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làm bừa thui,ai trên 11 điểm tích mình mình tích lại
Số số hạng là :
Có số cặp là :
50 : 2 = 25 ( cặp )
Mỗi cặp có giá trị là :
99 - 97 = 2
Tổng dãy trên là :
25 x 2 = 50
Đáp số : 50
1) \(\left(\left|x\right|-\frac{1}{8}\right)\left(-\frac{1}{8}\right)^5=\left(-\frac{1}{8}\right)^7\)
\(\left(\left|x\right|-\frac{1}{8}\right)\left(-\frac{1}{8}\right)^5=-\frac{1}{2097152}\)
\(\left(\left|x\right|-\frac{1}{8}\right)\left(-\frac{1}{32768}\right)=-\frac{1}{2097152}\)
\(\left(\left|x\right|-\frac{1}{8}\right)=\left(-\frac{1}{2097152}\right)\left(-32768\right)\)
\(\left|x\right|-\frac{1}{8}=\frac{1}{64}\)
\(\left|x\right|=\frac{1}{64}+\frac{1}{8}\)
\(x=\frac{9}{64}\)
\(\frac{1}{2}\left(\frac{4}{9}-x\right)-\frac{3}{2}\left(16-x\right)+\frac{1}{2}\left(5x+10\right)=0\)
\(\Leftrightarrow\frac{2}{9}-\frac{1}{2}x-24+\frac{3}{2}x+\frac{5}{2}x+5=0\)
\(\Leftrightarrow-\frac{169}{9}=\frac{7}{2}x\Leftrightarrow x=-\frac{338}{63}\)
Sai thì thông cảm cho mk nha
\(a,\Leftrightarrow\left|x+\dfrac{2}{5}\right|=\dfrac{7}{4}\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{2}{5}=\dfrac{7}{4}\left(x\ge-\dfrac{2}{5}\right)\\x+\dfrac{2}{5}=-\dfrac{7}{4}\left(x< -\dfrac{2}{5}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{27}{20}\left(tm\right)\\x=-\dfrac{43}{20}\left(tm\right)\end{matrix}\right.\)
\(b,\Leftrightarrow\left|x-\dfrac{13}{10}\right|=\dfrac{13}{10}\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{13}{10}=\dfrac{13}{10}\left(x\ge\dfrac{13}{10}\right)\\x-\dfrac{13}{10}=-\dfrac{13}{10}\left(x< \dfrac{13}{10}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{13}{5}\left(tm\right)\\x=0\left(tm\right)\end{matrix}\right.\)
\(c,\Leftrightarrow\left|\dfrac{3}{4}-\dfrac{1}{2}x\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{4}-\dfrac{1}{2}x=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\\\dfrac{1}{2}x-\dfrac{3}{4}=\dfrac{1}{2}\left(x>\dfrac{3}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{5}{2}\left(tm\right)\end{matrix}\right.\)
\(d,\Leftrightarrow\left|5-2x\right|=4\Leftrightarrow\left[{}\begin{matrix}5-2x=4\left(x\le\dfrac{5}{2}\right)\\2x-5=4\left(x>\dfrac{5}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(tm\right)\\x=\dfrac{9}{2}\left(tm\right)\end{matrix}\right.\)
\(đ,\Leftrightarrow\left\{{}\begin{matrix}x-3,5=0\\x-1,3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3,5\\x=1,3\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(e,\Leftrightarrow\left\{{}\begin{matrix}x-2021=0\\x-2022=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2021\\x=2022\end{matrix}\right.\left(vô.lí\right)\Leftrightarrow x\in\varnothing\)
\(f,\Leftrightarrow\left|x\right|=\dfrac{1}{3}-x\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}-x\left(x\ge0\right)\\x=x-\dfrac{1}{3}\left(x< 0\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{6}\left(tm\right)\\0x=-\dfrac{1}{3}\left(vô.lí\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{6}\)
\(g,\Leftrightarrow\left[{}\begin{matrix}x-2=x\left(x\ge2\right)\\2-x=x\left(x< 2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}0x=2\left(vô.lí\right)\\x=1\left(tm\right)\end{matrix}\right.\Leftrightarrow x=1\)
Tổng \(\left(x^2.y^2\right)+\left(x^3.y^3\right)+...+\left(x^{10}.y^{10}\right)\) có 9 số hạng (bạn tự tính nhá)
\(xy+\left(x^2.y^2\right)+\left(x^3.y^3\right)+...+\left(x^{10}.y^{10}\right)\) (1)
Thay x = -1 và y = 1 vào biểu thức (1) ta có :
\(-1.1+\left(\left(-1\right)^2.1^2\right)+\left(\left(-1^3\right).1^3\right)+...+\left(\left(-1\right)^{10}.1^{10}\right)\)
\(=-1+1+1+...+1\)
\(=-1+\left(1.9\right)\)
\(=8\)
Vậy 8 là giá trị của biểu thức \(xy+\left(x^2.y^2\right)+\left(x^3.y^3\right)+...+\left(x^{10}.y^{10}\right)\) tại x = -1 và y = 1
*\(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left[\frac{1}{14}+\frac{1}{7}-\left(-\frac{3}{35}\right)\right].\frac{4}{3}}=\frac{\left(\frac{18}{60}-\frac{16}{60}-\frac{21}{60}\right).\frac{5}{19}}{\left(\frac{5}{70}+\frac{10}{70}+\frac{6}{70}\right).\frac{4}{3}}=\frac{\frac{-19}{60}.\frac{5}{19}}{\frac{21}{70}.\frac{4}{3}}=\frac{\frac{-1}{12}}{\frac{14}{35}}=-\frac{1}{12}.\frac{35}{14}=\frac{-35}{168}\)
*\(\frac{\left(1+2+3+...+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(6,3.12-21.3,6\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(\frac{63}{10}.12-21.\frac{18}{5}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(\frac{378}{5}-\frac{378}{5}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)
=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).0}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}=0\)
a, \(\frac{3}{4}-x=\frac{1}{2}\Leftrightarrow x=\frac{3}{4}-\frac{1}{2}=\frac{1}{4}\)Vậy \(x=\frac{1}{4}\)
b, \(\left|x+\frac{2}{3}\right|=\frac{5}{6}\)
TH1 : \(x+\frac{2}{3}=\frac{5}{6}\Leftrightarrow x=\frac{5}{6}-\frac{2}{3}=\frac{1}{6}\)
TH2 : \(x+\frac{2}{3}=-\frac{5}{6}\Leftrightarrow x=-\frac{5}{6}-\frac{2}{3}=\frac{-9}{6}=\frac{-3}{2}\)
Vậy \(x=\left\{\frac{1}{6};-\frac{3}{2}\right\}\)
a,\(\frac{3}{4}-x=\frac{1}{2}\)
\(\Leftrightarrow x=\frac{3}{4}-\frac{1}{2}\)
\(\Leftrightarrow x=\frac{1}{4}\)
b,\(\left|x+\frac{2}{3}\right|=\frac{5}{6}\)
\(\Leftrightarrow x+\frac{2}{3}=\pm\frac{5}{6}\)
TH1:\(x+\frac{2}{3}=\frac{5}{6}\)
\(\Leftrightarrow x=\frac{5}{6}-\frac{2}{3}\)
\(\Leftrightarrow x=\frac{1}{6}\)
TH2:\(x+\frac{2}{3}=-\frac{5}{6}\)
\(\Leftrightarrow x=-\frac{5}{6}-\frac{2}{3}\)
\(\Leftrightarrow x=-\frac{3}{2}\)
bạn coi lại đề đi
ko có đề bài
lm sao mak lm đc
tìm x biết nha bạn