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a)\(2+\frac{3}{x-5}=1\)
\(\Rightarrow\frac{3}{x-5}=-1\)
\(\Rightarrow3=-x+5\)
\(\Leftrightarrow x+3=5\)
\(\Rightarrow x=2\)
a,\(\frac{2x+5}{3}-2=\frac{3x-7}{5}\)
\(\Rightarrow5\left(2x+5\right)-30=3\left(3x-7\right)\)
\(\Leftrightarrow10x+25-30=9x-27\)
\(\Leftrightarrow x=-22\)
vậy....................
\(b,\frac{x}{6}+x=\frac{2x+1}{2}\)
\(\Rightarrow2x+12x=6\left(2x+1\right)\)
\(\Leftrightarrow14x=12x+6\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
vậy.....................
c,\(\frac{x}{4}-\frac{2x-1}{3}=-\frac{5x}{12}\)
\(\Rightarrow3x-4\left(2x-1\right)=-5x\)
\(\Leftrightarrow3x-8x+4=-5x\)
\(\Leftrightarrow0x=-4\left(PTVN\right)\)
VẬY................
P/s : bạn chú ý \(\Rightarrow\)với \(\Leftrightarrow\)nha
\(\frac{25x-655}{95}-\frac{5\left(x-12\right)}{209}=\frac{89-3x-\frac{2\left(x-18\right)}{5}}{11}\)
\(< =>\frac{5x-131}{19}=\frac{1631-52x-\frac{38x-684}{5}}{209}\)
\(< =>\left(5x-131\right)209=\left(1631-52x-\frac{38x-684}{5}\right)19\)
\(< =>55x-1441=1631-52x-\frac{38x-684}{5}\)
\(< =>3072-107x=\frac{38x-684}{5}\)
\(< =>\left(3072-107x\right)5=38x-684\)
\(< =>15360-535x-38x-684=0\)
\(< =>14676=573x< =>x=\frac{14676}{573}=\frac{4892}{191}\)
nghệm xấu thế
\(\frac{8\left(x+22\right)}{45}-\frac{7x+149+\frac{6\left(x+12\right)}{5}}{9}=\frac{x+35+\frac{2\left(x+50\right)}{9}}{5}\)
\(< =>\frac{8x+176}{45}-\frac{41x+817}{45}=\frac{11x+415}{45}\)
\(< =>993-33x-11x-415=0\)
\(< =>578=44x< =>x=\frac{289}{22}\)
1) \(\frac{14}{3x-12}-\frac{2+x}{x-4}=\frac{3}{8-2x}-\frac{5}{6}\) (1)
ĐK: x \(\ne\)4
(1) <=> \(\frac{14}{3\left(x-4\right)}-\frac{2+x}{x-4}+\frac{3}{2\left(x-4\right)}=-\frac{5}{6}\)
<=> \(\frac{28-6\left(2+x\right)+9}{6\left(x-4\right)}=-\frac{5}{6}\)
<=> \(\frac{25-6x}{x-4}=-5\)
<=> 25 - 6x = - 5x + 20
<=> x = 5 ( thỏa mãn )
Vậy x = 5.
b) ĐK: x \(\ne\)1; -1
\(\left(1-\frac{x-1}{x+1}\right)\left(x+2\right)=\frac{x+1}{x-1}+\frac{x-1}{x+1}\)
<=> \(\frac{2\left(x+2\right)}{x+1}=\frac{2x^2+2}{\left(x-1\right)\left(x+1\right)}\)
<=> \(\frac{2\left(x+2\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{2x^2+2}{\left(x-1\right)\left(x+1\right)}\)
<=> \(2x^2+2x-4=2x^2+2\)
<=> \(x=3\)( thỏa mãn)
Vậy x = 3.
Nhìn sơ qua thì thấy bài 3, b thay -2 vào x rồi giải bình thường tìm m
Bài 2:
a) \(x+x^2=0\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x+1=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=0\\x=0-1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=0\\x=-1\end{cases}}\)
b) \(0x-3=0\)
\(\Leftrightarrow0x=3\)
\(\Rightarrow vonghiem\)
c) \(3y=0\)
\(\Leftrightarrow y=0\)
ĐKXĐ : \(x\ne-2,x\ne2\)
\(\frac{x+1}{x-2}+\frac{5}{2+x}=\frac{12}{x^2-4}+1\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{x^2-4}-\frac{5\left(x-2\right)}{x^2-4}=\frac{12}{x^2-4}+\frac{x^2-4}{x^2-4}\)
\(\Leftrightarrow x^2+3x+2-5x+10=8+x^2\)
\(\Leftrightarrow12-2x=8\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\left(ktm\right)\)
Vậy phương trình vô nghiệm
\(\frac{x+1}{x-2}+\frac{5}{2+x}=\frac{12}{x^2-4}+1\)
ĐKXĐ: \(x\ne\pm2\)
\(\Leftrightarrow\frac{x+1}{x-2}+\frac{-5}{x-2}=\frac{12}{\left(x-2\right)\left(x+2\right)}+1\)
\(\Leftrightarrow\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{-5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\frac{12}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-2\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)-5\left(x+2\right)=12+\left(x-2\right)\left(x+2\right)\)
\(\Leftrightarrow x^2+2x+x+2-5x-10=12+x^2+2x-2x-4\)
\(\Leftrightarrow x^2-2x-8=8+x^2\)
\(\Leftrightarrow x^2-2x-8-8-x^2=0\)
\(\Leftrightarrow-2x-16=0\)
\(\Leftrightarrow-2x=0+16\)
\(\Leftrightarrow-2x=16\)
\(\Leftrightarrow x=16:\left(-2\right)\)
\(\Leftrightarrow x=-8\)(TMĐKXĐ)
Vậy S = \(\left\{-8\right\}\)