\(\orbr{\begin{cases}\\\end{cases}}6+\left(\frac{1}{2}\right)^3-\left|-\frac{1}{2}\right|\orbr{\begin{cases}\\\end{cases}\frac{3}{2}}\)
Những câu hỏi liên quan
\(\hept{\begin{cases}\\\end{cases}\hept{\begin{cases}\\\end{cases}}\hept{\begin{cases}\\\end{cases}}\hept{\begin{cases}\\\\\end{cases}}_{ }_{ }_{ }^2^2^{ }\orbr{\begin{cases}\\\end{cases}}\orbr{\begin{cases}\\\end{cases}}\orbr{\begin{cases}\\\end{cases}}\orbr{\begin{cases}\\\end{cases}}\frac{ }{ }\frac{ }{ }\frac{ }{ }\frac{ }{ }\sqrt[]{}\sqrt{ }}\)
Thế thì đừng hỏi trong khi câu mình ko biết mà người khác cũng ko biết đi cho đỡ phức tạp nhe bạn nhen
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g) |9-7x|5x-3Vì |9-7x|ge0;forall xRightarrow5x-3ge0Rightarrow xgefrac{3}{5}Ta có: |9-7x|5x-3Leftrightarroworbr{begin{cases}9-7x5x-39-7x3-5xend{cases}}Leftrightarroworbr{begin{cases}-7x-5x-3-9-7x+5x3-9end{cases}}Leftrightarroworbr{begin{cases}-12x-12-2x-6end{cases}}Leftrightarroworbr{begin{cases}x1frac{3}{5}left(chonright)x3frac{3}{5}left(chonright)end{cases}}Vậy xinleft{1;3right}h) 8x-|4x+1|x+2Leftrightarrow|4x+1|7x+2Vì |4x+1|ge0;forall xRightarrow7x+2ge0Rightarrow xgefrac{-2}{7}Ta có: |4x+...
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g) \(|9-7x|=5x-3\)
Vì \(|9-7x|\ge0;\forall x\)
\(\Rightarrow5x-3\ge0\)
\(\Rightarrow x\ge\frac{3}{5}\)
Ta có: \(|9-7x|=5x-3\)
\(\Leftrightarrow\orbr{\begin{cases}9-7x=5x-3\\9-7x=3-5x\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-7x-5x=-3-9\\-7x+5x=3-9\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-12x=-12\\-2x=-6\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=1>\frac{3}{5}\left(chon\right)\\x=3>\frac{3}{5}\left(chon\right)\end{cases}}\)
Vậy \(x\in\left\{1;3\right\}\)
h) \(8x-|4x+1|=x+2\)
\(\Leftrightarrow|4x+1|=7x+2\)
Vì \(|4x+1|\ge0;\forall x\)
\(\Rightarrow7x+2\ge0\)
\(\Rightarrow x\ge\frac{-2}{7}\)
Ta có: \(|4x+1|=7x+2\)
\(\Leftrightarrow\orbr{\begin{cases}4x+1=7x+2\\4x+1=-7x-2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}-3x=1\\11x=-3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{3}< \frac{-2}{7}\left(loai\right)\\x=\frac{-3}{11}>\frac{-2}{7}\left(chon\right)\end{cases}}\)
Vậy \(x=\frac{-3}{11}\)
P/s : Mọi người đừng để ý ạ !
\(\left|4-x\right|+2x=3\)
\(\Leftrightarrow\orbr{\begin{cases}4-x+2x=3\\x-4+2x=3\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\3x-7=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{7}{3}\end{cases}}\)
Vậy \(x\in\left\{-1;\frac{7}{3}\right\}\)
Tính hợp lí :\(\orbr{\begin{cases}\\\end{cases}9-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right):\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)}\)
\(\left[9-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)\right]\div\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)\)
\(=\left[\left(1+1+1+...+1\right)-\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\right)\right]\div\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)\)
có 9 số 1 có 9 số hạng
\(=\left[\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{3}\right)+\left(1-\frac{1}{4}\right)+...+\left(1-\frac{1}{10}\right)\right]\div\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)\)
\(=\left[\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right]\div\left(\frac{1}{2}+\frac{2}{3}+\frac{3}{4}+...+\frac{9}{10}\right)\)
\(=1\)
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\(\orbr{\begin{cases}\\4\frac{1}{2}:2,5-40\%+\left(-\frac{1}{5}\right)\end{cases}}.\frac{5}{9}\)
\(\left[4\frac{1}{2}:2,5-40\%+\left(-\frac{1}{5}\right)\right]\cdot\frac{5}{9}\)
\(=\left[\frac{9}{2}\cdot\frac{5}{2}-\frac{2}{5}+\left(-\frac{1}{5}\right)\right]\cdot\frac{5}{9}\)
\(\left[\frac{45}{4}-\frac{2}{5}+\left(-\frac{1}{5}\right)\right]\cdot\frac{5}{9}\)
\(=\frac{213}{20}\cdot\frac{5}{9}=\frac{71}{12}\)
xin ti c k
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\(\left(0,125+40\%-\frac{3}{40}\right)\):\(\orbr{\begin{cases}\\\end{cases}11\frac{3}{7}+8\frac{1}{2}-\left(\frac{13}{12}-5\frac{4}{7}\right)}\)
mk là KHÁNH LINH.
Tính tổng :
\(S_n=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{2_n+1}{\orbr{\begin{cases}\\n\left(n+1\right)^2\end{cases}}}\)
\(\Rightarrow S=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+..+\frac{1}{n^2}-\frac{1}{n+1^2}\)
\(\Rightarrow S=1-\frac{1}{n+1}\)
\(\Rightarrow S+\frac{n}{n+1}\)
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\(\orbr{\begin{cases}\\\end{cases}\left(-25\right)}.\left(-27\right).\left(-x\right)\orbr{\begin{cases}\\\end{cases}}:y\)với a = 5 ; b = -3
rut gọn biểu thức sau
a)\(3x^2-2x\left(5+1.5x\right)+10\)
b)\(7x\left(4y-x\right)+4y\left(y-7x\right)-2\left(2y^2-3.5x\right)\)
c)\(\hept{\begin{cases}\\\end{cases}}2x-3\left(x-1\right)-5\orbr{\begin{cases}\\\end{cases}x-4\left(3-2x\right)+10\orbr{\begin{cases}\\\end{cases}}}\)