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1. biến đổi vế trái
= a2x2 + a2y2 + b2x2 + b2y2
= (ax -by)2 + (bx+ ay)2 - 2abxy + 2abxy
= (ax -by)2 + ( bx + ay)2 = vế phải( dpcm)
a) Đặt \(A=\left(3+1\right)\left(3^2+1\right)...\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=2.\left(3+1\right)\left(3^2+1\right)...\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)...\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)...\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(2A=\left(3^4-1\right)...\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(...\)
\(2A=\left(3^{32}-1\right)\left(3^{32}+1\right)\)
\(2A=3^{64}-1\)
\(A=\frac{3^{64}-1}{2}\)
1a)\(a^2+b^2+1\ge ab+a+b\)
\(\Leftrightarrow2\left(a^2+b^2+1\right)\ge2\left(ab+b+a\right)\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)(luôn đúng)
Dấu "=" xảy ra khi x=y=1
b)\(a^2+b^2+c^2\ge a\left(b+c\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2ac\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+b^2+c^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+b^2+c^2\ge0\)(luôn đúng)
Dấu "=" xảy ra khi a=b=c=0
3a2c2 + bd + 3abc + acd
= 3ac(ac + b) + d(ac + b)
= (ac + b)(3ac + d)
ab(a + b) - bc(a + c) + abc
= b(a2 + ab - ac - c2 + ac)
= b(a2 + ab - c2)
a(b2 + c2) + b(c2 + a2) + c(a2 + b2) + 2abc
= ab2 + ac2 + bc2 + a2b + c(a2 + 2ab + b2)
= c2(a + b) + ab(a + b) + c(a + b)2
= (a + b)(c2 + ab + ac + bc)
= (a + b)[c(b + c) + a(b + c)]
= (a + b)(a + c)(b + c)
bc(b + c) + ac(c - a) - ab(a + b)
= bc(b + c) + ac[(b + c) - (a + b)] - ab(a + b)
= bc(b + c) + ac(b + c) - ac(a + b) - ab(a + b)
= c(b + c)(a + b) - a(a + b)(b + c)
= (a + b)(b + c)(c - a)
B1:a)(3x-5)2-(3x+1)2=8
[(3x-5)+(3x+1)].[(3x-5)-(3x+1)]=8
(3x-5+3x+1)(3x-5-3x-1)=8
9x2-15x-9x2-3x-15x+25+15x+5+9x2-15x-9x2-3x+3x-5-3x-1=8
-36x+24=8
-36x=8-24=16
x=16:(-36)=\(\dfrac{-4}{9}\)
Bài 5:
a: \(=\left(xy-u^2v^3\right)\left(xy+u^2v^3\right)\)
b: \(=\left(2xy^2-3xy^2+1\right)\left(2xy^2+3xy^2-1\right)\)
\(=\left(1-xy^2\right)\left(5xy^2-1\right)\)
Bài 6:
a: \(\left(a+b+c-d\right)\left(a+b-c+d\right)\)
\(=\left(a+b\right)^2+\left(c-d\right)^2\)
\(=a^2+2ab+b^2+c^2-2cd+d^2\)
b: \(\left(a+b-c-d\right)\left(a-b+c-d\right)\)
\(=\left(a-d\right)^2-\left(b-c\right)^2\)
\(=a^2-2ad+d^2-b^2+2bc-c^2\)
\(c)\)
\(a^3+b^3+c^3-3abc\)
\(=a^3+3ab\left(a+b\right)+b^3+c^3-3abc-3ab\left(a+b\right)\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ab-ac+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(d)\)
\(\left(a+b+c\right)^3-a^3-b^3-c^3\)
\(=[\left(a+b\right)c]^3-a^3-b^3-c^3\)
\(=\left(a+b\right)^3+c^3+3\left(a+b\right)c\left(a+b+c\right)-a^3-b^3-c^3\)
\(=a^3+b^3+3ab\left(a+b\right)+c^3+3\left(a+b\right)c\left(a+b+c\right)-a^3-b^3-c^3\)
\(=3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=3\left(a+b\right)[a\left(b+c\right)+c\left(b+c\right)]\)
\(=3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Bài 1:Cách thông thường nhất là sos hoặc cauchy-Schwarz nhưng thôi ko làm:v Thử cách này cho nó mới dù rằng ko chắc
Giả sử \(a\ge b\ge c\Rightarrow c\le1\Rightarrow a+b=3-c\ge2\) và \(a\ge1\)
Ta có \(LHS=a^3.a+b^3.b+c^3.c\)
\(=\left(a^3-b^3\right)a+\left(b^3-c^3\right)\left(a+b\right)+c^3\left(a+b+c\right)\)
\(\ge\left(a^3-b^3\right).1+\left(b^3-c^3\right).2+3c^3\)
\(=a^3+b^3+c^3=RHS\)
Đẳng thức xảy ra khi a = b = c = 1
=a, a(b2+c2)+b(a2+c2)+c(a2+b2)+2abc
= ab2+ac2+ba2+bc2+ca2+cb2+2abc
= c2(a+b)+ab(a+b)+c(a2+b2+2ab)
= c2(a+b)+ab(a+b)+c(a+b)2
= (a+b)\(\left[c^2+ab+c\left(a+b\right)\right]\)
= (a+b)(c2+ab+ca+cb)
= (a+b)\(\left[c\left(a+c\right)+b\left(a+c\right)\right]\)
=(a+b)(a+c)(b+c)
b, a(b-c)3+b(c-a)3+c(a-b)3
= a(b-c)3-b\(\left[\left(b-c\right)+\left(a-b\right)\right]\)3+c(a-b)3
= a(b-c)3-b(b-c)3-3b(b-c)2(a-b)-3b(b-c)(a-b)2-b(a-b)3+c(a-b)3
= a(b-c)3-b(b-c)3-3b(b-c)(a-b)(b-c+a-b)-b(a-b)3+c(a-b)3
= a(b-c)3-b(b-c)3-3b(b-c)(a-b)(a-c)-b(a-b)3+c(a-b)3
= (b-c)3(a-b)-3b(b-c)(a-b)(a-c)-(a-b)3(b-c)
= (b-c)(a-b)\(\left[\left(b-c\right)^2-3b\left(a-c\right)-\left(a-b\right)^2\right]\)
=(b-c)(a-b)(b2-2bc+c2-3ab+3bc-a2+2ab-b2)
= (b-c)(a-b)(c2-a2+bc-ab)
= (b-c)(a-b)\(\left[\left(c-a\right)\left(c+a\right)+b\left(c-a\right)\right]\)
= (b-c)(a-b)(c-a)(c+a+b)
c, a2b2(a-b)+b2c2(b-c)+c2a2(c-a)
= a2b2(a-b)-b2c2\(\left[\left(a-b\right)+\left(c-a\right)\right]\)+c2a2(c-a)
= a2b2(a-b)-b2c2(a-b)-b2c2(c-a)+c2a2(c-a)
= b2(a-b)(a2-c2)+c2(c-a)(a2-b2)
= b2(a-b)(a-c)(a+c)-c2(a-c)(a-b)(a+b)
= (a-c)(a-b)\(\left[b^2\left(a+c\right)-c^2\left(a+b\right)\right]\)
= (a-c)(a-b)(b2a+b2c-c2a-c2b)
= (a-c)(a-b)\(\left[a\left(b^2-c^2\right)+bc\left(b-c\right)\right]\)
= (a-c)(a-b)\(\left[a\left(b-c\right)\left(b+c\right)+bc\left(b-c\right)\right]\)
= (a-c)(a-b)(b-c)\(\left[a\left(b+c\right)+bc\right]\)
= (a-c)(a-b)(b-c)(ab+ac+bc)
d, a4(b-c)+b4(c-a)+c4(a-b)
= a4(b-c)-b4[(b-c)+(a-b)]+c4(a-b)
= (b-c)(a4-b4)+(a-b)(c4-b4)
= (b-c)(a2-b2)(a2+b2)+(a-b)(c2-b2)(c2+b2)
= (b-c)(a-b)(a+b)(a^2+b^2)-(a-b)(b-c)(b+c)(b2+c2)
= (b-c)(a-b)(a3+ab2+ba2+b3-bc2-b3-cb2-c3)
= (b-c)(a-b)(a3+ab2+ba2-bc2-c3-cb2)
= (b-c)(a-b)(a3-c3)+b2(a-c)+b(a2-c2)
= (b-c)(a-b)(a-c)(a2+ac+c2)+b2(a-c)+b(a-c)(a+c)
= (b-c)(a-b)(a-c)(a2+ac+c2+b2+ab+ac)
= (a-b)(b-c)(c-a)(a2+b2+c2+ab+bc+ca)
bạn làm giỏi thế có phương pháo nào ko mách mk