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a) \(2\sqrt{x}+3=0\)
\(2\sqrt{x}=-3\)
\(\sqrt{x}=\frac{-3}{2}\)
\(x=\frac{9}{4}\)
vậy \(x=\frac{9}{4}\)
b) \(\frac{5}{12}\sqrt{x}-\frac{1}{6}=\frac{1}{3}\)
\(\frac{5}{12}\sqrt{x}=\frac{1}{3}+\frac{1}{6}\)
\(\frac{5}{12}\sqrt{x}=\frac{1}{2}\)
\(\sqrt{x}=\frac{1}{2}:\frac{5}{12}\)
\(\sqrt{x}=\frac{6}{5}\)
\(x=\frac{36}{25}\)
vậy \(x=\frac{36}{25}\)
c) \(\sqrt{x+3}+3=0\)
\(\sqrt{x+3}=-3\)
\(\Rightarrow x\in\varnothing\) vì ko thỏa mãn ĐKXĐ của căn thức \(x\ge0\)
hay nói khác đi căn thức \(\sqrt{x+3}\) ko có nghĩa
vậy \(x\in\varnothing\)
1.
a) \(x-4\sqrt{x}=0\)
\(\Rightarrow\sqrt{x}.\left(\sqrt{x}-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\\sqrt{x}=0+4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\\sqrt{x}=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=16\end{matrix}\right.\)
Vậy \(x\in\left\{0;16\right\}.\)
b) \(\left|\frac{3}{5}\sqrt{x}-\frac{1}{20}\right|-\frac{3}{4}=\frac{1}{5}\)
\(\Rightarrow\left|\frac{3}{5}\sqrt{x}-\frac{1}{20}\right|=\frac{1}{5}+\frac{3}{4}\)
\(\Rightarrow\left|\frac{3}{5}\sqrt{x}-\frac{1}{20}\right|=\frac{19}{20}.\)
\(\Rightarrow\left[{}\begin{matrix}\frac{3}{5}\sqrt{x}-\frac{1}{20}=\frac{19}{20}\\\frac{3}{5}\sqrt{x}-\frac{1}{20}=-\frac{19}{20}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\frac{3}{5}\sqrt{x}=1\\\frac{3}{5}\sqrt{x}=-\frac{9}{10}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\sqrt{x}=1:\frac{3}{5}\\\sqrt{x}=\left(-\frac{9}{10}\right):\frac{3}{5}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{x}=\frac{5}{3}\\\sqrt{x}=-\frac{3}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{25}{9}\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x=\frac{25}{9}.\)
Câu c) làm tương tự như câu b).
Chúc bạn học tốt!
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
a) \(\frac{1}{4}+\frac{1}{3}:2x=-5\)
\(\frac{1}{3}:2x=\frac{-21}{4}\)
\(2x=\frac{-4}{63}\)
\(x=\frac{2}{63}\)
b) \(\left(3x-\frac{1}{4}\right)\left(x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-1}{2}\end{cases}}\)
Vậy.........
a) \(3-\sqrt{x}=\)0
\(\sqrt{x}=0+3\)
\(\sqrt{x}=3\)
mà :\(\sqrt{9}=3\)
=> x = 9
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\)\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1 +\frac{x+349}{5}-4=0\)
\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
\(\Leftrightarrow\)\(x+329=0\) (vì 1/327 + 1/326 + 1/325 + 1/324 + 1/5 khác 0 )
\(\Leftrightarrow\)\(x=-329\)
Bài 1 :
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
\(\Leftrightarrow\)\(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)
\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)
\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Vì \(\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)
\(\Rightarrow\)\(x+329=0\)
\(\Rightarrow\)\(x=-329\)
Vậy \(x=-329\)
\(a,2\sqrt{x}+3=0\)
\(\Leftrightarrow2\sqrt{x}=-3\)
\(\Leftrightarrow\sqrt{x}=-\frac{3}{2}\)( loại )
\(b,\frac{5}{12}\sqrt{x}-\frac{1}{6}=\frac{1}{3}\Leftrightarrow\frac{5}{12}\sqrt{x}=\frac{1}{2}\Leftrightarrow\sqrt{x}=\frac{6}{5}\Leftrightarrow x=\frac{36}{25}\)
\(c,\sqrt{x+3}+3=0\Leftrightarrow\sqrt{x+3}=-3\)( loại )