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\(ĐKXĐ:\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)
\(\frac{2}{x+1}+\frac{1}{2-x}=\frac{3x-11}{x^2-x-2}\)
\(\Leftrightarrow\frac{2}{x+1}-\frac{1}{x-2}-\frac{3x-11}{\left(x+1\right)\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{2\left(x-2\right)-\left(x+1\right)-\left(3x-11\right)}{\left(x+1\right)\left(x-2\right)}=0\)
\(\Leftrightarrow2x-4-x-1-3x+11=0\)
\(\Leftrightarrow-2x+6=0\)
\(\Leftrightarrow x=3\)
Vậy tập nghiệm của phương trình là \(S=\left\{3\right\}\)
\(\frac{1}{2-x}+1=\frac{1}{x+2}-\frac{6-x}{3x^2-12}\)ĐKXĐ : \(x\ne\pm2\)
\(\Leftrightarrow\frac{-3\left(x+2\right)}{3\left(x-2\right)\left(x+2\right)}+\frac{3\left(x-2\right)\left(x+2\right)}{3\left(x-2\right)\left(x+2\right)}=\frac{3\left(x-2\right)}{3\left(x-2\right)\left(x+2\right)}+\frac{x-6}{3\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow\frac{-3x-6+3\left(x^2-4\right)}{3\left(x-2\right)\left(x+2\right)}-\frac{3x-6+x-6}{3\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{-3x-6+3x^2-12-3x+6-x+6}{3\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow\frac{-7x-6+3x^2}{3\left(x-2\right)\left(x+2\right)}=0\)
\(\Leftrightarrow3x^2-7x-6=0\)
\(\Leftrightarrow3x^2-9x+2x-6=0\)
\(\Leftrightarrow3x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=\frac{-2}{3}\end{cases}}\)( thỏa mãn )
Vậy....
a) 7x - 35 = 0
<=> 7x = 0 + 35
<=> 7x = 35
<=> x = 5
b) 4x - x - 18 = 0
<=> 3x - 18 = 0
<=> 3x = 0 + 18
<=> 3x = 18
<=> x = 5
c) x - 6 = 8 - x
<=> x - 6 + x = 8
<=> 2x - 6 = 8
<=> 2x = 8 + 6
<=> 2x = 14
<=> x = 7
d) 48 - 5x = 39 - 2x
<=> 48 - 5x + 2x = 39
<=> 48 - 3x = 39
<=> -3x = 39 - 48
<=> -3x = -9
<=> x = 3
\(\frac{3x-1-\frac{x-1}{2}}{3}-\frac{2x+\frac{1-2x}{3}}{2}=\frac{\frac{3x-1}{2}}{5}\)
\(\frac{6.\left(3x-1\right)-3x-1}{3}-\frac{12x+2\left(1-2x\right)}{2}=\frac{3\left(3x-1\right)}{5}\)
\(\Leftrightarrow\)\(\frac{18x-6-3x+3}{3}-\frac{12x+2-4x}{2}=\frac{9x-3}{5}\)
\(\Leftrightarrow\)\(\frac{15x-3}{3}-\frac{8x+2}{2}=\frac{9x-3}{5}\)
\(\Leftrightarrow\)\(10\left(15x-3\right)-15\left(8x+2\right)=6\left(9x-3\right)\)
\(\Leftrightarrow\)\(150x-30-120x-30=54x-18\)
\(\Leftrightarrow\)\(150x-120x-54x=-18+30+30\)
\(\Leftrightarrow-24x=42\)
\(\Leftrightarrow x=-\frac{7}{4}\)
\(\frac{2x-1}{3x^2+7x+2}+\frac{3}{9x^2+15x+4}-\frac{2x+7}{3x^2-5x-12}=\frac{5}{x+2}\)
\(\Leftrightarrow\frac{2x-1}{\left(3x+1\right)\left(x+2\right)}+\frac{3}{\left(3x+1\right)\left(3x+4\right)}-\frac{2x+7}{\left(4x+3\right)\left(x-3\right)}=\frac{5}{\left(x+2\right)}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{3x+1}+\frac{1}{3x+1}-\frac{1}{3x+4}+\frac{1}{3x+4}-\frac{1}{x-3}=\frac{5}{x+2}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x-3}=\frac{5}{x+2}\)
\(\Leftrightarrow\frac{x-3-x-2}{\left(x+2\right)\left(x-3\right)}=\frac{5\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}\)
\(\Leftrightarrow5x-3=-5\)
\(\Leftrightarrow x=-\frac{2}{5}\)
Chúc bạn học tốt !!!
\(\frac{1}{x+1}+\frac{1}{1-x}=\frac{3x-6}{1-x^2}\)
\(\frac{1-x+x+1}{1-x^2}=\frac{3x-6}{1-x^2}\)
\(2=3x-6\)
\(4=3x\)
\(x=\frac{4}{3}\)
\(\frac{1}{x+1}-\frac{1}{x-1}=\frac{3x-6}{1-x^2}\)
\(\Leftrightarrow\frac{1}{x+1}+\frac{1}{1-x}=\frac{3x-6}{\left(1-x\right)\left(x+1\right)}\)
Quy đồng rồi khử mẫu ta được:
\(1-x+x+1=3x-6\)
\(\Leftrightarrow-x+x-3x=-6-1-1\)
\(\Leftrightarrow-3x=-8\)
\(\Leftrightarrow x=\frac{8}{3}\)
Vậy ....
\(\text{GIẢI :}\)
ĐKXĐ : \(x\ne-1\)
\(\frac{3x}{2}+\frac{x}{x+1}=2\)
\(\Leftrightarrow\frac{3x\left(x+1\right)}{2\left(x+1\right)}+\frac{2x}{2\left(x+1\right)}=\frac{4\left(x+1\right)}{2\left(x+1\right)}\)
\(\Rightarrow3x\left(x+1\right)+2x=4\left(x+1\right)\)
\(\Leftrightarrow3x\left(x+1\right)+2x-4\left(x+1\right)=0\)
\(\Leftrightarrow3x^2+3x+2x-4x-4=0\)
\(\Leftrightarrow3x^2+x-4=0\)
\(\Leftrightarrow3x^2-3x+4x-4=0\)
\(\Leftrightarrow3x\left(x-1\right)+4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\3x+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{4}{3}\end{cases}}\)
2 ngiệm vừa tìm được đều thỏa mãn ĐKXĐ.
Vậy tập nghiệm của phương trình là \(S=\left\{1;-\frac{4}{3}\right\}.\)
3x/2 + x/x+1 = 2 <=> 3x(x+1)/2(x+1) + 2x/2(x+1) = 4(x+1)/2(x+1) \(\frac{3}{2}\). NHÂN PHÁ NGOẶC VÀ KHỬ MẪU TA ĐC:
<=> 3x2 + 3x + 2x = 4x + 4 <=> 3x2 + x - 4 = 0\(\Delta\)
Đen - ta (kí hiệu tam giác) = b2 - 4ac = 12 - 4.(-4).3 = 1 + 48 = 49 > 0 => Phương trình có 2 nghiệm phân biệt :
x1 = -b+ căn đen ta / 2a = -1 + căn 49 / 2.3 = 6/6 =1
x2 = -b - căn đen ta / 2a = -1 - căn 49 / 2.3 = -8/6
Vậy phương trình có 2 nghiệm phân biệt là : S\(\hept{\begin{cases}\\\end{cases}}1,-\frac{8}{6}\)