Bài 1:
1: Ta có: \(\left(x-3\right)^2-4x\left(5-x\right)\)
\(=x^2-6x+9-20x+4x^2\)
\(=5x^2-26x+9\)
2: Ta có: \(\left(2x-1\right)\left(2x+1\right)-\left(x+1\right)\left(x-3\right)\)
\(=4x^2-1-x^2+3x-x+3\)
\(=3x^2+2x+2\)
3: Ta có: \(\left(-2x\right)\left(x-3x^2+4\right)-\left(x+2\right)^2\)
\(=-2x^2+6x^3-8x-x^2-4x-4\)
\(=6x^3-3x^2-12x-4\)
4: Ta có: \(\left(3x-2\right)^2-\left(2x+5\right)\left(5-2x\right)\)
\(=\left(3x-2\right)^2+\left(2x+5\right)\left(2x-5\right)\)
\(=9x^2-12x+4+4x^2-25\)
\(=13x^2-12x-21\)
Bài 2:
1: Ta có: \(\left(x+1\right)^2-\left(x+2\right)\left(x-3\right)=4\)
\(\Leftrightarrow x^2+2x+1-x^2+3x-2x+6=4\)
\(\Leftrightarrow3x=-3\)
hay x=-1
2: Ta có: \(\left(x+2\right)\left(x-2\right)-x\left(x-7\right)=3x+1\)
\(\Leftrightarrow x^2-4-x^2+7x-3x=1\)
\(\Leftrightarrow4x=5\)
hay \(x=\dfrac{5}{4}\)
3: Ta có: \(\left(4-x\right)^2-\left(x^2+7x-3\right)=2x-5\)
\(\Leftrightarrow x^2-8x+16-x^2-7x+3-2x=-5\)
\(\Leftrightarrow-17x=-24\)
hay \(x=\dfrac{24}{17}\)
4: Ta có: \(2x\left(2x-1\right)-\left(2x-3\right)^2=0\)
\(\Leftrightarrow4x^2-2x-4x^2+12x-9=0\)
\(\Leftrightarrow10x=9\)
hay \(x=\dfrac{9}{10}\)
