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\(2.A=x\left(x^2-y\right)-x^2\left(x+y\right)+y\left(x^2-x\right)=x^3-xy-x^3-x^2y+x^2y-xy=-2xy\\ Thayx=\frac{1}{2};y=-100vàoAđược:A=-2.\frac{1}{2}.\left(-100\right)=100\)
\(3.x\left(5-2x\right)+2x\left(x-1\right)=15\Leftrightarrow5x-2x^2+2x^2-2x=15\Leftrightarrow3x=15\Leftrightarrow x=5\)
Bài 1:
a) \(\left(x-1\right)\left(x+1\right)\left(x+2\right)\)
\(=\left(x^2-1\right)\left(x+2\right)\)
\(=x^3+x-2\)
b) \(\dfrac{1}{2}x^2y^2\left(2x+y\right)\left(2x-y\right)\)
\(=\dfrac{1}{2}x^2y^2\cdot\left(4x^2-y^2\right)\)
\(=2x^4y^2-\dfrac{1}{2}x^2y^4\)
Bài 2:
a) \(2x\cdot\left(x-5\right)-x\left(2x+3\right)=26\)
\(\Rightarrow2x^2-10x-2x^2-3x=26\)
\(\Rightarrow-13x=26\)
\(\Rightarrow x=2\)
b) \(\left(3y^2-y+1\right)\cdot\left(y-1\right)+y^2\cdot\left(4-3y\right)-\dfrac{5}{2}=0\)
\(\Rightarrow3y^3-3y^2-y^2+y+y-1+4y^2-3y^3-\dfrac{5}{2}=0\)
\(\Rightarrow2y+\dfrac{7}{5}=0\)
\(\Rightarrow2y=-1,4\)
\(\Rightarrow y=-0,7\)
c) \(2x^2+3\left(x-1\right)\cdot\left(x+1\right)=5x\left(x+1\right)\)
\(\Rightarrow2x^2+3\left(x^2-1\right)=5x^2+5x\)
\(\Rightarrow2x^2+3x^2-3=5x^2+5x\)
\(\Rightarrow5x^2-5x^2-5x=3\)
\(\Rightarrow-5x=3\)
\(\Rightarrow x=0,6\)
b) \(=y^3-1+\frac{2}{3}x^3y-2xy+\frac{1}{3}x^2y^3-y^3\)
\(=\frac{2}{3}x^3y+\frac{1}{3}x^2y^3-2xy-1\)
a) \(\left(x^2-2x+3\right)\left(\dfrac{1}{2}x-5\right)=\dfrac{1}{2}x^3-5x^2-x^2+10x+\dfrac{3}{2}x-15\)
\(=\dfrac{1}{2}x^3-6x^2+\dfrac{32}{3}x-15\)
b) \(\left(x^2-2xy+y^2\right)\left(x-y\right)=\left(x-y\right)^2\left(x-y\right)=\left(x-y\right)^3\)
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a/ \(\dfrac{2}{3}a^2b\left(3ab-a^2+b\right)=6a^3b^2-\dfrac{2a^2}{3}+\dfrac{2a^2b^2}{3}\)
b/ \(\left(x+y\right)y+x\left(x-y\right)=xy+y^2+x^2-xy=x^2+y^2\)
c/ \(2x\left(x-1\right)+x\left(5-2x\right)=15\)
\(\Leftrightarrow2x^2-2x+5x-2x^2=15\)
\(\Leftrightarrow3x=15\)
\(\Leftrightarrow x=5\)
Vậy..