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23 tháng 3 2019
\(\frac{4x}{x^2+4x+3}-1=6\left(\frac{1}{x+3}-\frac{1}{2x+2}\right)\) \(ĐK:x\ne-1;x\ne-3\)
\(\Leftrightarrow\frac{4x}{x^2+4x+3}-\frac{x^2+4x+3}{x^2+4x+3}=6\left[\frac{2\left(x+1\right)}{2\left(x+3\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)\left(x+3\right)}\right]\)
\(\Leftrightarrow\frac{4x-x^2-4x-3}{x^2+4x+3}=6\left[\frac{2\left(x+1\right)-x-3}{2\left(x+3\right)\left(x+1\right)}\right]\)
\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=6\left[\frac{2x+2-x-3}{2\left(x^2+4x+3\right)}\right]\)
\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=\frac{6\left(x-1\right)}{2\left(x^2+4x+3\right)}\)
\(\Leftrightarrow\frac{-x^2-3}{x^2+4x+3}=\frac{3\left(x-1\right)}{x^2+4x+3}\)
\(\Leftrightarrow-x^2-3=3x-3\)
\(\Leftrightarrow-x^2-3x=0\)
\(\Leftrightarrow-x\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\left(loại\right)\end{cases}}\)
Vậy x = 0
23 tháng 3 2019
\(ĐK:x\ne\frac{-1}{2};x\ne\frac{-3}{2}\)
\(\frac{3}{2x+1}=\frac{6}{2x+3}+\frac{8}{4x^2+8x+3}\)
\(\Leftrightarrow\frac{3}{2x+1}-\frac{6}{2x+3}=\frac{8}{4x^2+8x+3}\)
\(\Leftrightarrow\frac{3\left(2x+3\right)-6\left(2x+1\right)}{\left(2x+1\right)\left(2x+3\right)}=\frac{8}{4x^2+8x+3}\)
\(\Leftrightarrow\frac{6x+9-12x-6}{4x^2+8x+3}=\frac{8}{4x^2+8x+3}\)
\(\Leftrightarrow-6x+3=8\)
\(\Leftrightarrow x=-\frac{5}{6}\)
Vậy ...
KS
2
21 tháng 7 2019
\(\frac{\left(x-2\right)^2}{3}-\frac{2x-1}{4}=4-\frac{\left(2x-3\right)^2}{6}.\)
\(\Rightarrow\frac{4\left(x-2\right)^2}{12}-\frac{3\left(2x-1\right)^2}{12}=\frac{48}{12}-\frac{2\left(2x-3\right)^2}{12}\)
\(\Rightarrow4\left(x^2-4x+4\right)-3\left(4x^2-4x+1\right)=48-2\left(4x^2-12x+9\right)\)
\(\Rightarrow4x^2-16x+16-12x^2+12x-3=48-8x^2+24x-18\)
\(\Rightarrow-16x+12x+16-3=24x+48-18\)
\(\Rightarrow28x=-17\Leftrightarrow x=-\frac{17}{28}\)
KS
1
21 tháng 7 2019
-------------------ko chép đề nha---------
\(\Leftrightarrow\frac{4\left(x^2-4x+4\right)-3\left(2x+1\right)}{12}=\frac{12-2\left(4x^2-12x+9\right)}{12}\)
\(\Rightarrow4x^2+16x+16-6x-3=12-8x^2+24x-18\)
\(\Leftrightarrow4x^2+10x+13=-8x^2+24x-6\)
\(\Leftrightarrow4x^2+8x^2+10x-24x+13+6=0\)
\(\Leftrightarrow12x-14x+19=0\)
Ta có :\(\Delta'=7^2-12.19=-179< 0\)
\(\Rightarrow\)phương trình vô nghiệm
H
25 tháng 2 2019
\(a,\frac{1}{2x-3}-\frac{3}{x\left(2x-3\right)}=\frac{5}{x}\) ĐKXĐ : \(x\ne0;x\ne\frac{3}{2}\)
\(\Leftrightarrow\frac{x}{x\left(2x-3\right)}-\frac{3}{x\left(2x-3\right)}=\frac{5\left(2x-3\right)}{x\left(2x-3\right)}\)
\(\Leftrightarrow x-3=10x-15\)
\(\Leftrightarrow x-10x=3-15\)
\(\Leftrightarrow-9x=-12\)
\(\Leftrightarrow x=\frac{-12}{-9}=\frac{4}{3}\)(TMĐKXĐ)
KL :....
H
25 tháng 2 2019
\(b,\frac{x+2}{x-2}-\frac{1}{x}=\frac{2}{x\left(x-2\right)}\) ĐKXĐ : \(x\ne0;2\)
\(\Leftrightarrow\frac{x\left(x+2\right)}{x\left(x-2\right)}-\frac{x-2}{x\left(x-2\right)}=\frac{2}{x\left(x-2\right)}\)
\(\Leftrightarrow x^2+2x-x+2=2\)
\(\Leftrightarrow x^2+x=2-2\)
\(\Leftrightarrow x\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)
KL ::
PV
2
BT
11 tháng 3 2017
<=> \(\frac{3\left(x+2\right)-5x}{5.3}=\frac{2x-5}{2}< =>\frac{3x+6-5x}{15}=\frac{2x-5}{2}\) <=> 2(6-2x)=15(2x-5)
<=> 12-4x=30x-75 => 34x=87 => x=\(\frac{87}{34}\)
11 tháng 3 2017
\(\frac{x+2}{5}-\frac{x}{3}-\frac{2x-5}{2}=0\)0
\(\Leftrightarrow\frac{6\left(x+2\right)-10x-15\left(2x-5\right)}{30}\)=0
\(\Leftrightarrow6x+12-10x-30x+75\)=0
\(\Leftrightarrow-34x=-87\)
\(\Leftrightarrow x=\frac{87}{34}\)
Vay S={\(\frac{87}{34}\)}
3 tháng 4 2019
a,\(\frac{2x-5}{3}-\frac{3x-1}{2}< \frac{3-x}{5}-\frac{2x-1}{4}\)
\(\Leftrightarrow\frac{\left(2x-5\right)20}{60}-\frac{\left(3x-1\right)30}{60}< \frac{\left(3-x\right)12}{60}-\frac{\left(2x-1\right)15}{60}\)
\(\Leftrightarrow40x-100-90x+30< 36-12x-30x+15\)
\(\Leftrightarrow40x-90x+12x+30x< 36+15+100-30\)
\(\Leftrightarrow-8x< 121\)
\(\Leftrightarrow x>-\frac{378}{25}\)
\(\frac{x}{2x-6}+\frac{x}{2x+2}=\frac{2x^2}{x^2+2x-3}\)
\(ĐKXĐ:x^2+2x-3=\left(x+1\right)\left(x-3\right)\\ \Rightarrow x\ne-1;x\ne3\)
\(\frac{x}{2x-6}+\frac{x}{2x+2}=\frac{2x^2}{\left(x-3\right)\left(x+1\right)}\)
\(\Leftrightarrow\frac{x}{2\left(x-3\right)}+\frac{x}{2\left(x+1\right)}=\frac{2x^2}{\left(x-3\right)\left(x+1\right)}\)
\(\Leftrightarrow\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}=\frac{2x^2}{\left(x-3\right)\left(x+1\right)}\)
\(\Rightarrow x\left(x+1\right)+x\left(x-3\right)=4x^2\)
\(\Leftrightarrow x^2+x+x^2-3x=4x^2\)
\(\Leftrightarrow2x^2-2x=4x^2\)
\(\Leftrightarrow2x^2-4x^2-2x=0\)
\(\Leftrightarrow-2x^2-2x=0\)
\(\Leftrightarrow2x\left(-x-1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}2x=0\\-x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\left(N\right)\\x=-1\left(L\right)\end{cases}}\)
Tự kết luận tập nghiệm bạn nhé!
x2+2x-3 = (x+1)(x-3)
vậy MSC = 2(X+1(X-3) qui đồng mẫu số r làm dc r, đk x khác 1; -3