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H9
HT.Phong (9A5)
CTVHS
12 tháng 8 2023
\(48^2-42^2+64-52^2\)
\(=\left(48^2-52^2\right)-\left(42^2-64\right)\)
\(=\left(48^2-52^2\right)-\left(42^2-8^2\right)\)
\(=\left(48-52\right)\left(48+52\right)-\left(42+8\right)\left(42-8\right)\)
\(=-4\cdot100-50\cdot34\)
\(=-8\cdot50-50\cdot34\)
\(=-50\cdot\left(8+34\right)\)
\(=-50\cdot42\)
\(=-2100\)
21 tháng 9 2020
482 + 522 + 52.96
= 482 + 2.48.52 + 522
= ( 48 + 52 )2
= 1002 = 10 000
NM
1
31 tháng 8 2021
\(C=48\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)=2\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)=2\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\left(5^{128}-1\right)=2.5^{128}-2\)
31 tháng 8 2021
c: Ta có: \(C=48\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\cdot\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^2-1\right)\left(5^2+1\right)\cdot\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^{16}-1\right)\cdot\left(5^{16}+1\right)\cdot\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^{32}-1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^{64}-1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^{128}-1\right)\)
\(=2\cdot5^{128}-2\)
NT
1
31 tháng 7 2022
a: \(A=\dfrac{\left(37+12\right)\left(37^2+12^2-37\cdot12\right)}{49-37\cdot12}\)
\(=\dfrac{49\cdot1069}{49-37\cdot12}\simeq-132.61\)
b: \(=\dfrac{\left(52-48\right)\left(52^2+48^2+52\cdot48\right)}{4+52\cdot48}\)
\(=\dfrac{4\cdot7504}{4+52\cdot48}=\dfrac{7504}{625}\)
CN
1
19 tháng 8 2021
a: Ta có: \(A=\dfrac{35^3+13^3}{48}-35\cdot13\)
\(=35^2-35\cdot13+13^2-35\cdot13\)
\(=\left(35-13\right)^2\)
\(=22^2=484\)
b: Ta có: \(B=\dfrac{68^3-52^3}{16}+68\cdot52\)
\(=68^2+68\cdot52+52^2+68\cdot52\)
\(=\left(68+52\right)^2=14400\)
\(52^2+2.52.48+48^2=\left(52+48\right)^2=100^2=10000\)
\(52^2+2.52.48+48^2=\left(52+48\right)^2\)
\(=100^2\)
\(=10000\)