Câu 1:
a) Ta có: \(2\left(x-1\right)^2+\left(x+3\right)^2=3\left(x-2\right)\left(x+1\right)\)
\(\Leftrightarrow2\left(x^2-2x+1\right)+x^2+6x+9=3\left(x^2-x-2\right)\)
\(\Leftrightarrow2x^2-4x+2+x^2+6x+9=3x^2-3x-6\)
\(\Leftrightarrow3x^2+2x+11-3x^2+3x+6=0\)
\(\Leftrightarrow5x+17=0\)
\(\Leftrightarrow5x=-17\)
hay \(x=-\frac{17}{5}\)
Vậy: \(x=-\frac{17}{5}\)
b) Ta có: \(\left(x+2\right)^2-2\left(x-3\right)=\left(x-1\right)^2\)
\(\Leftrightarrow x^2+4x+4-2x+6=x^2-2x+1\)
\(\Leftrightarrow x^2+2x+10-x^2+2x-1=0\)
\(\Leftrightarrow4x+9=0\)
\(\Leftrightarrow4x=-9\)
hay \(x=-\frac{9}{4}\)
Vậy: \(x=-\frac{9}{4}\)
c) Ta có: \(\left(x-1\right)^2+\left(x-2\right)^2=2\left(x+4\right)^2-\left(22x+27\right)\)
\(\Leftrightarrow x^2-2x+1+x^2-4x+4=2\left(x^2+8x+16\right)-22x-27\)
\(\Leftrightarrow2x^2-6x+5=2x^2+16x+32-22x-27\)
\(\Leftrightarrow2x^2-6x+5=2x^2-6x+5\)
\(\Leftrightarrow2x^2-6x+5-2x^2+6x-5=0\)
\(\Leftrightarrow0x=0\)
Vậy: \(x\in R\)
a) \(2.\left(x-1\right)^2+\left(x+3\right)^2=3.\left(x-2\right)\left(x+1\right)\)
\(\Rightarrow2.\left(x^2-2x1+1^2\right)+\left(x^2+2x3+3^2\right)=3.\left[x\left(x+1\right)-2.\left(x+1\right)\right]\)
\(\Rightarrow2x^2-4x+2+x^2+6x+9=3.\left(x.x+x-2x-2\right)\)
\(\Rightarrow2x^2-4x+2+x^2+6x+9=3.x^2+3x-6x-6\)
\(\Rightarrow2x^2-4x+2+x^2+6x+9-3x^2-3x+6x+6=0\)
\(\Rightarrow\left(2x^2+x^2-3x^2\right)+\left(-4x+6x-3x+6x\right)+\left(2+9+6\right)=0\)
\(\Rightarrow5x+17=0\)
\(\Rightarrow5x=0-17=-17\)
\(\Rightarrow x=-\frac{17}{5}\)
b) \(\left(x+2\right)^2-2\left(x-3\right)=\left(x-1\right)^2\)
\(\Rightarrow\left(x+2\right)^2-2\left(x-3\right)-\left(x-1\right)^2=0\)
\(\Rightarrow\left(x^2+2x2+2^2\right)-2x+6-\left(x^2-2x1+1^2\right)=0\)
\(\Rightarrow x^2+4x+4-2x+6-x^2+2x-1=0\)
\(\Rightarrow\left(x^2-x^2\right)+\left(4x-2x+2x\right)+\left(4+6-1\right)=0\)
\(\Rightarrow4x+9=0\)
\(\Rightarrow4x=0-9=-9\)
\(\Rightarrow x=-\frac{9}{4}\)
