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1) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)
2) Ta có: \(\left(\dfrac{1}{2}-x\right)^2:\dfrac{9}{11}=\dfrac{11}{4}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=\dfrac{11}{4}\cdot\dfrac{9}{11}=\dfrac{9}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=\dfrac{3}{2}\\x-\dfrac{1}{2}=\dfrac{-3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy: \(x\in\left\{2;-1\right\}\)
3) Ta có: \(\left(\dfrac{1}{2}+\dfrac{1}{3}\right):x^3=\dfrac{-5}{6}\)
\(\Leftrightarrow\dfrac{5}{6}:x^3=\dfrac{-5}{6}\)
\(\Leftrightarrow x^3=-1\)
hay x=-1
Vậy: x=-1
4) Ta có: \(\left(11x-\dfrac{3}{4}\right)^3+21\dfrac{9}{17}=29\dfrac{9}{17}\)
\(\Leftrightarrow\left(11x-\dfrac{3}{4}\right)^3=8\)
\(\Leftrightarrow11x-\dfrac{3}{4}=2\)
\(\Leftrightarrow11x=\dfrac{11}{4}\)
\(\Leftrightarrow x=\dfrac{1}{4}\)
Vậy: \(x=\dfrac{1}{4}\)
5) Ta có: \(1\dfrac{5}{27}-\left(3\left|x\right|-\dfrac{7}{9}\right)^3=\dfrac{24}{27}\)
\(\Leftrightarrow\left(3\left|x\right|-\dfrac{7}{9}\right)^3=1+\dfrac{5}{27}-\dfrac{24}{27}\)
\(\Leftrightarrow\left(3\left|x\right|-\dfrac{7}{9}\right)^3=\dfrac{8}{27}\)
\(\Leftrightarrow3\left|x\right|-\dfrac{7}{9}=\dfrac{2}{3}\)
\(\Leftrightarrow3\left|x\right|=\dfrac{13}{9}\)
\(\Leftrightarrow\left|x\right|=\dfrac{13}{27}\)
hay \(x\in\left\{\dfrac{13}{27};-\dfrac{13}{27}\right\}\)
Vậy: \(x\in\left\{\dfrac{13}{27};-\dfrac{13}{27}\right\}\)
6) Ta có: \(\dfrac{8}{9}x-\dfrac{2}{3}=\dfrac{1}{3}x+1\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{8}{9}x-\dfrac{1}{3}x=\dfrac{4}{3}+\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{5}{9}x=2\)
hay \(x=\dfrac{18}{5}\)
Vậy: \(x=\dfrac{18}{5}\)
7) Ta có: \(x-25\%\cdot x-\dfrac{1}{2}x=-1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{4}x=-\dfrac{5}{4}\)
hay x=-5
Vậy: x=-5
8) Ta có: \(x^3-\dfrac{4}{25}x=0\)
\(\Leftrightarrow x\left(x^2-\dfrac{4}{25}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-\dfrac{4}{25}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{5}\\x=-\dfrac{2}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\dfrac{2}{5};-\dfrac{2}{5}\right\}\)
9) Ta có: \(8\dfrac{1}{2}-\left|2x-\dfrac{3}{4}\right|=-1\dfrac{3}{4}\)
\(\Leftrightarrow\left|2x-\dfrac{3}{4}\right|=\dfrac{17}{2}+\dfrac{7}{4}=\dfrac{41}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-\dfrac{3}{4}=-\dfrac{41}{4}\\2x-\dfrac{3}{4}=\dfrac{41}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{-19}{2}\\2x=11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{19}{4}\\x=\dfrac{11}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{-\dfrac{19}{4};\dfrac{11}{2}\right\}\)
10) Ta có: \(\dfrac{x-3}{x-5}=\left(-\dfrac{3}{5}\right)^2\)
\(\Leftrightarrow\dfrac{x-3}{x-5}=\dfrac{9}{25}\)
\(\Leftrightarrow25\left(x-3\right)=9\left(x-5\right)\)
\(\Leftrightarrow25x-75=9x-45\)
\(\Leftrightarrow25x-9x=-45+75=30\)
\(\Leftrightarrow16x=30\)
hay \(x=\dfrac{15}{8}\)
Vậy: \(x=\dfrac{15}{8}\)
Bài 7:
Số học sinh trung bình là:
\(45\cdot\dfrac{7}{15}=21\)(bạn)
Số học sinh khá là:
\(\left(45-21\right)\cdot\dfrac{5}{8}=24\cdot\dfrac{5}{8}=15\)(bạn)
Số học sinh giỏi là:
45-21-15=30-21=9(bạn)
Bài 9:
a) 32 trang sách chiếm:
\(1-\dfrac{1}{3}-\dfrac{2}{5}=\dfrac{15}{15}-\dfrac{5}{15}-\dfrac{6}{15}=\dfrac{4}{15}\)
Số trang của quyển sách là:
\(32:\dfrac{4}{15}=32\cdot\dfrac{15}{4}=120\)(trang)
b) Ngày thứ nhất Nam đọc được:
\(120\cdot\dfrac{2}{5}=48\)(trang)
Ngày thứ hai Nam đọc được:
\(120\cdot\dfrac{1}{3}=40\)(trang)
