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1:
a: =>(|x|+4)(|x|-1)=0
=>|x|-1=0
=>x=1; x=-1
b: =>x^2-4>=0
=>x>=2 hoặc x<=-2
d: =>|2x+5|=2x-5
=>x>=5/2 và (2x+5-2x+5)(2x+5+2x-5)=0
=>x=0(loại)
b: \(=\dfrac{4x\left(x-1\right)\left(x+1\right)}{6x\left(x-1\right)}=\dfrac{2\left(x+1\right)}{3}\)
c: \(=\dfrac{\left(5-x-1\right)\left(5+x+1\right)}{\left(x+6\right)^2}=\dfrac{\left(4-x\right)\left(x+6\right)}{\left(x+6\right)^2}=\dfrac{4-x}{x+6}\)
d: \(=\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\dfrac{x+3}{x+2}\)
\(\dfrac{2x}{x-1}+\dfrac{4}{x^2+2x-3}=\dfrac{2x-5}{x+3}\)
\(\Leftrightarrow\dfrac{2x}{x-1}+\dfrac{4}{\left(x-1\right)\left(x+3\right)}=\dfrac{2x-5}{x+3}\)
\(ĐK:x\ne1;-3\)
\(\Leftrightarrow\dfrac{2x\left(x+3\right)+4}{\left(x-1\right)\left(x+3\right)}=\dfrac{\left(2x-5\right)\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}\)
\(\Leftrightarrow2x\left(x+3\right)+4=\left(2x-5\right)\left(x-1\right)\)
\(\Leftrightarrow2x^2+6x+4=2x^2-2x-5x+5\)
\(\Leftrightarrow13x=1\)
\(\Leftrightarrow x=\dfrac{1}{13}\left(tm\right)\)
f: Ta có: \(16x^2-9\left(x+1\right)^2=0\)
\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\)
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a) ĐKXĐ: x≠-5
Ta có: \(\dfrac{2x-5}{x+5}=4\)
\(\Leftrightarrow2x-5=4\left(x+5\right)\)
\(\Leftrightarrow2x-5=4x+20\)
\(\Leftrightarrow2x-5-4x-20=0\)
\(\Leftrightarrow-2x-25=0\)
\(\Leftrightarrow-2x=25\)
hay \(x=\dfrac{-25}{2}\)(nhận)
Vậy: \(S=\left\{-\dfrac{25}{2}\right\}\)
b) ĐKXĐ: x≠0
Ta có: \(\dfrac{x^2-4}{x}=\dfrac{2x+3}{2}\)
\(\Leftrightarrow2\left(x^2-4\right)=x\left(2x+3\right)\)
\(\Leftrightarrow2x^2-8=2x^2+3x\)
\(\Leftrightarrow2x^2-8-2x^2-3x=0\)
\(\Leftrightarrow-3x-8=0\)
\(\Leftrightarrow-3x=8\)
hay \(x=\dfrac{-8}{3}\)(nhận)
Vậy: \(S=\left\{-\dfrac{8}{3}\right\}\)
c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{2};-5\right\}\)
Ta có: \(\dfrac{2x+3}{2x-1}=\dfrac{x-3}{x+5}\)
\(\Leftrightarrow\left(2x+3\right)\left(x+5\right)=\left(2x-1\right)\left(x-3\right)\)
\(\Leftrightarrow2x^2+10x+3x+15=2x^2-6x-x+3\)
\(\Leftrightarrow2x^2+13x+15=2x^2-7x+3\)
\(\Leftrightarrow2x^2+13x+15-2x^2+7x-3=0\)
\(\Leftrightarrow20x+12=0\)
\(\Leftrightarrow20x=-12\)
hay \(x=-\dfrac{3}{5}\)(nhận)
Vậy: \(S=\left\{-\dfrac{3}{5}\right\}\)
d) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)
Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)
\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(x+7\right)\left(6x+1\right)\)
\(\Leftrightarrow6x^2-9x-4x+6=6x^2+x+42x+7\)
\(\Leftrightarrow6x^2-13x+6=6x^2+43x+7\)
\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)
\(\Leftrightarrow-56x-1=0\)
\(\Leftrightarrow-56x=1\)
hay \(x=-\dfrac{1}{56}\)(nhận)
Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)
\(a,\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)
\(\Leftrightarrow x^2-4x+4-\left(x^2-9\right)=6\)
\(\Leftrightarrow-4x+13=6\)
\(\Leftrightarrow-4x=-7\)
\(\Leftrightarrow x=\dfrac{7}{4}\)
\(b,\left(x+3\right)^2+\left(4+x\right)\left(4-x\right)=10\)
\(\Leftrightarrow x^2+6x+9+16-x^2=10\)
\(\Leftrightarrow6x+25=10\)
\(\Leftrightarrow6x=-15\)
\(\Leftrightarrow x=-\dfrac{5}{2}\)
\(c,\left(x+4\right)^2+\left(1-x\right)\left(1+x\right)=7\)
\(\Leftrightarrow x^2+8x+16+1-x^2=7\)
\(\Leftrightarrow8x+17=7\)
\(\Leftrightarrow8x=-10\)
\(\Leftrightarrow x=-\dfrac{5}{4}\)
\(d,\left(x-4\right)^2-\left(x-2\right)\left(x+2\right)=6\)
\(\Leftrightarrow x^2-8x+16-\left(x^2-4\right)=6\)
\(\Leftrightarrow-8x+20=6\)
\(\Leftrightarrow-8x=-14\)
\(\Leftrightarrow x=\dfrac{7}{4}\)
#\(Urushi\)