3{x-2}²/x{x-2}{x²+2x+4}=3{x-2}/x²+2x+4=-3/x+2 

nho tick minh nha

\(\frac{3x^2-12x+12}{x^4-8x}=\frac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}=\frac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}=\frac{3}{x\left(x^2+2x+4\right)}\)(điều kiện: \(x\ne\left\{0;2\right\}\)

ĐK: \(x\ne\left\{0;2\right\}\)

Ta có: \(3x^2-12x+12\)\(=3\left(x^2-4x+4\right)=3\left(x-2\right)^2\) (1)

\(x^4-8x\)\(=\left(x-2\right)\left(x^3+2x^2+4x\right)\) (chỗ này mình làm hơi tắt xíu,bạn tự giải ra chi tiết nha)

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

Từ (1) và (2),ta có: \(\frac{3x^2-12x+12}{x^4-8x}=\frac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}\)\(=\frac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)

\(a,=\dfrac{\left(x+1\right)\left(x+y\right)}{\left(x-y\right)\left(x+1\right)}=\dfrac{x+y}{x-y}\\ b,=\dfrac{\left(x-3\right)^2}{3x\left(x-3\right)}=\dfrac{x-3}{3x}\\ c,=\dfrac{\left(y-x\right)\left(y+x\right)}{xy\left(x-y\right)}=\dfrac{-x-y}{xy}\)

Lời giải:

a.

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

b.

\(\frac{x^2-6x+9}{3x^2-9x}=\frac{(x-3)^2}{3x(x-3)}=\frac{x-3}{3x}\)

c.

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

a)\(\dfrac{3x^2-12x+12}{x^4-8x}=\dfrac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}=\dfrac{3\left(x-2\right)^2}{x\left(x^3-2^3\right)}=\dfrac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)

a) 5x + 10 = 5(x + 2)

25x2 + 50x = 25x(x + 2)

⇒ Nhân tử chung của chúng là: 5(x + 2)

b)

a) Ta có: \(-7xy\cdot\sqrt{\dfrac{3}{xy}}\)

\(=\dfrac{-7xy\cdot\sqrt{3xy}}{xy}\)

\(=-7\sqrt{3}\cdot\sqrt{xy}\)

b) Ta có: \(ab+b\sqrt{a}+\sqrt{a}+1\)

\(=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)

\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)

$a)-7xy.\sqrt{\dfrac{3}{xy}}$

$=-7.\sqrt{x^2y^2.\dfrac{3}{xy}}(do \,x,y>0a\to xy>0)$

$=-7.\sqrt{\dfrac{xy}{3}}$

$b)ab+b\sqrt{a}+\sqrt{a}+1(a \ge 0)$

$=b\sqrt{a}(\sqrt{a}+1)+\sqrt{a}+1$

$=(\sqrt{a}+1)(b\sqrt{a}+1)$

a) \(-7xy.\sqrt{\dfrac{3}{xy}}=-7xy.\dfrac{\sqrt{3xy}}{xy}=-7\sqrt{3xy}\)

b) \(ab+b\sqrt{a}+\sqrt{a}+1=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)

a: \(-7xy\cdot\sqrt{\dfrac{3}{xy}}=-7xy\cdot\dfrac{\sqrt{3}}{\sqrt{xy}}=-7\sqrt{3xy}\)

b: \(ab+b\sqrt{a}+\sqrt{a}+1\)

\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)

a, \(=\left(x+y\right)^2-2^2=\left(x+y-2\right)\left(x+y+2\right)\)

b, =  \(y^2-4x^2+4x^2=y^2\)

Thay y = 10 vào BT trên, ta có:

\(y^2=10^2=100\)

Vậy giá trị của BT là 100

 xác định khi:

(x – 2y)(x + 2y) ≠ 0 ⇔  ⇒ x  ≠   ± 2y