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Tham khảo:
a) \((45{x^5} - 5{x^4} + 10{x^2}):5{x^2}\)\( = 9{x^3} - {x^2} + 2\)
b) \((9{t^2} - 3{t^4} + 27{t^5}):3t = (27{t^5} - 3{t^4} + 9{t^2}):3t\\=(27t^5):(3t) - (3t^4):(3t)+(9t^2):(3t) = 9{t^4} - {t^3}+3t\)
\(\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{8.9.10}\right)x=\frac{22}{45}\)
=> \(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{8.9.10}\right)=\frac{22}{45}\)
=> \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{8.9}-\frac{1}{9.10}\right)=\frac{22}{45}\)
\(\Rightarrow\left(\frac{1}{1.2}-\frac{1}{9.10}\right)x=\frac{22}{45}:\frac{1}{2}\)
\(\Rightarrow\left(\frac{1}{2}-\frac{1}{90}\right)x=\frac{44}{45}\)
=> \(\frac{44}{45}x=\frac{44}{45}\)
=> x = 1
Vậy x = 1
`3x+20=0`
`=>3x=0-20`
`=>3x=-20`
`=>x=-20/3`
`---`
`2(-4x+9)=0`
`=>-4x+9=0`
`=>-4x=-9`
`=>x=9/4`
`---`
`2x(x-45)=0`
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x-45=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=45\end{matrix}\right.\)
`---`
`-5x(2x+47)=0`
\(\Rightarrow\left[{}\begin{matrix}-5x=0\\2x+47=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{47}{2}\end{matrix}\right.\)
`----`
`x^2 -912=0`
`=>x^2=912`
`=>x∈∅`
1)
`3x+20=0`
`<=>3x=-20`
`<=>x=-20/3`
2)
`2(-4x+9)=0`
<=>-4x+9=0`
`<=>-4x=-9`
`<=>x=9/4`
3)
`2x(x-45)=0`
\(< =>\left[{}\begin{matrix}2x=0\\x-45=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=45\end{matrix}\right.\)
4)
`-x(2x+47)=0`
\(< =>\left[{}\begin{matrix}-x=0\\2x+47=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=-\dfrac{47}{2}\end{matrix}\right.\)
5)
`x^2 -912=0`
`<=>x^2=912`
câu 5 xem lại nhé
3.(\(x\) - 2)4 = 45
( \(x\) - 2)4 = 45: 3
(\(x\) - 2)4 = 15
\(\left[{}\begin{matrix}x-2=\sqrt[4]{15}\\x-2=-\sqrt[4]{15}\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=2+\sqrt[4]{15}\\x=2-\sqrt[4]{15}\end{matrix}\right.\)