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ĐKXĐ : \(\hept{\begin{cases}x^2+x-6\ne0\\x^2+4x+3\ne0\\2x-1\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}\left(x+3\right)\left(x-2\right)\ne0\\\left(x+1\right)\left(x+3\right)\ne0\\x\ne\frac{1}{2}\end{cases}\Rightarrow\hept{\begin{cases}x\ne2;-3\\x\ne-1;-3\\x\ne\frac{1}{2}\end{cases}}}}\)
TXĐ : \(x\ne\left\{-3;-1;\frac{1}{2};2\right\}\)
\(pt\Leftrightarrow\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)
\(\Leftrightarrow\frac{5\left(x+1\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)
\(\Leftrightarrow\frac{3x+9}{\left(x-2\right)\left(x+1\right)\left(x+3\right)}=\frac{-3}{2x-1}\)
\(\Leftrightarrow\frac{3}{\left(x-2\right)\left(x+1\right)}=\frac{-3}{2x-1}\)
\(\Leftrightarrow\frac{1}{x^2-x-2}=\frac{1}{1-2x}\)
\(\Leftrightarrow x^2-x-2-1+2x=0\)
\(\Leftrightarrow x^2+x-3=0\)
\(\Leftrightarrow\left(x^2+2.\frac{1}{2}.x+\frac{1}{4}\right)-\frac{13}{4}=0\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)^2-\left(\frac{\sqrt{13}}{2}\right)^2=0\)
\(\Leftrightarrow\left(x+\frac{1-\sqrt{13}}{2}\right)\left(x+\frac{1+\sqrt{13}}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{13}-1}{2}\\x=\frac{-\sqrt{13}-1}{2}\end{cases}}\)
\(\frac{5}{x^2+x-6}-\frac{2}{x^2+4+3}=-\frac{3}{2x-1}\)
<=> \(\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{2}{\left(x+1\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{5\left(x+1\right)-2\left(x-2\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{5x+5-2x+4}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{3x+9}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{3\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=-\frac{3}{2x-1}\)
<=> \(\frac{1}{x-2}=-\frac{1}{2x-1}\)
<=> x-2=1-2x <=> 3x=3
=> x=1
Đáp số: x=1
Giải phương trình chứa ẩn ở mẫu:a) 4x 2/3x-6-x/2-x=1 3x/2x-4b) x-3/x 3-x 3/x-3=3/x2-9Các bạn hãy giúp mik với:))
a:=>x^2-1-x=2x-1
=>x^2-x-1=2x-1
=>x^2-3x=0
=>x=0(loại) hoặc x=3(nhận)
b:=>x+2=0 hoặc 5-3x=0
=>x=-2 hoặc x=5/3
c:=>20(1-2x)+6x=9(x-5)-24
=>20-40x+6x=9x-45-24
=>-34x+20=9x-69
=>-43x=-89
=>x=89/43
d: =>x^2+4x+4-x^2-2x+3=2x^2+8x-4x-16-3
=>2x^2+4x-19=-2x+7
=>2x^2+6x-26=0
=>x^2+3x-13=0
=>\(x=\dfrac{-3\pm\sqrt{61}}{2}\)
e: =>(2x-3)(2x-3-x-1)=0
=>(2x-3)(x-4)=0
=>x=4 hoặc x=3/2
a) x(4x + 2) = 4x2 - 14
⇔ 4x2 + 2x = 4x2 - 14
⇔ 4x2 - 4x2 + 2x = -14
⇔ 2x = -14
⇔ x = -7
Vậy tập nghiệm S = ......
b) (x2 - 9)(2x - 1) = 0
⇔ x2 - 9 = 0 hoặc 2x - 1 = 0
⇔ x2 = 9 hoặc 2x = 1
⇔ x = 3 hoặc -3 hoặc x = \(\dfrac{1}{2}\)
Vậy .......
c) \(\dfrac{3}{x-2}\) + \(\dfrac{4}{x+2}\) = \(\dfrac{x-12}{x^2-4}\)
⇔ \(\dfrac{3}{x-2}\) + \(\dfrac{4}{x+2}\) = \(\dfrac{x-12}{\left(x-2\right)\left(x+2\right)}\)
ĐKXĐ: x - 2 ≠ 0 và x + 2 ≠ 0
⇔ x ≠ 2 và x ≠ -2MSC (mẫu số chung): (x - 2)(x + 2)Quy đồng mẫu hai vế và khử mẫu ta được:3x + 6 + 4x - 8 = x - 12⇔ 3x + 4x - x = 8 - 6 - 12⇔ 6x = -10⇔ x = \(-\dfrac{5}{3}\) (nhận)Vậy ........
Ta có: \(\dfrac{2x}{x^2-x+1}-\dfrac{x}{x^2+x+1}=\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{2x\left(x^2+x+1\right)-x\left(x^2-x+1\right)}{\left(x^2-x+1\right)\left(x^2+x+1\right)}=\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{2x^3+2x^2+2x-x^3+x^2-x}{\left(x^2-x+1\right)\left(x^2+x+1\right)}=\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{x^3+3x^2+x}{\left(x^2+1\right)^2-x^2}=\dfrac{5}{3}\)
\(\Leftrightarrow3x^3+9x^2+3x=5\left(x^4+2x^2+1-x^2\right)\)
\(\Leftrightarrow3x^3+9x^2+3x=5x^4+5x^2+5\)
\(\Leftrightarrow5x^4+5x^2+5-3x^3-9x^2-3x=0\)
\(\Leftrightarrow5x^4-3x^3-4x^2-3x+5=0\)
\(\Leftrightarrow5x^4-5x^3+2x^3-2x^2-2x^2+2x-5x+5=0\)
\(\Leftrightarrow5x^3\left(x-1\right)+2x^2\left(x-1\right)-2x\left(x-1\right)-5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x^3+2x^2-2x-5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x^3-5x^2+7x^2-7x+5x-5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[5x^2\left(x-1\right)+7x\left(x-1\right)+5\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(5x^2+7x+5\right)=0\)
mà \(5x^2+7x+5>0\forall x\)
nên x-1=0
hay x=1
\(\dfrac{x+3}{6}-3=1-\dfrac{2x}{x^2+x}\)
=>\(\dfrac{x-15}{6}=1-\dfrac{2x}{x^2+x}=\dfrac{x^2+x-2x}{x^2+x}=\dfrac{x-1}{x+1}\)
=>x^2-14x-15=6x-6
=>x^2-20x-9=0
=>x=10+căn 109 hoặc x=10-căn 109
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