1. \(x^2-4x+4=\left(x-2\right)^2\)
Thế x = -1 ta có \(\left(-1-2\right)^2=9\)
2. \(a=\left(x-1\right)^2+\left(x+1\right)\left(3-x\right)\)
\(\Leftrightarrow a=x^2-2x+1+3x-x^2+3-x\)
\(\Leftrightarrow a=4\)
Vậy biểu thức a không phụ thuộc vào giá trị biến
3. a) \(2x-\left(x-3\right)^2=5-x\)
\(\Leftrightarrow2x-x^2+6x-9=5-x\)
\(\Leftrightarrow-x^2+9x-14=0\)
\(\Leftrightarrow-x^2+2x+7x-14=0\)
\(\Leftrightarrow-x\left(x-2\right)+7\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(-x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-2=0\\-x+7=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=7\end{matrix}\right.\)
Vậy..
b) \(\left(5x-2x\right)^2-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(5x-2x+x+2\right)\left(5x-2x-x-2\right)=0\)
\(\Leftrightarrow\left(4x+2\right)\left(2x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}4x+2=0\\2x-2=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=1\end{matrix}\right.\)
Vậy...
Bài 1:
Tại \(x=-1\) thì giá trị biểu thức là:
\(x^2-4x+4=\left(x-2\right)^2=\left(-1-2\right)^2=9\)
Bài 2:
\(A=\left(x-1\right)^2+\left(x+1\right)\left(3-x\right)\)
\(=x^2-2x+1+3\left(x+1\right)-x\left(x+1\right)\)
\(=x^2-2x+1+3x+3-x^2-x\)
\(=4\). Vậy giá trị biểu thức A ko phụ thuộc vào biến
Bài 3:
a)\(2x-\left(x-3\right)^2=5-x\)
\(\Leftrightarrow2x-\left(x-3\right)^2-5+x=0\)
\(\Leftrightarrow3x-5-\left(x-3\right)^2=0\)
\(\Leftrightarrow3x-5-x^2+6x-9=0\)
\(\Leftrightarrow-x^2+9x-14=0\)
\(\Leftrightarrow\left(7-x\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}7-x=0\\x-2=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=7\\x=2\end{matrix}\right.\)
b)\(\left(5x-2x\right)^2-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(5x-2x+x+2\right)\left[\left(5x-2x\right)-\left(x+2\right)\right]=0\)
\(\Leftrightarrow2\left(2x+1\right)\left[\left(5x-2x\right)-\left(x+2\right)\right]=0\)
\(\Leftrightarrow4\left(2x+1\right)\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x-1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\)
1) \(x^2-4x+4\)
\(=x^2-2.x.2+2^2=\left(x-2\right)^2\)
Tại x=-1, ta có:
\(\left(x-2\right)^2=\left(1-2\right)^2=\left(-1\right)^2=1\)
2) \(A=\left(x-1\right)^2+\left(x+1\right)\left(3-x\right)\)
\(=x^2-2x+1+3x-x^2+3-x\)
\(=3\)
Vậy biểu thức sau không phụ thuộc vào biến x
