giải pt
\(\dfrac{1}{x^2-2x+2}+\dfrac{2}{x^2-2x+3}=\dfrac{6}{x^2-2x+4}\)
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9 tháng 8 2023
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)
TA
1
11 tháng 9 2021
Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)
\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)
\(\Leftrightarrow x+8+20x-12=0\)
\(\Leftrightarrow x=\dfrac{4}{21}\)
22 tháng 4 2021
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: \(S=\varnothing\)
22 tháng 4 2021
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}
11 tháng 7 2023
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
1
16 tháng 3 2018
ĐKXD: ∀x
Ta có \(\dfrac{x^{2^{ }}+2x+1}{x^2+2x+2}\) + \(\dfrac{x^2+2x+2}{x^2+2x+3}\) = \(\dfrac{7}{6}\)
Đặt x2 + 2x + 2 là a (a ∈ Q) Ta có phương trình mới ẩn a:
\(\dfrac{a-1}{a}+\dfrac{a}{a+1}\) = \(\dfrac{7}{6}\)
⇔ \(\dfrac{6\left(a-1\right)\left(a+1\right)}{6a\left(a+1\right)}\)+\(\dfrac{6a^2}{6a\left(a+1\right)}\) = \(\dfrac{7}{6}\)
⇔\(\dfrac{6\left(a^2-1\right)+6a^2}{6a\left(a+1\right)}\) = \(\dfrac{7a\left(a+1\right)}{6a\left(a+1\right)}\)
Suy ra: 6a2 - 6 + 6a2 = 7a2 + 7a
⇔ 12a2 - 6 - 7a2 - 7a
⇔ 5a2 - 7a - 6 = 0
⇔5a2 - 10a + 3a - 6 = 0
⇔5a( a - 2 ) + 3( a - 2 ) = 0
⇔ (5a + 3)(a - 2) = 0
⇔\(\left[{}\begin{matrix}a-2=0\\5a+3=0\end{matrix}\right.\) ⇔\(\left[{}\begin{matrix}a=2\\a=-0,6\end{matrix}\right.\)
Với a = 2 thì:
x2 + 2x + 2 = 2 ⇔ x2 + 2x = 0
⇔ x(x + 2) = 0 ⇔ \(\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Với a = -0,6 thì:
x2 + 2x + 2 = -0,6 ⇔ x2 + 2x + 1 = -1,6
⇔ (x + 1)2 = -1,6 (Vô lí vì (x + 1)2 ≥ 0)
Vậy S ∈ \(\left\{0;-2\right\}\)
10 tháng 11 2021
\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
Đặt \(u=x^2-2x+2\)
=> Pt tương đương :
\(\dfrac{1}{u}+\dfrac{2}{u+1}=\dfrac{6}{u+2}\)
\(\Leftrightarrow\dfrac{\left(u+1\right)\left(u+2\right)+2u\cdot\left(u+2\right)}{u\left(u+1\right)\left(u+2\right)}=\dfrac{6u\left(u+1\right)}{u\left(u+1\right)\left(u+2\right)}\)
\(\Leftrightarrow\left(u+1\right)\left(u+2\right)+2u\left(u+2\right)=6u\left(u+1\right)\)
\(\Leftrightarrow u^2+3u+2+2u^2+4u=6u^2+6u\)
\(\Leftrightarrow-3u^2+u+2=0\)
\(\Rightarrow\left[{}\begin{matrix}u=1\\u=-\dfrac{2}{3}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x^2-2x+2=1\\x^2-2x+2=-\dfrac{2}{3}\end{matrix}\right.\Rightarrow x=1\)
Kết luận \(x=1\)
\(pt\Leftrightarrow\dfrac{1}{\left(x-1\right)^2+1}+\dfrac{2}{\left(x-1\right)^2+2}=\dfrac{6}{\left(x-1\right)^2+3}\)
Đặt: \(\left(x-1\right)^2=t\ge0\)
\(pt\Leftrightarrow\dfrac{1}{t+1}+\dfrac{2}{t+2}=\dfrac{6}{t+3}\)
\(\Rightarrow\dfrac{t+2+2\left(t+1\right)}{\left(t+1\right)\left(t+2\right)}=\dfrac{6}{t+3}\)
\(\Rightarrow\dfrac{t+2+2t+2}{\left(t+1\right)\left(t+2\right)}=\dfrac{6}{t+3}\)
\(\Rightarrow\dfrac{3t+4}{\left(t+1\right)\left(t+2\right)}=\dfrac{6}{t+3}\)
\(\Rightarrow\left(3t+4\right)\left(t+3\right)=6\left(t+1\right)\left(t+2\right)\)
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