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a: \(\Leftrightarrow-\dfrac{4}{9}-\dfrac{2}{9}+\dfrac{2}{3}< =x< =\dfrac{11}{7}+\dfrac{3}{7}+\dfrac{2}{5}-\dfrac{7}{5}\)
=>0<=x<=2-1=1
hay \(x\in\left\{0;1\right\}\)
b: \(\Leftrightarrow-\dfrac{8}{13}+\dfrac{21}{13}+\dfrac{7}{17}< =x< =\dfrac{-9}{14}-\dfrac{5}{14}+3\)
=>24/17<=x<=2
hay x=2
Ta có : \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)
\(B=-\dfrac{5}{6}+\dfrac{17}{7}-\dfrac{3}{7}-\dfrac{1}{6}=2-\dfrac{6}{6}=1\)
mà 99/100 < 1 hay A < B
bạn đăng tách cho mn giúp nhé
Bài 6 :
\(\Rightarrow30-3y=xy\Leftrightarrow xy+3y=30\Leftrightarrow y\left(x+3\right)=30\)
\(\Rightarrow x+3;y\inƯ\left(30\right)=\left\{\pm1;\pm2;\pm3;\pm5;\pm6;\pm10;\pm15;\pm30\right\}\)
x + 3 | 1 | -1 | 2 | -2 | 3 | -3 | 5 | -5 | 6 | -6 | 10 | -10 | 15 | -15 | 30 | -30 |
y | 30 | -30 | 15 | -15 | 10 | -10 | 6 | -6 | 5 | -5 | 3 | -3 | 2 | -2 | 1 | -1 |
x | -2 | -4 | -1 | -5 | 0 | -6 | 2 | -8 | 3 | -9 | 7 | -13 | 12 | -18 | 27 | -33 |
a: \(\Leftrightarrow0< =x< =1\)
hay \(x\in\left\{0;1\right\}\)
b: \(\Leftrightarrow\dfrac{18}{17}< =x< =2\)
hay x=2
Bài 3:
\(a,\left(5x-2\right)+\left(-3x+1\right)=\left(-42\right)-\left(-91\right)\\ \Rightarrow5x-2+\left(-3x\right)+1=50\\ \Rightarrow2x-1=49\\ \Rightarrow2x=50\\ \Rightarrow x=25\\ b,\left(3-x\right)\left(9+3x\right)=0\\ \Rightarrow\left[{}\begin{matrix}3-x=0\\9+3x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\3x=-9\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(c,5x^2-\left(-6\right)=\left(-33\right)-\left(-44\right)\\ \Rightarrow5x^2+6=11\\ \Rightarrow5x^2=5\\ \Rightarrow x^2=1\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
\(d,2\left(2x-4\right)^2-77=-45\\ \Rightarrow2\left(2x-4\right)^2=32\\ \Rightarrow\left(2x-4\right)^2=16\\ \Rightarrow\left[{}\begin{matrix}2x-4=-4\\2x-4=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=0\\2x=8\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
a: \(=\left(6+\dfrac{1}{3}+2+\dfrac{2}{3}\right)+\left(-3-\dfrac{2}{5}-1-\dfrac{3}{5}\right)+4=9-5+4=8\)
b: \(=\dfrac{5}{6}-\dfrac{1}{9}+\dfrac{4}{5}-\dfrac{4}{6}-\dfrac{7}{9}+\dfrac{3}{5}+\dfrac{3}{5}-\dfrac{1}{9}-\dfrac{1}{6}\)
=-1+2=1