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a) \(A=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(5x+5\right)^2\)
\(A=\left[\left(3x+1\right)-\left(5x+5\right)\right]^2\)
\(A=\left(-2x-4\right)^2\)
A = (3x + 1)2 - 2(3x + 1)(5x + 5) + (5x + 5)2
= [(3x + 1)-(5x + 5)]2
= (3x + 1 - 5x - 5)2
= [(-2x) - 4]2
B = (3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)
=> (3 - 1)B = (3 - 1)(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)
=>2B = (32 - 1)(32 + 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)
= (34 - 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)
= (38 - 1)(38 + 1)(316 +1)(332 + 1)
= (316 - 1)316 +1)(332 + 1)
= (332 - 1)(332 + 1)
= 364 - 1
vì 2B = 364 - 1
=> B = \(\dfrac{3^{64}-1}{2}\)
C = a2 + b2 + c2 + 2ab - 2ac - 2bc + a2 + b2 + c2 - 2ab + 2ac - 2bc - 2( b2 - 2bc + c2)
= 2a2 + 2b2 + 2c2 - 4bc - 2b2 + 4bc - 2c2
= 2a2
a,b,c,f tìm cách áp dụng HĐT vào nhé! động não tí xem :)
d) Sửa đề :\(100^2-99^2+98^2-97^2+...+2^2-1^2\)
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(=199+195+...+3\)
Khi đó tổng sẽ là:
\(\dfrac{\left(199+3\right)\left[\dfrac{\left(199-3\right)}{4}+1\right]}{2}=5050.\)
e) \(3\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)+...+\left(2^{64}+1\right)+1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)+...+\left(2^{64}+1\right)+1\)
\(=2^{128}-1+1\)
\(=2^{128}.\)
\(A=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)
\(=\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\left(a^2+b^2-c^2-a^2+b^2-c^2\right)-4a^2b^2\)
\(=2a^2.2b^2-4a^2b^2=0\)
\(C=\left(2-6x\right)^2+\left(2-5x\right)^2+2\left(6x-2\right)\left(2-5x\right)\)
\(=\left[\left(2-6x\right)+\left(2-5x\right)\right]^2\)
\(=\left[4-11x\right]^2\)
\(=16-88x+121x^2\)
chúc bn học tốt
2 ) b )
\(a+b+c+d=0\)
\(\Leftrightarrow a+b=-\left(c+d\right)\)
\(\Leftrightarrow\left(a+b\right)^3=-\left(c+d\right)^3\)
\(\Leftrightarrow a^3+b^3+3a^2b+3b^2a=-c^3-3c^2d-3d^2c-d^3\)
\(\Leftrightarrow a^3+b^3+3a^2b+3b^2a+c^3+3c^2d+3d^2c+d^3=0\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3a^2b-3b^2a-3c^2d-3d^2c\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=-3ab\left(a+b\right)-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3ab\left(c+d\right)-3cd\left(c+d\right)\)
\(\Leftrightarrow a^3+b^3+c^3+d^3=3\left(ab-cd\right)\left(c+d\right)\) \(\left(đpcm\right)\)
a: \(=ab\left(a+b\right)-bc\left(b+a\right)-bc\left(c-a\right)-ac\left(c-a\right)\)
\(=\left(a+b\right)\left(ab-bc\right)+\left(a-c\right)\left(bc-ac\right)\)
\(=\left(a+b\right)\cdot b\left(a-c\right)+\left(a-c\right)\cdot c\left(b-a\right)\)
\(=\left(a-c\right)\left(ab+b^2+cb-ac\right)\)
b: \(=ab^2+ac^2+bc^2+a^2b+a^2c+b^2c+2abc\)
\(=ab\left(a+b\right)+c^2\left(a+b\right)+c\left(a+b\right)^2\)
\(=\left(a+b\right)\left(ab+c^2+ac+cb\right)\)
\(=\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
d: \(=a^3\left(b-c\right)-b^3\left(b-c+a-b\right)+c^3\left(a-b\right)\)
\(=a^3\left(b-c\right)-b^3\left(b-c\right)-b^3\left(a-b\right)+c^3\left(a-b\right)\)
\(=\left(b-c\right)\left(a-b\right)\left(a^2+ab+b^2\right)-\left(a-b\right)\left(b-c\right)\left(b^2+bc+c^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a^2+ab+b^2-b^2-bc-c^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a^2+ab-bc-c^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\cdot\left[\left(a-c\right)\left(a+c\right)+b\left(a-c\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(a+b+c\right)\)
nhiều thế, đăng ít một thôi bạn
a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)