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\(\text{GIẢI :}\)
ĐKXĐ : \(\hept{\begin{cases}x\ne0\\x\ne-1\end{cases}}\).
\(\frac{1}{x}\left(\frac{x-1}{x+1}+\frac{2}{x+1}\right)=\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{x}\cdot\frac{x-1+2}{x+1}\)
\(\Leftrightarrow\frac{x+1}{x\left(x+1\right)}=\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{x}=\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{x}-\frac{2}{3}=0\)
\(\Leftrightarrow\frac{3}{3x}-\frac{2x}{3x}=0\)
\(\Rightarrow\text{ }3-2x=0\)
\(\Leftrightarrow\text{ }2x=3\text{ }\Leftrightarrow\text{ }x=\frac{3}{2}\) (thỏa mãn ĐKXĐ)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{3}{2}\right\}\).
\(\frac{1}{x}\left(\frac{x-1}{x+1}+\frac{2}{x+1}\right)=\frac{2}{3}\)\(\left(đk:x\ne0;-1\right)\)
\(< =>\frac{1}{x}.\frac{x-1+2}{x+1}=\frac{2}{3}\)
\(< =>\frac{x+1}{x^2+x}=\frac{2}{3}\)
\(< =>3\left(x+1\right)=2\left(x^2+x\right)\)
\(< =>3x+3=2x^2+2x\)
\(< =>2x^2-x-3=0\)
Ta có : \(\Delta=\left(-1\right)^2-4.\left(2\right).\left(-3\right)=1+24=25\)
Vì delta > 0 nên phương trình có 2 nghiệm phân biệt
\(x_1=\frac{1+\sqrt{25}}{4}=\frac{1+5}{4}=\frac{3}{2}\)
\(x_2=\frac{1-\sqrt{25}}{4}=\frac{1-5}{4}=\frac{4}{4}=1\)
Vậy tập nghiệm của phương trình trên là \(\left\{1;\frac{3}{2}\right\}\)
\(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\left(x\ne3;x\ne-1\right)\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}-\frac{2x\cdot2}{2\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{x^2+x+x^2-3x-4x}{2\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{2x^2-6x}{2\left(x-3\right)\left(x+1\right)}=0\)
\(\Leftrightarrow\frac{2x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}=0\)
=> 2x=0
<=> x=0
Vậy x=0
+ Ta có: \(\frac{x}{2.\left(x-3\right)}+\frac{x}{2.\left(x+1\right)}=\frac{2x}{\left(x+1\right).\left(x-3\right)}\)\(\left(ĐKXĐ: x\ne-1, x\ne3\right)\)
\(\Leftrightarrow\frac{x.\left(x+1\right)+x.\left(x-3\right)}{2.\left(x-3\right).\left(x+1\right)}=\frac{4x}{2.\left(x-3\right).\left(x+1\right)}\)
\(\Rightarrow x^2+x+x^2-3x=4x\)
\(\Leftrightarrow\left(x^2+x^2\right)+\left(x-3x-4x\right)=0\)
\(\Leftrightarrow2x^2-6x=0\)
\(\Leftrightarrow2x.\left(x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\left(TM\right)\\x=6\left(TM\right)\end{cases}}\)
Vậy \(S=\left\{0,6\right\}\)
+ Ta có: \(\frac{1}{x-1}+\frac{2}{x^2+x+1}=\frac{3x^2}{x^3-1}\)\(\left(ĐKXĐ:x\ne1,x^2+x+1\ne0\right)\)
\(\Leftrightarrow\frac{\left(x^2+x+1\right)+2.\left(x-1\right)}{\left(x-1\right).\left(x^2+x+1\right)}=\frac{3x^2}{\left(x-1\right).\left(x^2+x+1\right)}\)
\(\Rightarrow x^2+x+1+2x-2=3x^2\)
\(\Leftrightarrow\left(x^2-3x^2\right)+\left(x+2x\right)+\left(1-2\right)=0\)
\(\Leftrightarrow-2x^2+3x-1=0\)
\(\Leftrightarrow2x^2-3x+1=0\)
\(\Leftrightarrow\left(2x^2-2x\right)-\left(x-1\right)=0\)
\(\Leftrightarrow2x.\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(2x-1\right).\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=1\\x=1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\left(TM\right)\\x=1\left(L\right)\end{cases}}\)
Vậy \(S=\left\{\frac{1}{2}\right\}\)
\(\text{GIẢI :}\)
ĐKXĐ : \(x\ne\pm1\)
\(\frac{2}{x+1}+\frac{x}{x-1}=\frac{\left[1\frac{1}{6}\cdot\frac{6}{7}+\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\right]x+1}{x^2-1}\)
\(\Leftrightarrow\frac{2}{x+1}+\frac{x}{x-1}=\frac{x+1}{x^2-1}\)
\(\Leftrightarrow\frac{2}{x+1}+\frac{x}{x-1}-\frac{x+1}{x^2-1}=0\)
\(\Leftrightarrow\frac{2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{x+1}{\left(x+1\right)\left(x-1\right)}=0\)
\(\Rightarrow\text{ }2\left(x-1\right)+x\left(x+1\right)-(x+1)=0\)
\(\Leftrightarrow\text{ }2\left(x-1\right)+\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2+x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x-1\text{ (loại)}\\x=-3\text{ (Chọn)}\end{cases}}}\)
Vậy tập nghiệm của phương trình là \(S=\left\{-3\right\}\).
\(\frac{2}{x+1}+\frac{x}{x-1}=\frac{\left[1\frac{1}{6}.\frac{6}{7}+\left(\frac{1}{2}-\frac{1}{3}-\frac{1}{6}\right)\right]x+1}{x^2-1}\)\(đk:x\ne\pm1\)
\(< =>\frac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{x\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{\left[\frac{7}{6}.\frac{6}{7}+\left(1\right)\right]x+1}{x^2-1}\)
\(< =>\frac{2x-2+x^2+x}{x^2+x-x-1}=\frac{2x+1}{x^2-1}\)\(< =>\frac{x^2+3x-2}{x^2-1}=\frac{2x-1}{x^2-1}\)
\(< =>x^2+2x-2=2x-1\)\(< =>x^2+2x-2x-2+1=0\)
\(< =>x^2-1=0< =>x^2=1\)\(< =>x=\pm1\)\(\left(ktmđk\right)\)
Vậy phương trình trên vô nghiệm
ĐKXĐ : \(x\ne2,x\ne4\)
Pt \(\Leftrightarrow\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-4}-12\left(\frac{x-2}{x-4}\right)^2=0\) (2)
Đặt \(\frac{x+1}{x-2}=a,\frac{x-2}{x-4}=b\Rightarrow ab=\frac{x+1}{x-4}\)
Khi đó pt (2) trở thành :
\(a^2+ab-12b=0\)
\(\Leftrightarrow a^2-3ab+4ab-12b=0\)
\(\Leftrightarrow a\left(a-3b\right)+4b\left(a-3b\right)=0\)
\(\Leftrightarrow\left(a-3b\right)\left(a+4b\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=3b\\a=-4b\end{cases}}\)
Bạn thay vào tính, được nghiệm là \(S=\left\{3,\frac{4}{3}\right\}\)
\(\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x}{\left(x+1\right)\left(x-3\right)}\left(x\ne3;x\ne-1\right)\)
\(\Leftrightarrow\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}-\frac{2x}{\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{2\left(x+1\right)\left(x-3\right)}+\frac{x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}-\frac{2x\cdot2}{2\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{x^2+x}{2\left(x+1\right)\left(x-3\right)}+\frac{x^2-3x}{2\left(x+1\right)\left(x-3\right)}-\frac{4x}{2\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{x^2+x+x^2-3x-4x}{2\left(x+1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{2x^2-6x}{2\left(x+1\right)\left(x-3\right)}=\frac{2x\left(x-3\right)}{2\left(x+1\right)\left(x-3\right)}=\frac{2x}{2\left(x+1\right)}=0\)
=> 2x=0
=> x=0(tmđk)
Vậy x=0 là nghiệm của phương trình
a) \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
<=> 1 - x + 3(x + 1) = 2x + 3
<=> 1 - x + 3x + 3 = 2x + 3
<=> 1 - x + 3x + 3 - 2x = 3
<=> 4 = 3 (vô lý)
=> pt vô nghiệm
b) ĐKXĐ: \(x\ne1;x\ne2\)
\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
<=> (x - 2)(2 - x) - 5(x + 1)(2 - x) = 15(x - 2)
<=> 2x - x2 - 4 + 2x - 5x - 5x2 + 10 = 15x - 30
<=> -x + 4x2 - 14 = 15x - 30
<=> x - 4x2 + 14 = 15x - 30
<=> x - 4x2 + 14 + 15x - 30 = 0
<=> 16x - 4x2 - 16 = 0
<=> 4(4x - x2 - 4) = 0
<=> -x2 + 4x - 4 = 0
<=> x2 - 4x + 4 = 0
<=> (x - 2)2 = 0
<=> x - 2 = 0
<=> x = 2 (ktm)
=> pt vô nghiệm
c) xem bài 4 ở đây: Câu hỏi của gjfkm
d) ĐKXĐ: \(x\ne1;x\ne2;x\ne3\)
\(\frac{x+4}{x^2-3x+2}+\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)
<=> \(\frac{x+4}{\left(x-1\right)\left(x-2\right)}+\frac{x+1}{\left(x-1\right)\left(x-3\right)}=\frac{2x+5}{\left(x-1\right)\left(x-3\right)}\)
<=> (x + 4)(x - 3) + (x + 1)(x - 2) = (2x + 5)(x - 2)
<=> x2 - 3x + 4x - 12 + x2 - 2x + x - 2 = 2x2 - 4x + 5x - 10
<=> 2x2 - 14 = 2x2 + x - 10
<=> 2x2 - 14 - 2x2 = x - 10
<=> -14 = x - 10
<=> -14 + 10 = x
<=> -4 = x
<=> x = -4
17) \(ĐKXĐ:x\ne1\)
\(\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}\)
\(\Leftrightarrow\frac{x^2+x+1-3x^2-2x^2+2x}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow-4x^2+3x+1=0\)
\(\Leftrightarrow-\left(x-1\right)\left(4x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\4x+1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\left(ktm\right)\\x=-\frac{1}{4}\left(tm\right)\end{cases}}\)
Vậy tập nghiệm của phương trình là \(S=\left\{-\frac{1}{4}\right\}\)
18) \(ĐKXĐ:x\ne1\)
\(\frac{1}{x-1}+\frac{2x^2-5}{x^3-1}=\frac{4}{x^2+x+1}\)
\(\Leftrightarrow\frac{x^2+x+1+2x^2-5-4x+4}{\left(x-1\right)\left(x^2+x+1\right)}=0\)
\(\Leftrightarrow3x^2-3x=0\)
\(\Leftrightarrow3x\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=1\left(ktm\right)\end{cases}}\)
Vậy tập nghiệm của phương trình là \(S=\left\{0\right\}\)
19) \(ĐKXĐ:\hept{\begin{cases}x\ne2\\x\ne3\\x\ne\frac{1}{2}\end{cases}}\)
\(\frac{x+4}{2x^3-5x+2}+\frac{x+1}{2x^2-7x+3}=\frac{2x+5}{2x^2-7x+3}\)
\(\Leftrightarrow\frac{x+4}{\left(2x-1\right)\left(x-2\right)}+\frac{x+1}{\left(2x-1\right)\left(x-3\right)}-\frac{2x+5}{\left(2x-1\right)\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{x^2+x-12+x^2-x-2-2x^2-x+10}{\left(x-2\right)\left(x-3\right)\left(2x-1\right)}=0\)
\(\Leftrightarrow-x-4=0\)
\(\Leftrightarrow x=-4\)(TM)
Vậy tập nghiệm của phương trình là \(S=\left\{-4\right\}\)
20) \(ĐKXĐ:x\ne0\)
\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\)
\(\Leftrightarrow\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}-\frac{3}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}=0\)
\(\Leftrightarrow\frac{x\left(x+1\right)\left(x^2-x+1\right)-x\left(x-1\right)\left(x^2+x+1\right)-3}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}=0\)
\(\Leftrightarrow x^4+x-x^4+x-3=0\)
\(\Leftrightarrow2x-3=0\)
\(\Leftrightarrow x=\frac{3}{2}\)(TM)
Vậy tập nghiệm của phương trình là \(S=\left\{\frac{3}{2}\right\}\)
a) 0,75x(x + 5) = (x + 5)(3 - 1,25x)
<=> 0,75x(x + 5) - (x + 5)(3 - 1,25x) = (x + 5)(3 - 1,25x) - (x + 5)(3 - 1,25x)
<=> 0,75x(x + 5) - (x + 5)(3 - 1,25x) = 0
<=> (x + 5)(0,75 + 1,25x - 3) = 0
<=> (x + 5)(2x - 3) = 0
<=> x + 5 = 0 hoặc 2x - 3 = 0
<=> x = -5 hoặc x = 3/2
b) 4/5 - 3 = 1/5x(4x - 15)
<=> -11/5 = x(4x - 15)/5
<=> -11 = x(4x - 15)
<=> -11 = 4x2 - 15x
<=> 11 + 4x2 - 15x = 0
<=> 4x2 - 4x - 11x + 11 = 0
<=> 4x(x - 1) - 11(x - 1) = 0
<=> (4x - 11)(x - 1) = 0
<=> 4x - 11 = 0 hoặc x - 1 = 0
<=> x = 11/4 hoặc x = 1
c) \(\left(x-3\right)-\frac{\left(x-3\right)\left(2x-5\right)}{6}=\frac{\left(x-3\right)\left(3-x\right)}{4}\)
<=> 12x - 36 - 2(x - 3)(2x - 5) = 3(x - 3)(3 - x)
<=> 12x - 36 - 4x2 + 10x + 12x - 30 = 9x - 3x2 - 27 + 9x
<=> 34x - 66 - 4x2 = 18x - 3x2 - 27
<=> 34x - 66 - 4x2 - 18x + 3x2 + 27 = 0
<=> 16x - 39x - x2 = 0
<=> x2 - 16x + 39x = 0
<=> (x - 3)(x - 13) = 0
<=> x - 3 = 0 hoặc x - 13 = 0
<=> x = 3 hoặc x = 13
d) \(\frac{\left(3x+1\right)\left(3x-2\right)}{3}+5\left(3x+1\right)=\frac{2\left(2x+1\right)\left(3x+1\right)}{3}+2x\left(3x+1\right)\)
<=> (3x + 1)(3x - 2) + 15(3x + 1) = 2(2x + 1)(3x + 1) + 6x(3x + 1)
<=> 9x2 - 6x + 3x - 2 + 45x + 15 = 12x3 + 4x + 6x + 2 + 18x2 + 6x
<=> 9x2 + 42x + 13 = 30x2 + 16x + 2
<=> 9x2 + 42x + 13 - 30x2 - 16x - 2 = 0
<=> -21x2 + 26x + 11 = 0
<=> 21x2 - 26x - 11 = 0
<=> 21x2 + 7x - 33x - 11 = 0
<=> 7x(3x + 1) - 11(3x + 1) = 0
<=> (7x - 11)(3x + 1) = 0
<=> 7x - 11 = 0 hoặc 3x + 1 = 0
<=> x = 11/7 hoặc x = -1/3
Giải :
\(\frac{x-1}{2}+\frac{x-1}{4}=1-\frac{2\left(x-1\right)}{3}\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{2}+\frac{1}{4}+\frac{2}{3}\right)=1\)
\(\Leftrightarrow\left(x-1\right)\frac{17}{12}=1\)
\(\Leftrightarrow x-1=\frac{12}{17}\)
\(\Leftrightarrow x=\frac{29}{17}\)
Vậy phương trình có tập nghiệm là \(S=\left\{\frac{29}{17}\right\}\).
\(\frac{x-1}{2}+\frac{x-1}{4}=1-\frac{2\left(x-1\right)}{3}\)
\(\Leftrightarrow\frac{x-1}{2}+\frac{x-1}{4}=1-\frac{2x-2}{3}\)
\(\Leftrightarrow\frac{6\left(x-1\right)}{12}+\frac{3\left(x-1\right)}{12}=\frac{12}{12}-\frac{4\left(2x-2\right)}{12}\)
\(\Leftrightarrow6\left(x-1\right)+3\left(x-1\right)=12-4\left(2x-2\right)\)
\(\Leftrightarrow6x-6+3x-3=12-8x+8\)
\(\Leftrightarrow6x+3x+8x=12+8+6+3\)
\(\Leftrightarrow17x=29\)
\(\Leftrightarrow x=\frac{29}{17}\)
Vậy phương trình có nghiệm x=29/17