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a) \(f\left(x\right)=5x^3-7x^2+2x+5\)
\(\Rightarrow f\left(1\right)=5.1^3-7.1^2+2.1+5\)
\(\Rightarrow f\left(1\right)=5.1-7.1+2+5\)
\(\Rightarrow f\left(1\right)=5-7+7\)
\(\Rightarrow f\left(1\right)=5\)
Vậy f(1) = 5.
\(g\left(x\right)=7x^3-7x^2+2x+5\)
\(\Rightarrow g\left(\frac{1}{2}\right)=7.\left(\frac{1}{2}\right)^3-7.\left(\frac{1}{2}\right)^2+2.\frac{1}{2}+5\)
\(\Leftrightarrow g\left(\frac{1}{2}\right)=7.\frac{1}{8}-7.\frac{1}{4}+1+5\)
\(\Leftrightarrow g\left(\frac{1}{2}\right)=\frac{7}{8}-\frac{14}{8}+6\)
\(\Leftrightarrow g\left(\frac{1}{2}\right)=\frac{-7}{8}+\frac{48}{8}\)
\(\Leftrightarrow g\left(\frac{1}{2}\right)=\frac{41}{8}\)
Vậy \(g\left(\frac{1}{2}\right)=\frac{41}{8}\)
\(h\left(x\right)=2x^3+4x+1\)
\(\Rightarrow h\left(0\right)=2.0^3+4.0+1\)
\(\Rightarrow h\left(0\right)=0+0+1\)
\(\Rightarrow h\left(0\right)=1\)
Vậy \(h\left(0\right)=1\)
\(=-\dfrac{7}{2}x+1+\dfrac{5}{4}x-3-\dfrac{1}{2}x\left(2x^2+x-2x-1\right)\)
\(=\dfrac{-9}{4}x-2-x^3-\dfrac{1}{2}x^2+x^2+\dfrac{1}{2}x\)
\(=-x^3+\dfrac{1}{2}x^2-\dfrac{7}{4}x-2\)
Ta có: \(\frac{2x-2}{3}=\frac{7x+3}{2-1}\)
\(\Leftrightarrow\frac{2x-2}{3}=7x+3\)
\(\Leftrightarrow2x-2=21x+9\)
\(\Leftrightarrow19x=-11\)
\(\Rightarrow x=-\frac{11}{19}\)
\(\left|3x-2\right|=x+1\left(1\right)\)
\(ĐK:x+1\ge0\Rightarrow x\ge-1\)
Với \(x\ge-1\) thì \(\left(1\right)\Rightarrow\left[{}\begin{matrix}3x-2=x+1\\3x-2=-\left(x+1\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-x=1+2\\3x-2=-x-1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=3\\3x+x=-1+2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1,5\\4x=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1,5\\x=0,25\end{matrix}\right.\)
Vậy \(x\in\left\{1,5;0,25\right\}\)
\(\left|2x-1\right|-3+5x=7x-2\)
\(\Rightarrow\left|2x-1\right|=7x-2+3-5x\)
\(\Rightarrow\left|2x-1\right|=\left(7x-5x\right)-2+3\)
\(\Rightarrow\left|2x-1\right|=2x+1\left(2\right)\)
\(ĐK:2x+1\ge0\Rightarrow2x\ge-1\Rightarrow x\ge\dfrac{-1}{2}\)
Với \(x\ge\dfrac{-1}{2}\) thì
\(\left(2\right)\Rightarrow\left[{}\begin{matrix}2x-1=2x+1\\2x-1=-\left(2x+1\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x-2x=1+1\\2x-1=-2x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}0=2(loại)\\2x+2x=-1+1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}0=2\left(loại\right)\\4x=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}0=2\left(loại\right)\\x=0\left(tm\right)\end{matrix}\right.\)
Vậy \(x=0\)
`@` `\text {Ans}`
`\downarrow`
`7x*(1/7x - 2x^2 + 1)`
`= 7x*1/7x + 7x*(-2x^2) + 7x`
`= x^2 - 14x^3 + 7x`
`= -14x^3 + x^2 + 7x`