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a)\(\left(x^2+1\right)\left(x^2-4x+4\right)=0\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x^2-4x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=-1\left(vn\right)\\\left(x-2\right)^2=0\end{cases}\Rightarrow}x=2}\)
b)\(\left(3x-2\right)\left(\frac{2x+6}{7}-\frac{4x-3}{5}\right)=0\\ \Rightarrow\left(3x-2\right)\left(\frac{10x+30-28x+21}{35}\right)=0\)
\(\Rightarrow\left(3x-2\right)\left(\frac{-18x+51}{35}\right)=0\Rightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{17}{6}\end{cases}}\)
c)\(\left(3,3-11x\right)\left(\frac{21x+6+10-30x}{15}\right)=0\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{10}\\x=\frac{16}{9}\end{cases}}\)
b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
Giải:
a) \(8\left(3x-2\right)-13x=5\left(12-3x\right)+7x\)
\(\Leftrightarrow24x-16-13x=60-15x+7x\)
\(\Leftrightarrow24x-13x+15x-7x=60+16\)
\(\Leftrightarrow19x=76\)
\(\Leftrightarrow x=\dfrac{76}{19}=4\)
Vậy ...
b) \(\dfrac{5x}{x+2}-\dfrac{3}{x-2}+\dfrac{3x^2+6}{\left(x-2\right)\left(x+2\right)}=0\) (1)
ĐKXĐ: \(x\ne\pm2\)
\(\left(1\right)\Leftrightarrow\dfrac{5x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{3x^2+6}{\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow5x\left(x-2\right)-3\left(x+2\right)+3x^2+6=0\)
\(\Leftrightarrow5x^2-10x-3x-6+3x^2+6=0\)
\(\Leftrightarrow8x^2-13x=0\)
\(\Leftrightarrow x\left(8x-13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\8x-13=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(TM\right)\\x=\dfrac{13}{8}\left(TM\right)\end{matrix}\right.\)
Vậy ...
c) \(\dfrac{x}{2\left(x-3\right)}+\dfrac{x}{2x+2}=\dfrac{2x}{\left(x+1\right)\left(x-3\right)}\) (2)
ĐKXĐ: \(x\ne-1;x\ne3\)
\(\left(2\right)\Leftrightarrow\dfrac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\dfrac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\dfrac{4x}{2\left(x+1\right)\left(x-3\right)}\)
\(\Leftrightarrow x\left(x+1\right)+x\left(x-3\right)=4x\)
\(\Leftrightarrow x\left(x+1+x-3\right)=4x\)
\(\Leftrightarrow x\left(2x-2\right)=4x\)
\(\Leftrightarrow2x-2=4\)
\(\Leftrightarrow x=3\)
Vậy ...
b: =>(2x-1)(2x-1+4-2x)=0
=>3(2x-1)=0
=>2x-1=0
=>x=1/2
c: =>(x+1)(x^2-x+1)-x(x+1)=0
=>(x+1)(x-1)^2=0
=>x=1 hoặc x=-1
e: =>(2x-1)(2x+1)=0
=>x=1/2 hoặc x=-1/2
h: =>x[(x^2-5)^2-4]=0
=>x(x^2-7)(x^2-3)=0
=>\(x\in\left\{0;\pm\sqrt{7};\pm\sqrt{3}\right\}\)
k: =>(x-1)(5x+3-3x+8)=0
=>(x-1)(2x+11)=0
=>x=1 hoặc x=-11/2
l: =>x^2(x+1)+(x+1)=0
=>(x+1)(x^2+1)=0
=>x+1=0
=>x=-1
a) \(\left(4x-10\right)\left(24+5x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{10}{4}=\dfrac{5}{2}\\x=-\dfrac{24}{5}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{24}{5};\dfrac{5}{2}\right\}\)
b) \(\left(3.5-7x\right)\left(0.1x+2.3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3.5-7x=0\\0.1x+2.3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3.5}{7}=\dfrac{1}{2}\\x=-\dfrac{2.3}{0.1}=-23\end{matrix}\right.\)
Vậy \(S=\left\{-23;\dfrac{1}{2}\right\}\)