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Câu 3:
a: \(49^2=2401\)
b: \(51^2=2601\)
c: \(99\cdot100=9900\)
a) Sử dụng công thức bình phương của tổng với số hạng thứ nhất là a + b và số hạng thứ hai là c.
Biến đổi thu được A = a 2 + b 2 + c 2 + 2ab + 2bc + 2 ac;
b) a 2 + b 2 + c 2 - 2ab + 2bc - 2 ac.
\(A=\left(100-99\right)\left(100+99\right)+\left(99-98\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\\ A=100+99+99+98+...+2+1\\ A=\left(100+1\right)\left(100-1+1\right):2=5050\)
\(B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^1-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\\ B=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\\ B=\left(2^{64}-1\right)\left(2^{64}+1\right)+1=2^{128}-1+1=2^{128}\)
\(C=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2+2ab-2ac-2bc-2a^2-4ab-2b^2\\ C=2c^2\)
a) ( a - b + c ) 2 = a2 + b2 + c2 + 2ab - 2ac - 2bc
b ) ( a - b - c )2 = a² + b² + c² - 2ab + 2bc - 2ca
Giải
a/\(\left(a-b+c\right)^2=a^2+b^2+c^2+2ab-2ac-2bc\)
b/\(\left(a-b-c\right)^2=a^2+b^2+c^2-2ab+2bc-2ca\)
a: \(A=\left(100-99\right)\left(100+99\right)+\left(98+97\right)\left(98-97\right)+....+\left(2+1\right)\left(2-1\right)\)
\(=100+99+98+97+...+2+1\)
=5050
b: \(B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^4-1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^8-1\right)\left(2^8+1\right)\cdot...\cdot\left(2^{64}+1\right)+1\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\)
\(=\left(2^{64}-1\right)\cdot\left(2^{64}+1\right)+1\)
\(=2^{128}-1+1=2^{128}\)
a. \(A=100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=199+195+...+3\)
\(=\dfrac{\left(199+3\right)\left(\dfrac{199-3}{4}+1\right)}{2}=5050\)
b. \(B=3\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)
\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1^2\)
\(=2^{128}-1+1=2^{128}\)
c) \(C=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2+2ab-2ac-2bc-2a^2-2b^2-4ab\)
\(=2c^2\)
Bài 3:
a) \(4x^2+4x+1=\left(2x+1\right)^2\)
b) \(9x^2-12x+4=\left(3x-2\right)^2\)
c) \(ab^2+\dfrac{1}{4}a^2b^4+1=\left(\dfrac{1}{2}ab^2+1\right)^2\)
1. \(A=\left(a+b\right)^2+\left(a+b\right)^2\)
\(\Leftrightarrow A=2\left(a^2+2ab+b^2\right)\)
\(\Leftrightarrow A=2a^2+4ab+2b^2\)
2. \(B=\left(a+b\right)^2-\left(a-b\right)^2\)
\(\Leftrightarrow B=\left(a+b+a-b\right)\left(a+b-a+b\right)\)
\(\Leftrightarrow B=2a.2b=4ab\)
1) \(A=\left(a+b\right)^2+\left(a+b\right)^2\)
\(A=a^2+2ab+b^2+a^2+2ab+b^2\)
\(A=2a^2+4ab+2b^2\)
2) \(B=\left(a+b\right)^2-\left(a-b\right)^2\)
\(B=a^2+2ab+b^2-a^2+2ab-b^2\)
\(B=4ab\)