\(B=\left(\dfrac{a-b}{a^2+ab}-\dfrac{a}{b^2+ab}\right):\left(\dfrac{b^3}{a^3-ab^2}+\dfrac{1}{a+b}\right)\)

    \(=\left(\dfrac{a-b}{a\left(a+b\right)}-\dfrac{a}{b\left(a+b\right)}\right):\left(\dfrac{b^3}{a\left(a-b\right)\left(a+b\right)}+\dfrac{1}{a+b}\right)\)

    \(=\dfrac{b\left(a-b\right)-a^2}{ab\left(a+b\right)}:\dfrac{b^3+a\left(a-b\right)}{a\left(a-b\right)\left(a+b\right)}\)

    \(=\dfrac{ab-b^2-a^2}{ab\left(a+b\right)}\cdot\dfrac{a\left(a-b\right)\left(a+b\right)}{a^2-ab+b^3}\)

    \(=\dfrac{\left(a-b\right)\left(ab-b^2-a^2\right)}{b\left(a^2-ab+b^3\right)}\)

    \(=\dfrac{-\left(a-b\right)\left(a^2-ab+b^2\right)}{b\left(a^2-ab+b^3\right)}\)

Đề lỗi rồi chứ mình ko rút gọn đc nữa

1) a) \(\left(a-b\right)^2-\left(a+b\right)^2=\left(a-b-a-b\right)\left(a-b+a+b\right)\)

\(=-2b\left(2a\right)=-4ab\)

b) ta có : \(\left(a+2b\right)^2+\left(b-a\right)^2-\left(a-b\right)^2=\left(a+2b\right)^2+\left(b-a\right)-\left(b-a\right)^2\)

\(=\left(a+2b\right)^2\)

2) ta có : \(\left(a-b\right)^2=\left(-\left(b-a\right)\right)^2=\left(b-a\right)^2\left(đpcm\right)\)

3) \(\left(a-b\right)^4=\left(a-b\right)^2\left(a-b\right)^2=\left(a^2-2ab+b^2\right)\left(a^2-2ab+b^2\right)\)

\(=a^4-2a^3b+a^2b^2-2a^3b+4a^2b^2-2ab^3+b^2a^2-2ab^3+b^4\)

\(=a^4-4a^3b+6a^2b^2-4ab^3+b^4\)

a)  Điều kiện :  \(a\ne-b;b\ne1;a\ne-1\)

\(P=\frac{a^2\left(1+a\right)-b^2\left(1-b\right)-a^2b^2\left(a+b\right)}{\left(a+b\right)\left(1-b\right)\left(1+a\right)}\)

\(P=\frac{a^3+a^2+b^3-b^2-a^2b^2\left(a+b\right)}{\left(a+b\right)\left(1-b\right)\left(1+a\right)}\)

\(P=\frac{\left(a+b\right)\left(a^2-ab+b^2\right)+\left(a+b\right)\left(a-b\right)-a^2b^2\left(a+b\right)}{\left(a+b\right)\left(1-b\right)\left(1+a\right)}\)

\(P=\frac{\left(a+b\right)\left(a^2-ab+b^2+a-b-a^2b^2\right)}{\left(a+b\right)\left(1-b\right)\left(1+a\right)}\)

\(P=\frac{a^2+b^2-a^2b^2+a-b-ab}{\left(1-b\right)\left(1+a\right)}\)

\(P=\frac{a^2\left(1-b^2\right)-\left(1-b^2\right)+a\left(1-b\right)+\left(1-b\right)}{\left(1-b\right)\left(1+a\right)}\)

\(P=\frac{\left(1-b\right)\left(a^2+a^2b-1-b+a+1\right)}{\left(1-b\right)\left(1+a\right)}\)

\(P=\frac{a^2+a^2b+a-b}{1+a}\)

\(P=\frac{a\left(a+1\right)+b\left(a-1\right)\left(a+1\right)}{1+a}\)

\(P=\frac{\left(a+1\right)\left(a+ab-b\right)}{1+a}\)

P = a + ab - b

b)

P = 3

<=>  a + ab - b = 3

<=>  a(b+1) - (b+1) +1 - 3 = 0

<=>   (b+1)(a-1)  = 2

Ta có bảng sau với a, b nguyên

Vậy (a;b) \(\in\){ (3; 0) ; (0; -3)}

a) \(A=\left[\left(\frac{1}{5}\right)^2\right]^{\frac{-3}{2}}-\left[2^{-3}\right]^{\frac{-2}{3}}=5^3-2^2=121\)

b) \(B=6^2+\left[\left(\frac{1}{5}\right)^{\frac{3}{4}}\right]^{-4}=6^2+5^3=161\)

c) \(C=\frac{a^{\sqrt{5}+3}.a^{\sqrt{5}\left(\sqrt{5}-1\right)}}{\left(a^{2\sqrt{2}-1}\right)^{2\sqrt{2}+1}}=\frac{a^{\sqrt{5}+3}.a^{5-\sqrt{5}}}{a^{\left(2\sqrt{2}\right)^2-1^2}}\)

                              \(=\frac{a^{\sqrt{5}+3+5-\sqrt{5}}}{a^{8-1}}=\frac{a^8}{a^7}=a\)

d) \(D=\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)^2:\left(b-2b\sqrt{\frac{b}{a}}+\frac{b^2}{a}\right)\)

        \(=\left(\sqrt{a}-\sqrt{b}\right)^2:b\left[1-2\sqrt{\frac{b}{a}}+\left(\sqrt{\frac{b}{a}}\right)^2\right]\)

        \(=\left(\sqrt{a}-\sqrt{b}\right)^2:b\left(1-\sqrt{b}a\right)^2\)

Bài giải:

a) (a + b)² – (a – b)² = (a² + 2ab + b²) – (a² – 2ab + b²)

= a² + 2ab + b² – a² + 2ab - b² = 4ab

Hoặc (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

b) (a + b)³ – (a – b)³ – 2b³

= (a³ + 3a²b + 3ab² + b³) – (a³ – 3a²b + 3ab² – b³) – 2b³

= a³ + 3a²b + 3ab² + b³ – a³ + 3a²b - 3ab² + b³ – 2b³

= 6a²b

Hoặc (a + b)³ – (a – b)³ – 2b³ = [(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) (x + y + z)² – 2(x + y + z)(x + y) + (x + y)²

= x² + y² + z²+ 2xy + 2yz + 2xz – 2(x² + xy + yx + y² + zx + zy) + x² + 2xy + y²

= 2x² + 2y² + z² + 4xy + 2yz + 2xz – 2x² – 4xy – 2y² – 2xz – 2yz = z²