\(x^2.\left(2x-6\right)-2x^2=0\\ \Leftrightarrow2x^2.\left(x-3-1\right)-0\\ \Leftrightarrow2x^2\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(x^2.\left(2x-6\right)-2x^2=0\\ \Leftrightarrow2x^2.\left(x-3-1\right)-0\\ \Leftrightarrow2x^2\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
Tìm x biết:
\(a.\left|2x-3\right|=x-2\\ b.\left(2x+3\right)^6-\left(2x+3\right)^4=0\)
\(\left(2x+3\right)^2+\left(3x-2\right)^4=0\) vì \(\left(2x+3\right)^2\ge0;\left(3x-2\right)^4\ge0\) nên\(\Rightarrow\hept{\begin{cases}\left(2x+3\right)^2=0\\\left(3x-2\right)^4=0\end{cases}\Rightarrow\hept{\begin{cases}2x+3=0\\3x-2=0\end{cases}}}\) \(\Rightarrow\hept{\begin{cases}x=-\frac{3}{2}\\x=\frac{2}{3}\end{cases}}\)
Tìm x và y biết:
d)\(-1\frac{2}{3}-\left(\left|2x\right|+\frac{5}{6}\right)=\)\(-2\)e)\(\left(-\frac{1}{2}+\frac{1}{3}\right):\left|1-2x\right|-1\frac{1}{4}:\left(-\frac{5}{8}\right).\left(-\frac{1}{2}\right)^2=\frac{1}{3}\)
c)\(\left|2x-1\right|+\left|2y+1\right|+\left|2x-y\right|=0\)b)\(\left|2x-1\right|=2x-1\)
a)\(\left|x-3\right|=x+4\)
\(2\left|2x-6\right|=\dfrac{5}{6}-\left|x-3\right|\)
2:\(\left|x+2013\right|+\left|x+2014\right|+\left|x+2045\right|=2\)
3:\(\left|2x-1\right|=\left|x+1\right|\)
4:\(\sqrt{\left(x+\sqrt{5}\right)}+\sqrt{\left(y-\sqrt{3}\right)^2}+\left|x-y-z\right|=0\)
a)\(2\cdot\text{|}3-2x\text{|}+\dfrac{1}{2}=\dfrac{5}{2}\) b)\(x^2\cdot\left(2^x-6\right)-2x^3=0\)
Bài 2: Tìm x,y,z biết:
a)\(\left(x-1\right)\)\(:\)\(\dfrac{2}{3}\)=\(\dfrac{-2}{5}\)
b) \(\left|x-\dfrac{1}{2}\right|-\dfrac{1}{3}=0\)
c) \(\left|4x+2\right|=\left|6+2x\right|\)
Bài 2: Cho đa thức: \(A\left(x\right)=2x^2+3x+6\)
\(B\left(x\right)=2x^2+2x+3\)
a) Tính: \(P\left(x\right)=A\left(x\right)-B\left(x\right)\)
b) Tính: \(P\left(x\right)\) tại \(x=-3\) ; \(x=2\)
Tìm \(x:\)\(\left(2x-6\right)\left(x^2-1\right)-\left(3x-9\right)\left(x+1\right)^2=0\)
Tính giá trị biểu thức
A=\(2x+2y+3xy\left(x+y\right)+5\left(x^3y_{ }^2+x^2y^3\right)\)
tại x+y=0
B=\(3xy\left(x+y\right)+2x^3y+2x^2y^2\)
tại x+y=0