Bài 1:

b: \(B=1:\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{x-1}\right)\)

\(=1:\left(\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\right)\)

\(=1:\frac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=1:\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=1:\frac{x-\sqrt{x}}{\left.\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\right.}\)

\(=\frac{\left(\sqrt{x}\right)-1)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)

Bài 2:

a: \(\left(\frac{6a+1}{a^2-6a}+\frac{6a-1}{a^2+6a}\right)\cdot\frac{a^2-36}{a^2+1}\)

\(=\left(\frac{6a+1}{a\left(a-6\right)}+\frac{6a-1}{a\left(a+6\right)}\right)\cdot\frac{\left(a-6\right)\left(a+6\right)}{a^2+1}\)

\(=\frac{\left(6a+1\right)\left(a+6\right)+\left(6a-1\right)\left(a-6\right)}{a\left(a-6\right)\left(a+6\right)}\cdot\frac{\left(a-6\right)\left(a+6\right)}{a^2+1}\)

\(=\frac{6a^2+37a+6+6a^2-37a+6}{a}\cdot\frac{1}{a^2+1}=\frac{12a^2+12}{a\left(a^2+1\right)}\)

\(=\frac{12\left(a^2+1\right)}{a\left(a^2+1\right)}=\frac{12}{a}\)

b: \(\left(\frac{1}{x}+\frac{1}{y}\right)\cdot\frac{1}{x+2\sqrt{xy}+y}+\frac{2}{\left.\left(\sqrt{x}+\sqrt{y}\right)^3\right.}\cdot\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\)

\(=\frac{x+y}{xy}\cdot\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}\cdot\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)

\(=\frac{x+y}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)^2}=\frac{x+y+2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}=\frac{1}{xy}\)

\(\frac{\sqrt{x}-\sqrt{y}}{xy\cdot\sqrt{xy}}:\left\lbrack\left(\frac{1}{x}+\frac{1}{y}\right)\cdot\frac{1}{x+2\sqrt{xy}+y}+\frac{2}{\left.\left(\sqrt{x}+\sqrt{y}\right)^3\right.}\cdot\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\right\rbrack\)

\(=\frac{\sqrt{x}-\sqrt{y}}{xy\cdot\sqrt{xy}}:\frac{1}{xy}\)

\(=\frac{\sqrt{x}-\sqrt{y}}{xy\cdot\sqrt{xy}}\cdot xy=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)

Bài 1:

b: \(B=1:\left(\frac{x+2}{x\sqrt{x}-1}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{\sqrt{x}+1}{x-1}\right)\)

\(=1:\left(\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}+1}{x+\sqrt{x}+1}-\frac{1}{\sqrt{x}-1}\right)\)

\(=1:\frac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=1:\frac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=1:\frac{x-\sqrt{x}}{\left.\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\right.}\)

\(=\frac{\left(\sqrt{x}\right)-1)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}=\frac{x+\sqrt{x}+1}{\sqrt{x}}\)

Bài 2:

a: \(\left(\frac{6a+1}{a^2-6a}+\frac{6a-1}{a^2+6a}\right)\cdot\frac{a^2-36}{a^2+1}\)

\(=\left(\frac{6a+1}{a\left(a-6\right)}+\frac{6a-1}{a\left(a+6\right)}\right)\cdot\frac{\left(a-6\right)\left(a+6\right)}{a^2+1}\)

\(=\frac{\left(6a+1\right)\left(a+6\right)+\left(6a-1\right)\left(a-6\right)}{a\left(a-6\right)\left(a+6\right)}\cdot\frac{\left(a-6\right)\left(a+6\right)}{a^2+1}\)

\(=\frac{6a^2+37a+6+6a^2-37a+6}{a}\cdot\frac{1}{a^2+1}=\frac{12a^2+12}{a\left(a^2+1\right)}\)

\(=\frac{12\left(a^2+1\right)}{a\left(a^2+1\right)}=\frac{12}{a}\)

b: \(\left(\frac{1}{x}+\frac{1}{y}\right)\cdot\frac{1}{x+2\sqrt{xy}+y}+\frac{2}{\left.\left(\sqrt{x}+\sqrt{y}\right)^3\right.}\cdot\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\)

\(=\frac{x+y}{xy}\cdot\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2}{\left(\sqrt{x}+\sqrt{y}\right)^3}\cdot\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}}\)

\(=\frac{x+y}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}+\frac{2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)^2}=\frac{x+y+2\sqrt{xy}}{xy\left(\sqrt{x}+\sqrt{y}\right)^2}=\frac{1}{xy}\)

\(\frac{\sqrt{x}-\sqrt{y}}{xy\cdot\sqrt{xy}}:\left\lbrack\left(\frac{1}{x}+\frac{1}{y}\right)\cdot\frac{1}{x+2\sqrt{xy}+y}+\frac{2}{\left.\left(\sqrt{x}+\sqrt{y}\right)^3\right.}\cdot\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)\right\rbrack\)

\(=\frac{\sqrt{x}-\sqrt{y}}{xy\cdot\sqrt{xy}}:\frac{1}{xy}\)

\(=\frac{\sqrt{x}-\sqrt{y}}{xy\cdot\sqrt{xy}}\cdot xy=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{xy}}\)

Bài 3:

a: \(A=\left(\frac{x}{x-y}-\frac{y}{x+y}\right):\left(\frac{x+y}{x-y}-\frac{2xy}{x^2-y^2}\right)\)

\(=\frac{x\left(x+y\right)-y\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}:\frac{\left(x+y\right)^2-2xy}{\left(x-y\right)\left(x+y\right)}\)

\(=\frac{x^2+xy-xy+y^2}{\left(x-y\right)\left(x+y\right)}\cdot\frac{\left(x-y\right)\left(x+y\right)}{x^2+2xy+y^2-2xy}=\frac{x^2+y^2}{x^2+y^2}=1\)

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

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

\(=\left(a+\sqrt{a}+1+\sqrt{a}\right)\cdot\frac{1}{\left(\sqrt{a}+1\right)^2}=\frac{\left(\sqrt{a}+1\right)^2}{\left(\sqrt{a}+1\right)^2}=1\)

Bài 3:

\(a,A=\dfrac{x^2+xy-xy+y^2}{\left(x-y\right)\left(x+y\right)}:\dfrac{x^2+2xy+y^2-2xy}{\left(x-y\right)\left(x+y\right)}\\ A=\dfrac{x^2+y^2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{\left(x-y\right)\left(x+y\right)}{x^2+y^2}=1\\ b,=\left[\dfrac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}+\sqrt{a}\right]\left[\dfrac{1-\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}\right]^2\\ =\left(a+2\sqrt{a}+1\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2\\ =\left(\sqrt{a}+1\right)^2\cdot\dfrac{1}{\left(\sqrt{a}+1\right)^2}=1\)

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

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

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

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

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

b: \(C=\frac{a}{b-2}-\left\lbrack\frac{\left(a^2+2a+1\right)}{b^2-4}\right\rbrack\cdot\frac{b+2}{a+1}\)

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

\(=-\frac{1}{b-2}\)

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

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

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

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

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

b: \(C=\frac{a}{b-2}-\left\lbrack\frac{\left(a^2+2a+1\right)}{b^2-4}\right\rbrack\cdot\frac{b+2}{a+1}\)

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

\(=-\frac{1}{b-2}\)

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

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

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

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

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

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

Đề lỗi rồi chứ mình ko rút gọn đc nữa

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

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

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

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

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

b: \(C=\frac{a}{b-2}-\left\lbrack\frac{\left(a^2+2a+1\right)}{b^2-4}\right\rbrack\cdot\frac{b+2}{a+1}\)

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

\(=-\frac{1}{b-2}\)

\(A=\frac{\left(a+2\right)^2\left(5a-15a^2\right)}{\left(a-3\right)\left(4a-a^3\right)}=\frac{\left(a+2\right)^2.5a.\left(1-3a\right)}{\left(a-3\right).a.\left(2-a\right)\left(a+2\right)}\)

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

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

\(=\dfrac{\left(a+b+c\right)\left[\left(a+b\right)^2-c\left(a+b\right)+c^2\right]-3ab\left(a+b+c\right)}{a^2+b^2+c^2-ab-bc-ac}\)

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

=a+b+c

b: 

Sửa đề: \(=\dfrac{x^3-y^3+z^3+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)

\(=\dfrac{\left(x-y\right)^3+z^3+3xy\left(x-y\right)+3xyz}{\left(x+y\right)^2+\left(y+z\right)^2+\left(z-x\right)^2}\)
\(=\dfrac{\left(x-y+z\right)\left(x^2-2xy+y^2-xz+yz+z^2\right)+3xy\left(x-y+z\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)

\(=\dfrac{\left(x-y+z\right)\left(x^2+y^2+z^2+xy-xz+yz\right)}{2\left(x^2+y^2+z^2+xy+yz-xz\right)}\)

\(=\dfrac{x-y+z}{2}\)

a) \(\dfrac{a^3+b^3+c^3-3abc}{a^2+b^2+c^2-ab-bc-ca}\)

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

\(=a+b+c\)

\(A=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\\ A=\dfrac{2\left(\dfrac{1}{2}-2\right)}{\dfrac{1}{2}+2}=\dfrac{2\left(-\dfrac{3}{2}\right)}{\dfrac{5}{2}}=\left(-3\right)\cdot\dfrac{2}{5}=-\dfrac{6}{5}\)

\(B=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}=\dfrac{-5}{-5+10}=\dfrac{-5}{5}=-1\)