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a. x ( x - 1 ) = 2
=> x thuộc 1,2,-1,-2
Thử ta thấy ko có x thỏa mãn
Câu b tương tự
\(3x^2+5x-2=0\)
\(\Leftrightarrow3x^2-x+6x-2=0\)
\(\Leftrightarrow x\left(3x-1\right)+2\left(3x-1\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=-2\end{cases}}}\)
Vậy ...
\(3x^2+5x-2=0\)
\(\Leftrightarrow3x^2+6x-x-2=0\)
\(\Leftrightarrow3x\left(x+2\right)-\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\3x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=\frac{1}{3}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{1}{3}\end{cases}}\)
`x^2-5x-2x^3+x^4+1 + (-5x^3) - 3+8x^4+x^2`
`= ( x^4 + 8x^4 ) - ( 2x^3 + 5x^3 ) + ( x^2 + x^2 ) - 5x + ( 1 - 3 )`
`= 9x^4 - 7x^3 + 2x^2 - 5x - 2`
= x2-5x-2x3+x4+1+(-5x3)-3+8x4+x2
=(x2+x2)+(-2x3-5x3)+(x4+8x4)-5x+(1-3)
=2x2+(-7x3)+9x4-5x+(-2)
\(\dfrac{\left(-0.25\right)^{-5}\cdot9^4\cdot\left(-2\right)^{-3}-2^{-2}\cdot6^9}{2^9\cdot3^6+6^6\cdot40}\)
\(=\dfrac{2^7\cdot3^8-2^8\cdot3^9}{2^9\cdot3^6+2^9\cdot3^6\cdot5}\)
\(=\dfrac{2^7\cdot3^8\cdot\left(-5\right)}{2^9\cdot3^6\cdot6}\)
\(=\dfrac{1}{4}\cdot9\cdot\dfrac{-5}{6}=\dfrac{-45}{24}=\dfrac{-15}{8}\)
\(\frac{4}{3}\left(x-\frac{1}{4}\right)=\frac{3}{2}\left(2x-1\right)\)
\(\Leftrightarrow\frac{4}{3}x-\frac{1}{3}=3x-\frac{3}{2}\)
\(\Leftrightarrow\frac{4}{3}x-3x=\frac{1}{3}-\frac{3}{2}\)
\(\Leftrightarrow-\frac{5}{3}x=-\frac{7}{6}\)
\(\Leftrightarrow x=\left(-\frac{7}{6}\right)\div\left(-\frac{5}{3}\right)\)
\(\Leftrightarrow x=\frac{7}{10}\)
\(\frac{4}{3}\left(x-\frac{1}{4}\right)=\frac{3}{2}\left(2x-1\right)\)
\(\Rightarrow\frac{4}{3}x-\frac{1}{3}-3x+\frac{3}{2}=0\)
\(\Rightarrow-\frac{5}{3}x+\frac{7}{6}=0\)
\(\Rightarrow-\frac{5}{3}x=-\frac{7}{6}\)
\(\Rightarrow x=\frac{7}{10}\)
Study well
th1; x-1/2 = -1,5
x = 1/2 - 1,5 = -1
th2; x -1/2 = 1,5
x = 1/2 +1,5 = 2