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3a-b=1/2(a+b)
=>6a-2b=a+b
=>5a=3b
=>a/3=b/5=k
=>a=3k; b=5k
\(A=\dfrac{a^{2022}+3^{2022}}{b^{2022}+5^{2022}}\)
\(=\dfrac{3^{2022}\left(k^{2022}+1\right)}{5^{2022}\left(k^{2022}+1\right)}=\left(\dfrac{3}{5}\right)^{2022}\)
A = 7 - 8 + 9 -10 + 11 - 12 +...+ 2009 - 2010
A = (7-8) + (9 - 10) + ( 11 - 12) +...+ ( 2009 - 2010)
Xét dãy số: 7; 9; 11;...; 2009
Dãy số trên là dãy số cách đều với khoảng cách là: 9 - 7 = 2
Dãy số trên có số số hạng là: (2009 - 7) : 2 + 1 = 1002
Vậy tổng A có 1002 nhóm mỗi nhóm có giá trị là: 7 - 8 = -1
A = -1 \(\times\) 1002 = - 1002
B = 1 - 2 - 3 - 4 -...- 2022 - 2023
B = 1 - ( 2 + 3 + 4 +...+ 2022 + 2023)
B = 1 - (2 + 2023).{ ( 2023 - 2): 1 + 1}: 2 = -2047274
Ta có: \(B=\frac12+\frac13-\frac14+\frac15-\frac16+\cdots-\frac{1}{2022}+\frac{1}{2023}\)
=>\(B=\frac12+\frac13+\frac14+\frac15+\frac16+\cdots+\frac{1}{2022}+\frac{1}{2023}-2\left(\frac14+\frac16+\cdots+\frac{1}{2022}\right)\)
\(=\frac12+\frac13+\frac14+\frac15+\cdots+\frac{1}{2022}+\frac{1}{2023}-\frac12-\frac13-\cdots-\frac{1}{1011}\)
\(=\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2022}+\frac{1}{2023}\)
=C
=>B-C=0
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=>a=bk; c=dk
\(\frac{a^{2022}+b^{2022}}{c^{2022}+d^{2022}}=\frac{\left(bk\right)^{2022}+b^{2022}}{\left(dk\right)^{2022}+d^{2022}}=\frac{b^{2022}\left(k^{2022}+1\right)}{d^{2022}\left(k^{2022}+1\right)}=\frac{b^{2022}}{d^{2022}}\)
\(\frac{\left(a+b\right)^{2022}}{\left(c+d\right)^{2022}}=\frac{\left(bk+b\right)^{2022}}{\left(dk+d\right)^{2022}}=\frac{\left\lbrack b\left(k+1\right)\right\rbrack^{2022}}{\left\lbrack d\left(k+1\right)\right\rbrack^{2022}}=\frac{b^{2022}}{d^{2022}}\)
Do đó: \(\frac{a^{2022}+b^{2022}}{c^{2022}+d^{2022}}=\frac{\left(a+b\right)^{2022}}{\left(c+d\right)^{2022}}\)
\(\dfrac{a}{2022}=\dfrac{b}{2021}=\dfrac{c}{2020}=\dfrac{c-a}{-2}=\dfrac{c-b}{-1}=\dfrac{b-a}{-1}\\ \Rightarrow c-a=2\left(c-b\right)=2\left(b-a\right)\\ \Rightarrow\left(c-a\right)^3=\left[2\left(c-b\right)\right]^3=8\left(c-b\right)^2\left(c-b\right)=8\left(c-b\right)^2\left(b-a\right)\)
Ta có: \(S=1-\frac12+\frac13-\frac14+\cdots+\frac{1}{2021}-\frac{1}{2022}\)
\(=1+\frac12+\frac13+\frac14+\cdots+\frac{1}{2021}+\frac{1}{2022}-2\left(\frac12+\frac14+\cdots+\frac{1}{2022}\right)\)
\(=1+\frac12+\cdots+\frac{1}{2022}-1-\frac12-\cdots-\frac{1}{1011}\)
\(=\frac{1}{1012}+\frac{1}{1013}+\cdots+\frac{1}{2022}=P\)
=>S-P=0
=>\(\left(S-P\right)^{2022}=0\)
TA có: \(3a-b=\frac12\left(a+b\right)\)
=>2(3a-b)=a+b
=>6a-2b=a+b
=>5a=3b
=>a=0,6b
\(C=\frac{a^{2022}+3^{2022}}{b^{2022}+5^{2022}}\)
\(=\frac{\left(0,6b\right)^{2022}+3^{2022}}{b^{2022}+5^{2022}}\)
\(=\frac{\left(0,6\right)^{2022}\left(b^{2022}+5^{2022}\right)}{b^{2022}+5^{2022}}=\left(0,6\right)^{2022}=\left(\frac35\right)^{2022}\)