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\(A=7+7^2+7^3+7^4+7^5+7^6+7^7+7^8\)
\(A=\left(7+7^3\right)+\left(7^2+7^4\right)+\left(7^5+7^7\right)+\left(7^6+7^8\right)\)
\(A=7\cdot\left(7+7^2\right)+7^2\cdot\left(1+7^2\right)+7^5\cdot\left(1+7^2\right)+7^6\cdot\left(1+7^2\right)\)
\(A=7\cdot50+7^2\cdot50+7^5\cdot50+7^6\cdot50\)
\(A=50\cdot\left(7+7^2+7^5+7^6\right)\)
\(A=5\cdot10\cdot\left(7+7^2+7^5+7^6\right)\)
Ta có: 5 ⋮ 5
⇒ \(A=5\cdot10\cdot\left(7+7^2+7^5+7^6\right)\) ⋮ 5 (đpcm)
A = 7 + 72 + 73 + 74 + 75 + 76 + 77 + 78
A = (7 + 73) + (72+ 74) + (75 + 77) + (76 + 78)
A = 7.(1 + 72) + 72.(1 + 72) + 75.(1 + 72) + 76.(1 + 72)
A = 7.( 1 + 49) + 72.( 1 + 49) + 75.(1 + 49) + 76. (1 + 49)
A = 7.50 + 72.50 + 75.50 + 76.50
A = 50.(7 + 72 + 75 + 76)
Vì 50 ⋮ 5 nên A = 50.(7 + 72 + 76) ⋮ 5 đpcm
a) Ta có : 3 > 2 và 300 > 200
\(\Rightarrow3^{300}>2^{200}\)
b) Ta có : 1000 > 999
\(\Rightarrow5^{1000}>5^{999}\)
c) Ta có : \(243^5=\left(3^5\right)^5=3^{25}\)
\(3.243^5=3.\left(3^5\right)^5=3.3^{25}=3^{26}\)
\(3.27^8=3.\left(3^3\right)^8=3.3^{24}=3^{25}\)
mà 25 = 25 < 26
\(\Rightarrow3^{25}=3^{25}< 3^{26}\)
\(\Rightarrow243^5=3.27^8< 3.243^5\)
d) Ta có : \(125^5=\left(5^3\right)^5=5^{15}\)
\(25^7=\left(5^2\right)^7=5^{14}\)
mà 15 > 14
\(\Rightarrow5^{15}>5^{14}\)
\(\Rightarrow125^5>25^7\)
\(\dfrac{1}{15}+\dfrac{1}{35}+\dfrac{1}{63}+\dfrac{1}{99}+\dfrac{1}{143}+\dfrac{1}{195}\)
\(=\dfrac{1}{3.5}+\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}=\dfrac{4}{15}\)
\(P=\left(a^2+b\right)-\left(2a^2+b\right)+2\left(ab+2021\right)\)
\(P=a^2+b-2a^2+b+2ab+4042\)
\(P=-a^2+2ab+4042\)
\(P=-a\left(a-2b\right)+4042\)
Để cho: \(a-2b=2021\)
\(\Rightarrow P=-a.2021+4042\)
\(P=-2021a+4042\)
Vậy: \(P=-2021a+4042\)
$A = (a + b) - (a + b) + (a - c) - (a + c)$
$A=a+b-a-b+a-c-a-c$
$A=-2c$
(a - b + c) - (a + c) = a - b + c - a - c = -b (đpcm)
(a + b) - (b - a) + c = a + b - b + a + c = 2a + c (đpcm)
-(a + b - c) + (a - b - c) = -a - b + c + a - b - c = -2b (đpcm)
a.(b + c) - a.(b + d) = a.(b + c - b - d) = a.(c - d) (đpcm)
a.(b - c) + a.(d + c) = a.(b - c + d + c) = a.(b + d) (đpcm)
