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1 tháng 5 2017

ĐKXĐ: \(\left\{{}\begin{matrix}x+2\ne0\\2-x\ne0\\x^2-4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)

Pt \(\Leftrightarrow\) \(\dfrac{\left(x-2\right)}{x^2-4}+\dfrac{-5\left(x+2\right)}{x^2-4}=\dfrac{2x-3}{x^2-4}\)

\(\Leftrightarrow x-2-5x-10=2x-3\)

\(\Leftrightarrow x-5x-2x=10+2-3\)

\(\Leftrightarrow-6x=9\)

\(\Leftrightarrow x=\dfrac{-3}{2}\) ( thỏa mãn)

Vậy nghiệm của pt là \(x=\dfrac{-3}{2}\)

1: Sửa đề: 2/x+2

\(\dfrac{2x+1}{x^2-4}+\dfrac{2}{x+2}=\dfrac{3}{2-x}\)

=>\(\dfrac{2x+1+2x-4}{x^2-4}=\dfrac{-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)

=>4x-3=-3x-6

=>7x=-3

=>x=-3/7(nhận)

2: \(\Leftrightarrow\dfrac{\left(3x+1\right)\left(3-x\right)+\left(3+x\right)\left(1-3x\right)}{\left(1-3x\right)\left(3-x\right)}=2\)

=>9x-3x^2+3-x+3-9x+x-3x^2=2(3x-1)(x-3)

=>-6x^2+6=2(3x^2-10x+3)

=>-6x^2+6=6x^2-20x+6

=>-12x^2+20x=0

=>-4x(3x-5)=0

=>x=5/3(nhận) hoặc x=0(nhận)

3: \(\Leftrightarrow x\cdot\dfrac{8}{3}-\dfrac{2}{3}=1+\dfrac{5}{4}-\dfrac{1}{2}x\)

=>x*19/6=35/12

=>x=35/38

a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)

\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)

\(\Leftrightarrow x^2-2x+12-8-x^2=0\)

\(\Leftrightarrow-2x+4=0\)

\(\Leftrightarrow-2x=-4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

b) Ta có: \(\left|2x+6\right|-x=3\)

\(\Leftrightarrow\left|2x+6\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)

Vậy: S={-3}

=>4x-6(2x+1)=2x-3x

=>4x-12x-6+x=0

=>-7x=6

hay x=-6/7

25 tháng 1 2022

\(\dfrac{x}{3}-\dfrac{2x+1}{2}=\dfrac{x}{6}-\dfrac{x}{4}\)

\(\Leftrightarrow\dfrac{4x}{12}-\dfrac{6\left(2x+1\right)}{12}=\dfrac{2x}{12}-\dfrac{3x}{12}\)

\(\Leftrightarrow4x-6\left(2x+1\right)=2x-3x\)

\(\Leftrightarrow4x-12x-6=-x\)

\(\Leftrightarrow4x-12x-6+x=0\)

\(\Leftrightarrow-7x-6=0\)

\(\Leftrightarrow x=-\dfrac{6}{7}\)

NV
21 tháng 4 2021

ĐKXĐ: \(x\ne\left\{2;4\right\}\)

Đặt \(\left\{{}\begin{matrix}\dfrac{x+1}{x-2}=a\\\dfrac{x-2}{x-4}=b\end{matrix}\right.\) \(\Rightarrow\dfrac{x+1}{x-4}=ab\)

Phương trình trở thành:

\(a^2-12b^2+ab=0\)

\(\Leftrightarrow a^2+4ab-3ab-12b^2=0\)

\(\Leftrightarrow a\left(a+4b\right)-3b\left(a+4b\right)=0\)

\(\Leftrightarrow\left(a-3b\right)\left(a+4b\right)=0\Leftrightarrow\left[{}\begin{matrix}a-3b=0\\a+4b=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x+1}{x-2}-\dfrac{3\left(x-2\right)}{x-4}=0\\\dfrac{x+1}{x-2}+\dfrac{4\left(x-2\right)}{x-4}=0\end{matrix}\right.\)

Bạn tự quy đồng và hoàn thành phần còn lại nhé

22 tháng 4 2021

e cảm ơn ạ

 

a) Sửa đề: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

ĐKXĐ: \(x\notin\left\{3;\dfrac{1}{5}\right\}\)

Ta có: \(\dfrac{3}{5x-1}+\dfrac{2}{3-x}=\dfrac{4}{\left(1-5x\right)\left(x-3\right)}\)

\(\Leftrightarrow\dfrac{3\left(3-x\right)}{\left(5x-1\right)\left(3-x\right)}+\dfrac{2\left(5x-1\right)}{\left(3-x\right)\left(5x-1\right)}=\dfrac{4}{\left(5x-1\right)\left(3-x\right)}\)

Suy ra: \(9-3x+10x-2=4\)

\(\Leftrightarrow7x+7=4\)

\(\Leftrightarrow7x=-3\)

hay \(x=-\dfrac{3}{7}\)

Vậy: \(S=\left\{-\dfrac{3}{7}\right\}\)

b: \(\Leftrightarrow\dfrac{20}{x}-\dfrac{20}{x+20}=\dfrac{1}{6}\)

=>\(\dfrac{20x+400-20x}{x\left(x+20\right)}=\dfrac{1}{6}\)

=>x*(x+20)=400*6=2400

=>x^2+20x-2400=0

=>(x+60)(x-40)=0

=>x=-60 hoặc x=40

c: \(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}=\dfrac{8}{4x^2-1}\)

=>(2x+1)^2-(2x-1)^2=8

=>4x^2+4x+1-4x^2+4x-1=8

=>8x=8

=>x=1(nhận)

9 tháng 8 2023

câu b sai đề rồi anh ơi và câu a đâu rồi ạ

NV
24 tháng 4 2021

ĐKXĐ: ...

\(\left(\dfrac{x-1}{x+2}\right)^2-4\left(\dfrac{x+2}{x-3}\right)^2+3\left(\dfrac{x-1}{x-3}\right)=0\)

Đặt \(\left\{{}\begin{matrix}\dfrac{x-1}{x+2}=a\\\dfrac{x+2}{x-3}=b\end{matrix}\right.\)

\(\Rightarrow a^2-4b^2+3ab=0\Leftrightarrow\left(a-b\right)\left(a+4b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a-b=0\\a+4b=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x-1}{x+2}-\dfrac{x+2}{x-3}=0\\\dfrac{x-1}{x+2}+\dfrac{4x+8}{x-3}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)\left(x-3\right)-\left(x+2\right)^2=0\\\left(x-\right)\left(x-3\right)+4\left(x+2\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow...\)

11 tháng 8 2021

\(1,\dfrac{4x-4}{3}=\dfrac{7-x}{5}\\ \Leftrightarrow5\left(4x-4\right)=3\left(7-x\right)\\ \Leftrightarrow20x-20=21-3x\\ \Leftrightarrow17x=41\Leftrightarrow x=\dfrac{41}{17}\)

\(2,\dfrac{3x-9}{5}=\dfrac{3-x}{2}\\ \Leftrightarrow6x-18=15-5x\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=3\)

\(3,\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\\ \Leftrightarrow\dfrac{6x-3-15+5x}{15}=1\\ \Leftrightarrow11x-18=1\\ \Leftrightarrow x=\dfrac{19}{11}\)

\(4,\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\\ \Leftrightarrow2x-10+9x+12=5x+2\\ \Leftrightarrow6x=0\Leftrightarrow x=0\)

\(5,\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\\ \Leftrightarrow5x-15+4x+6=2x+5\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\)

Tick nha

2: Ta có: \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)

\(\Leftrightarrow6x-18=15-5x\)

\(\Leftrightarrow11x=33\)

hay x=3

8 tháng 5 2017

Đặt t=x2-2x+3(t\(\ge\)2)

PTTT: \(\dfrac{1}{t-1}+\dfrac{1}{t}=\dfrac{9}{2\left(t+1\right)}\)

<=>2t2+2t+2t2-2=9t2-9

<=>5t2-2t-7=0

<=>(t+1)(5t-7)=0

Do t\(\ge\)2

=>t+1>0 5t-7>0

Vậy pt vô nghiệm

9 tháng 5 2017

\(\dfrac{1}{x^2-2x+2}+\dfrac{1}{x^2-2x+3}=\dfrac{9}{2\left(x^2-2x+4\right)}\)

Đặt \(t=x^2-2x+2=\left(x-1\right)^2+1\ge1\)

Thì ta có:

\(PT\Leftrightarrow\dfrac{1}{t}+\dfrac{1}{t+1}=\dfrac{9}{2\left(t+2\right)}\)

\(\Leftrightarrow5t^2-t-4=0\)

\(\Leftrightarrow\left(5t^2-5t\right)+\left(4t-4\right)=0\)

\(\Leftrightarrow\left(t-1\right)\left(5t+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5t+4=0\\t-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}t=-\dfrac{4}{5}\left(l\right)\\t=1\end{matrix}\right.\)

\(\Rightarrow x^2-2x+2=1\)

\(\Leftrightarrow x=1\)

Vậy PT có 1 nghiệm là x = 1