Ta có: \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)+4\left(x-7\right)\)
\(=x^2-4x+4-x^2+3^2+4x-28\\ =\left(x^2-x^2\right)-\left(4x-4x\right)+\left(4+3^2-28\right)\\=-15 \)
⇒ Gía trị của biểu thức \(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)+4\left(x-7\right)\) không phụ thuộc vào biến \(x\) \(\left(đcpcm\right)\)
3
a,\(\left(x-5\right)\left(x+5\right)-x\left(x-4\right)=0\)
\(\Leftrightarrow x^2-5^2-x^2+4x=0\\ \Leftrightarrow\left(x^2-x^2\right)+4x-5^2\\ \Leftrightarrow4x=25\\ \Leftrightarrow x=\dfrac{25}{4}\)
Vậy....
b,\(\left(2x-1\right)\left(4x^2+2x+1\right)-\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow8x^3-1-\left(x^2-1^2\right)=0\\ \Leftrightarrow8x^3-x^2-\left(1-1\right)=0\\ \Leftrightarrow8x^3-x^2=0\\ \Leftrightarrow x^2\left(8x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x^2=0\\8x-1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{8}\end{matrix}\right.\)
Vậy...
c,\(\left(x-3\right)^2+x\left(2x-3\right)=3x^2-6x+9\)
\(\Leftrightarrow x^2-6x+9+2x^2-3x=3x^2-6x+9\\ \Leftrightarrow x^2+2x^2-3x^2-6x+6x-3x=9-9\\ \Leftrightarrow-3x=0\\ \Leftrightarrow x=0\)
Vậy...
5,
a. \(\left(x+y+z\right)^2\)
\(=\left[\left(x+y\right)+z\right]^2\\ =\left(x+y\right)^2+2\left(x+y\right)z+z^2\\ =x^2+2xy+y2+\left(2x+2y\right)z+z^2\\ =x^2+y^2+z^2+2xy+2yz+2zx\)
Giải giúp mik nhé!
