1) \(\left|5x-4\right|=\left|x+2\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-4=x+2\\5x-4=-\left(x+2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-4=x+2\\5x-4=-x-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-x=2+4\\5x+x=-2+4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=6\\6x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy \(x=\dfrac{1}{3}\) hoặc \(x=\dfrac{3}{2}\)

2) \(\left|x+15\right|=\left|3x-4\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}x+15=3x-4\\x+15=-\left(3x-4\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+15=3x-4\\x+15=-3x+4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3x=-4-15\\x+3x=4-15\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x=-19\\4x=-11\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{19}{2}\\x=-\dfrac{11}{4}\end{matrix}\right.\)

Vậy \(x=-\dfrac{11}{4}\) hoặc \(x=\dfrac{19}{2}\)

3) \(\left|\dfrac{5}{4}x-\dfrac{7}{2}\right|-\left|\dfrac{5}{8}x+\dfrac{3}{5}\right|=0\)

\(\Leftrightarrow\left|\dfrac{5}{4}x-\dfrac{7}{2}\right|=\left|\dfrac{5}{8}x+\dfrac{3}{5}\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{4}x-\dfrac{7}{2}=\dfrac{5}{8}x+\dfrac{3}{5}\\\dfrac{5}{4}x-\dfrac{7}{2}=-\left(\dfrac{5}{8}x+\dfrac{3}{5}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{4}x-\dfrac{7}{2}=\dfrac{5}{8}x+\dfrac{3}{5}\\\dfrac{5}{4}x-\dfrac{7}{2}=-\dfrac{5}{8}x-\dfrac{3}{5}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{4}x-\dfrac{5}{8}x=\dfrac{3}{5}+\dfrac{7}{2}\\\dfrac{5}{4}x+\dfrac{5}{8}x=-\dfrac{3}{5}+\dfrac{7}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{8}x=\dfrac{41}{10}\\\dfrac{15}{8}x=\dfrac{29}{10}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{164}{25}\\x=\dfrac{116}{75}\end{matrix}\right.\)

Vậy \(x=\dfrac{116}{75}\) hoặc \(x=\dfrac{164}{25}\)

4) \(\left|2x-6\right|-\left|x+3\right|=0\)

\(\Leftrightarrow\left|2x-6\right|=\left|x+3\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-6=x+3\\2x-6=-\left(x+3\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-6=x+3\\2x-6=-x-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3+6\\2x+x=-3+6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\3x=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=9\\x=1\end{matrix}\right.\)

Vậy \(x=1\) hoặc \(x=9\)

@@ nhiều z @@

\(1\text{)} \left(x-3\right)-9.\left(2x-7\right)=16:2\\ x-3-18x+63=8\\ -17x=-52\\ x=\dfrac{52}{17}\)

\(2\text{)} x-\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\\ -\dfrac{1}{6}x=\dfrac{5}{12}\\ x=-2,5\)

\(3\text{)} -2.\left(x-1\right)+5.\left(x-2\right)=x-14\\ -2x+2+5x-10=x-14\\ 2x=-6\\ x=-3\)

\(4\text{)} x.\left(x+1\right)-x-1=0\\ \left(x+1\right)\left(x-1\right)=0\\ x^2-1=0\\ x^2=1\\ x=\pm1\)

Cảm ơn rất nhiều!

1. |x| - x = 0
<=> |x| = x
<=> \(\left[{}\begin{matrix}x=x\\x=-x\end{matrix}\right.\) (thỏa mãn)
@Phan Đức Gia Linh

2. |x| + x = 0
<=> |x| = -x
Do |x| \(\ge\) 0, mà -x < 0 => không tồn tại x thỏa mãn
@Phan Đức Gia Linh

1. x³ - \(\dfrac{4}{25}\)x = 0
<=> x(x² - \(\dfrac{4}{25}\)) = 0
<=> \(\left[{}\begin{matrix}x=0\\x^2-\dfrac{4}{25}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2=\dfrac{4}{25}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\end{matrix}\right.\) (thỏa mãn)
Vậy x = 0; 2/5
@Phan Đức Gia Linh

1 ) \(x^3-\dfrac{4}{25}x=0\)

\(\Leftrightarrow x\left(x^2-\dfrac{4}{25}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{4}{25}=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\left[{}\begin{matrix}x-\dfrac{2}{5}=0\\x+\dfrac{2}{5}=0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm\dfrac{2}{5}\end{matrix}\right.\)

Vậy .............

2 ) \(3^{4x+4}=9^{x+2}\)

\(\Leftrightarrow3^{4x+4}=\left(3^2\right)^{x+2}\)

\(\Leftrightarrow4x+4=2x+4\)

\(\Leftrightarrow2x=0\Leftrightarrow x=0.\)

3 ) \(3\left(\dfrac{1}{5.8}+\dfrac{1}{8.11}+\dfrac{1}{11.14}+...+\dfrac{1}{97.100}\right)=\dfrac{319}{100}\) ( thiếu đề hay sao )

4 ) \(\left(6-x\right)^{2014}=\left(6-x\right)^{2015}\)

\(\Leftrightarrow\left(6-x\right)^{2014}-\left(6-x\right)^{2015}=0\)

\(\Leftrightarrow\left(6-x\right)^{2014}\left(1-6+x\right)=0\)

\(\Leftrightarrow\left(6-x\right)^{2014}\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(6-x\right)^{2014}=0\\x-5=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=5\end{matrix}\right.\)

Vậy ......

5) \(2+4+6+...+2x=210\)

\(\Leftrightarrow2.1+2.2+2.3+...+2.x=210\)

\(\Leftrightarrow2\left(1+2+3+...+x\right)=210\)

\(\Leftrightarrow1+2+3+...+x=105\)

\(\Leftrightarrow\dfrac{\left(x+1\right).x}{2}=105\)

\(\Leftrightarrow x\left(x+1\right)=210\)

Ta lại có : \(x\left(x+1\right)=14\left(14+1\right)\)

\(\Leftrightarrow x=14\)

Vậy ......

6 ) \(\dfrac{1}{21}+\dfrac{1}{28}+\dfrac{1}{36}+..+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{1}{3.7}+\dfrac{1}{4.7}+\dfrac{1}{4.7}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{2}{2.3.7}+\dfrac{2}{2.4.7}+\dfrac{2}{2.4.9}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{2}{6.7}+\dfrac{2}{8.7}+\dfrac{2}{8.9}+...+\dfrac{2}{x\left(x+1\right)}=\dfrac{2}{9}\)

\(\Leftrightarrow2\left(\dfrac{1}{6.7}+\dfrac{1}{8.7}+\dfrac{1}{8.9}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{2}{9}\)

\(\Leftrightarrow2.\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{7}+\dfrac{1}{8}-\dfrac{1}{9}+...+\dfrac{1}{\dfrac{x-1}{x+1}}\right)=\dfrac{2}{9}\)

\(\Leftrightarrow\dfrac{1}{6}+\dfrac{1}{x+1}=\dfrac{1}{9}\)

\(\Leftrightarrow\dfrac{1}{x+1}=\dfrac{1}{18}\)

\(\Leftrightarrow x=17.\)

Vậy ...........

\(\)

\(\left|x-2\right|+\left|2y-5\right|=0\)

\(\left\{{}\begin{matrix}\left|x-2\right|\ge0\forall x\\\left|2y-5\right|\ge0\forall y\end{matrix}\right.\)

\(\left|x-2\right|+\left|2y-5\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|x-2\right|=0\Rightarrow x=2\\\left|2y-5\right|=0\Rightarrow2y=5\Rightarrow y=\dfrac{5}{2}\end{matrix}\right.\)

\(\left|3y-2\right|+\left|xy-6\right|=0\)

\(\left\{{}\begin{matrix} \left|3y-2\right|\ge0\forall y\\\left|xy-6\right|\ge0\forall x;y\end{matrix}\right.\)

\(\Rightarrow\left|3y-2\right|+\left|xy-6\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|3y-2\right|=0\Rightarrow3y=2\Rightarrow y=\dfrac{3}{2}\\\left|xy-6\right|=0\Rightarrow\dfrac{3}{2}x=6\Rightarrow x=4\end{matrix}\right.\)

\(\left|x-\dfrac{1}{2}\right|+\left|2y-\dfrac{1}{3}\right|+\left|4z-5\right|\le0\)

\(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|\ge0\forall x\\\left|2y-\dfrac{1}{3}\right|\ge0\forall y\\ \left|4z-5\right|\ge0\forall z\end{matrix}\right.\)

\(\Rightarrow\left|x-\dfrac{1}{2}\right|+\left|2y-\dfrac{1}{3}\right|+\left|4z-5\right|\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left|x-\dfrac{1}{2}\right|+\left|2y-\dfrac{1}{3}\right|+\left|4z-5\right|\ge0\\\left|x-\dfrac{1}{2}\right|+\left|2y-\dfrac{1}{3}\right|+\left|4z-5\right|\le0\end{matrix}\right.\)

\(\Rightarrow\left|x-\dfrac{1}{2}\right|+\left|2y-\dfrac{1}{3}\right|+\left|4z-5\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|=0\Rightarrow x=\dfrac{1}{2}\\\left|2y-\dfrac{1}{3}\right|=0\Rightarrow2y=\dfrac{1}{3}\Rightarrow y=\dfrac{1}{6}\\\left|4z-5\right|=0\Rightarrow4z=5\Rightarrow z=\dfrac{5}{4}\end{matrix}\right.\)

THANKS, THANKS!

a) Ta có: \(\dfrac{1}{7}+x=-\dfrac{2}{3}\)

\(\Leftrightarrow x=-\dfrac{2}{3}-\dfrac{1}{7}=\dfrac{-14}{21}-\dfrac{3}{21}\)

hay \(x=-\dfrac{17}{21}\)

Vậy: \(x=-\dfrac{17}{21}\)

b) Ta có: \(\dfrac{-2}{3}:x=\dfrac{-5}{6}\)

\(\Leftrightarrow x=\dfrac{-2}{3}:\dfrac{-5}{6}=\dfrac{-2}{3}\cdot\dfrac{6}{-5}=\dfrac{-12}{-15}=\dfrac{4}{5}\)

Vậy: \(x=\dfrac{4}{5}\)

c) Ta có: \(\left(\dfrac{3}{5}-2x\right)\cdot\dfrac{5}{8}=1\)

\(\Leftrightarrow\left(\dfrac{3}{5}-2x\right)=1:\dfrac{5}{8}=\dfrac{8}{5}\)

\(\Leftrightarrow-2x=\dfrac{8}{5}-\dfrac{3}{5}=1\)

hay \(x=-\dfrac{1}{2}\)

Vậy: \(x=-\dfrac{1}{2}\)

d) Ta có: \(\dfrac{3}{4}+\dfrac{2}{5}x=\dfrac{29}{60}\)

\(\Leftrightarrow x\cdot\dfrac{2}{5}=\dfrac{29}{60}-\dfrac{3}{4}=\dfrac{29}{60}-\dfrac{45}{60}=\dfrac{-16}{60}=\dfrac{-4}{15}\)

hay \(x=\dfrac{-4}{15}:\dfrac{2}{5}=\dfrac{-4}{15}\cdot\dfrac{5}{2}=\dfrac{-20}{30}=-\dfrac{2}{3}\)

Vậy: \(x=-\dfrac{2}{3}\)

e) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}:x=\dfrac{2}{5}\)

\(\Leftrightarrow\dfrac{1}{4}:x=\dfrac{2}{5}-\dfrac{3}{4}=\dfrac{8}{20}-\dfrac{15}{20}=\dfrac{-7}{20}\)

hay \(x=-\dfrac{1}{4}:\dfrac{7}{20}=\dfrac{-1}{4}\cdot\dfrac{20}{7}=\dfrac{-20}{28}=\dfrac{-5}{7}\)

Vậy: \(x=-\dfrac{5}{7}\)

f) Ta có: \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)

\(\Leftrightarrow-x+\dfrac{11}{12}-\dfrac{2}{5}-\dfrac{2}{3}=0\)

\(\Leftrightarrow-x+\dfrac{55}{60}-\dfrac{24}{60}-\dfrac{40}{60}=0\)

\(\Leftrightarrow-x-\dfrac{9}{60}=0\)

\(\Leftrightarrow-x=\dfrac{9}{60}=\dfrac{3}{20}\)

hay \(x=-\dfrac{3}{20}\)

Vậy: \(x=-\dfrac{3}{20}\)

g) Ta có: \(\left|x+\dfrac{1}{3}\right|-4=\dfrac{-1}{2}\)

\(\Leftrightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{-1}{2}+4=\dfrac{-1}{2}+\dfrac{8}{2}=\dfrac{7}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{7}{2}\\x+\dfrac{1}{3}=-\dfrac{7}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{21}{6}-\dfrac{2}{6}=\dfrac{19}{6}\\x=-\dfrac{7}{2}-\dfrac{1}{3}=\dfrac{-21}{6}-\dfrac{2}{6}=\dfrac{-23}{6}\end{matrix}\right.\)

Vậy: \(x\in\left\{\dfrac{19}{6};-\dfrac{23}{6}\right\}\)

cảm ơn cậu cutee gì đó ơi nha

1: \(x^2\left(2-x\right)\le0\)

\(\Leftrightarrow2-x\le0\)

hay x>=2

2: \(\left(x-7\right)\left(x+3\right)< 0\)

=>x+3>0 và x-7<0

=>-3<x<7

3: \(\left(x+4\right)\left(x-3\right)>0\)

=>x-3>0 hoặc x+4<0

=>x>3 hoặc x<-4

Lời giải:

a.

 \(\frac{10}{x+2}=\frac{60}{6(x+2)}=\frac{60(x-2)}{6(x+2)(x-2)}=\frac{60(x-2)}{6(x^2-4)}\)

\(\frac{5}{2x-4}=\frac{15(x+2)}{6(x-2)(x+2)}=\frac{15(x+2)}{6(x^2-4)}\)

\(\frac{1}{6-3x}=\frac{x+2}{3(2-x)}=\frac{2(x+2)^2}{6(2-x)(2+x)}=\frac{-2(x+2)^2}{6(x^2-4)}\)

b.

\(\frac{1}{x+2}=\frac{x(2-x)}{x(x+2)(2-x)}=\frac{x(2-x)}{x(4-x^2)}\)

\(\frac{8}{2x-x^2}=\frac{8(x+2)}{(x+2)x(2-x)}=\frac{8(x+2)}{x(4-x^2)}\)

c.

\(\frac{4x^2-3x+5}{x^3-1}\)

\(\frac{1-2x}{x^2+x+1}=\frac{(1-2x)(x-1)}{(x-1)(x^2+x+1)}=\frac{-2x^2+3x-1}{x^3-1}\)

\(-2=\frac{-2(x^3-1)}{x^3-1}\)

\(a,x=\dfrac{1}{5}+\dfrac{-3}{7}\)

   \(x=\dfrac{7}{35}+\dfrac{-15}{35}\)

   \(x=-\dfrac{8}{35}\)

\(b,\dfrac{3}{5}-\dfrac{4}{7}:x=\dfrac{-9}{10}\)

           \(\dfrac{4}{7}:x=\dfrac{3}{5}-\dfrac{-9}{10}\)

           \(\dfrac{4}{7}:x=\dfrac{3}{2}\)

                 \(x=\dfrac{4}{7}:\dfrac{3}{2}\)

                 \(x=\dfrac{4}{7}\times\dfrac{2}{3}\)

                 \(x=\dfrac{8}{21}\)

\(c,x-\left(\dfrac{-3}{4}\right)=\dfrac{-2}{3}-\dfrac{1}{2}\)

   \(x+\dfrac{3}{4}=\dfrac{-4}{6}-\dfrac{3}{6}\)

   \(x+\dfrac{3}{4}=-\dfrac{7}{6}\)

           \(x=-\dfrac{7}{6}-\dfrac{3}{4}\)

           \(x=-\dfrac{23}{12}\)

\(d,\dfrac{-5}{9}-x=\dfrac{1}{3}+\dfrac{7}{18}\)

    \(\dfrac{-5}{9}-x=\dfrac{6}{18}+\dfrac{7}{18}\)

     \(\dfrac{-5}{9}-x=\dfrac{13}{18}\)

                \(x=\dfrac{-5}{9}-\dfrac{13}{18}\)

                \(x=\dfrac{-10}{18}-\dfrac{13}{18}\)

                \(x=-\dfrac{23}{18}\)