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1 tháng 3 2020
\(x-\frac{\frac{x}{2}-\frac{3+x}{4}}{2}=3-\frac{\left(1-\frac{6-x}{3}\right).\frac{1}{2}}{2}\)
\(\Leftrightarrow2x-\frac{x}{2}+\frac{3+x}{4}=6-\frac{1}{2}+\frac{6-x}{6}\)
\(\Leftrightarrow24x-6x+9+3x=72-6+12-2x\)
\(\Leftrightarrow23x=69\)
\(\Leftrightarrow x=3\)
Vậy nghiệm của pt x=3
T
0
P
22 tháng 4 2017
\(a,\Leftrightarrow5\left(x-2\right)-15x\le9+10\left(x+1\right)\)
\(\Leftrightarrow5x-10-15x\le9+10x+10\)
\(\Leftrightarrow-20x\le29\)
\(\Leftrightarrow x\ge-1,45\)
Vậy ...........
\(b,\Rightarrow\left(x+2\right)-3\left(x-3\right)=5\left(x-2\right)\)
\(\Leftrightarrow x+2-3x+9-5x+10=0\)
\(\Leftrightarrow-7x+21=0\)
\(\Leftrightarrow x=3\)
Vậy ..............
23 tháng 4 2017
\(\frac{x-2}{6}-\frac{x}{2}\le\frac{3}{10}+\frac{x+1}{3}\Leftrightarrow\frac{5\left(x-2\right)}{30}-\frac{15x}{30}\le\frac{9}{30}+\frac{10\left(x+1\right)}{30}\)
\(\Leftrightarrow5x-10-15x-9-10x-10\le0\)
\(\Leftrightarrow-20x-29\le0\Leftrightarrow\left(-20x\right)\cdot\frac{-1}{20}\ge29\cdot-\frac{1}{20}\)
\(\Leftrightarrow x\ge-\frac{29}{20}\)
TH
23 tháng 3 2019
a) \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}=\frac{x-4}{5}+\frac{x-5}{6}\)
\(\left(\frac{x-1}{2}+1\right)+\left(\frac{x-2}{3}+3\right)+\left(\frac{x-3}{4}+1\right)=\left(\frac{x-4}{5}+1\right)+\left(\frac{x-5}{6}+1\right)\)
\(\frac{x-1}{2}+\frac{x-1}{3}+\frac{x-1}{4}=\frac{x-1}{5}+\frac{x-1}{6}\)
\(\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\)=0
\(x-1=0\)
\(x=1\)
8 tháng 4 2018
\(\frac{x+2}{x+1}-\frac{3}{2-x}=\frac{-3}{\left(x+1\right)\left(x-2\right)}+2\)(1)
ĐKXĐ : \(x\ne-1;x\ne\pm2\)
Quy đồng và khử mẫu phương trình (1) , ta được :
\(\left(x+2\right)\left(2-x\right)\left(x-2\right)-3\left(x+1\right)\left(x-2\right)=-3\left(2-x\right)+2\left(x+1\right)\left(x-2\right)\left(2-x\right)\)
\(\Leftrightarrow-\left(x+2\right)\left(x-2\right)^2-3\left(x^2-x-2\right)=-6+3x-2\left(x+1\right)\left(x^2-4x+4\right)\)
\(\Leftrightarrow-\left(x-2\right)\left(x^2-4\right)-3x^2+3x+6=-6+3x-2\left(x^3-3x^2+4\right)\)
\(\Leftrightarrow-x^3+2x^2+4x-8-3x^2+3x+6=-6+3x-2x^3+6x^2-8\)
\(\Leftrightarrow-x^3-x^2+7x-2+6-3x+2x^3-6x^2+8=0\)
\(\Leftrightarrow x^3-7x^2+4x+12=0\)
\(\Leftrightarrow x^3-2x^2-5x^2+10x-6x+12=0\)
\(\Leftrightarrow x^2\left(x-2\right)-5x\left(x-2\right)-6\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2+x-6x-6\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x\left(x+1\right)-6\left(x+1\right)\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-6\right)\left(x+1\right)=0\)
\(\Leftrightarrow x=2\)(loại) ; \(x=6\)(chọn ) ; \(x=-1\)(loại).
Vậy S={6}.
HP
3
8 tháng 1 2019
\(x+\frac{x+1}{2}+\frac{x+2}{3}+\frac{x+3}{4}=1\)
\(\Rightarrow\frac{12x}{12}+\frac{6x+6}{12}+\frac{4x+8}{12}+\frac{3x+9}{12}=\frac{12}{12}\)
\(\Rightarrow25x+23=12\)
\(\Rightarrow x=\frac{-11}{25}\)
DD
1
21 tháng 4 2017
phương trình tương đương với 1+\(\frac{1}{x}+1+\frac{1}{x+3}\)=1+\(\frac{1}{x+1}+1+\frac{1}{x+2}\)\(\Leftrightarrow\frac{1}{x}+\frac{1}{x+3}=\frac{1}{x+2}+\frac{1}{x+1}\)
\(\Leftrightarrow\frac{2x+3}{x\left(x+3\right)}=\frac{2x+3}{\left(x+1\right)\left(x+2\right)}\)\(\Leftrightarrow\left(2x+3\right)\left(\frac{1}{x\left(x+3\right)}-\frac{1}{\left(x+1\right)\left(x+2\right)}\right)\)=0
\(\Leftrightarrow\left(2x+3\right)\left(\frac{\left(x+1\right)\left(x+2\right)-x\left(x+3\right)}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(\frac{2}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\right)=0\)\(\Leftrightarrow2x+3=0\Leftrightarrow x=\frac{-3}{2}\)
ĐK: x \(\ne\)-1; x \(\ne\)2
\(\frac{x+2}{x+1}+\frac{3}{x-2}=\frac{3}{x^2-x-2}+1\)
<=> \(\frac{\left(x+2\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}+\frac{3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{3}{\left(x+1\right)\left(x-2\right)}+\frac{\left(x+1\right)\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}\)
<=> x2 - 4 + 3x + 3 = 3 + x2 - x - 2
<=> x2 + 3x - x2 + x = 1 + 1
<=> 4x = 2
<=> x = 1/2
Vậy S = {1/2}