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a: \(\Leftrightarrow0< =x< =1\)
hay \(x\in\left\{0;1\right\}\)
b: \(\Leftrightarrow\dfrac{18}{17}< =x< =2\)
hay x=2
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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 |
![](https://rs.olm.vn/images/avt/0.png?1311)
a: =>2x-3-3x+1=34
=>-x-2=34
=>-x=36
hay x=-36
b: =>x+1=0 hoặc 5-5x=0
=>x=-1 hoặc x=1
c: \(\Leftrightarrow\left(2x+1\right)^3=45+19=64\)
=>2x+1=4
=>2x=3
hay x=3/2
d: \(\Leftrightarrow4x^2=16\)
=>x=2 hoặc x=-2
a) \(\left(2x-3\right)+\left(-3x+1\right)=34\)
\(2x-3-3x+1=34\)
\(-x=36\)
\(x=-36\)
b) \(\left(x+1\right)\left(5-5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
c) \(\left(2x+1\right)^3=49-\left(-19\right)=64=4^3\)
\(2x+1=4\)
\(x=\dfrac{3}{2}\)
d) \(4x^2-3=13\)
\(4x^2=16\)
\(x^2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 3:
\(a,\left(5x-2\right)+\left(-3x+1\right)=\left(-42\right)-\left(-91\right)\\ \Rightarrow5x-2+\left(-3x\right)+1=50\\ \Rightarrow2x-1=49\\ \Rightarrow2x=50\\ \Rightarrow x=25\\ b,\left(3-x\right)\left(9+3x\right)=0\\ \Rightarrow\left[{}\begin{matrix}3-x=0\\9+3x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\3x=-9\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)
\(c,5x^2-\left(-6\right)=\left(-33\right)-\left(-44\right)\\ \Rightarrow5x^2+6=11\\ \Rightarrow5x^2=5\\ \Rightarrow x^2=1\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=1\end{matrix}\right.\)
\(d,2\left(2x-4\right)^2-77=-45\\ \Rightarrow2\left(2x-4\right)^2=32\\ \Rightarrow\left(2x-4\right)^2=16\\ \Rightarrow\left[{}\begin{matrix}2x-4=-4\\2x-4=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=0\\2x=8\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 9:
\(a,\left(2n+1\right)⋮\left(n-1\right)\\
\Rightarrow\left[\left(2n-2\right)+3\right]⋮\left(n-1\right)\\
\Rightarrow\left[2\left(n-1\right)+3\right]⋮\left(n-1\right)\)
Mà \(2\left(n-1\right)⋮\left(n-1\right)\Rightarrow3⋮\left(n-1\right)\Rightarrow n-1\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)
Ta có bảng:
n-1 | -3 | -1 | 1 | 3 |
n | -2(loại) | 0(tm) | 2(tm) | 4(tm) |
Vậy \(n\in\left\{0;2;4\right\}\)
b, c, d bạn làm tương tự nhé
Bài 10:
a: Gọi a=UCLN(n+1;2n+3)
\(\Leftrightarrow2n+3-2\left(n+1\right)⋮a\)
\(\Leftrightarrow1⋮a\)
=>a=1
Vậy: n+1/2n+3 là phân số tối giản
b: Gọi a=UCLN(3n+2;5n+3)
\(\Leftrightarrow5\left(3n+2\right)-3\left(5n+3\right)⋮a\)
\(\Leftrightarrow1⋮a\)
=>a=1
Vậy: 3n+2/5n+3 là phân số tối giản
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
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\(a)(-3/5)*x=-1/20+1/2=9/20=>x=9/20:(-3/5)=-3/4\)
Các câu kia làm tương tự nhé, chúc em học giỏi
a: =>-3/5x=-1/20+1/2=-1/20+10/20=-9/20
=>x=3/4
b: =>-1/15x-2/15=3/5
=>-1/15x=6/15+2/15=8/15
=>x=-8
c: \(\Leftrightarrow\left(2x-1\right)\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{\dfrac{1}{2};3;-3\right\}\)