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\(\left(m-n\right)\left(m+n\right)\)
\(\Leftrightarrow\left(m-n\right).m+\left(m-n\right).n\)
\(=m^2-nm+mn-n^2\)
\(=\left(-nm+mn\right)+\left(m^2-n^2\right)\)
\(=0+\left(m^2-n^2\right)\)
\(=m^2-n^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x\left(y+z\right)-y\left(x-z\right)=xy+xz-yx+yz\)
\(=xy-xy+\left(zx+zy\right)\)
\(=\left(x+y\right)z\)
b, \(\left(m-n\right)\left(m+n\right)=m^2+mn-nm-n^2\)
\(=m^2-n^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có: VT = -(-a + b - 17) + (-3b + a - 13) - 20
= a - b + 17 - 3b + a - 13 - 20 = 2a - 4b - 16 = 2(a - 2b - 8)
VP = -2(2b - a + 1) + (-14)
= -4b + 2a - 2 - 14 = 2(a - 2b - 8)
=> VT = VP
b) Ta có: 3n + 8 \(⋮\)n - 1
=> 3(n - 1) + 11 \(⋮\)n - 1
Do 3(n - 1) \(⋮\)n - 1 => 11 \(⋮\)n - 1
=> n - 1 \(\in\)Ư(11) = {1; -1; 11; -11}
Lập bảng:
n - 1 | 1 | -1 | 11 | -11 |
n | 2 | 0 | 12 | -10 |
Vậy ...
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Lời giải:
\(a^{n+1}-1=(a^{n+1}-a^n)+(a^n-a^{n-1})+.....+(a-1)\)
\(=a^n(a-1)+a^{n-1}(a-1)+...+(a-1)=(a-1)(a^n+a^{n-1}+...+1)\)
Ta có đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\)
Vì \(\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{1}{n}-\dfrac{1}{n+1}\)
\(\Rightarrow\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\left(đpcm\right)\)
Chứng minh
\(\dfrac{1}{n\left(n+1\right)}\)=\(\dfrac{1}{n}-\dfrac{1}{n+1}\)
Ta có:VP=\(\dfrac{1}{n}-\dfrac{1}{n+1}\)=
\(\dfrac{n+1}{n\left(n+1\right)}-\dfrac{n}{n\left(n+1\right)}\)
=\(\dfrac{n+1-n}{n\left(n+1\right)}=\dfrac{1}{n\left(n+1\right)}=VT\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(m\left(n-p\right)-n\left(m+p\right)=-p\left(m+n\right)\)
\(vt=mn-mp-nm-np\)
\(=-mp-np\)
\(=-p\left(m+n\right)=vp\)
vâyj đẳng thức được chứng minh
![](https://rs.olm.vn/images/avt/0.png?1311)
1 -a.(c-d)-d.(a+c)
=-a.c+a.d-d.a-d.c
=(a.d-a.d)-a.c-d.c
=0-(a.c+d.c)
=-(a.c+d.c)
=-[c.(a+d)]
=-c.(a+d)
![](https://rs.olm.vn/images/avt/0.png?1311)
câu 1 :
\(VT=A\left(B+C\right)=A\cdot B+A\cdot C=VP\)
Vậy \(A\left(B+C\right)=A\cdot B+A\cdot C\)đpcm
câu 2 :
\(\left(x-5\right)\left(2-y\right)=7\)
\(\left(x-5\right)\left(2-y\right)=1\cdot7\)
\(\Rightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x-5=1\\2-y=7\end{matrix}\right.\\\left\{{}\begin{matrix}x-5=7\\2-y=1\end{matrix}\right.\end{matrix}\right.\\ \left\{{}\begin{matrix}\left\{{}\begin{matrix}x=6\\y=-5\end{matrix}\right.\\\left\{{}\begin{matrix}x=12\\y=1\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow x=12;y=1\) ( \(y=-5\) ko phải là số nguyên )
Vậy........................................