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Giải:
a) \(A=\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3+1\right)\)
\(\Leftrightarrow2A=2\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3+1\right)\)
\(\Leftrightarrow2A=\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3+1\right)\left(3-1\right)\)
\(\Leftrightarrow2A=\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3^2-1\right)\)
\(\Leftrightarrow2A=\left(3^8+1\right)\left(3^4+1\right)\left(3^4-1\right)\)
\(\Leftrightarrow2A=\left(3^8+1\right)\left(3^8-1\right)\)
\(\Leftrightarrow2A=3^{16}-1\)
\(\Leftrightarrow A=\dfrac{3^{16}-1}{2}\)
Vậy ...
b) \(B=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{32}+1\right)\)
\(\Leftrightarrow2B=2.12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{32}+1\right)\)
\(\Leftrightarrow2B=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{32}+1\right)\)
\(\Leftrightarrow2B=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{32}+1\right)\)
\(\Leftrightarrow2B=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{32}+1\right)\)
...
\(\Leftrightarrow2B=\left(5^{32}-1\right)\left(5^{32}+1\right)\)
\(\Leftrightarrow2B=5^{64}-1\)
\(\Leftrightarrow B=\dfrac{5^{64}-1}{2}\)
Vậy ...
Rút gọn biểu thức
A= 12(52 +1)(54 +1)(58 +1)(516 +1)
Chứng minh
(a+b+c)3 = a3+b3+c3+3(a+b)(b+c)(c+a)
Bài 1:
A = \(12.\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
=> \(\left(5^2-1\right)A\) = \(12\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
=> 24A = \(12\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
=> A = \(\dfrac{12}{24}.\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
=> A = \(\dfrac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
=> A = \(\dfrac{1}{2}\left(5^{32}-1\right)\)
Bài 2:
Ta có: \(\left(a+b+c\right)^3=\left[\left(a+b\right)+c\right]^3\)
= \(\left(a+b\right)^3+c^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2\)
= \(a^3+b^3+3ab\left(a+b\right)+c^3+3\left(a+b\right)\left(ac+bc+c^2\right)\)
= \(a^3+b^3+c^3+3\left(a+b\right)\left(ab+bc+ca+c^2\right)\)
= \(a^3+b^3+c^3+3\left(a+b\right)\left[b\left(a+c\right)+c\left(a+c\right)\right]\)
= \(a^3+b^3+c^3+3\left(a+b\right)\left(a+c\right)\left(b+c\right)\) => đpcm
3. ( 22 + 1 ).( 24 + 1 ).( 28 + 1 )......( 264 + 1 ) + 1
= ( 22 - 1 ).( 22 + 1 ).( 24 + 1 ).( 28 + 1 )....( 264 + 1 ) + 1
= ( 24 - 1 ).( 24 + 1 ).( 28 + 1 )......( 264 + 1 ) + 1
= ( 28 + 1 ).....( 264 + 1 ) + 1
= ( 264 - 1 ).( 264 + 1 ) + 1
= 2128 - 1 + 1
= 2128
8.( 32 + 1 ).( 34 + 1 ).( 38 + 1 )....( 3128 + 1 ) + 1
= ( 32 - 1 ).( 32 + 1 ).( 34 + 1 ).( 38 + 1 )....( 3128 + 1 ) + 1
= ( 34 - 1 ).( 34 + 1 ).( 38 + 1 )....( 3128 + 1 ) + 1
= ( 38 - 1 ).( 38 + 1 )....( 3128 + 1 ) + 1
= ( 316 - 1 )......( 3128 + 1 ) + 1
= ( 3128 - 1 ).( 3128 + 1 ) + 1
= 3256 - 1 + 1
= 3256
Giải:
1) B = 272 - 252 = (27 - 25)(27 + 25) = 20.52
Suy ra A<B, vì 202<20.52
2) D = 20032 - 1 = 20032 - 12 = (2003 - 1)(2003 + 1) = 2002.2004
Suy ra C = D.
3) Nhân (2-1) vào E, ta đươc: E = (2-1)(2+1)(22+1)(24+1)(28+1)(216+1)
Áp dụng lân lượt hằng đẳng thức số 3 (Hiệu hai bình phương) vào E, ta được kế quả:
E = 232-1
Suy ra E<F
4) Nhân (3-1) vào G, ta đươc: 2G = (3-1)(3+1)(32+1)(34+1)(38+1)(316+1)
Áp dụng lân lượt hằng đẳng thức số 3 (Hiệu hai bình phương) vào G, ta được kế quả:
2G = 332-1
Suy ra G = (332-1)/2
Mà (332-1)/2 < 332/2
Suy ra G<H
5)
Nhân 2 vào I, ta đươc: 2I = 2.12(52+1)(54+1)(58+1)...(532+1)
Áp dụng lân lượt hằng đẳng thức số 3 (Hiệu hai bình phương) vào I, ta được kế quả:
2I = 564-1
Suy ra I = (564-1)/2
Mà (564-1)/2 < 564-1
Suy ra I<K.
Chúc chị học tốt!
a)\(\left(2x-y\right)[\left(2x\right)^2-2.2x.y+y^2]\)
\(=\left(2x-y\right)^3\)
b)\(2x^2-3xy+5y^2\)
c)\(2x^3-10x^2-11x^2+55x+12x-60\)
\(=2x^2\left(x-5\right)-11x\left(x-5\right)+12\left(x-5\right)\)
\(=\left(x+5\right)\left(2x^2-11x+12\right)\)
\(\Leftrightarrow(2x^3-21x^2+67x-60)/\left(x-5\right)=2x^2-11x+12\)
\(\frac{3^{30}-5.3^{16}.3^{12}+4.3^{16}.3^8}{41.3^{24}}=\frac{3^{24}\left(3^6-5.3^4+4\right)}{41.3^{24}}=\frac{3^4\left(3^{2-5}\right)+4}{41}=\frac{4\left(3^4+1\right)}{41}=\frac{4.82}{41}=8\)
a,\(5^3:\left(-5\right)^2=5^3:5^2=5^{3-2}=5\)
b,\(\left(\dfrac{3}{4}\right)^5:\left(\dfrac{3}{4}\right)^3=\left(\dfrac{3}{4}\right)^{5-3}=\left(\dfrac{3}{4}\right)^2\)
c,\(\left(-12\right)^3:8^3=\left(\left(-12\right):8\right)^3=\left(-\dfrac{3}{2}\right)^3\)