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a) \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)
\(\Leftrightarrow\left(x^2+6x+9\right)-\left(x^2+4x-32\right)-1=0\)
\(\Leftrightarrow2x=-40\)
\(\Rightarrow x=-20\)
b) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)
\(\Leftrightarrow x^3+27-x^3+4x=15\)
\(\Leftrightarrow4x=-12\)
\(\Rightarrow x=-3\)
c) \(\left(x-2\right)^2-\left(x+3\right)^2-4\left(x+1\right)=5\)
\(\Leftrightarrow\left(x^2-4x+4\right)-\left(x^2+6x+9\right)-\left(4x+4\right)=5\)
\(\Leftrightarrow-14x=14\)
\(\Rightarrow x=-1\)
d) \(\left(2x-3\right)\left(2x+3\right)-\left(x-1\right)^2-3x\left(x-5\right)=-44\)
\(\Leftrightarrow4x^2-9-\left(x^2-2x+1\right)-\left(3x^2-15x\right)=-44\)
\(\Leftrightarrow17x=-34\)
\(\Rightarrow x=-2\)
e) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)
\(\Leftrightarrow24x=24\)
\(\Rightarrow x=1\)
1/ \(1+\frac{2}{x-1}+\frac{1}{x+3}=\frac{x^2+2x-7}{x^2+2x-3}\)
ĐKXĐ: \(\hept{\begin{cases}x-1\ne0\\x+3\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne-3\end{cases}}\)
<=> \(1+\frac{2\left(x+3\right)+x-1}{\left(x-1\right)\left(x+3\right)}=\frac{x^2+2x-3-5}{x^2+2x-3}\)
<=> \(1+\frac{2x+6+x-1}{x^2+2x-3}=1-\frac{5}{x^2+2x-3}\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=1-1\)
<=> \(\frac{3x+5}{x^2+2x-3}+\frac{5}{x^2+2x-3}=0\)
<=> \(\frac{3x+10}{x^2+2x-3}=0\)
<=> \(3x+10=0\)
<=> \(x=-\frac{10}{3}\)
a: ĐKXĐ: x∉{3;-1}
\(\frac{2}{x+1}-\frac{1}{x-3}=\frac{3x-11}{x^2-2x-3}\)
=>\(\frac{2}{x+1}-\frac{1}{x-3}=\frac{3x-11}{\left(x-3\right)\left(x+1\right)}\)
=>\(\frac{2\left(x-3\right)-x-1}{\left(x-3\right)\left(x+1\right)}=\frac{3x-11}{\left(x-3\right)\left(x+1\right)}\)
=>3x-11=2(x-3)-x-1
=>3x-11=2x-6-x-1=x-7
=>3x-x=-7+11
=>2x=4
=>x=2(nhận)
b: ĐKXĐ: x<>0; x<>2
\(\frac{3}{x-2}+\frac{1}{x}=\frac{-2}{x\left(x-2\right)}\)
=>\(\frac{3x+x-2}{x\left(x-2\right)}=\frac{-2}{x\left(x-2\right)}\)
=>\(\frac{4x-2}{x\left(x-2\right)}=\frac{-2}{x\left(x-2\right)}\)
=>4x-2=-2
=>4x=0
=>x=0(loại)
c: ĐKXĐ: x<>3; x<>-3
\(\frac{x-3}{x+3}-\frac{2}{x-3}=\frac{3x+1}{9-x^2}\)
=>\(\frac{\left(x-3\right)^2-2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{-3x-1}{\left(x-3\right)\left(x+3\right)}\)
=>\(\left(x-3\right)^2-2\left(x+3\right)=-3x-1\)
=>\(x^2-6x+9-2x-6+3x+1=0\)
=>\(x^2-5x+4=0\)
=>(x-1)(x-4)=0
=>x=1(nhận) hoặc x=4(nhận)
d: ĐKXĐ: x<>2; x<>-1
\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-5}{x^2-x-2}\)
=>\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-5}{\left(x-2\right)\left(x+1\right)}\)
=>\(\frac{2\left(x-2\right)-x-1}{\left(x-2\right)\left(x+1\right)}=\frac{3x-5}{\left(x-2\right)\left(x+1\right)}\)
=>3x-5=2x-4-x-1=x-5
=>2x=0
=>x=0(nhận)
e: ĐKXĐ: x<>2; x<>-2
\(\frac{x-2}{x+2}+\frac{3}{x-2}=\frac{x^2-11}{x^2-4}\)
=>\(\frac{\left(x-2\right)^2+3\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}=\frac{x^2-11}{\left(x-2\right)\left(x+2\right)}\)
=>\(\left(x-2\right)^2+3\left(x+2\right)=x^2-11\)
=>\(x^2-4x+4+3x+6=x^2-11\)
=>-x+10=-11
=>-x=-21
=>x=21(nhận)
f: ĐKXĐ: x<>-1;x<>0
\(\frac{x+3}{x+1}+\frac{x-2}{x}=2\)
=>\(\frac{x\left(x+3\right)+\left(x-2\right)\left(x+1\right)}{x\left(x+1\right)}=2\)
=>2x(x+1)=x(x+3)+(x-2)(x+1)
=>\(2x^2+2x=x^2+3x+x^2-x-2=2x^2+2x-2\)
=>0=-2(vô lý)
=>Phương trình vô nghiệm
g: ĐKXĐ: x<>5; x<>-5
\(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\)
=>\(\frac{\left(x+5\right)^2-\left(x-5\right)^2}{\left(x+5\right)\left(x-5\right)}=\frac{20}{\left(x-5\right)\left(x+5\right)}\)
=>\(\left(x+5\right)^2-\left(x-5\right)^2=20\)
=>\(x^2+10x+25-x^2+10x-25=20\)
=>20x=20
=>x=1
h: ĐKXĐ: x<>1; x<>-1
\(\frac{x+4}{x+1}+\frac{x}{x-1}=\frac{2x^2}{x^2-1}\)
=>\(\frac{\left(x+4\right)\left(x-1\right)+x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{2x^2}{\left(x-1\right)\left(x+1\right)}\)
=>\(\left(x+4\right)\left(x-1\right)+x\left(x+1\right)=2x^2\)
=>\(x^2+3x-4+x^2+x=2x^2\)
=>4x-4=0
=>4x=4
=>x=1(loại)
i: ĐKXĐ: x<>1; x<>-1
\(\frac{x+1}{x-1}-\frac{1}{x+1}=\frac{x^2+2}{x^2-1}\)
=>\(\frac{\left(x+1\right)^2-\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{x^2+2}{\left(x-1\right)\left(x+1\right)}\)
=>\(\left(x+1\right)^2-\left(x-1\right)=x^2+2\)
=>\(x^2+2x+1-x+1=x^2+2\)
=>x+2=2
=>x=0(nhận)