`5x - 12 = x`

`5x - x    = 12`

`4x           = 12`

`   x           = 12 ÷ 4`

`   x           = 3`

Vậy `x = 3`

Mình cx thấy nek!

\(\dfrac{12}{7}-\dfrac{4}{9}\)

\(=\dfrac{108}{63}-\dfrac{28}{63}\)

\(=\dfrac{80}{63}\)

Dòng cuối là \(4.12+4=52\) nhá

Sửa đề:  \(100-96+92-88+84-80+...+12-8+4\)

Số số hạng dãy số đó là :

\(\left(100-4\right):4+1=25\) (số hạng)

Vì trong dãy số đó có 1 số 4 nên có số số hạng (có nhóm) trong tổng đó là : \(25-1=24\) (số hạng) 

Các cặp trong dãy là : \(24:2=12\) (nhóm)

Ta có:

\(\left(100-96\right)+\left(92-88\right)+\left(84-80\right)+...+\left(12-8\right)+4\)

\(=4+4+4+...+4+4\)

\(=4.12+1=52\)

Đáp số :...

\(a,\left(y-24\right):28=20\)

     \(y-24\)         \(=20\times28\)

     \(y-24\)         \(=560\)

     \(y\)                 \(=560+24\)

     \(y\)                 \(=584\)

\(b,13\times\left(y-6\right)=4\times y-6\)

   \(13y-78\)       \(=4y-6\)

   \(13y-4y\)       \(=78-6\)

        \(9y\)            \(=72\)

          \(y\)            \(=72:9\)

          \(y\)            \(=8\)

@Mira ghi TK vào bài ak!

Bài 1: 

\(a,A=\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{24.25}\)

\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

\(=\dfrac{1}{5}-\dfrac{1}{25}=>\dfrac{5}{25}-\dfrac{1}{25}\)

\(=\dfrac{4}{25}\)

\(b,B=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)

       \(=1.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\right)\)

       \(=1.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

       \(=1.\left(1-\dfrac{1}{101}\right)\)

       \(=\dfrac{100}{101}\)

\(c,K=\dfrac{4}{11.16}+\dfrac{4}{16.21}+\dfrac{4}{21.26}+...+\dfrac{4}{61.66}\)

        \(=\dfrac{4}{5}.\left(\dfrac{1}{11.16}+\dfrac{1}{16.21}+\dfrac{1}{21.26}+...+\dfrac{1}{61.66}\right)\)

        \(=\dfrac{4}{5}.\left(\dfrac{1}{11}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{21}+...+\dfrac{1}{61}-\dfrac{1}{66}\right)\)

        \(=\dfrac{4}{5}.\left(\dfrac{1}{11}-\dfrac{1}{66}\right)\)

       \(=\dfrac{4}{5}.\dfrac{5}{66}=>4.\dfrac{1}{66}\)

       \(=\dfrac{4}{66}=\dfrac{2}{33}\)

\(d,N=\dfrac{4}{1.3}+\dfrac{4}{3.5}+\dfrac{4}{5.7}+...+\dfrac{4}{99.101}\)

        \(=2.\left(\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\right)\)

        \(=2.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

        \(=2.\left(1-\dfrac{1}{101}\right)\)

        \(=2.\dfrac{100}{101}\)

        \(=\dfrac{200}{101}\)

Bài 2:

\(K=\dfrac{5}{3.7}+\dfrac{5}{7.11}+\dfrac{5}{11.15}+...+\dfrac{5}{81.85}+\dfrac{5}{85.89}\)

    \(=\dfrac{5}{4}.\left(\dfrac{1}{3.7}+\dfrac{1}{7.11}+\dfrac{1}{11.15}+...+\dfrac{1}{81.85}+\dfrac{1}{85.89}\right)\)

    \(=\dfrac{5}{4}.\left(\dfrac{1}{3}-\dfrac{1}{7}+...+\dfrac{1}{85}-\dfrac{1}{89}\right)\)

    \(=\dfrac{5}{4}.\left(\dfrac{1}{3}-\dfrac{1}{89}\right)\)

    \(=\dfrac{5}{4}.\dfrac{86}{267}\)

    \(=\dfrac{215}{534}\)

Bài 3:

\(A=\dfrac{1}{25.24}+\dfrac{1}{24.23}+...+\dfrac{1}{7.6}+\dfrac{1}{6.5}\)

    \(=\dfrac{1}{5.6}+\dfrac{1}{6.7}+...+\dfrac{1}{23.24}+\dfrac{1}{24.25}\)

    \(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{24}-\dfrac{1}{25}\)

    \(=\dfrac{1}{5}-\dfrac{1}{25}\)

    \(=\dfrac{4}{25}\)

Bài 4 :

\(A=\dfrac{5}{3.6}+\dfrac{5}{6.9}+\dfrac{5}{9.12}+...+\dfrac{5}{99.102}\)

    \(=\dfrac{5}{3}.\left(\dfrac{1}{3.6}+\dfrac{1}{6.9}+\dfrac{1}{9.12}+...+\dfrac{1}{99.102}\right)\)

    \(=\dfrac{5}{3}.\left(\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{9}+...+\dfrac{1}{99}-\dfrac{1}{102}\right)\)

    \(=\dfrac{5}{3}.\left(\dfrac{1}{3}-\dfrac{1}{102}\right)\)

    \(=\dfrac{5}{3}.\dfrac{11}{34}\)

    \(=\dfrac{55}{102}\)

Bài 5 :

Sửa đề :\(a,E=\dfrac{1}{7}+\dfrac{1}{91}+\dfrac{1}{247}+\dfrac{1}{475}+\dfrac{1}{775}+\dfrac{1}{1147}\)

        \(=\dfrac{1}{1.7}+\dfrac{1}{7.13}+\dfrac{1}{13.19}+\dfrac{1}{19.25}+\dfrac{1}{25.31}+\dfrac{1}{31.37}\)

        \(=\dfrac{1}{6}.\left(\dfrac{1}{1.7}+\dfrac{1}{7.13}+\dfrac{1}{13.19}+\dfrac{1}{19.25}+\dfrac{1}{25.31}+\dfrac{1}{31.37}\right)\)

        \(=\dfrac{1}{6}.\left(1-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{13}+...+\dfrac{1}{31}-\dfrac{1}{37}\right)\)

        \(=\dfrac{1}{6}.\left(1-\dfrac{1}{37}\right)\)

        \(=\dfrac{1}{6}.\dfrac{36}{37}\)

        \(=\dfrac{6}{37}\)

\(b,C=\dfrac{2}{15}+\dfrac{2}{35}+\dfrac{2}{63}+\dfrac{2}{99}+\dfrac{2}{143}\)

       \(=\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+\dfrac{2}{11.13}\)

       \(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{11}-\dfrac{1}{13}\)

       \(=\dfrac{1}{3}-\dfrac{1}{13}\)

       \(=\dfrac{10}{39}\)

Bài 6 :

\(a,\dfrac{3}{5.7}+\dfrac{3}{7.9}+\dfrac{3}{9.11}+...+\dfrac{3}{x\left(x+2\right)}=\dfrac{24}{35}\) 

    \(\dfrac{3}{2}\left(\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}+...+\dfrac{2}{x\left(x+2\right)}\right)=\dfrac{24}{35}\)

    \(\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{x+2}\right)=\dfrac{24}{35}\)

          \(\dfrac{1}{5}-\dfrac{1}{x+2}=\dfrac{24}{35}:\dfrac{3}{2}\)

          \(\dfrac{1}{5}-\dfrac{1}{x+2}=\dfrac{16}{35}\)

                  \(\dfrac{1}{x+2}=\dfrac{1}{5}-\dfrac{16}{35}\)

                  \(\dfrac{1}{x+2}=-\dfrac{9}{35}\)

                  \(-9\left(x+2\right)=1.35\)

                  \(-9\left(x+2\right)=35\)

                         \(x+2=35:-9\)

                        \(x+2=\dfrac{-35}{9}\)

                        \(x\)        \(=\dfrac{-35}{9}-2\)

                        \(x\)        \(=\dfrac{-53}{9}\)

Vậy \(x=\dfrac{-53}{9}\)

\(b,\dfrac{2}{4.7}+\dfrac{2}{7.10}+\dfrac{2}{10.13}+...+\dfrac{2}{x\left(x+3\right)}=\dfrac{1}{9}\)

   \(\dfrac{2}{3}.\left(\dfrac{3}{4.7}+\dfrac{3}{7.10}+\dfrac{3}{10.13}+...+\dfrac{3}{x\left(x+3\right)}\right)=\dfrac{1}{9}\)

   \(\dfrac{2}{3}.\left(\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{1}{9}\)

   \(\dfrac{2}{3}.\left(\dfrac{1}{4}-\dfrac{1}{x+3}\right)\)                                        \(=\dfrac{1}{9}\)

   \(\dfrac{1}{6}-\dfrac{2}{3.\left(x+3\right)}\)                                            \(=\dfrac{1}{9}\)

           \(\dfrac{2}{3.\left(x+3\right)}\)                                            \(=\dfrac{1}{6}-\dfrac{1}{9}\)

           \(\dfrac{2}{3.\left(x+3\right)}\)                                            \(=\dfrac{1}{18}\)

           \(\dfrac{2}{3.\left(x+3\right)}\)                                            \(=\dfrac{2}{36}\)

      ⇒   \(3.\left(x+3\right)=36\)

                 \(x+3=36:3\)

                 \(x+3=12\) 

                 \(x\)       \(=12-3\)

                 \(x\)       \(=9\)

Vậy \(x=9\)

Bài 7:

\(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=1\dfrac{1989}{1991}\)

\(=>\dfrac{2}{2}+\dfrac{2}{6}+\dfrac{2}{12}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{3980}{1991}\)

\(=>\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{x.\left(x+1\right)}=\dfrac{3980}{1991}\)

\(=>2.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}...+\dfrac{1}{x.\left(x+1\right)}\right)=\dfrac{3980}{1991}\)

\(=>2.\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{3980}{1991}\)

\(=>2.\left(1-\dfrac{1}{x+1}\right)=\dfrac{3980}{1991}\)

             \(1-\dfrac{1}{x+1}=\dfrac{3980}{1991}:2\)

             \(1-\dfrac{1}{x+1}=\dfrac{1990}{1991}\)

                    \(\dfrac{1}{x+1}=1-\dfrac{1990}{1991}\)

                    \(\dfrac{1}{x+1}=\dfrac{1}{1991}\)

           \(=>x+1=1991\)

                  \(x\)       \(=1991-1\)

                  \(x\)       \(=1990\)

Vậy \(x=1990\)

`5/13 + -5/17 + -21/41 + 8/13 + -20/41`

`= (5/13 + 8/13) + (-21/41 + -20/41) + -5/17`

`= 1 + -1 + -5/17`

`= 0 + -5/17`

`= -5/17`

$\color{#87CEFA}{\text{@Ann}}$

\(2,\)

\(a,\dfrac{-5}{2}:\dfrac{5}{8}\)

\(=\dfrac{-5}{2}.\dfrac{8}{5}\)

\(=\dfrac{-1}{1}.\dfrac{4}{1}\left(-1.4\right)\)

\(=-4\)

\(b,4\dfrac{1}{5}:\left(-2\dfrac{4}{5}\right)\)

\(=\dfrac{21}{5}:\left(-\dfrac{14}{5}\right)\)

\(=\dfrac{21}{5}.-\dfrac{5}{14}\)

\(=\dfrac{3}{1}.-\dfrac{1}{2}\)

\(=-\dfrac{3}{2}\)

\(c,7:\left(-3,5\right)=-2\)

\(d,-1\dfrac{4}{5}:\left(-\dfrac{3}{4}\right)\)

\(=-\dfrac{9}{5}:\left(-\dfrac{3}{4}\right)\)

\(=-\dfrac{9}{5}.-\dfrac{4}{3}\)

\(=-\dfrac{3}{5}.-\dfrac{4}{1}\)

\(=\dfrac{12}{5}\)

\(e,4,2:\dfrac{-15}{12}\)

\(=\dfrac{42}{10}:\dfrac{-15}{12}\)

\(=\dfrac{42}{10}.-\dfrac{12}{15}\)

\(=\dfrac{42}{5}.-\dfrac{6}{15}\)

\(=8,4.-0,4=-3,36\)

\(g,6\dfrac{9}{11}:\left(-3\right)\)

\(=\dfrac{75}{11}:\left(-3\right)\)

\(=\dfrac{75}{11}.\left(-\dfrac{1}{3}\right)\)

\(=\dfrac{25}{11}.\left(-\dfrac{1}{1}\right)\)

\(=-\dfrac{25}{11}\)

\(#T-T\)