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a) \(3,6-\left|x-0,4\right|=0\)
\(\Leftrightarrow\left|x-0,4\right|=3,6\)
\(\Leftrightarrow\left[{}\begin{matrix}x-0,4=3,6\\x-0,4=-3,6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3,2\end{matrix}\right.\)
Vậy \(x\in\left\{4;-3,2\right\}\)
b) Ta có:
\(\frac{x}{2}=y=\frac{z}{3}=\frac{2y}{2}=\frac{x-2y+z}{2-2+3}=\frac{210}{3}=70\)
\(\Rightarrow\left\{{}\begin{matrix}\frac{x}{2}=70\\y=70\\\frac{z}{3}=70\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=140\\y=70\\z=210\end{matrix}\right.\)
Vậy \(x=140\); \(y=70\); \(z=210\)
c)\(\left|x+0,25\right|-4=\frac{1}{4}\)
\(\Leftrightarrow\left|x+\frac{1}{4}\right|=\frac{17}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{1}{4}=\frac{17}{4}\\x+\frac{1}{4}=\frac{-17}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\frac{-9}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{4;\frac{-9}{2}\right\}\)
d) \(x:\left(0,25\right)^4=\left(0,5\right)^2\)
\(\Leftrightarrow x=\left(0,25\right)^4.\left(0,5\right)^2\)
\(\Leftrightarrow x=\left(0,5\right)^8.\left(0,5\right)^2\)
\(\Leftrightarrow x=\left(0,5\right)^{10}=\left(\frac{1}{2}\right)^{10}=\frac{1}{2^{10}}=\frac{1}{1024}\)
Vậy \(x=\frac{1}{1024}\)
e) \(3^{x-1}+5.3^{x-1}=162\)
\(\Leftrightarrow6.3^{x-1}=162\)
\(\Leftrightarrow3^{x-1}=27\)
\(\Leftrightarrow3^{x-1}=3^3\)
\(\Leftrightarrow x-1=3\)
\(\Leftrightarrow x=4\)
f) \(\frac{x}{-25}=\frac{2}{5}\)
\(\Leftrightarrow x=\left(-25\right).\frac{2}{5}=-10\)
Vậy \(x=-10\)
g) \(\left|x+\frac{3}{4}\right|-\frac{3}{4}=\sqrt{\frac{1}{9}}\)
\(\Leftrightarrow\left|x+\frac{3}{4}\right|-\frac{3}{4}=\frac{1}{3}\)
\(\Leftrightarrow\left|x+\frac{3}{4}\right|=\frac{13}{12}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{3}{4}=\frac{13}{12}\\x+\frac{3}{4}=-\frac{13}{12}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=-\frac{11}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{3};-\frac{11}{6}\right\}\)
a) \(3,6-\left|x-0,4\right|=0\)
\(\Rightarrow\left|x-0,4\right|=3,6-0\)
\(\Rightarrow\left|x-0,4\right|=3,6.\)
\(\Rightarrow\left[{}\begin{matrix}x-0,4=3,6\\x-0,4=-3,6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3,6+0,4\\x=\left(-3,6\right)+0,4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-3,2\end{matrix}\right.\)
Vậy \(x\in\left\{4;-3,2\right\}.\)
c) \(\left|x+0,25\right|-4=\frac{1}{4}\)
\(\Rightarrow\left|x+\frac{1}{4}\right|=\frac{1}{4}+4\)
\(\Rightarrow\left|x+\frac{1}{4}\right|=\frac{17}{4}.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{1}{4}=\frac{17}{4}\\x+\frac{1}{4}=-\frac{17}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{17}{4}-\frac{1}{4}\\x=\left(-\frac{17}{4}\right)-\frac{1}{4}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-\frac{9}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{4;-\frac{9}{2}\right\}.\)
d) \(x:\left(0,25\right)^4=\left(0,5\right)^2\)
\(\Rightarrow x:\left(0,25\right)^4=0,25\)
\(\Rightarrow x=\left(0,25\right).\left(0,25\right)^4\)
\(\Rightarrow x=\left(0,25\right)^5\)
\(\Rightarrow x=\frac{1}{1024}\)
Vậy \(x=\frac{1}{1024}.\)
Chúc bạn học tốt!
a)\(\left|x+\frac{1}{5}\right|-4=-2\)
\(\Rightarrow\left|x+\frac{1}{5}\right|=2\)
\(\Rightarrow x+\frac{1}{5}=2\) hoặc \(-2\)
Xét \(x+\frac{1}{5}=2\Leftrightarrow x=\frac{9}{5}\)
Xét \(x+\frac{1}{5}=-2\Leftrightarrow x=-\frac{11}{5}\)
a) | x - 1/5 | + | 2x - 0,4 | >= 0 với mọi x
theo đề : | x - 1/5 | + | 2x - 0,4 | = 0
=> \(\hept{\begin{cases}x-\frac{1}{5}=0\\2x-0,4=0\end{cases}}\)=> \(\hept{\begin{cases}x=\frac{1}{5}\\x=\frac{1}{5}\end{cases}}\)=> x = \(\frac{1}{5}\)
Thử lại ta thấy đúng
Vậy x = \(\frac{1}{5}\)
b) Ta có : | x + 1/4 | + | 2x + 0,5 | >= 0 với mọi x
theo đề : | x + 1/4 | + | 2x + 0,5 | = 0
Giải tương tự câu a ta được x = -1/4
nhớ nha
1.
a) \(x\in\left\{4;5;6;7;8;9;10;11;12;13\right\}\)
b) x=0
d) \(x=\frac{-1}{35}\) hoặc \(x=\frac{-13}{35}\)
e) \(x=\frac{2}{3}\)
a, \(\frac{3}{5}\left(2x-\frac{1}{3}\right)+\frac{4}{15}=\frac{12}{30}\)
\(\Leftrightarrow\frac{3}{5}\left(2x-\frac{1}{3}\right)=\frac{2}{15}\)
\(\Leftrightarrow2x-\frac{1}{3}=\frac{2}{9}\)
\(\Leftrightarrow2x=\frac{5}{9}\)
\(\Leftrightarrow x=\frac{5}{18}\)
b,\(\left(-0,2\right)^x=\frac{1}{25}\)
\(\Leftrightarrow\left(\frac{-1}{5}\right)^x=\left(\frac{-1}{5}\right)^2\)
\(\Leftrightarrow x=2\)
c,\(\left|x-1\right|-\frac{3}{12}=\left(-\frac{1}{2}\right)^2\)
\(\Leftrightarrow\left|x-1\right|-\frac{3}{12}=\frac{1}{4}\)
\(\Leftrightarrow\left|x-1\right|=\frac{1}{2}\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=\frac{1}{2}\\x-1=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{1}{2}\end{cases}}\)
\(a,\frac{3}{5}\left(2x-\frac{1}{3}\right)=\frac{12}{30}-\frac{4}{15}\)
\(\frac{3}{5}\left(2x-\frac{1}{3}\right)=\frac{2}{15}\)
\(2x-\frac{1}{3}=\frac{2}{9}\)
\(x=\frac{5}{18}\)
\(b,\left(-0,2\right)^x=\frac{1}{25}\)
\(\left(-0,2\right)^x=\left(-\frac{1}{5}\right)^2\)
\(\left(-0,2\right)^x=\left(-0,2\right)^2\)
\(x=2\)
c,/x-1/=1/2
Nếu
\(x-1\ge0\)
\(x\ge1\)
suy ra x-1=1/2
x=3/2(thỏa mãn điều kiện )
nếu \(x-1\le0\)
\(x\le1\)
suy ra x-1=-1/2
x=1/2 (thỏa mãn điều kiện )
Vậy ...
nha !!!
a)\(-x^2\left(x^2-4\right)=-25\left(x^2-4\right)\)
\(\Leftrightarrow-x^2=-25\)
\(\Leftrightarrow x^2=25\)
\(\Leftrightarrow x=\pm5\)
a/ \(4^2-x=36^4:12^4\)
\(\Rightarrow16-x=\left(36:12\right)^4\)
\(\Leftrightarrow16-x=3^4\)
\(\Leftrightarrow16-x=81\)
\(\Leftrightarrow x=-65\)
b/ \(3,6-\left|x-0,4\right|=15-\left|-5\right|\)
\(\Leftrightarrow3,6-\left|x-0,4\right|=15-5\)
\(\Leftrightarrow3,6-\left|x-0,4\right|=10\)
\(\Leftrightarrow\left|x-0,4\right|=-6,4\) (vô lí)
=> x thuộc rỗng