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a) ĐKXĐ: \(x\ne-1;x\ne2\)
Ta có: \(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
⇔\(\frac{1}{x+1}-\frac{5}{x-2}+\frac{15}{\left(x+1\right)\left(x-2\right)}=0\)
⇔\(\frac{x-2}{\left(x+1\right)\left(x-2\right)}-\frac{5\left(x+1\right)}{\left(x-2\right)\left(x+1\right)}+\frac{15}{\left(x+1\right)\left(x-2\right)}=0\)
⇔\(x-2-5x-5+15=0\)
⇔\(-4x+8=0\)
⇔\(-4x=-8\)
⇔\(x=\frac{-8}{-4}=2\)(loại)
Vậy: x không có giá trị
b) ĐKXĐ: \(x\ne0;x\ne\frac{3}{2}\)
Ta có: \(\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\)
⇔\(\frac{x}{\left(2x-3\right)\cdot x}-\frac{3}{x\left(2x-3\right)}-\frac{5\left(2x-3\right)}{x\left(2x-3\right)}=0\)
⇔\(x-3-10x+15=0\)
⇔\(-9x+12=0\)
⇔\(-9x=-12\)
⇔\(x=\frac{-12}{-9}=\frac{4}{3}\)
Vậy: \(x=\frac{4}{3}\)
c) ĐKXĐ:\(x\ne3;x\ne1\)
Ta có: \(\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2x-6}\)
⇔\(\frac{6}{x-1}-\frac{4}{x-3}=\frac{8}{2\left(x-3\right)}\)
⇔\(\frac{6}{x-1}-\frac{4}{x-3}=\frac{4}{x-3}\)
⇔\(\frac{6}{x-1}-\frac{4}{x-3}-\frac{4}{x-3}=0\)
⇔\(\frac{6}{x-1}-\frac{8}{x-3}=0\)
⇔\(\frac{6\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}-\frac{8\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}=0\)
⇔\(6\left(x-3\right)-8\left(x-1\right)=0\)
⇔6x-18-8x+8=0
⇔-2x-10=0
⇔-2(x+5)=0
Vì 2≠0 nên x+5=0
hay x=-5
Vậy: x=-5
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
a)
ĐKXĐ: \(x\neq 0; x\neq -10\)
\(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\)
\(\Leftrightarrow \frac{x+10+x}{x(x+10)}=\frac{1}{12}\)
\(\Leftrightarrow \frac{2x+10}{x(x+10)}=\frac{1}{12}\)
\(\Rightarrow 12(2x+10)=x(x+10)\)
\(\Leftrightarrow x^2-14x-120=0\)
\(\Leftrightarrow (x+6)(x-20)=0\Rightarrow \left[\begin{matrix} x=-6\\ x=20\end{matrix}\right.\) (đều thỏa mãn)
b)
ĐKXĐ: \(x\neq 0; x\neq 3\)
PT\(\Leftrightarrow \frac{(x+3).x-(x-3)}{x(x-3)}=\frac{3}{x(x-3)}\)
\(\Leftrightarrow \frac{x^2+2x+3}{x(x-3)}=\frac{3}{x(x-3)}\)
\(\Rightarrow x^2+2x+3=3\)
\(\Leftrightarrow x^2+2x=0\Leftrightarrow x(x+2)=0\Rightarrow \left[\begin{matrix} x=0\\ x=-2\end{matrix}\right.\) . Kết hợp với đkxđ suy ra $x=-2$
c)
ĐKXĐ: \(x\neq \pm 2\)
\(\frac{3}{x+2}-\frac{2}{x-2}+\frac{8}{x^2-4}=0\)
\(\Leftrightarrow \frac{3(x-2)-2(x+2)}{(x+2)(x-2)}+\frac{8}{x^2-4}=0\)
\(\Leftrightarrow \frac{x-10}{x^2-4}+\frac{8}{x^2-4}=0\)
\(\Leftrightarrow \frac{x-2}{x^2-4}=0\Leftrightarrow \frac{1}{x+2}=0\) (vô lý)
Vậy pt vô nghiệm.
d)
ĐKXĐ: \(x\neq -2; x\neq 3\)
PT \(\Leftrightarrow \frac{3(x-3)-2(x+2)}{(x+2)(x-3)}=\frac{8}{(x-3)(x+2)}\)
\(\Leftrightarrow \frac{x-13}{(x+2)(x-3)}=\frac{8}{(x-3)(x+2)}\)
\(\Rightarrow x-13=8\Rightarrow x=21\) (thỏa mãn)
Vậy..........
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
a) ĐKXĐ: x≠-5
Ta có: \(\frac{2x-5}{x+5}=3\)
\(\Leftrightarrow\frac{2x-5}{x+5}-3=0\)
\(\Leftrightarrow\frac{2x-5}{x+5}-\frac{3\left(x+5\right)}{x+5}=0\)
\(\Leftrightarrow2x-5-3\left(x+5\right)=0\)
\(\Leftrightarrow2x-5-3x-15=0\)
\(\Leftrightarrow-x-20=0\)
\(\Leftrightarrow-x=20\)
\(\Leftrightarrow x=-20\)(tmđk)
Vậy: x=-20
b) ĐKXĐ: x≠1;x≠-1
Ta có: \(\frac{2}{x-1}=\frac{6}{x+1}\)
\(\Leftrightarrow\frac{2}{x-1}-\frac{6}{x+1}=0\)
\(\Leftrightarrow\frac{2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{6\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Leftrightarrow2\left(x+1\right)-6\left(x-1\right)=0\)
\(\Leftrightarrow2x+2-6x+6=0\)
\(\Leftrightarrow-4x+8=0\)
\(\Leftrightarrow-4x=-8\)
\(\Leftrightarrow x=2\)(tmđk)
Vậy: x=2
c) ĐKXĐ: x≠1;x≠-1
Ta có: \(\frac{2x+1}{x-1}=\frac{5\left(x-1\right)}{x+1}\)
\(\Leftrightarrow\frac{2x+1}{x-1}-\frac{5\left(x-1\right)}{x+1}=0\)
\(\Leftrightarrow\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{5\left(x-1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x+1\right)-5\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow2x^2+2x+x+1-5x^2+10x-5=0\)
\(\Leftrightarrow-3x^2+13x-4=0\)
\(\Leftrightarrow-3x^2+x+12x-4=0\)
\(\Leftrightarrow x\left(-3x+1\right)+4\left(3x-1\right)=0\)
\(\Leftrightarrow x\left(1-3x\right)-4\left(1-3x\right)=0\)
\(\Leftrightarrow\left(1-3x\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}1-3x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=1\\x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=4\end{matrix}\right.\)(thỏa mãn điều kiện)
Vậy: \(x\in\left\{\frac{1}{3};4\right\}\)
d) ĐKXĐ: x≠1;x≠-1
Ta có: \(\frac{x}{x-1}-\frac{2x}{x^2-1}=0\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\frac{2x}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x\left(x+1\right)-2x=0\)
\(\Leftrightarrow x^2+x-2x=0\)
\(\Leftrightarrow x^2-x=0\)
\(\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=1\left(ktm\right)\end{matrix}\right.\)
Vậy: x=0
e) ĐKXĐ: x≠2
Ta có: \(\frac{1}{x-2}+3=\frac{x-3}{2-x}\)
⇔\(\frac{1}{x-2}+3-\frac{x-3}{2-x}=0\)
⇔\(\frac{1}{x-2}+3+\frac{x-3}{x-2}=0\)
⇔\(\frac{1}{x-2}+\frac{3\left(x-2\right)}{x-2}+\frac{x-3}{x-2}=0\)
\(\Leftrightarrow1+3\left(x-2\right)+x-3=0\)
\(\Leftrightarrow1+3x-6+x-3=0\)
\(\Leftrightarrow4x-8=0\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\)(không thỏa mãn)
Vậy: x∈∅
f) ĐKXĐ: \(x\ne\pm2\)
Ta có: \(\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\)
⇔\(\frac{x+1}{x-2}+\frac{x-1}{x+2}-\frac{2\left(x^2+2\right)}{x^2-4}=0\)
⇔\(\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{2x^2+4}{\left(x+2\right)\left(x-2\right)}=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)+\left(x-1\right)\left(x-2\right)-2x^2-4=0\)
\(\Leftrightarrow x^2+2x+x+2+x^2-2x-x+2-2x^2-4=0\)
\(\Leftrightarrow0=0\)
Vậy: x∈R
g) ĐKXĐ: \(x\ne\pm2\)
Ta có: \(\frac{x+2}{x-2}+\frac{1}{x+2}=\frac{x\left(x-5\right)}{x^2-4}\)
⇔\(\frac{x+2}{x-2}+\frac{1}{x+2}-\frac{x\left(x-5\right)}{\left(x-2\right)\left(x+2\right)}=0\)
⇔\(\frac{\left(x+2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x+2\right)\left(x-2\right)}-\frac{x^2-5x}{\left(x-2\right)\left(x+2\right)}=0\)
⇔\(\left(x+2\right)^2+x-2-x^2+5x=0\)
\(\Leftrightarrow x^2+4x+4+x-2-x^2+5x=0\)
\(\Leftrightarrow10x-2=0\)
\(\Leftrightarrow10x=2\)
\(\Leftrightarrow x=\frac{2}{10}=\frac{1}{5}\)(thỏa mãn)
Vậy: \(x=\frac{1}{5}\)
cảm ơn bạn nha