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a: ĐK của A là x<>-3; x<>2
ĐKXĐ của B là x<>3
DKXĐ của C là x<>0; x<>4/3
ĐKXĐ của D là x<>-2
ĐKXĐ của E là x<>2; x<>-2
ĐKXĐ của F là x<>2
b,c:
\(A=\dfrac{2\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\dfrac{2}{x-2}\)
Để A=0 thì 2=0(loại)
\(B=\dfrac{\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x-3\right)}=\dfrac{x+3}{x-3}\)
Để B=0 thì x+3=0
=>x=-3
\(C=\dfrac{\left(3x-4\right)\left(3x+4\right)}{x\left(3x-4\right)}=\dfrac{3x+4}{x}\)
Để C=0 thì 3x+4=0
=>x=-4/3
\(D=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}=\dfrac{x+2}{2}\)
Để D=0 thì x+2=0
=>x=-2(loại)
\(E=\dfrac{x\left(2-x\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{-x}{x+2}\)
Để E=0 thì x=0
\(F=\dfrac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3}{x-2}\)
Để F=0 thì 3=0(loại)
1: \(B=\left(\dfrac{4x}{x+2}-\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\dfrac{4\left(x^2-2x+4\right)}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{16}{x+2}\cdot\dfrac{\left(x+2\right)\left(x+1\right)}{x^2+x+1}\)
\(=\left(\dfrac{4x}{x+2}-\dfrac{4\left(x^2+2x+4\right)}{\left(x+2\right)^2}\right)\cdot\dfrac{x+2}{16}\cdot\dfrac{\left(x+2\right)\left(x+1\right)}{x^2+x+1}\)
\(=\dfrac{4x^2+8x-4x^2-8x-16}{\left(x+2\right)^2}\cdot\dfrac{\left(x+2\right)^2\cdot\left(x+1\right)}{16\left(x^2+x+1\right)}\)
\(=\dfrac{-\left(x+1\right)}{x^2+x+1}\)
2: Để B=0 thì -x-1=0
hay x=-1(nhận)
a: \(=\dfrac{4x^3+8x^2-11x+3-\left(x^2-5\right)\left(2x-1\right)-2x^3-5x^2+x+1}{\left(2x-1\right)^3}\)
\(=\dfrac{2x^3+3x^2-10x+4-2x^3+x^2+10x-5}{\left(2x-1\right)^3}\)
\(=\dfrac{4x^2-1}{\left(2x-1\right)^3}=\dfrac{2x+1}{\left(2x-1\right)^2}\)
b: \(=\dfrac{1+x+1-x}{1-x^2}+\dfrac{2}{1+x^2}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{2+2x^2+2-2x^2}{1-x^4}+\dfrac{4}{1+x^4}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{4+4x^4+4-4x^4}{1-x^8}+\dfrac{8}{1+x^8}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{8+8x^8+8-8x^8}{1-x^{16}}+\dfrac{16}{1+x^{16}}\)
\(=\dfrac{32}{1+x^{32}}\)
a: \(\Leftrightarrow-12x-4=8x-2-8-6x\)
=>-12x-4=2x-10
=>-14x=-6
hay x=3/7
b: \(\Leftrightarrow3\left(5x-3\right)-2\left(5x-1\right)=-4\)
=>15x-9-10x+2=-4
=>5x-7=-4
=>5x=3
hay x=3/5(loại)
c: \(\Leftrightarrow x^2-4+3x+3=3+x^2-x-2\)
\(\Leftrightarrow x^2+3x-1=x^2-x+1\)
=>4x=2
hay x=1/2(nhận)
a)
\(Q=\left(\dfrac{2x-x^2}{2x^2+8}-\dfrac{2x^2}{x^3-2x^2+4x-8}\right)\left(\dfrac{2}{x^2}+\dfrac{1-x}{x}\right)\\ =\left(\dfrac{-x^3-4x}{2\left(x^2+4\right)\left(x-2\right)}\right)\left(\dfrac{2+x-x^2}{x^2}\right)\\ =\dfrac{x\left(x-2\right)^2\left(x+2\right)\left(x+1\right)}{2x^2\left(x^2+4\right)\left(x-2\right)}\)
\(=\dfrac{\left(x^2-4\right)\left(x+1\right)}{2x\left(x^2+4\right)}\)
a, \(\dfrac{x^2-x}{x-2}+\dfrac{4-3x}{x-2}\)
\(=\dfrac{x^2-x+4-3x}{x-2}=\dfrac{x^2-4x+4}{x-2}\)
c) \(\dfrac{2}{x^2-9}+\dfrac{1}{x+3}\)
Ta có: \(\dfrac{1}{x+3}=\dfrac{1\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-3}{x^2-9}\)
\(\Rightarrow\dfrac{2}{x^2-9}+\dfrac{1}{x+3}=\dfrac{2}{x^2-9}+\dfrac{x-3}{x^2-9}=\dfrac{2+x-3}{x^2-9}=\dfrac{x-1}{x^2-9}\)
a: \(B=\left(\dfrac{4x}{x+2}-\dfrac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\cdot\dfrac{4\left(x^2-2x+4\right)}{\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{x+2}{16}\cdot\dfrac{\left(x+2\right)\left(x+1\right)}{x^2+x+1}\)
\(=\left(\dfrac{4x}{x+2}-\dfrac{4\left(x^2+2x+4\right)}{\left(x+2\right)^2}\right)\cdot\dfrac{x+2}{16}\cdot\dfrac{\left(x+2\right)\left(x+1\right)}{x^2+x+1}\)
\(=\dfrac{4x^2+8x-4x^2-8x-16}{\left(x+2\right)^2}\cdot\dfrac{\left(x+2\right)^2\cdot\left(x+1\right)}{16\left(x^2+x+1\right)}\)
\(=\dfrac{-16}{16\left(x^2+x+1\right)}\cdot\left(x+1\right)=-\dfrac{x+1}{x^2+x+1}\)
b: \(B=\dfrac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x+2}{x^2+x+1}\)
\(P=A+B=\dfrac{-x-1+x+2}{x^2+x+1}=\dfrac{1}{x^2+x+1}=\dfrac{1}{\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}}< =1:\dfrac{3}{4}=\dfrac{4}{3}\)
Dấu = xảy ra khi x=-1/2