$\color{#B0E0E6}{\text{1/5 . 11/16 + 1/5 . 5/16 + 4/5}}$

$\color{#B0E0E6}{\text{ = 1/5. ( 11/16 + 5/16) + 4/5}}$

$\color{#B0E0E6}{\text{ = 1/5 . 1 + 4/5 }}$

$\color{#B0E0E6}{\text{ = 1/5 + 4/5}}$

$\color{#B0E0E6}{\text{1}}$

$\color{#87CEFA}{\text{@456}}$

$\color{#B0E0E6}{\text{ What your favourite room?}}$

$\color{#87CEFA}{\text{= > What room is your favourite spot?}}

(sửa lại)

$\color{#B0E0E6}{\text{ What your favourite room?}}$

$\color{#87CEFA}{\text{= > What room do you like?}}

(không biết có đúng k)

\(\left(2\times x-17\right):12-0,5=75\%\)

\(\left(2\times x-17\right):12-0,5=0,75\)

\(\left(2\times x-17\right):12\)           \(=0,75+0,5\)

\(\left(2\times x-17\right):12\)           \(=1,25\)

\(2\times x-17\)                     \(=1,25\times12\)

\(2\times x-17\)                     \(=15\)

\(2\times x\)                              \(=15+17\)

\(2\times x\)                               \(=32\)

       \(x\)                              \(=32:2\)

       \(x\)                              \(=16\)

Vậy \(x=16\)

\(205-\left[1200-\left(4^2-2.3\right)^2\right]:40\)

\(=205-\left[1200-\left(16-2.3\right)^2\right]:40\)

\(=205-\left[1200-\left(16-6\right)^2\right]:40\)

\(=205-\left[1200-10^2\right]:40\)

\(=205-\left[1200-100\right]:40\)

\(=205-1100:40\)

\(=205-27,5\)

\(=177,5\)

\(\dfrac{3}{1\times2}+\dfrac{3}{2\times3}+...+\dfrac{3}{99\times100}\)

\(=3\times\left(\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+...+\dfrac{1}{99\times100}\right)\)

\(=3\times\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)

\(=3\times\left(1-\dfrac{1}{100}\right)\)

\(=3\times\dfrac{99}{100}\)

\(=\dfrac{297}{100}\)

\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{2000}-\dfrac{5}{6}\)

\(=\dfrac{5}{6}+\dfrac{1}{2000}-\dfrac{5}{6}\)

\(=\left(\dfrac{5}{6}-\dfrac{5}{6}\right)+\dfrac{1}{2000}\)

\(=0+\dfrac{1}{2000}\)

\(=\dfrac{1}{2000}\)

\(a,32< 2^x< 128\)

\(=2^5< 2^x< 2^7\)

\(=>2^x=2^6< =>x=6\)

Vậy...

\(b,2.16\ge2^x>4\)

\(=2^5\ge2^x>2^2\)

\(=>x\in\left\{3;4;5\right\}\)

Vậy...

\(c,9.27\le3^x\le243\)

\(=>3^2.3^3\le3^x\le3^5\)

\(=>3^5\le3^x\le3^5\)

\(=>x\in\left\{5\right\}\)

Vậy ...

\(d,x^{2019}=x\)

\(=>x^{2019}-x=0\)

\(=>x.\left(x^{2018}-1\right)=0\)

\(=>\left[{}\begin{matrix}x=0\\x^{2018}-1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x^{2018}=1\end{matrix}\right.=>\left[{}\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)

Vậy \(x\in\left\{0;\pm1\right\}\)

\(\dfrac{1}{2}+\dfrac{2}{4}\)

\(=\dfrac{1}{2}+\dfrac{1}{2}\)

\(=1\)

\(\dfrac{8}{9}-\dfrac{5}{9}\)

\(=\dfrac{8-5}{9}\)

\(=\dfrac{3}{9}=\dfrac{1}{3}\)

\(\dfrac{2}{3}\times\dfrac{3}{4}\)

\(=\dfrac{1}{1}\times\dfrac{1}{2}\)

\(=\dfrac{1}{2}\)

\(\dfrac{5}{6}:\dfrac{4}{3}\)

\(=\dfrac{5}{6}\times\dfrac{3}{4}\)

\(=\dfrac{5}{2}\times\dfrac{1}{4}\)

\(=\dfrac{5}{8}\)

$\color{#87CEFA}{\text{@456}}$$\color{#B0E0E6}{\text{@Aoi}}$

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