Rút gọn biểu thức:
\(2x\left(3x^3-x\right)-4x^2\left(x-x^2+1\right)+\left(x-3x^2\right)x\)
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a: \(2x\left(2x-1\right)^2-3x\left(x+3\right)\left(x-3\right)-4x\left(x+1\right)^2\)
\(=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)\)
\(=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x\)
\(=x^3-16x^2+25x\)
\(a,\left(3x+5\right)^2+\left(3x-5\right)^2-\left(3x+2\right)\left(3x-2\right)=9x^2+30x+25+9x^2-30x+25-9x^2+4=9x^2+54\)
\(b,BT=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x=x^3-16x^2+25x\)
\(c,BT=\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2=\left(x+y-z-x-y\right)^2=z^2\)
\(2x\left(2x-1\right)^2-3x\left(x+3\right)\left(x-3\right)-4x\left(x+1\right)^2\)
\(=x\left[2\left(2x-1\right)^2-3\left(x^2-9\right)-4\left(x+1\right)^2\right]\)
\(=x\left(8x^2-8x+1-3x^2+27-4x^2-8x-4\right)\)
\(=x\left(x^2-16x+28\right)=x\left(x-2\right)\left(x-14\right)\)
Rút gọn biểu thức:
\(\left(x+3\right)^2+\left(2x+1\right)\left(3x-5\right)-2x\left(3-x\right)+4x+25\)
a)
\(A=\left(x+3\right)\left(x^2-3x+9\right)-\left(54+x^3\right)\)
\(=x^3-3x^2+9x+3x^2-9x+27-54-x^3\)
\(=-27\)
or
\(A=x^3+27-54-x^3=-27\)
b)
\(B=\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3=2y^3\)
c)
\(C=\left(2x+1\right)^2+\left(1-3x\right)^2+2\left(2x+1\right)\left(3x-1\right)\)
\(=\left(2x+1+3x-1\right)^2=\left(5x\right)^2=25x^2\)
d)
\(D=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=x^3-8-\left(x-1\right)^3+3\left(x-1\right)\left(x+1\right)\)
\(=6x^2-3x-10\)
\(C=\left\lbrack\frac{1}{1+x}+\frac{2x}{1-x^2}\right\rbrack:\left(\frac{1}{x}-1\right)\)
\(=\frac{1-x+2x}{\left(1-x\right)\left.\right.\left(1+x\right)}:\frac{1-x}{x}\)
\(=\frac{1+x}{\left(1-x\right)\left(1+x\right)}\cdot\frac{x}{1-x}=\frac{x}{\left(1-x\right)^2}\)
\(D=\frac{x^2-y^2}{x+y}\cdot\frac{\left(x+y\right)^2}{x}+\frac{y^2}{x+y}\cdot\frac{\left(x+y\right)^2}{x}\)
\(=\frac{\left(x^2-y^2\right)\left(x+y\right)}{x}+\frac{y^2\left(x+y\right)}{x}=\frac{\left(x+y\right)\cdot x^2}{x}=x\left(x+y\right)\)
\(E=\frac{\left|x-3\right|}{x^2-9}\left(x^2-6x+9\right)\)
\(=\frac{\left|x-3\right|}{\left.\left(x-3\right)\left(x+3\right)\right.}\cdot\left(x-3\right)^2=\frac{\left|x-3\right|\cdot\left(x-3\right)}{x+3}\)
\(F=\frac{\sqrt{x}}{\sqrt{x}-5}-\frac{10\sqrt{x}}{x-25}-\frac{5}{\sqrt{x}+5}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+5\right)-10\sqrt{x}-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\frac{x-5\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\frac{\sqrt{x}-5}{\sqrt{x}+5}\)
a: ĐKXĐ: x<>1; x<>-1
TH1: \(x^2-1>0\)
=>\(x^2>1\)
=>x>1 hoặc x<-1
\(A=\frac{x+2}{\left|x^2-1\right|}+\frac{x^2}{x+1}\)
\(=\frac{x+2}{\left(x^2-1\right)}+\frac{x^2}{x+1}\)
\(=\frac{x+2+x^2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x^3-x^2+x+2}{\left(x-1\right)\left(x+1\right)}\)
TH2: \(x^2-1<0\)
=>-1<x<1
\(A=\frac{x+2}{\left|x^2-1\right|}+\frac{x^2}{x+1}\)
\(=\frac{-\left(x+2\right)}{\left(x-1\right)\left(x+1\right)}+\frac{x^2}{x+1}\)
\(=\frac{-\left(x+2\right)+x^2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x^3-x^2-x-2}{\left(x-1\right)\left(x+1\right)}\)
b: \(B=2x:\frac12x+\left(x+1\right)^2\)
\(=\left(2:\frac12\right)\cdot\left(\frac{x}{x}\right)+x^2+2x+1\)
\(=x^2+2x+1+4=x^2+2x+5\)
c: \(C=\left\lbrack\frac{1}{1+x}+\frac{2x}{1-x^2}\right\rbrack:\left(\frac{1}{x}-1\right)\)
\(=\frac{1-x+2x}{\left(1-x\right)\left.\right.\left(1+x\right)}:\frac{1-x}{x}\)
\(=\frac{1+x}{\left(1-x\right)\left(1+x\right)}\cdot\frac{x}{1-x}=\frac{x}{\left(1-x\right)^2}\)
d: \(D=\frac{x^2-y^2}{x+y}\cdot\frac{\left(x+y\right)^2}{x}+\frac{y^2}{x+y}\cdot\frac{\left(x+y\right)^2}{x}\)
\(=\frac{\left(x^2-y^2\right)\left(x+y\right)}{x}+\frac{y^2\left(x+y\right)}{x}=\frac{\left(x+y\right)\cdot x^2}{x}=x\left(x+y\right)\)
e: \(E=\frac{\left|x-3\right|}{x^2-9}\left(x^2-6x+9\right)\)
\(=\frac{\left|x-3\right|}{\left.\left(x-3\right)\left(x+3\right)\right.}\cdot\left(x-3\right)^2=\frac{\left|x-3\right|\cdot\left(x-3\right)}{x+3}\)
f: \(F=\frac{\sqrt{x}}{\sqrt{x}-5}-\frac{10\sqrt{x}}{x-25}-\frac{5}{\sqrt{x}+5}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+5\right)-10\sqrt{x}-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\frac{x-5\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)
\(=\frac{\sqrt{x}-5}{\sqrt{x}+5}\)
Ta có: \(2x\left(3x^3-x\right)-4x^2\left(x-x^2+1\right)+\left(x-3x^2\right)x\)
=\(6x^4-2x^2-4x^3+4x^4-4x^2+x^2-3x^3\)
=\(10x^4-7x^3-5x^2\)
2x.3x3-2x.x-4x2.x-4x2.(-x2)-4x2.1+x.x-3x2.x
6x4-2x2-4x3+4x4-4x2+x2-3x3
(6x4+4x4)+(-4x3-3x3)+(-2x2-4x2+x2)
10x4-7x3-5x2
-
2x(3x3-x)-4x2(x-x2+1)+(x-3x2)x
=6x4-2x2-4x3+4x4-4x2+x2-3x3
=10x4-7x3-5x2