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a) \(\left(4x-1\right)^2-\left(3x+2\right)\left(3x-2\right)=\left(7x-1\right)\left(x+2\right)+\left(2x+1\right)^2-\left(4x^2+7\right)\)(1)
\(\Leftrightarrow\left(16x^2-8x+1\right)-\left(9x^2-4\right)=\left(7x^2+14x-x-2\right)+\left(4x^2+4x+1\right)-\left(4x^2+7\right)\)
\(\Leftrightarrow16x^2-8x+1-9x^2+4=7x^2+13x-2+4x^2+4x+1-4x^2-7\)
\(\Leftrightarrow7x^2-8x+5=7x^2+17x-8\)
\(\Leftrightarrow7x^2-8x-7x^2-17x=-8-5\)
\(\Leftrightarrow-25x=-13\)
\(\Leftrightarrow x=\dfrac{13}{25}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{\dfrac{13}{25}\right\}\)
\(A=\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3-2\right)\)
\(\Rightarrow A=\left(x^3+8\right)-\left(x^3-2\right)\)
\(\Rightarrow A=x^3+8-x^3+2\)
\(\Rightarrow A=\left(x^3-x^3\right)+\left(8+2\right)\)
\(\Rightarrow A=10\)
\(A=\left(x+2\right)\left(x^2-2x+4\right)-\left(x^3-2\right)\)
\(=x^3+8-x^3+2\)
\(=10\)
\(B=\left(x+2\right)\left(x-2\right)\left(x^2+2x+4\right)\left(x^2-2x+4\right)\)
\(=\left(x+2\right)\left(x^2-2x+4\right)\left(x-2\right)\left(x^2+2x+4\right)\)
\(=\left(x^3+8\right)\left(x^3-8\right)\)
\(=x^6-64\)
\(C=\left(x^2+3x+1\right)^2+\left(3x-1\right)^2-2\left(x^2+3x+1\right)\left(3x-1\right)\)
\(=\left(x^2+3x+1\right)^2-2\left(x^2+3x+1\right)\left(3x-1\right)+\left(3x-1\right)^2\)
\(=\left(x^2+3x+1-3x+1\right)^2\)
\(=\left(x^2+2\right)^2\)
\(D=\left(3x^3+3x+1\right)\left(3x^3-3x+1\right)-\left(3x^3+1\right)^2\)
\(=\left(3x^3+1+3x\right)\left(3x^3+1-3x\right)-\left(3x^3+1\right)^2\)
\(=\left(3x^3+1\right)^2-9x^2-\left(3x^3+1\right)^2\)
\(=-9x^2\)
\(E=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)-\left(2x^2+1\right)^2\)
\(=\left(2x^2+1+2x\right)\left(2x^2+1-2x\right)-\left(2x^2+1\right)^2\)
\(=\left(2x^2+1\right)^2-4x^2-\left(2x^2+1\right)^2\)
\(=-4x^2\)
1,\(2x\left(x-5\right)-\left(x-2\right)^2-\left(x+3\right)\left(x-3\right)\)
\(=2x^2-10x-x^2+4x-4-x^2+9\)
\(=\left(2x^2-x^2-x^2\right)+\left(-10x+4x\right)+\left(-4+9\right)\)
\(=-6x+5\)
2,\(\left(x+1\right)^2-3\left(x-5\right)\left(x+5\right)-\left(2x-1\right)^2\)
\(=x^2+2x+1-3\left(x^2-25\right)-\left(4x^2-4x+1\right)\)
\(=x^2+2x+1-3x^2+75-4x^2+4x-1\)
\(=-6x^2+6x+75\)
3,\(\left(x-1\right)^3-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=\left(x-1\right)^3-\left(x^3-27\right)\)
\(=x^3-3x^2+3x-1-x^3+27\)
\(=-3x^2+3x+26\)
4,\(\left(x+5\right)\left(x^2-5x+25\right)-\left(x+2\right)^3\)
\(=\left(x^3+125\right)-\left(x^3+6x^2+12x+8\right)\)
\(=x^3+125-x^3-6x^2-12x-8\)
\(=-6x^2-12x+117\)
5,\(2x\left(x-7\right)-\left(x+3\right)\left(x-2\right)^2+\left(x+1\right)^2\)
\(=2x^2-14x-\left(x+3\right)\left(x^2-4x+4\right)+x^2+2x+1\)
=\(2x^2-14x-x^3+4x^2-4x-3x^2+12x-12+x^2+2x+1\)
\(=-x^3+4x^2-4x+1\)
6,\(\left(2x+5\right)\left(x-3\right)-\left(x+5\right)\left(x-1\right)-\left(x-4\right)^2\)
\(=2x^2-6x+5x-15-x^2+x-5x+5-x^2+8x-16\)
\(=3x-26\)
7,\(\left(x+5\right)\left(x-5\right)\left(x+2\right)-\left(x+2\right)^3\)
=\(\left(x^2-25\right)\left(x+2\right)-x^3-6x^2-12x-8\)
\(=x^3+2x^2-25x-50-x^3-6x^2-12x-8\)
\(=-4x^2-27x-58\)
Nếu đúng thì tick cho mk nha ^_^
Bài 2: a) \(3x^3-3x=0\Leftrightarrow3x\left(x^2-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)
b) \(x^2-x+\frac{1}{4}=0\Leftrightarrow x^2-2.\frac{1}{2}+\left(\frac{1}{2}\right)^2=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
23x+2=4x+5
23x+2=22.(x+5)
23x+2=22x+10
=>3x+2=2x+10
=>3x-2x=-2+10
=>(3-2)x=8
=>x=8
Chúc bn học tốt !!!