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\(a.\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)
\(\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-4x-4=5\)
\(\left(-4x-6x\right)+\left(4-9\right)-4x-4=5\)
\(-10x-5-4x-4=5\)
\(-14x-9=5\)
\(-14x=14\Rightarrow x=-1\)
\(b.\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)
\(4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)
\(4x^2-9-x^2+2x-1-3x^2+15x=-44\)
\(17x-10=-44\)
\(17x=-34\Rightarrow x=-2\)
\(c.\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
\(25x^2+10x+1-\left(25x^2-9\right)=30\)
\(10x+10=30\)
\(10x=20\Rightarrow x=2\)
\(d.\left(x+3\right)^2+\left(x-2\right)\left(x+2\right)-2\left(x-1\right)^2=7\)
\(\left(x^2+6x+9\right)+\left(x^2-4\right)-2\left(x^2-2x+1\right)=7\)
\(2x^2+6x+5-2x^2+4x-2=7\)
\(10x+3=7\)
\(10x=4\Rightarrow x=\frac{4}{10}=\frac25\)
\(f.\left(3x-8\right)^2=0\)
\(3x-8=0\Rightarrow x=\frac83\)
\(e.6\left(x+1\right)^2-2\left(x+1\right)+2\left(x-1\right)\left(x^2+x+1\right)=0\)
\(6\left(x^2+2x+1\right)-2x-2+2\left(x^3-1\right)=0\)
\(6x^2+12x+6-2x-2+2x^3-2=0\)
\(2x^3+6x^2+10x+2=0\)
\(\Rightarrow x\approx-0,23\)
(8 - 5x)(x + 2) + 4(x - 2)(x + 1) + 2(x - 2)(x + 2) = 0
=> 8(x + 2) - 5x(x + 2) + 4[x(x + 1) - 2(x + 1)] + 2(x2 - 4) = 0
=> 8x + 16 - 5x2 - 10x + 4(x2 + x - 2x - 2) + 2x2 - 8 = 0
=> 8x + 16 - 5x2 - 10x + 4x2 + 4x - 8x - 8 + 2x2 - 8 = 0
=> (8x - 10x + 4x - 8x) + (16 - 8 - 8) + (-5x2 + 4x2 + 2x2) = 0
=> 0 + x2 = 0
=> x2 = 0 => x = 0
\(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\)
\(-5x^2-2x+16+4\left(x^2-x-2\right)+2\left(x^2-4\right)=0\)
\(-5x^2-2x+16+4x^2-4x-8+2x^2-8=0\)
\(x^2-6x=0\)
\(x\left(x-6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=6\end{cases}}\)
Câu a:
(x + 2)(x -3) - (x -2)(x + 5) = 0
x^2 + 2x - 3x - 6 - x^2 - 5x + 2x + 10 = 0
(x^2 - x^2) + (2x - 3x - 5x + 2x) + (10 - 6) = 0
0 + (-x - 5x + 2x) + 4 = 0
-6x + 2x + 4 = 0
-4x + 4 = 0
4x = 4
x = 4 : 4
x = 1
Vậy x = 1
Câu b:
(2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)
2x^2 - 8x + 3x - 12+ x^2 - 2x - 5x + 10 = 3x^2 - 12x - 5x + 20
(2x^2 + x^2) - (8x+5x-3x +2x) - (12 - 10) = 3x^2 - (12x + 5x) + 20
3x^2 - (13x - 3x + 2x) - 2 = 3x^2 - 17x + 20
3x^2 - (10x + 2x) - 2 = 3x^2 - 17x + 20
3x^2 - 12x - 2 = 3x^2 - 17x + 20
3x^2 - 12x - 2 - 3x^2 + 17x - 20 = 0
(3x^2 - 3x^2) + (-12x + 17x) - (2 + 20) = 0
0 + 5x - 22 = 0
5x = 22
x = 22/5
Vậy x = 22/5