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a: \(A=\dfrac{1.3-2.6}{2.6}-\dfrac{5}{6}:2=\dfrac{-1}{2}-\dfrac{5}{12}=\dfrac{-11}{12}\)
\(B=\left(\dfrac{47}{8}-\dfrac{9}{4}-\dfrac{1}{2}\right):\dfrac{75}{26}\)
\(=\dfrac{47-18-4}{8}\cdot\dfrac{26}{75}=\dfrac{25}{75}\cdot\dfrac{26}{8}=\dfrac{1}{3}\cdot\dfrac{13}{4}=\dfrac{13}{12}\)
b: Để A<x<B thì \(\dfrac{-11}{12}< x< \dfrac{13}{12}\)
mà x là số nguyên
nên \(x\in\left\{0;1\right\}\)
Chả hỉu gì cả.
Đăng từng bài một thôi bạn!
1)\(\left(-\dfrac{5}{13}\right)^{2017}.\left(\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).\left(-\dfrac{5}{13}\right)^{2016}.\left(\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).\left(\dfrac{5}{13}\right)^{2016}.\left(\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).\left(\dfrac{5}{13}.\dfrac{13}{5}\right)^{2016}\)
\(=\left(-\dfrac{5}{13}\right).1^{2016}\)
\(=-\dfrac{5}{13}\)
\(M=\left|x-2002\right|+\left|x-2001\right|\)\(=\left|x-2002\right|+\left|2001-x\right|\ge\left|x-2002+2001-x\right|=\left|-2002+2001\right|=1\)
tức \(M\ge1\) \(\Leftrightarrow\left[{}\begin{matrix}x-2001=0\\x-2002=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2001\\x=2002\end{matrix}\right.\)
Vậy MinM = - 1 \(\Leftrightarrow\left[{}\begin{matrix}x=2001\\x=2002\end{matrix}\right.\)
\(\left(\dfrac{-5}{13}\right)^{2017}\cdot\left(\dfrac{13}{5}\right)^{2016}=\left(\dfrac{-5}{13}\right)\cdot\left(-\dfrac{5}{13}\right)^{2016}\cdot\left(\dfrac{13}{5}\right)^{2016}=\left(\dfrac{-5}{13}\right)\cdot\left(\dfrac{5}{13}\right)^{2016}\cdot\left(\dfrac{13}{5}\right)^{2016}=\left(-\dfrac{5}{13}\right)\cdot\left[\left(\dfrac{5}{13}\right)^{2016}\cdot\left(\dfrac{13}{5}\right)^{2016}\right]=\left(-\dfrac{5}{13}\right)\cdot1^{2016}=\left(-\dfrac{5}{13}\right)\cdot1=-\dfrac{5}{13}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\) \(\begin{cases} a = bk \\ c = dk \end{cases}\)
Ta có: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(1\right)\)
\(\dfrac{a.c}{b.d}=\dfrac{bk.dk}{b.d}=\dfrac{k^2.b.d}{b.d}=k^2\left(2\right)\)
Từ (1) và (2) suy ra: \(\dfrac{a.c}{b.d}=\dfrac{a^2+c^2}{b^2+d^2}\) \(\rightarrow đpcm\).
Bài 1:
1: \(M=\left|x-1\right|+x+2\)
Trường hợp 1: x>=1
M=x-1+x+2=2x+1
Trường hợp 2: x<1
M=1-x+x+2=3
2: \(N=x-3+\left|x-3\right|\)
Trường hợp 1: x>=3
\(N=x-3+x-3=2x-6\)
Trường hợp 2: x<3
\(N=x-3+3-x=0\)
3: \(P=2x-1-\left|x-2\right|\)
Trường hợp 1: x<2
\(P=2x-1-\left(2-x\right)=2x-1-2+x=3x-3\)
TRường hợp 2: x>=2
\(P=2x-1-x+2=x+1\)
a: \(A=\dfrac{1.3-2.6}{2.6}-\dfrac{5}{6}:2=\dfrac{-1}{2}-\dfrac{5}{12}=\dfrac{-11}{12}\)
\(B=\left(\dfrac{47}{8}-\dfrac{9}{4}-\dfrac{1}{2}\right):\dfrac{75}{26}=\dfrac{47-18-4}{8}\cdot\dfrac{26}{75}=\dfrac{25}{75}\cdot\dfrac{26}{8}=\dfrac{13}{12}\)
b: Để A<x<B thì -11/12<x<13/12
mà x là số nguyên
nên \(x\in\left\{0;1\right\}\)
a) \(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}+\dfrac{2}{3}-1\dfrac{15}{17}\)
\(=\left(\dfrac{15}{34}+\dfrac{19}{34}\right)+\left(\dfrac{7}{21}+\dfrac{2}{3}\right)-1\dfrac{15}{17}\)
\(=1+1-1\dfrac{15}{17}=\dfrac{2}{17}\)
\(x^2+5x< 0\)
\(x\left(x+5\right)< 0\)
\(\Leftrightarrow x< 0\)
\(\Leftrightarrow x+5>0\Leftrightarrow x>-5\)
\(-5< x< 0\)
\(x\in\left\{-4;-3;-2;-1\right\}\)
\(\Leftrightarrow x>0\)
\(\Leftrightarrow x-5< 0\Leftrightarrow x< 5\)
\(0< x< 5\)
\(x\in\left\{1;2;3;4\right\}\)
Vậy.......