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\(a^2-2a+1+b^2-4b+4=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-2\right)^2=0\)
=>a=1 và b=2
\(a^{27}+b^2+2022=1^{27}+2^2+2022=2022+4+1=2027\)
\(A=\left(1+2+3+...+100\right)^2\cdot\left(a+2b\right)^4\cdot\left(a+3b\right)^5\cdot\left(\dfrac{4}{5}-\dfrac{4}{5}\right)^6\)
=0
a: \(A=\left(\dfrac{1}{99}+1\right)+\left(\dfrac{2}{98}+1\right)+...+\left(\dfrac{98}{2}+1\right)+1\)
\(=\dfrac{100}{99}+\dfrac{100}{98}+...+\dfrac{100}{2}+\dfrac{100}{100}\)
\(=100\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)\)=100B
=>B/A=1/100
b: \(A=\left(\dfrac{1}{49}+1\right)+\left(\dfrac{2}{48}+1\right)+\left(\dfrac{3}{47}+1\right)+...+\left(\dfrac{48}{2}+1\right)+\left(1\right)\)
\(=\dfrac{50}{49}+\dfrac{50}{48}+....+\dfrac{50}{2}+\dfrac{50}{50}\)
\(=50\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)
\(B=\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{49}+\dfrac{2}{50}\)
\(=2\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)\)
=>A/B=25
\(\frac{a}{3}=\frac{4b}{5},\frac{10b}{3}=\frac{5c}{2}\)
\(\Rightarrow\frac{a}{3}=\frac{b}{\frac{5}{4}},\frac{b}{\frac{3}{10}}=\frac{c}{\frac{2}{5}}\)
\(\Rightarrow\frac{a}{\frac{9}{10}}=\frac{b}{\frac{3}{8}},\frac{b}{\frac{3}{8}}=\frac{c}{\frac{1}{2}}\)
\(=>\frac{a}{\frac{9}{10}}=\frac{b}{\frac{3}{8}}=\frac{c}{\frac{1}{2}}\)
áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\frac{a}{\frac{9}{10}}=\frac{b}{\frac{3}{8}}=\frac{c}{\frac{1}{2}}=\frac{a+b+c}{\frac{71}{40}}=\frac{142}{\frac{71}{40}}=80\)
\(\frac{a}{\frac{9}{10}}=80\Rightarrow a=72\)
\(\frac{b}{\frac{3}{8}}=80\Rightarrow b=30\)
\(\frac{c}{\frac{1}{2}}=80\Rightarrow c=40\)
vậy a=72,b=30,c=40
A chia hết cho 2 sẵn rồi
CM A chia hết cho 30:
\(2+2^2+2^3+...+2^{100}\)
\(=\left(2+2^2+2^3+2^4\right)+2^4\left(2+2^2+2^3+2^4\right)+....+2^{96}\left(2+2^2+2^3+2^4\right)\)
\(=30.\left(1+2^4+...+2^{96}\right)⋮30\)
Gợi ý;
B chia hết cho 5 sắn rồi
chia hết cho 6 nhóm 2 số vào
Chi hết cho 31 nhóm 3 số vào
Ta có 2A = 22+23 +......+2100+2101
=> 2A - A = A = 22+23+.......+2101 - 2 - 22 - ....... -2100
= 2101 - 2
5B = 52 + 53 + ... + 5101
=> 4B = 5B - B = 52 + 53 + .....+5101 - 5 - 52 - 5100
= 5101 - 5
Vậy A + 4B = 5101 + 2101 - 7
Ta có: A= 2 + 2^2 +...+ 2^100
Suy ra 2A= 2^2 + 2^3 +...+ 2^101
2A - A= (2^2 + 2^3 +...+ 2^101) - (2 + 2^2 +...+ 2^100)
A= 2^2 + 2^3 +...+ 2^101 - 2 - 2^2 -...- 2^100
A= 2^101 - 2
Và B= 5+ 5^2 + ..... + 5^100
Suy ra 5B= 5^2+ 5^3 + ..... + 5^101
5B - B= (5^2 + 5^3 +...+ 5^101) - (5 + 5^2 +...+ 5^100)
4B= 5^2 + 5^3 +...+ 5^101 - 5 - 5^2 -...- 5^100
4B= 5^101 - 5
Suy ra A + 4B = 2^101 - 2 + 5^101 -5
A + 4B = 2^101 + 5^101 -7
Vậy A + 4B = 2^101 + 5^101 -7
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