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\(\text{Bn hỏi từ từ từng câu 1 thôi}\)
\(\text{Bn hỏi thế ai mà dám làm}\)
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Chí lí
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sọ ghi 2 hàng khoogn đc tích tăng lê hiều hàng
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Bài 1:
Ta có:
\(3^{x+1}+3^{x+2}+3^{x+3}+...+3^{x+100}\)
\(=\left(3^{x+1}+3^{x+2}+3^{x+3}+3^{x+4}\right)+...+\left(3^{x+97}+3^{x+98}+3^{x+99}+3^{x+100}\right)\)
\(=3^x.\left(3+3^2+3^3+3^4\right)+...+3^{x+96}.\left(3+3^2+3^3+3^4\right)\)
\(=3^x.120+3^{x+4}.120+...+3^{x+96}.120\)
\(=120.\left(3^x+3^{x+4}+...+3^{x+96}\right)\)
Vì \(120⋮120.\)
\(\Rightarrow120.\left(3^x+3^{x+4}+...+3^{x+96}\right)⋮120\)
\(\Rightarrow3^{x+1}+3^{x+2}+3^{x+3}+...+3^{x+100}⋮120\left(\forall x\in N\right)\left(đpcm\right).\)
Chúc bạn học tốt!
Bài 2:
Vì \(f\left(x_1.x_2\right)=f\left(x_1\right).f\left(x_2\right)\)
\(\Rightarrow f\left(4\right)=f\left(2.2\right)=f\left(2\right).f\left(2\right)=10.10=100\)
\(\Rightarrow f\left(16\right)=f\left(4.4\right)=f\left(4\right).f\left(4\right)=100.100=10000.\)
\(\Rightarrow f\left(32\right)=f\left(16.2\right)=f\left(16\right).f\left(2\right)=10000.10=100000.\)
Vậy \(f\left(32\right)=100000.\)
Chúc bạn học tốt!
Bài 4:
x O y z m n
Giải:
Vì Om là tia phân giác của góc xOz nên:
mOz = 1/2.xOz
Vì On là tia phân giác của góc zOy nên:
zOn = 1/2 . zOy
Ta có: xOz + zOy = 180o ( kề bù )
=> 1/2(xOz + zOy) = 1/2 . 180o
=> 1/2.xOz + 1/2.zOy = 90o
=> mOz + zOn = 90o
=> mOn = 90o (đpcm)
Bài 2:
7^6 + 7^5 - 7^4 = 7^4.( 7^2 + 7 - 1 ) = 7^4 . 55 chia hết cho 55
Vậy 7^6 + 7^5 - 7^4 chia hết cho 55
A = 1 + 5 + 5^2 + ... + 5^50
=> 5A = 5 + 5^2 + 5^3 + ... + 5^51
=> 5A - A = ( 5 + 5^2 + 5^3 + ... + 5^51 ) - ( 1 + 5 + 5^2 + ... + 5^50 )
=> 4A = 5^51 - 1
=> A = ( 5^51 - 1 )/4
2.
\(3xy+x-y=1\)
\(\Rightarrow9xy+3x-3y=3\)
\(\Rightarrow\left(9xy+3x\right)-3y-1=3-1\)
\(\Rightarrow\left(9xy+3x\right)-\left(3y+1\right)=2\)
\(\Rightarrow3x.\left(3y+1\right)-\left(3y+1\right)=2\)
\(\Rightarrow\left(3y+1\right).\left(3x-1\right)=2\)
Vì \(x,y\in Z\Rightarrow\left\{{}\begin{matrix}3y+1\in Z\\3x-1\in Z\end{matrix}\right.\)
\(\Rightarrow3y+1\inƯC\left(2\right);3x-1\inƯC\left(2\right)\)
\(\Rightarrow3y+1\in\left\{1;2;-1;-2\right\};3x-1\in\left\{1;2;-1;-2\right\}.\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3y+1=1\\3x-1=2\end{matrix}\right.\\\left\{{}\begin{matrix}3y+1=2\\3x-1=1\end{matrix}\right.\\\left\{{}\begin{matrix}3y+1=-1\\3x-1=-2\end{matrix}\right.\\\left\{{}\begin{matrix}3y+1=-2\\3x-1=-1\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}3y=0\\3x=3\end{matrix}\right.\\\left\{{}\begin{matrix}3y=1\\3x=2\end{matrix}\right.\\\left\{{}\begin{matrix}3y=-2\\3x=-1\end{matrix}\right.\\\left\{{}\begin{matrix}3y=-3\\3x=0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=0\\x=1\end{matrix}\right.\left(TM\right)\\\left\{{}\begin{matrix}y=\frac{1}{3}\\x=\frac{2}{3}\end{matrix}\right.\left(KTM\right)\\\left\{{}\begin{matrix}x=-\frac{2}{3}\\x=-\frac{1}{3}\end{matrix}\right.\left(KTM\right)\\\left\{{}\begin{matrix}y=-1\\x=0\end{matrix}\right.\left(TM\right)\end{matrix}\right.\)
Vậy cặp số nguyên \(\left(x;y\right)\) thỏa mãn đề bài là: \(\left(1;0\right),\left(0;-1\right).\)
Chúc bạn học tốt!
a, \(A=\frac{2^{12}\cdot3^5-4^6\cdot9^2}{(2^2\cdot3)^6+8^4\cdot3^5}-\frac{5^{10}\cdot7^3-25^5\cdot49^2}{(125\cdot7)^3+5^9\cdot14^3}\)
\(A=\frac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\frac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot2^3\cdot7^3}\)
\(A=\frac{2^{12}\cdot3^4(3-1)}{2^{12}\cdot3^5(3+1)}-\frac{5^{10}\cdot7^3(1-7)}{5^9\cdot7^3(1+2^3)}\)
\(A=\frac{2^{12}\cdot3^4\cdot2}{2^{12}\cdot3^5\cdot4}-\frac{5^{10}\cdot7^3\cdot(-6)}{5^9\cdot7^3\cdot9}=\frac{1}{6}-\frac{-10}{3}=\frac{7}{2}\)
b,\(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=(3^{n+2}+3^n)-(2^{n+2}-2^n)\)
\(=(3^n\cdot3^2+3^n)-(2^n\cdot2^2-2^n)\)
\(=3^n\cdot(3^2+1)-2^n\cdot(2^2+1)\)
\(=3^n\cdot9+1-2^n\cdot4+1\)
\(=3^n\cdot10-2^n\cdot5\)
Vì \(2\cdot5⋮10\Rightarrow2^n\cdot5⋮10\)
\(3^n\cdot10⋮10\)
Vậy : ....
5. \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
\(\Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{x+10}\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{x+10}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\\left(x-7\right)^{x+10}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)
Vậy ...
4/ Ta có :
\(3^{n+2}-2^{n+2}+3^n-2^n=3^n\left(3^2+1\right)-2^n\left(2^2+1\right)\)
\(=3^n.10-2^n.5\)
\(=3^n.10-2^{n-1}.10\)
\(=10\left(3^n-2^{n-1}\right)⋮10\left(đpcm\right)\)
3/ Ta có :
\(A=1+5+5^2+......+5^{49}+5^{50}\)
\(\Leftrightarrow5A=5+5^2+5^3+.......+5^{50}+5^{51}\)
\(\Leftrightarrow5A-A=\left(5+5^2+.........+5^{51}\right)-\left(1+5+5^2+....+5^{50}\right)\)
\(\Leftrightarrow4A=5^{51}-1\)
\(\Leftrightarrow A=\frac{5^{51}-1}{4}\)
Vậy..
2.
\(7^6+7^5-7^4\)
\(=7^4.\left(7^2+7-1\right)\)
\(=7^4.\left(49+7-1\right)\)
\(=7^4.55\)
Vì \(55⋮55.\)
\(\Rightarrow7^4.55⋮55\)
\(\Rightarrow7^6+7^5-7^4⋮55\left(đpcm\right).\)
4.