Cho Pt \(\dfrac{x+a}{a-x}+\dfrac{x-a}{a+x}=\dfrac{a\left(3a+1\right)}{a^2-x^2}\)
a) GPT với a = -3
b) Tìm a biết x=\(\dfrac{1}{2}\)
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13 tháng 1 2018
minh giai phan d, nha bn :
x-a/b+c + x-b/c+a + x-c/a+b=3
=> (x-a/b+c - 1)+(x-b/a+c - 1 )+(x-c/a+b - 1) = 3-3=0
=>x-a-b-c/b+c + x-a-b-c/a+c + x-a-b-c/a+b =0
=>(x-a-b-c)(1/b+c + 1/a+c + 1/a+b )=0
Vi 1/b+c + 1/a+c + 1/a+b luon lon hon 0=>x-a-b-c=0
=>x=a+b+c
13 tháng 5 2022
a: Khi a=-3 thì phương trình sẽ là:
\(\dfrac{x+3}{x-3}-\dfrac{x-3}{x+3}+\dfrac{3\cdot9-3}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow x^2+6x+9-x^2+6x-9+24=0\)
=>12x=-24
hay x=-2
b: Khi a=1 thì phương trình trở thành:
\(\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}+\dfrac{4}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Leftrightarrow x^2-2x+1-x^2-2x-1+4=0\)
=>-4x+4=0
hay x=1(loại)
10 tháng 3 2017
\(\Leftrightarrow A=\dfrac{\left(x-a\right)^2-\left(x+a\right)^2+3a^2+a}{\left(x-a\right)\left(x+a\right)}\)
\(\Leftrightarrow A=\dfrac{-4ax+3a^2+a}{\left(x-a\right)\left(x+a\right)}\Leftrightarrow\left\{{}\begin{matrix}\left|x\right|\ne a\\4ax=a\left(3a+1\right)\left(1\right)\end{matrix}\right.\)
a) với a=-3
\(\left(1\right)\Leftrightarrow4x=3.\left(-3\right)+1\Rightarrow x=-2\)(NHAN)
b)với a=-1
\(\left(1\right)\Leftrightarrow4x=3.\left(-1\right)+1\Rightarrow x=-\dfrac{2}{4}=-\dfrac{1}{2}\)(NHẬN)
c)
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}a\ne0\\x=\dfrac{3a+1}{4}=0,5\Rightarrow a=\dfrac{1}{3}\left(nhan\right)\end{matrix}\right.\)
19 tháng 5 2022
a: \(E=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x\left(x-1\right)}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)
b: |x-3|=2
=>x-3=2 hoặc x-3=-2
=>x=5(nhận) hoặc x=1(loại)
Khi x=5 thì \(E=\dfrac{5^2}{5-1}=\dfrac{25}{4}\)
c: Để E=1/2 thì \(\dfrac{x^2}{x-1}=\dfrac{1}{2}\)
\(\Leftrightarrow2x^2-x+1=0\)
hay \(x\in\varnothing\)
19 tháng 5 2022
f) \(A=\dfrac{x^2}{x-1}=\dfrac{x^2-x+x-1+1}{x-1}=\dfrac{x\left(x-1\right)+x-1+1}{x-1}=x+1+\dfrac{1}{x-1}=x-1+\dfrac{1}{x-1}+2\ge2\sqrt{\left(x-1\right).\dfrac{1}{x-1}}+2=4\)\(A=4\Leftrightarrow x=2\)
-Vậy \(A_{min}=4\)
19 tháng 11 2018
a, \(\dfrac{x^2-x}{x-2}+\dfrac{4-3x}{x-2}\)
\(=\dfrac{x^2-x+4-3x}{x-2}=\dfrac{x^2-4x+4}{x-2}\)
19 tháng 11 2018
c) \(\dfrac{2}{x^2-9}+\dfrac{1}{x+3}\)
Ta có: \(\dfrac{1}{x+3}=\dfrac{1\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-3}{x^2-9}\)
\(\Rightarrow\dfrac{2}{x^2-9}+\dfrac{1}{x+3}=\dfrac{2}{x^2-9}+\dfrac{x-3}{x^2-9}=\dfrac{2+x-3}{x^2-9}=\dfrac{x-1}{x^2-9}\)
9 tháng 12 2022
2: \(\left(\dfrac{7}{a+7}+\dfrac{a^2+49}{a^2-49}-\dfrac{7}{a-7}\right):\dfrac{a+1}{2}\)
\(=\dfrac{7a-49+a^2+49-7a-49}{\left(a-7\right)\left(a+7\right)}\cdot\dfrac{2}{a+1}\)
\(=\dfrac{a^2-49}{\left(a-7\right)\left(a+7\right)}\cdot\dfrac{2}{a+1}=\dfrac{2}{a+1}\)
3: \(=\dfrac{x^4-4x^2+4x^2}{x^2-4}\cdot\left(\dfrac{x+2}{x-4}+\dfrac{2-3x}{x\left(x^2-4\right)}\cdot\dfrac{x^2-4}{x-2}\right)\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\left(\dfrac{x+2}{x-4}+\dfrac{2-3x}{x\left(x-2\right)}\right)\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x^2-4\right)+\left(2-3x\right)\left(x-4\right)}{x\left(x-2\right)\left(x-4\right)}\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-4x+2x-8-3x^2+12x}{x\left(x-2\right)\left(x-4\right)}\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-3x^2+10x-8}{x\left(x-2\right)\left(x-4\right)}\)
\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-x^2-2x^2+2x+8x-8}{x\left(x-2\right)\left(x-4\right)}\)
\(=\dfrac{x^3\left(x-1\right)\left(x^2-2x+8\right)}{\left(x-2\right)^2\cdot\left(x+2\right)\left(x-4\right)}\)
7 tháng 7 2021
Đk:\(a\ne\pm x\)
Pt \(\Leftrightarrow\dfrac{\left(a+x\right)^2-\left(x-a\right)\left(a-x\right)}{\left(a-x\right)\left(a+x\right)}=\dfrac{a\left(3a+1\right)}{a^2-x^2}\)
\(\Leftrightarrow\dfrac{2\left(a^2+x^2\right)}{a^2-x^2}=\dfrac{a\left(3a+1\right)}{a^2-x^2}\)
\(\Leftrightarrow2a^2+2x^2=3a^2+a\)
\(\Leftrightarrow a^2+a-2x^2=0\) (1)
Thay \(x=\dfrac{1}{2}\) vào (1) ta được:
\(a^2+a-2\left(\dfrac{1}{2}\right)^2=0\)
\(\Leftrightarrow a^2+a-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{-1+\sqrt{3}}{2}\\a=\dfrac{-1-\sqrt{3}}{2}\end{matrix}\right.\) (tm)
Vậy...
\(\dfrac{x+a}{a-x}+\dfrac{x-a}{a+x}=\dfrac{a\left(3a+1\right)}{a^2-x^2}\)
\(\Leftrightarrow\dfrac{\left(x+a\right)\left(a+x\right)}{\left(a-x\right)\left(a+x\right)}+\dfrac{\left(x-a\right)\left(a-x\right)}{\left(a+x\right)\left(a-x\right)}=\dfrac{a\left(3a+1\right)}{a^2-x^2}\)
\(\Leftrightarrow\dfrac{\left(x+a\right)\left(a+x\right)+\left(x-a\right)\left(a-x\right)}{\left(a-x\right)\left(a+x\right)}=\dfrac{a\left(3a+1\right)}{a^2-x^2}\)
\(\Leftrightarrow\dfrac{xa+x^2+a^2+ax+xa-x^2-a^2+ax}{\left(a-x\right)\left(a+x\right)}=\dfrac{a\left(3a+1\right)}{\left(a-x\right)\left(a+x\right)}\)
\(\Rightarrow4ax=a\left(3a+1\right)\)
<=> 4ax-a(3a+1)=0
<=> 4ax-3a2-a=0
<=> a(4x-3a-1)=0 (*)
a) Thay a=-3 vào phương trình ta có :
\(\dfrac{x-3}{-3-x}+\dfrac{x-3}{-3+x}=\dfrac{-3\left[3.\left(-3\right)+1\right]}{\left(-3\right)^2-x^2}\)
ĐKXĐ : \(x\ne\pm3\)
(*) <=> -3[4x-3.(-3)-1]=0
<=> -3(4x+8)=0
<=> (-3).4x+(-3).8=0
<=> -12x-24=0
<=> -12x=24
<=> x=-2
Vậy phương trình có nghiệm x=-2
b) Thay x=1/2 vào phương trình ta có :
(*) \(\Leftrightarrow a\left(4.\dfrac{1}{2}-3a-1\right)=0\)
\(\Leftrightarrow a\left(2-3a-1\right)=0\)
<=> a(1-3a)=0
\(\Leftrightarrow\left[{}\begin{matrix}a=0\\1-3a=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=0\\a=\dfrac{1}{3}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{0;\dfrac{1}{3}\right\}\)