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a) Ta có: (a+b)2 - (a-b)2
= (a+b+a-b)(a+b-a+b)
= 2a.2b
= 4ab
b) Ta có: (a+b)3 - (a-b)3 - 2b3
= a3 + 3a2b + 3ab2 + b3 - a3 + 3a2b - 3ab2 + b3 - 2b3
= 6a2b
c) Ta có: (x+y+z)2 - 2(x+y+z)(x+y) + (x+y)2
= (x+y+z-x-y)2
= z2
\(a,=a^3+3a^2b+3ab^2+b^3-a^3+3a^2b-3ab^2+b^3-2b^3=6a^2b\\ b,=\left(6x+1-6x+1\right)^2=2^2=4\)
a)(x+y+z)2 - 2(x+y+z)(x+y)+(x+y)2
=[(x+y+z)-(x-y)]2
=(x+y+z-x-y)2
=z2
b) (a+b)3 - (a - b)3 - 2b3
=[(a+b)-(a-b)][(a+b)2+(a+b)(a-b)+(a-b)2]-2b3
=(a+b-a+b)(a2+2ab+b2+a2-b2+a2-2ab+b2)-2b3
=2b(3a2+b2)-2b3
=6a2b+2b3-2b3
=6a2b
c) (a + b)2 - (a - b)2=[a+b+(a-b)][a+b-(a-b)]=(a+b+a-b)(a+b-a+b)
=2a.2b=4ab
a: Ta có: \(\frac{1}{2a-b}-\frac{a^2-1}{2a^3-b+2a-a^2b}\)
\(=\frac{1}{2a-b}-\frac{a^2-1}{a^2\left(2a-b\right)+\left(2a-b\right)}\)
\(=\frac{1}{2a-b}-\frac{a^2-1}{\left(2a-b\right)\left(a^2+1\right)}=\frac{a^2+1-a^2+1}{\left(2a-b\right)\left(a^2+1\right)}=\frac{2}{\left(2a-b\right)\left(a^2+1\right)}\)
\(\frac{4a+2b}{a^3b+ab}-\frac{2}{a}\)
\(=\frac{4a+2b}{ab\left(a^2+1\right)}-\frac{2}{a}=\frac{4a+2b-2b\left(a^2+1\right)}{ab\left(a^2+1\right)}\)
\(=\frac{4a-2a^2b}{ab\left(a^2+1\right)}=\frac{2a\left(2-ab\right)}{ab\cdot\left(a^2+1\right)}=\frac{2\left(2-ab\right)}{b\left(a^2+1\right)}\)
Ta có: \(A=\left(\frac{1}{2a-b}-\frac{a^2-1}{2a^3-b+2a-a^2b}\right):\left(\frac{4a+2b}{a^3b+ab}-\frac{2}{a}\right)\)
\(=\frac{2}{\left(2a-b\right)\left(a^2+1\right)}:\frac{2\left(2-ab\right)}{b\left(a^2+1\right)}=\frac{2b\left(a^2+1\right)}{2\left(2-ab\right)\left(2a-b\right)\left(a^2+1\right)}=\frac{b}{\left(2-ab\right)\left(2a-b\right)}\)
b:
Sửa đề: b>a>0
\(4a^2+b^2=5ab\)
=>\(4a^2-5ab+b^2=0\)
=>\(4a^2-4ab-ab+b^2=0\)
=>(a-b)(4a-b)=0
TH1: a-b=0
=>a=b
mà a>b
nên Loại
TH2: 4a-b=0
=>b=4a(nhận)
\(A=\frac{b}{\left(2-ab\right)\left(2a-b\right)}\)
\(=\frac{4a}{\left(2-a\cdot4a\right)\left(2a-4a\right)}=\frac{4a}{\left(2-4a^2\right)\left(-2a\right)}\)
\(=\frac{4a}{-2a\cdot\left(-2\right)\left(2a^2-1\right)}=\frac{1}{2a^2-1}\)
Bài 1:
a) \(\left(a+b\right)^2-\left(a-b\right)^2\)
\(=\left(a+b+\left(a-b\right)\right).\left(a+b-\left(a-b\right)\right)\)
\(=2a.2b\)
\(=4ab\)
Câu 1:
a) (a +b )2 - ( a -b )2
=a2+b2-a2+b2
=2b2
b) (a + b )3- ( a - b )3 - 2b3
=a3+b3-a+b3-2b3
=a3-a
c) ( x+y+z)2 - 2(x+y+z)(x+y) + (x + y )2
=x2+xy+xz+xy+y2+yz+xz+yz+z2-2.(x2+xy+xz+xy+y2+yz)+x2+xy+xy+y2
=x2+y2+z2+2xy+2xz+2yz-2x2-2y2-4xy-2xz-2yz+x2+2xy+y2
=0
\(a,\left(a+b\right)^2-\left(a-b\right)^2\)
\(=a^2+2ab+b^2-a^2+2ab-b^2\)
\(=4ab\)
\(b,\left(a+b\right)^3-\left(a-b\right)-\left(2b\right)^3\)
\(=a^3+3a^2b+3ab^2+b^3-a+b-8b^3\)
a) \(\left(a+b\right)^2-\left(a-b\right)^2\)
\(\left(a+b-a+b\right)\left(a+b+a-b\right)\)
\(\left(2b\right)\left(2a\right)\)
\(4ab\)
b) \(\left(a+b\right)^3-\left(a-b\right)-\left(2b\right)^3\)
\(a^3+3a^2b+3ab^2+b^3-a+b-8b^3\)
\(a\left(a^2-1\right)+3\left(a^2b+ab^2\right)+b\left(b^2+1-8b^2\right)\)
\(a\left(a-1\right)\left(a+1\right)+3\left[ab\left(a+b\right)\right]+b\left(-7b^2+1\right)\)