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23 tháng 10 2018

a)Với x>=0

 \(\frac{5}{11}\sqrt{x}=\frac{1}{6}+\frac{1}{3}=\frac{1}{2}\)

\(\sqrt{x}=\frac{1}{2}:\frac{5}{11}=\frac{11}{10}\)

\(x=\frac{11^2}{10^2}=\frac{121}{100}\)(thỏa mãn)

b) x=0

c) \(x=\pm\sqrt{3}\)vì x<0 => \(x=-\sqrt{3}\)

d) x=1 hoặc -1

e) \(x=\pm\sqrt{2}\)

23 tháng 10 2018

\(a,\frac{5}{11}\sqrt{x}-\frac{1}{3}=\frac{1}{6}.\)

\(\frac{5}{11}\sqrt{x}=\frac{1}{6}+\frac{1}{3}\)

\(\frac{5}{11}\sqrt{x}=\frac{1}{2}\)

\(\sqrt{x}=\frac{1}{2}:\frac{5}{11}\)

\(\sqrt{x}=\frac{11}{10}\)

\(\Rightarrow x=\frac{121}{100}\)

\(b.x^2=0\)

\(\Leftrightarrow x=0\)

\(c.x^2=3\left(x< 0\right)\)

\(\Leftrightarrow x=-\sqrt{3}\)

\(d.x^2=1\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-1\end{cases}}\)

\(e.x^2=2\)

\(\Leftrightarrow x=\sqrt{2}\)

28 tháng 11 2016

Bài 2:

a) \(\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|-6x=0\)

\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=6x\)

Ta có: \(\left|x+1\right|\ge0;\left|x+2\right|\ge0;\left|x+4\right|\ge0;\left|x+5\right|\ge0\)

\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|\ge0\)

\(\Rightarrow6x\ge0\)

\(\Rightarrow x\ge0\)

\(\Rightarrow\left|x+1\right|+\left|x+2\right|+\left|x+4\right|+\left|x+5\right|=x+1+x+2+x+4+x+5=6x\)

\(\Rightarrow4x+12=6x\)

\(\Rightarrow2x=12\)

\(\Rightarrow x=6\)

Vậy x = 6

b) Giải:

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x-2}{2}=\frac{y-3}{3}=\frac{z-3}{4}=\frac{2y-6}{6}=\frac{3z-9}{12}=\frac{x-2-2y+6+3z-9}{2-6+12}=\frac{\left(x-2y+3z\right)-\left(2-6+9\right)}{8}\)

\(=\frac{14-5}{8}=\frac{9}{8}\)

+) \(\frac{x-2}{2}=\frac{9}{8}\Rightarrow x-2=\frac{9}{4}\Rightarrow x=\frac{17}{4}\)

+) \(\frac{y-3}{3}=\frac{9}{8}\Rightarrow y-3=\frac{27}{8}\Rightarrow y=\frac{51}{8}\)

+) \(\frac{z-3}{4}=\frac{9}{8}\Rightarrow z-3=\frac{9}{2}\Rightarrow z=\frac{15}{2}\)

Vậy ...

c) \(5^x+5^{x+1}+5^{x+2}=3875\)

\(\Rightarrow5^x+5^x.5+5^x.5^2=3875\)

\(\Rightarrow5^x.\left(1+5+5^2\right)=3875\)

\(\Rightarrow5^x.31=3875\)

\(\Rightarrow5^x=125\)

\(\Rightarrow5^x=5^3\)

\(\Rightarrow x=3\)

Vậy x = 3

28 tháng 11 2016

@@ good :D

1) Tìm x:

a) \(\frac{11}{12}-\frac{5}{12}.\left(\frac{2}{5}+x\right)=\frac{2}{3}\)

\(\Leftrightarrow\frac{5}{12}.\left(\frac{2}{5}+x\right)=\frac{11}{12}-\frac{2}{3}=\frac{1}{4}\)

\(\Leftrightarrow\frac{2}{5}+x=\frac{1}{4}:\frac{5}{12}=\frac{3}{5}\)

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

b) \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)

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

\(\Leftrightarrow x=-\frac{7}{20}:\frac{1}{4}=\frac{-7}{5}\)

7 tháng 7 2019

a) \(\frac{11}{12}-\frac{5}{12}\left(\frac{2}{5}+x\right)=\frac{2}{3}\)

\(\Leftrightarrow\frac{11}{12}-\frac{5}{12}.\frac{2}{5}-\frac{5}{12}x=\frac{2}{3}\)

\(\Leftrightarrow\frac{11}{12}-\frac{1}{6}-\frac{5}{12}x=\frac{2}{3}\)

\(\Leftrightarrow\frac{-5}{12}x=\frac{2}{3}-\frac{11}{12}+\frac{1}{6}\)

\(\Leftrightarrow-\frac{5}{12}x=\frac{8}{12}-\frac{11}{12}+\frac{2}{12}=-\frac{1}{12}\)

\(\Leftrightarrow x=\frac{-1}{12}:\left(-\frac{5}{12}\right)=-\frac{1}{12}.\left(-\frac{12}{5}\right)=\frac{1}{5}\)

Vậy x = 1/5

b) \(\frac{3}{4}+\frac{1}{4}:x=\frac{2}{5}\)

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

\(\Leftrightarrow x=\frac{1}{4}:\left(-\frac{7}{20}\right)=\frac{1}{4}.\left(-\frac{20}{7}\right)=-\frac{5}{7}\)

Vậy x = -5/7

c) \(2x\left(x-\frac{1}{7}\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-\frac{1}{7}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{7}\end{matrix}\right.\)

d) \(\left(x+1\right)\left(x-2\right)< 0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1< 0\\x-2>0\end{matrix}\right.\\\left\{{}\begin{matrix}x+1>0\\x-2< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x>-1\\x< 2\end{matrix}\right.\end{matrix}\right.\)

Ta thấy x <-1 và x >2 vô lí

Do đó: x >-1 và x <2

Vậy -1 < x <2

e) \(\left(x-2\right)\left(x+\frac{2}{3}\right)>0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-2>0\\x+\frac{2}{3}>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-2< 0\\x+\frac{2}{3}< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>2\\x>-\frac{2}{3}\end{matrix}\right.\\\left\{{}\begin{matrix}x< 2\\x< -\frac{2}{3}\end{matrix}\right.\end{matrix}\right.\)

Vậy x > 2 hoặc x < -2/3