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6 tháng 7 2016
a) Xem lại đề
b) \(\Leftrightarrow3x^4-2x^2-1=0\Leftrightarrow\left(3x^2+1\right)\left(x^2-1\right)=0\Leftrightarrow x^2=1\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)
PA
28 tháng 11 2016
\(8-12x+6x^2-x^3\)
\(=\left(2-x\right)^3\)
\(125x^3-75x^2+15x-1\)
\(=\left(5x-1\right)^3\)
\(x^2-xz-9y^2+3yz\)
\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y-z\right)\)
\(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
\(x^3+2x^2-6x-27\)
\(=x^3+5x^2+9x-3x^2-15x-27\)
\(=x\left(x^2+5x+9\right)-3\left(x^2+5x+9\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
\(12x^3+4x^2-27x-9\)
\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(4x^2-9\right)\)
\(=\left(3x+1\right)\left(2x-3\right)\left(2x+3\right)\)
\(4x^4+4x^3-x^2-x\)
\(=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=x\left(x+1\right)\left(4x^2-1\right)\)
\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)
20 tháng 6 2019
a) \(\left(\frac{1}{81}\right)^x\cdot27^{2x}=\left(-9\right)^4\)
\(\Leftrightarrow\frac{1}{3^{4x}}\cdot3^{6x}=9^4\)
\(\Leftrightarrow\frac{3^{6x}}{3^{4x}}=3^8\)
\(\Leftrightarrow3^{2x}=3^8\)
\(\Leftrightarrow2x=8\)
\(\Leftrightarrow x=4\)
b) \(5^x\cdot\left(5^3\right)^2=625\)
\(\Leftrightarrow5^{x+6}=5^4\)
\(\Leftrightarrow x+6=4\)
\(\Leftrightarrow x=-2\)
c) \(\left(4x-1\right)^{30}=\left(4x-1\right)^{20}\)
\(\Leftrightarrow\left(4x-1\right)^{30}-\left(4x-1\right)^{20}=0\)
\(\Leftrightarrow\left(4x-1\right)^{20}\cdot\left[\left(4x-1\right)^{10}-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-1=0\\\left(4x-1\right)^{10}=1=\left(\pm1\right)^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{4}\\x=\frac{1}{2}\\x=0\end{matrix}\right.\)
Vậy....
MA
15 tháng 8 2016
a) \(3\left(x+\frac{1}{2}\right)=8\)
\(\Rightarrow x+\frac{1}{2}=\frac{8}{3}\)
\(\Rightarrow x=\frac{8}{3}-\frac{1}{2}\)
\(\Rightarrow x=\frac{13}{6}\)
b) \(4x+\frac{1}{2}^3=-7\)
\(\Rightarrow4x+\frac{1}{8}=-7\)
\(\Rightarrow4x=-7-\frac{1}{8}\)
\(\Rightarrow4x=-\frac{57}{8}\)
\(\Rightarrow x=-\frac{57}{8}:4\)
\(\Rightarrow x=\frac{1}{32}\)
c) \(4\left(x-\frac{1}{2}\right)+\left(-3\right)^2=9\)
\(\Rightarrow4\left(x-\frac{1}{2}\right)+9=9\)
\(\Rightarrow4\left(x-\frac{1}{2}\right)=0\)
\(\Rightarrow x-\frac{1}{2}=0\Rightarrow x=\frac{1}{2}\)
tíc mình nha
XO
10 tháng 7 2019
a) \(\frac{2}{5}:\left(2x+\frac{3}{4}\right)=-\frac{7}{10}\)
=> \(2x+\frac{3}{4}=-\frac{7}{10}:\frac{2}{5}\)
=> \(2x+\frac{3}{4}=-\frac{7}{4}\)
=> \(2x=\frac{-7}{4}-\frac{3}{4}\)
=> \(2x=-\frac{5}{2}\)
=> \(x=\frac{-5}{2}:2\)
=> \(x=\frac{-5}{4}\)
b) \(\frac{x+1}{3}=\frac{2-x}{2}\)
\(\Rightarrow2\left(x+1\right)=3\left(2-x\right)\)
\(\Rightarrow2x+2=6-3x\)
\(\Rightarrow2x-3x=6-2\)
\(\Rightarrow-x=4\)
\(\Rightarrow x=4\)
XO
10 tháng 7 2019
c) \(\left|x-\frac{3}{5}\right|.\frac{1}{2}-\frac{1}{5}=0\)
\(\Rightarrow\left|x-\frac{3}{5}\right|.\frac{1}{2}=\frac{1}{5}\)
\(\Rightarrow\left|x-\frac{3}{5}\right|=\frac{1}{5}:\frac{1}{2}\)
\(\Rightarrow\left|x-\frac{3}{5}\right|=\frac{2}{5}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{3}{5}=\frac{2}{5}\\x-\frac{3}{5}=-\frac{2}{5}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=\frac{3}{5}+\frac{2}{5}\\x=\frac{3}{5}+-\frac{2}{5}\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=1\\x=\frac{1}{5}\end{cases}}\)
d) \(x^2-4x=0\)
Ta có : \(x^2-4x=0\)
\(\Rightarrow xx-4x=0\)
\(\Rightarrow x\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-4=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=0+4\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=0\\x=4\end{cases}}\)