a)\(Q\left(x\right)=5x^3+2x^4-2x^2+4x^2-2x^3-x^4+1-3x^3\)
\(=x^4+2x^2+1\ge1\forall x\)nên đa thức này vô nghiệm(số mũ chẵn mà:>)
b)\(S\left(x\right)=x^2+x+1\)
\(=x^2+2.\frac{1}{2}.x+\frac{1}{4}+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\) nên đa thức này vô nghiệm(cái này phải dùng HĐT nhé,xem sau vở á)
c)\(T\left(x\right)=x^2-x+\frac{1}{2}=x^2-2.\frac{1}{2}.x+\frac{1}{4}+\frac{1}{4}=\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\ge\frac{1}{4}\forall x\)
nên PT này vô nghiệm
chắc bạn chx hiểu chô này nhỉ:
\(x^2+x+\frac{1}{4}=x^2+\frac{1}{2}x+\frac{1}{2}x+\frac{1}{4}=x.\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{2}\right)=\left(x+\frac{1}{2}\right).\left(x+\frac{1}{2}\right)\)
\(=\left(x+\frac{1}{2}\right)^2\) cái câu B cx tương tự nhưng à dấu trừ nha
đây là HĐT lớp 8
O x y z t
Ta có: Góc xOz + góc yOz = 180o (2 góc kề bù)
Góc xOz = 3 . góc yOz
=> Góc xOz = 180o : (3 + 1) . 3 = 135o và góc yOz = 180o - 135o = 45o
Mà góc xOz = góc tOy (2 góc đối đỉnh) và góc xOt = góc yOz (2 góc đối đỉnh)
=> Góc xOz = góc tOy = 135o và góc xOt = góc yOz = 45o
@kiềuanh2k8
O x y z t
^xOz + ^zOy = 180 (2 góc kề bù)
^xOz + ^xOt = 180 (2 góc kề bù)
-> ^xOt = ^yOz
\(\dfrac{1}{5}:\left(x-3\right)+\dfrac{1}{2}=1\dfrac{1}{8}\\ \Rightarrow\dfrac{1}{5}:\left(x-3\right)=\dfrac{9}{8}-\dfrac{1}{2}\\ \Rightarrow\dfrac{1}{5}:\left(x-3\right)=\dfrac{9}{8}-\dfrac{4}{8}\\ \Rightarrow\dfrac{1}{5}:\left(x-3\right)=\dfrac{5}{8}\\ \Rightarrow x-3=\dfrac{1}{5}:\dfrac{5}{8}\\ \Rightarrow x-3=\dfrac{8}{25}\\ \Rightarrow x=\dfrac{8}{25}+3\\ \Rightarrow x=\dfrac{83}{25}\)
\(\dfrac{1}{5}:\left(x-3\right)+\dfrac{1}{2}=1\dfrac{1}{8}\)
\(\dfrac{1}{5}:\left(x-3\right)+\dfrac{1}{2}=\dfrac{9}{8}\)
\(\dfrac{1}{5}:\left(x-3\right)=\dfrac{9}{8}-\dfrac{1}{2}=\dfrac{9}{8}-\dfrac{4}{8}\)
\(\dfrac{1}{5}:\left(x-3\right)=\dfrac{5}{8}\)
\(x-3=\dfrac{1}{5}:\dfrac{5}{8}=\dfrac{1}{5}.\dfrac{8}{5}\)
\(x-3=\dfrac{8}{25}\)
\(x=\dfrac{8}{25}+3=\dfrac{8}{25}+\dfrac{75}{25}\)
\(x=\dfrac{83}{25}\)
\(\dfrac{1}{5}:\left(x+3\right)+\dfrac{1}{2}=1\dfrac{1}{8}\\ \dfrac{1}{5}:\left(x+3\right)=\dfrac{5}{8}\\ x+3=\dfrac{8}{25}\\ x=\dfrac{-67}{25}\)
\(\dfrac{1}{5}:\left(x-3\right)+\dfrac{1}{2}=1\dfrac{1}{8}\)
\(\Rightarrow\dfrac{1}{5}:\left(x-3\right)=\dfrac{9}{8}-\dfrac{1}{2}\)
\(\Rightarrow\dfrac{1}{5}:\left(x-3\right)=\dfrac{5}{8}\)
\(\Rightarrow x-3=\dfrac{5.5}{8}=\dfrac{25}{8}\)
\(\Rightarrow x=\dfrac{25}{8}+3\Rightarrow x=\dfrac{49}{8}\)
`@` `\text {Ans}`
`\downarrow`
\(\dfrac{1}{5} \div (x - 3) + \dfrac{1}{2} = 1 \dfrac{1}{8}\)
\(\Rightarrow \dfrac{1}{5} \div (x - 3) = 1\dfrac{1}{8} - \dfrac{1}{2}\)
\(\Rightarrow \dfrac{1}{5} \div (x - 3) = \dfrac{5}{8}\)
\(\Rightarrow x - 3 = \dfrac{1}{5} \div \dfrac{5}{8}\)
\(\Rightarrow x - 3 =\dfrac{8}{25}\)
\(\Rightarrow x = \dfrac{8}{25} +3\)
\(\Rightarrow x = \dfrac{83}{25}\)
\(\dfrac{83}{25}\)nha
51:(x−3)+21=181⇒51:(x−3)=89−21⇒51:(x−3)=89−84⇒51:(x−3)=85⇒x−3=51:85⇒x−3=258⇒x=258+3⇒x=2583
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