1. = \(\dfrac{x+y}{x-y}\)
2. = \(\dfrac{x}{x+3}\)

`@` `\text {Ans}`

`\downarrow`

\((x+y)(x-y)+(xy^4-x^3y^2) \div (xy^2) \)

`= x(x-y) + y(x-y) + xy^4 \div xy^2 - x^3y^2 \div xy^2`

`= x^2 - xy + xy - y^2 + y^2 - x^2`

`= (x^2 - x^2) + (-xy + xy) + (-y^2 + y^2)`

`= 0`

Trường hợp 1: với thì tương lai, hiện tại đơn, hiện tại tiếp diễn:

C1:S+be+Vp2+ that +S-V

C2:S+be+VP2+to+ V

Trường hợp 2: với các thì Hiện tại honaf thành, quá khứ

C1:S+be+VP2+ that +S-V

C2:S+be+VP2+to have +VP2

\(B=9x^4-\left(2x+1\right)^2-\left(9x^4+6x^2+1\right)\\ =9x^4-4x^2-4x-1-9x^4-6x^2-1\\ =-10x^2-4x-2\)

\(B=\left(3x^2+1-2x\right)\left(3x^2+1+2x\right)-\left(3x^2+1\right)^2\\ B=\left(3x^2+1\right)^2-4x^2-\left(3x^2+1\right)^2=-4x^2\)

\(-\dfrac{2}{3}xy^2z.9x^2y^2=-6x^3y^4z\)

6xy4z

6xy4z

\(\dfrac{2sin8a-sin16a}{2sin8a+sin16a}=\dfrac{2sin8a-2sin8a.cos8a}{2sin8a+2sin8a.cos8a}=\dfrac{2sin8a\left(1-cos8a\right)}{2sin8a\left(1+cos8a\right)}=\dfrac{1-cos8a}{1+cos8a}=\dfrac{1-\left(1-2sin^24a\right)}{1+\left(1-2sin^24a\right)}=\dfrac{2sin^24a}{2-2sin^24a}=\dfrac{sin^24a}{1-sin^24a}=\dfrac{sin^24a}{cot^24a}=tan^24a\)

\(=\dfrac{2sin8a-2sin8a.cos8a}{2sin8a+2sin8a.cos8a}=\dfrac{2sin8a\left(1-cos8a\right)}{2sin8a\left(1+cos8a\right)}=\dfrac{1-cos8a}{1+cos8a}\)

\(=\dfrac{1-\left(1-2sin^24a\right)}{1+\left(2cos^24a-1\right)}=\dfrac{2sin^24a}{2cos^24a}=tan^24a\)

a) (2x^2 +2xy - xy -y^2 ) / (2x^2 - 2xy - xy +y^2)

= 2x(x+y) - y(x+y)  /  2x(x-y) - y(x-y)

= (2x-y)(x+y)  /  (2x-y)(x-y)

= x+y/x-y

Rút gọn cái sau:

\(\frac{32x+4x^2+2x^3}{x^3+64}\)

\(=\frac{2x\left(x^2+2x+16\right)}{\left(x+4\right)\left(x^2-4x+16\right)}\)

Đề có vẻ sai sai ? 

a) \(\sqrt{0,64.a^2}\left(a>0\right)=0,8.\left|a\right|=0,8a\)

b) \(\sqrt{a^2\left(a-2\right)^2}\left(a>2\right)=\left|a\left(a-2\right)\right|=a\left(a-2\right)=a^2-2a\)

c) \(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}\left(a\ge0,a\ne1\right)=\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}=1+\sqrt{a}+a\)