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

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

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

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

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

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

Đề lỗi rồi chứ mình ko rút gọn đc nữa

1)

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

\(=\left(a^2+2a.2b+\left(2b\right)^2\right)+\left(b^2-2ba+a^2\right)-\left(a^2-2ab+b^2\right)\)

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

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

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

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

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

ai tích mình mình tích lại cho

nhiều thế, đăng ít một thôi bạn

a/ \(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(2A=\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)

\(\Rightarrow2A=3^{128}-1\Rightarrow A=\dfrac{3^{128}-1}{2}\)

e) ta dể dàng thấy được : \(a^2+b^2=\left(a+b\right)^2-2ab\)

\(\Rightarrow E=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)

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

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

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

g) củng sử dụng cái trên ta có : \(G=\left(a+b+c+d\right)^2+\left(a+b-c-d\right)^2+\left(a+c-b-d\right)^2+\left(a+d-b-c\right)^2\)

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

\(=4\left(a+b\right)^2+4\left(a-b\right)^2-2\left(\left(a+b\right)^2-\left(c+d\right)^2\right)-2\left(\left(a-b\right)^2-\left(c-d\right)^2\right)\)

\(=4\left(\left(a+b\right)^2+\left(a-b\right)^2\right)-2\left(\left(a+b\right)^2+\left(a-b\right)^2\right)+2\left(\left(c+d\right)^2+\left(c-d\right)^2\right)\)

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

\(=2\left(4a^2-2\left(a^2-b^2\right)\right)+2\left(4c^2-2\left(c^2-d^2\right)\right)\)

\(=2\left(2a^2+2b^2\right)+2\left(2c^2+2d^2\right)=4\left(a^2+b^2+c^2+d^2\right)\)

bn đăng nhiều quá nên mk làm câu nào hay câu đó nha

mà nè mấy câu a;b;c;d hình như trên mạng có bn lên đó tìm nha .