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a: \(=x^3-3x^2+3x-1-x^3-64+3x^2-3x\)
=-65
b \(=8x^3+27y^3-8x^3+27y^3-54y^3+27\)
=27
c: \(=y\left(x^4-y^4\right)-y\left(x^4-y^4\right)=0\)
d: \(=x^3-3x^2+3x-1-x^3+1-3x\left(1-x\right)\)
\(=-3x^2+3x-3x+3x^2=0\)
a: \(-3x^2\cdot\left(\dfrac{4}{3}x^2+\dfrac{2}{3}x^2-\dfrac{1}{3}\right)\)
\(=-4x^4-2x^4+x^2\)
b: \(\left(x-3y\right)\left(3x^2+5xy+4y^2\right)\)
\(=3x^3+5x^2y+4xy^2-9x^2y-15xy^2-12y^3\)
\(=3x^3-4x^2y-11xy^2-12y^3\)
c: \(\left(x+8\right)^2-2\left(x+8\right)\left(x-2\right)+\left(x-2\right)^2\)
\(=\left(x+8-x+2\right)^2\)
\(=10^2=100\)
d: \(x\left(x-4\right)\left(x+4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x\left(x^2-16\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x^3-16x-x^4+1\)
a: \(=\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}:\left(\dfrac{1}{x+1}+\dfrac{x}{x-1}+\dfrac{2}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{x-1+x^2+x+2}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{x^2+2x+1}=\dfrac{4x}{x^2+2x+1}\)
b: \(=\dfrac{x+2}{-\left(x-2\right)}\cdot\dfrac{\left(x-2\right)^2}{4x^2}\cdot\left(\dfrac{2}{2-x}-\dfrac{4}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\dfrac{x^2-2x+4}{2-x}\right)\)
\(=\dfrac{-\left(x+2\right)\left(x-2\right)}{4x^2}\cdot\left(\dfrac{2}{2-x}-\dfrac{4}{\left(x+2\right)\left(2-x\right)}\right)\)
\(=\dfrac{-\left(x+2\right)\left(x-2\right)}{4x^2}\cdot\dfrac{2x+4-4}{\left(2-x\right)\left(x+2\right)}\)
\(=\dfrac{2x}{4x^2}=\dfrac{1}{2x}\)
\(A=\left(x-4\right)^2-\left(x+4\right)^2-16\left(x-2\right)\)
\(=x^2-8x+16-x^2-8x-16-16x+32\)
\(=-32x+32\)
Biểu thức phụ thuộc vào giá trị của biến
c) \(\left(3x+5\right)^2-2\left(2x+3\right)\left(3x+5\right)+\left(2x+3\right)^2=\left(x+2\right)^3\)
\(\Leftrightarrow\left[\left(3x+5\right)-\left(2x+3\right)\right]^2=\left(x+2\right)^3\)
\(\Leftrightarrow\left(3x+5-2x-3\right)^2=\left(x+2\right)^3\)
\(\Leftrightarrow\left(x+2\right)^2=\left(x+2\right)^3\)
\(\Leftrightarrow\left(x+2\right)^3-\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)^2.\left(x+2-1\right)=0\)
\(\Leftrightarrow\left(x+2\right)^2.\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-1\end{cases}}\)
Vậy tập nghiệm của phương trình là: \(S=\left\{-2;-1\right\}\)
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) \(\left(x+2\right)^2-2\left(x-3\right)=\left(x-1\right)^2\)
\(\Rightarrow\left(x+2\right)^2-2\left(x-3\right)-\left(x-1\right)^2=0\)
\(\Rightarrow\left(x^2+2x2+2^2\right)-2x+6-\left(x^2-2x1+1^2\right)=0\)
\(\Rightarrow x^2+4x+4-2x+6-x^2+2x-1=0\)
\(\Rightarrow\left(x^2-x^2\right)+\left(4x-2x+2x\right)+\left(4+6-1\right)=0\)
\(\Rightarrow4x+9=0\)
\(\Rightarrow4x=0-9=-9\)
\(\Rightarrow x=-\frac{9}{4}\)
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: \(x\in R\)