e)1/2 , 2/3 , 5/6 ta được ;.../... , .../... và ... / ... |
g)3/4 , 1/3 ,1/6 ta được ;.../... , .../... và .../... mọi người quy đồng ra nha |
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![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(B=\dfrac{5}{2}-\dfrac{7}{2}+\dfrac{3}{8}+\dfrac{6}{8}+\dfrac{-6}{11}-\dfrac{5}{11}=-2-1+\dfrac{9}{8}=\dfrac{9}{8}-3=-\dfrac{15}{8}\)
c: \(C=\left(\dfrac{4}{3}+\dfrac{7}{3}+\dfrac{1}{3}\right)+\left(\dfrac{2}{5}+\dfrac{3}{5}\right)=4+1=5\)
d: \(D=\dfrac{4}{19}\left(\dfrac{-5}{6}-\dfrac{7}{12}\right)-\dfrac{40}{57}\)
\(=\dfrac{4}{19}\cdot\dfrac{-17}{12}-\dfrac{40}{57}=-1\)
e: \(E=\dfrac{1}{3}\left(\dfrac{4}{5}-\dfrac{9}{5}\right)+\dfrac{2}{3}=\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(2x+5⋮x+1\)
=>\(2x+2+3⋮x+1\)
=>\(3⋮x+1\)
=>\(x+1\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{0;-2;2;-4\right\}\)
b: \(5x+9⋮x+2\)
=>\(5x+10-1⋮x+2\)
=>\(-1⋮x+2\)
=>\(x+2\in\left\{1;-1\right\}\)
=>\(x\in\left\{-1;-3\right\}\)
c: \(2x+11⋮x+3\)
=>\(2x+6+5⋮x+3\)
=>\(5⋮x+3\)
=>\(x+3\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{-2;-4;2;-8\right\}\)
d: \(4x+9⋮2x+1\)
=>\(4x+2+7⋮2x+1\)
=>\(7⋮2x+1\)
=>\(2x+1\in\left\{1;-1;7;-7\right\}\)
=>\(2x\in\left\{0;-2;6;-8\right\}\)
=>\(x\in\left\{0;-1;3;-4\right\}\)
e: \(6x+7⋮3x+1\)
=>\(6x+2+5⋮3x+1\)
=>\(5⋮3x+1\)
=>\(3x+1\in\left\{1;-1;5;-5\right\}\)
=>\(3x\in\left\{0;-2;4;-6\right\}\)
=>\(x\in\left\{0;-\dfrac{2}{3};\dfrac{4}{3};-2\right\}\)
g: \(10x+13⋮5x+1\)
=>\(10x+2+11⋮5x+1\)
=>\(11⋮5x+1\)
=>\(5x+1\in\left\{1;-1;11;-11\right\}\)
=>\(5x\in\left\{0;-2;10;-12\right\}\)
=>\(x\in\left\{0;-\dfrac{2}{5};2;-\dfrac{12}{5}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(E=5+\left(1-x\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=5-\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=5-\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\)
\(=5-\left(x^2+5x-6\right)\left(x^2+5x-6\right)\)
Đặt \(t=x^2+6x\)
\(\Rightarrow E=5+\left(t-6\right)\left(t+6\right)\)
\(=5+t^2-36\)
\(=t^2-31\)
Mà \(t^2\ge0\Rightarrow t^2-31\ge-31\)
\(\Rightarrow E\ge-31\)
Dấu "=" xảy ra \(\Leftrightarrow t^2=0\Leftrightarrow t=0\Leftrightarrow x^2+6x=0\Leftrightarrow x\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
\(E=5+\left(1-x\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\\ E=5-\left[\left(x-1\right)\left(x+6\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]\\ E=5-\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
Cách 1: \(E=5-\left(x^2+5x\right)^2+36=-\left(x^2+5x\right)^2+41\le41\)
\(E_{max}=41\Leftrightarrow x^2+5x=0\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=0\end{matrix}\right.\)
Cách 2: Đặt \(x^2+5x=t\)
\(\Leftrightarrow E=5-\left(t+6\right)\left(t-6\right)=5-t^2+36=-t^2+41\le41\\ E_{max}=41\Leftrightarrow t=0\Leftrightarrow x^2+5x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(4\dfrac{3}{8}+5\dfrac{2}{3}\)
\(=\dfrac{35}{8}+\dfrac{17}{3}\)
\(=\dfrac{105}{24}+\dfrac{136}{24}\)
\(=\dfrac{241}{24}\)
b) \(2\dfrac{3}{8}+1\dfrac{1}{4}+3\dfrac{6}{7}\)
\(=\dfrac{19}{8}+\dfrac{5}{4}+\dfrac{27}{7}\)
\(=\dfrac{29}{8}+\dfrac{27}{7}\)
\(=\dfrac{419}{56}\)
c) \(2\dfrac{3}{8}-1\dfrac{1}{4}+5\dfrac{1}{3}\)
\(=\dfrac{19}{8}-\dfrac{5}{4}+\dfrac{16}{3}\)
\(=\dfrac{9}{8}+\dfrac{16}{3}\)
\(=\dfrac{155}{24}\)
d) \(\left(\dfrac{5}{2}+\dfrac{1}{3}\right):\left(1-\dfrac{1}{2}\right)\)
\(=\dfrac{17}{6}:\dfrac{1}{2}\)
\(=\dfrac{17}{6}\cdot2\)
\(=\dfrac{17}{3}\)
e) \(\left(\dfrac{5}{2}-\dfrac{1}{3}\right)\cdot\dfrac{9}{2}-\dfrac{6}{7}\)
\(=\dfrac{13}{6}\cdot\dfrac{9}{2}-\dfrac{6}{7}\)
\(=\dfrac{39}{4}-\dfrac{6}{7}\)
\(=\dfrac{249}{28}\)
a: =4+3/8+5+2/3
=9+9/24+16/24
=9+25/24
=216/24+25/24=241/24
b: \(=\dfrac{19}{8}+\dfrac{5}{4}+\dfrac{27}{7}=\dfrac{19+10}{8}+\dfrac{27}{7}\)
=27/7+29/8
=419/56
c: =2+3/8-1-1/4+5+1/3
=6+3/8-1/4+1/3
=6+3/8+1/12
=144/24+9/24+2/24
=155/24
d: =(15/6+2/6):1/2
=17/6*2
=17/3
e: =(15/6-2/6)*9/2-6/7
=13/6*9/2-6/7
=117/12-6/7
=249/28