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tại sao không ai làm nhỉ?
(2x+1)(x+6) = ((X-2)(2X-3)
2x2 +12x+x +6 = 2x2 -3x-4x +6
20x =0
x =0
( 1 chữ cũng là thầy, nửa chữ cũng là thầy)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4\left(18-5x\right)-12\left(3.7\right)=15\left(2x-16\right)-6\left(2+14\right)\)
\(\Leftrightarrow72-20x-252=30x-240-96\)
\(\Leftrightarrow-20x-180=30x-336\)
\(\Leftrightarrow-50x=-156\)
\(\Leftrightarrow x=3,12\)
#H
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\(\left|x+3\right|+2\left|6+2x\right|+2016=4016\)
\(\left|x+3\right|+2\left|6+2x\right|=4016-2016\)
\(\left|x+3\right|+2\times2\left|3+x\right|=2000\)
\(1\left|x+3\right|+4\left|x+3\right|=2000\)
\(\left|x+3\right|\left(1+4\right)=2000\)
\(5\left|x+3\right|=2000\)
\(\left|x+3\right|=2000:5\)
\(\left|x+3\right|=400\)
Vậy x=-403 hoặc 397
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|x-3|-6=2x
<=> |x-3|-2x=6
Nếu \(x-3\ge0\Leftrightarrow x\ge3\). Ta có |x-3| = x-3
|x-3|-2x=6 <=> x-3-2x=6 <=> -x - 3 = 6 <=> -x = 9 <=> x = -9 < 3 ( loại )
Nếu \(x-3<0\Leftrightarrow x<3\). Ta có |x-3| = -x+3
|x-3|-2x=6 <=> -x+3-2x=6 <=> -3x + 3 = 6 <=> -3x=3 <=> x = -1 < 3 (thỏa mãn )
Vậy x = -1
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\(\frac{x}{4}=\frac{y}{3}\Rightarrow\frac{x}{12}=\frac{y}{9}\)
\(\frac{x}{3}=\frac{z}{5}\Rightarrow\frac{x}{12}=\frac{z}{20}\)
\(\Rightarrow\frac{x}{12}=\frac{y}{9}=\frac{z}{20}=\frac{2x-3y+z}{2\cdot12-3\cdot9+20}=\frac{6}{17}\)
\(\Rightarrow x=\frac{72}{17};y=\frac{54}{17};z=\frac{120}{17}\)
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a) \(\Leftrightarrow\left|2x-3\right|=\frac{1}{4}\Leftrightarrow\orbr{\begin{cases}x\ge\frac{3}{2}\mid:2x-3=\frac{1}{4}\Rightarrow2x=\frac{13}{4}\Rightarrow x=\frac{13}{8}\left(TM\right)\\x< \frac{3}{2}\mid:3-2x=\frac{1}{4}\Rightarrow2x=\frac{11}{4}\Rightarrow x=\frac{11}{8}\left(TM\right)\end{cases}.}\)
b) \(\Leftrightarrow\left|x-1\right|=\frac{3}{4}\Leftrightarrow\orbr{\begin{cases}x\ge1\mid:x-1=\frac{3}{4}\Rightarrow x=\frac{7}{4}\left(TM\right)\\x< 1\mid:1-x=\frac{3}{4}=>x=\frac{1}{4}\left(TM\right)\end{cases}}\)
c) \(\frac{3}{5\left(x-\frac{5}{6}\right)}-\frac{1}{2\left(\frac{3}{2}-1\right)}=-\frac{1}{4}\Leftrightarrow\frac{3}{\frac{5\left(6x-5\right)}{6}}-\frac{1}{2\cdot\frac{1}{2}}=-\frac{1}{4}\Leftrightarrow\frac{18}{5\left(6x-5\right)}=-\frac{1}{4}+1\)
\(\Leftrightarrow\frac{18}{5\left(6x-5\right)}=\frac{3}{4}\Leftrightarrow6x-5=\frac{24}{5}\Leftrightarrow6x=\frac{49}{5}\Leftrightarrow x=\frac{49}{30}\)
d) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Leftrightarrow\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(\Leftrightarrow2\cdot\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)
\(\Leftrightarrow2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2015}{2016}\Leftrightarrow2\cdot\frac{x+1-2}{2\left(x+1\right)}=\frac{2015}{2016}\Leftrightarrow\frac{x-1}{x+1}=\frac{2015}{2016}\)
\(\Leftrightarrow2016x-2016=2015x+2015\Leftrightarrow x=2015+2016=4031\)
Vậy x = 4031.
![](https://rs.olm.vn/images/avt/0.png?1311)
Vì GTTĐ luôn lớn hơn hoặc bằng 0, mà theo đề bài
=> 6 - 2x = 0 và x - 13 = 0
2x = 6 x = 13
x = 3
Vậy,................