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1:

a: \(\left(2x-5\right)^2-4x\left(x+3\right)\)

\(=4x^2-20x+25-4x^2-12x\)

=-32x+25

b: \(\left(x-2\right)^3-6\left(x+4\right)\left(x-4\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

\(=x^3-6x^2+12x-8-\left(x^3-8\right)-6\left(x^2-16\right)\)

\(=-6x^2+12x-6x^2+96=-12x^2+12x+96\)

c: \(\left(x-1\right)^2-2\left(x-1\right)\left(x+2\right)+\left(x+2\right)^2+5\left(2x-3\right)\)

\(=\left(x-1-x-2\right)^2+5\left(2x-3\right)\)

\(=\left(-3\right)^2+5\left(2x-3\right)\)

\(=9+10x-15=10x-6\)

2: 

a: \(\left(2-3x\right)^2-5x\left(x-4\right)+4\left(x-1\right)\)

\(=9x^2-12x+4-5x^2+20x+4x-4\)

\(=4x^2+12x\)

b: \(\left(3-x\right)\left(x^2+3x+9\right)+\left(x-3\right)^3\)

\(=27-x^3+x^3-9x^2+27x-27\)

\(=-9x^2+27x\)

c: \(\left(x-4\right)^2\left(x+4\right)-\left(x-4\right)\left(x+4\right)^2+3\left(x^2-16\right)\)

\(=\left(x-4\right)\left(x+4\right)\left(x-4-x-4\right)+3\left(x^2-16\right)\)

\(=\left(x^2-16\right)\left(-8\right)+3\left(x^2-16\right)\)

\(=-5\left(x^2-16\right)=-5x^2+80\)

bạn viết thế này khó nhìn quá

nhìn hơi đau mắt nhá bạn hoa mắt quá

Bài 1.

a)

\((x-2)(2x-1)-(2x-3)(x-1)-2\\=2x^2-x-4x+2-(2x^2-2x-3x+3)-2\\=2x^2-5x+2-(2x^2-5x+3)-2\\=2x^2-5x+2-2x^2+5x-3-2\\=(2x^2-2x^2)+(-5x+5x)+(2-3-2)\\=-3\)

b)

\(x(x+3y+1)-2y(x-1)-(y+x+1)x\\=x^2+3xy+x-2xy+2y-xy-x^2-x\\=(x^2-x^2)+(3xy-2xy-xy)+(x-x)+2y\\=2y\)

Bài 2.

a)

\((14x^3+12x^2-14x):2x=(x+2)(3x-4)\\\Leftrightarrow 14x^3:2x+12x^2:2x-14x:2x=3x^2-4x+6x-8\\ \Leftrightarrow 7x^2+6x-7=3x^2+2x-8\\\Leftrightarrow (7x^2-3x^2)+(6x-2x)+(-7+8)=0\\\Leftrightarrow 4x^2+4x+1=0\\\Leftrightarrow (2x)^2+2\cdot 2x\cdot 1+1^2=0\\\Leftrightarrow (2x+1)^2=0\\\Leftrightarrow 2x+1=0\\\Leftrightarrow 2x=-1\\\Leftrightarrow x=\frac{-1}2\)

b)

\((4x-5)(6x+1)-(8x+3)(3x-4)=15\\\Leftrightarrow 24x^2+4x-30x-5-(24x^2-32x+9x-12)=15\\\Leftrightarrow 24x^2-26x-5-(24x^2-23x-12)=15\\\Leftrightarrow 24x^2-26x-5-24x^2+23x+12=15\\\Leftrightarrow -3x+7=15\\\Leftrightarrow -3x=8\\\Leftrightarrow x=\frac{-8}3\\Toru\)

Bài 8:

Ta có: \(A=-x^2+2x+4\)

\(=-\left(x^2-2x-4\right)\)

\(=-\left(x^2-2x+1-5\right)\)

\(=-\left(x-1\right)^2+5\le5\forall x\)

Dấu '=' xảy ra khi x=1

Bài 2: 

a) Ta có: \(\left|x-2\right|=\left|4-x\right|\)

\(\Leftrightarrow x-2=4-x\)

\(\Leftrightarrow2x=6\)

hay x=3

b) Ta có: \(\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)+\left(-5\right)=6\)

\(\Leftrightarrow\left(\left|2x-1\right|-3\right)\cdot\left(-2\right)=11\)

\(\Leftrightarrow\left|2x-1\right|-3=\dfrac{-11}{2}\)

\(\Leftrightarrow\left|2x-1\right|=\dfrac{-11}{2}+\dfrac{6}{2}=\dfrac{-5}{2}\)(Vô lý)

\(\left(x+4\right)^2-81=0\Leftrightarrow\left(x+4\right)^2-9^2=0\)

\(\Leftrightarrow\left(x+4+9\right)\times\left(x+4-9\right)=0\)

\(\Leftrightarrow\left(x+13\right)\times\left(x-5\right)=0\)

\(\left[{}\begin{matrix}x+13=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=5\end{matrix}\right.\)