\(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{7}{6x+30}\)
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5 tháng 4 2020
ĐKXĐ: x∉{-5;5}
Ta có: \(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{7}{6x+30}\)
\(\Leftrightarrow\frac{3}{4\left(x-5\right)}-\frac{15}{2\left(x-5\right)\left(x+5\right)}-\frac{7}{6\left(x+5\right)}=0\)
\(\Leftrightarrow\frac{9\left(x+5\right)}{12\left(x-5\right)\left(x+5\right)}-\frac{90}{12\left(x-5\right)\left(x+5\right)}-\frac{14\left(x-5\right)}{12\left(x+5\right)\left(x-5\right)}=0\)
\(\Leftrightarrow9\left(x+5\right)-90-14\left(x-5\right)=0\)
\(\Leftrightarrow9x+45-90-14x+70=0\)
\(\Leftrightarrow-5x+25=0\)
\(\Leftrightarrow-5x=-25\)
hay x=5(ktm)
Vậy: x∈∅
28 tháng 3 2020
a) ĐKXĐ: x khác +2
\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)
<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)
<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22
<=> x^2 - 7x - 2 = 2x - 22
<=> x^2 - 7x - 2 - 2x + 22 = 0
<=> x^2 - 9x + 20 = 0
<=> (x - 4)(x - 5) = 0
<=> x - 4 = 0 hoặc x - 5 = 0
<=> x = 4 hoặc x = 5
làm nốt đi
26 tháng 3 2020
1) Ta có: \(5\left(x-2\right)=3x+10\)
\(\Leftrightarrow5x-10-3x-10=0\)
\(\Leftrightarrow2x-20=0\)
\(\Leftrightarrow2\left(x-10\right)=0\)
Vì 2>0
nên x-10=0
hay x=10
Vậy: x=10
2) Ta có: \(x^2\left(x-5\right)-4x+20=0\)
\(\Leftrightarrow x^2\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\\x=-2\end{matrix}\right.\)
Vậy: x∈{-2;2;5}
3) Ta có: \(\frac{3x+1}{4}+\frac{8x-21}{20}=\frac{3\left(x+2\right)}{5}-2\)
\(\Leftrightarrow\frac{5\left(3x+1\right)}{20}+\frac{8x-21}{20}-\frac{12\left(x+2\right)}{20}+\frac{40}{20}=0\)
\(\Leftrightarrow15x+5+8x-21-12\left(x+2\right)+40=0\)
\(\Leftrightarrow15x+5-8x-21-12x-24+40=0\)
\(\Leftrightarrow-5x=0\)
hay x=0
Vậy: x=0
4) ĐKXĐ: x≠5; x≠-5
Ta có: \(\frac{3}{4x-20}+\frac{7}{6x+30}=\frac{15}{2x^2-50}\)
\(\Leftrightarrow\frac{3}{4\left(x-5\right)}+\frac{7}{6\left(x+5\right)}-\frac{15}{2\left(x-5\right)\left(x+5\right)}=0\)
\(\Leftrightarrow\frac{9\left(x+5\right)}{12\left(x-5\right)\left(x+5\right)}+\frac{14\left(x-5\right)}{12\left(x+5\right)\left(x-5\right)}-\frac{180}{12\left(x-5\right)\left(x+5\right)}=0\)
\(\Leftrightarrow9x+45+14x-70-180=0\)
\(\Leftrightarrow23x-205=0\)
\(\Leftrightarrow23x=205\)
hay \(x=\frac{205}{23}\)(tm)
Vậy: \(x=\frac{205}{23}\)
\(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{7}{6x+30}\)đkxđ \(x\ne\pm5\)
\(\Leftrightarrow45+9x-90-14x+70=0\)
\(\Leftrightarrow25-5x=0\)
\(\Leftrightarrow-5\left(x-5\right)=0\)
\(\Leftrightarrow x-5=0\)
\(\Leftrightarrow x=5\left(\inđkxđ\right)\)
\(\Leftrightarrow x\in\varnothing\)
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