a; ( 2+2^2+2^3+........+2^60) chia hết cho 3; 7; 15
b; (1+3+3^2+3^3+......+3^1991) chia hết cho 13; 41
c; ( 3+3^2+3^3+........+3^1998) chia hết cho 12; 39
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Đỗ Thanh Hải
CTVVIP
5 tháng 9 2021
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NL
Nguyễn Lê Phước Thịnh
CTVHS
28 tháng 10 2025
1: \(A=\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\cdot\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)
ĐKXĐ: a>0; a<>1
Ta có: \(\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}\)
\(=\frac{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}+1\right)}\)
\(=\frac{a+\sqrt{a}+1-\left(a-\sqrt{a}+1\right)}{\sqrt{a}}=\frac{2\sqrt{a}}{\sqrt{a}}=2\)
Ta có: \(\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)
\(=\frac{\left(\sqrt{a}+1\right)^2+\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=\frac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{a-1}\)
\(=\frac{2a+2}{a-1}\)
Ta có: \(A=\frac{a\sqrt{a}-1}{a-\sqrt{a}}-\frac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\frac{1}{\sqrt{a}}\right)\cdot\left(\frac{\sqrt{a}+1}{\sqrt{a}-1}+\frac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)
\(=2+\frac{a-1}{\sqrt{a}}\cdot\frac{2a+2}{a-1}=2+\frac{2a+2}{\sqrt{a}}=\frac{2a+2\sqrt{a}+2}{\sqrt{a}}\)
2: A=7
=>\(\frac{2a+2\sqrt{a}+2}{\sqrt{a}}=7\)
=>\(2a+2\sqrt{a}+2=7\sqrt{a}\)
=>\(2a-5\sqrt{a}+2=0\)
=>\(\left(2\sqrt{a}-1\right)\left(\sqrt{a}-2\right)=0\)
=>\(\left[\begin{array}{l}2\sqrt{a}-1=0\\ \sqrt{a}-2=0\end{array}\right.\Rightarrow\left[\begin{array}{l}\sqrt{a}=\frac12\\ \sqrt{a}=2\end{array}\right.\Rightarrow\left[\begin{array}{l}a=\frac14\left(nhận\right)\\ a=4\left(nhận\right)\end{array}\right.\)
3: A>6
=>\(\frac{2a+2\sqrt{a}+2}{\sqrt{a}}>6\)
=>\(\frac{a+\sqrt{a}+1}{\sqrt{a}}>3\)
=>\(\frac{a+\sqrt{a}+1-3\sqrt{a}}{\sqrt{a}}>0\)
=>\(\frac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}}>0\) (luôn đúng với mọi a thỏa mãn ĐKXĐ)
3 tháng 11 2015
a, a=0 hoặc a=2
b, b=0
c, Vì a=0 nhung a:a=0:0 không được
\(\Rightarrow\)a=1
a: Đặt \(A=2+2^2+\cdots+2^{60}\)
Ta có: \(A=2+2^2+\cdots+2^{60}\)
\(=\left(2+2^2\right)+\left(2^3+2^4\right)+\cdots+\left(2^{59}+2^{60}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+\cdots+2^{59}\left(1+2\right)\)
\(=3\left(2+2^3+\cdots+2^{59}\right)\) ⋮3
Ta có: \(A=2+2^2+\cdots+2^{60}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+\cdots+\left(2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+\cdots+2^{58}\left(1+2+2^2\right)\)
\(=7\left(2+2^4+\cdots+2^{58}\right)\) ⋮7
TA có: \(A=2+2^2+\cdots+2^{60}\)
\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+\ldots+\left(2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(=2\left(1+2+2_{}^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+\cdots+2^{57}\left(1+2+2^2+2^3\right)\)
\(=15\left(1+2^5+\cdots+2^{57}\right)\) ⋮15
b: Ta có: \(B=1+3+3^2+\cdots+3^{1991}\)
\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+\cdots+\left(3^{1989}+3^{1990}+3^{1991}\right)\)
\(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+\cdots+3^{1989}\left(1+3+3^2\right)\)
\(=13\left(1+3^3+\cdots+3^{1989}\right)\) ⋮13
c: Ta có: \(C=3+3^2+3^3+\cdots+3^{1998}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+\cdots+\left(3^{1997}+3^{1998}\right)\)
\(=\left(3+3^2\right)+3^2\left(3+3^2\right)+\cdots+3^{1996}\left(3+3^2\right)\)
\(=12\left(1+3^2+\cdots+3^{1996}\right)\) ⋮12
Ta có: \(C=3+3^2+3^3+\cdots+3^{1998}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\cdots+\left(3^{1996}+3^{1997}+3^{1998}\right)\)
\(=\left(3+3^2+3^3\right)+3^3\left(3+3^2+3^3\right)+\cdots+3^{1995}\left(3+3^2+3^3\right)\)
\(=39\left(1+3^3+\cdots+3^{1995}\right)\) ⋮39