Ôn tập chương 1

Hà Nguyễn Thanh Hải

a)\(3,6-\left|x-0,4\right|=0\)

b)\(\frac{x}{2}=y=\frac{z}{3}\)\(x-2y+z=210\)

c)\(\left|x+0,25\right|-4=\frac{1}{4}\)

d)\(x:\left(0,25\right)^4=\left(0,5\right)^2\)

e)\(3^{x-1}+5.3^{x-1}=162\)

f)\(\frac{x}{-25}=\frac{2}{5}\)

g)\(\left|x+\frac{3}{4}\right|-\frac{3}{4}=\sqrt[]{\frac{1}{9}}\)

Hưng Nguyễn Lê Việt
10 tháng 12 2019 lúc 20:03

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 \(x\in\left\{\frac{1}{3};-\frac{11}{6}\right\}\)

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Vũ Minh Tuấn
10 tháng 12 2019 lúc 20:57

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!

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