B1: a)\(xy\left(3x-2y\right)-2xy^2=3x^2y-2y^2x-2xy^2=3x^2y-4xy^2\)

b) \(\left(x^2+4x+4\right):\left(x+2\right)=\left(x+2\right)^2:\left(x+2\right)=\left(x+2\right)\)

\(\dfrac{2\left(x-1\right)}{x^2}.\dfrac{x}{\left(x-1\right)}=\dfrac{2\left(x-1\right)x}{x^2\left(x-1\right)}=\dfrac{2}{x}\)

B2:

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

b)\(x^2-y^2+3x-3y=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)

Mấy bài này là mấy bài rất rất rất cơ bản, học sinh TB cũng phải tự làm được, mấy bài kiểu này đừng nên đăng lên hỏi nha:vv

a.

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

\(=\dfrac{\left(x-1\right)\left(x^2+x+1\right)}{x-1}-\dfrac{\left(x-1\right)\left(x+1\right)}{x+1}\)

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

b.

\(\dfrac{x+y}{2\left(x-y\right)}-\dfrac{x-y}{2\left(x+y\right)}+\dfrac{2y^2}{x^2-y^2}\)

\(=\dfrac{\left(x+y\right)^2}{2\left(x-y\right)\left(x+y\right)}-\dfrac{\left(x-y\right)^2}{2\left(x-y\right)\left(x+y\right)}+\dfrac{4y^2}{2\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{\left(x+y\right)^2-\left(x-y\right)^2+4y^2}{2\left(x-y\right)\left(x+y\right)}\)

\(=\dfrac{4xy+4y^2}{2\left(x-y\right)\left(x+y\right)}=\dfrac{4y\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}\)

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

c.

\(\dfrac{x+5}{2x-4}.\dfrac{4-2x}{x+2}=\dfrac{x+5}{2x-4}.\dfrac{-\left(2x-4\right)}{x+2}=-\dfrac{x+5}{x+2}\)

d.

\(\dfrac{8}{x^2+2x-3}+\dfrac{2}{x+3}+\dfrac{1}{x-1}=\dfrac{8}{\left(x-1\right)\left(x+3\right)}+\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x+3\right)}+\dfrac{x+3}{\left(x-1\right)\left(x+3\right)}\)

\(=\dfrac{8+2\left(x-1\right)+x+3}{\left(x-1\right)\left(x+3\right)}=\dfrac{3x+9}{\left(x-1\right)\left(x+3\right)}\)

\(=\dfrac{3\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}=\dfrac{3}{x-1}\)

Bài 1 . Chia :( x³ + 5x² - 4x - 20) cho ( x² + 3x - 10) ta được x+ 2

Chia :( x³ + 5x² - 4x - 20) cho ( x² + 7x + 10) ta được x - 2

Do đó , ta có :

\(\dfrac{1}{x^2+3x-10}=\dfrac{x+2}{\left(x^2+3x-10\right)\left(x+2\right)}=\dfrac{x+2}{x^3+5x^2-4x-20}\)

Và : \(\dfrac{x}{x^2+7x+10}=\dfrac{x\left(x-2\right)}{\left(x^2+7x+10\right)\left(x-2\right)}=\dfrac{x^2-2x}{x^3+5x^2-4x-20}\)

Bài 2 . a) Ta có :

\(\dfrac{x-1}{x^3+1}\)( giữ nguyên)

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

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

b) Ta có MTC = x²( y - z)²

Ta có :

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

\(\dfrac{y}{x^2\left(y-z\right)^2}\)( giữ nguyên )

\(\dfrac{z}{x^2}=\dfrac{z\left(y-z\right)^2}{x^2\left(y-z\right)^2}\)

1.

Áp dụng BĐT Cauchy-Schwarz:

\(\dfrac{a}{2a+a+b+c}=\dfrac{a}{25}.\dfrac{\left(2+3\right)^2}{2a+a+b+c}\le\dfrac{a}{25}\left(\dfrac{2^2}{2a}+\dfrac{3^2}{a+b+c}\right)=\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{a}{a+b+c}\)

Tương tự:

\(\dfrac{b}{3b+a+c}\le\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{b}{a+b+c}\)

\(\dfrac{c}{a+b+3c}\le\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{c}{a+b+c}\)

Cộng vế:

\(VT\le\dfrac{6}{25}+\dfrac{9}{25}.\dfrac{a+b+c}{a+b+c}=\dfrac{3}{5}\)

Dấu "=" xảy ra khi \(a=b=c\)

2.

Đặt \(\dfrac{x}{x-1}=a;\dfrac{y}{y-1}=b;\dfrac{z}{z-1}=c\)

Ta có: \(\dfrac{x}{x-1}=a\Rightarrow x=ax-a\Rightarrow a=x\left(a-1\right)\Rightarrow x=\dfrac{a}{a-1}\)

Tương tự ta có: \(y=\dfrac{b}{b-1}\) ; \(z=\dfrac{c}{c-1}\)

Biến đổi giả thiết:

\(xyz=1\Rightarrow\dfrac{abc}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}=1\)

\(\Rightarrow abc=\left(a-1\right)\left(b-1\right)\left(c-1\right)\)

\(\Rightarrow ab+bc+ca=a+b+c-1\)

BĐT cần chứng minh trở thành:

\(a^2+b^2+c^2\ge1\)

\(\Leftrightarrow\left(a+b+c\right)^2-2\left(ab+bc+ca\right)\ge1\)

\(\Leftrightarrow\left(a+b+c\right)^2-2\left(a+b+c-1\right)\ge1\)

\(\Leftrightarrow\left(a+b+c-1\right)^2\ge0\) (luôn đúng)

\(a,VP=\dfrac{x^2+4x+3}{x^2+6x+9}=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x+3\right)^2}=\dfrac{x+1}{x+3}=VT\)

Vậy ta có đpcm 

b, \(VP=\dfrac{3x\left(x+y\right)^2}{9x^2\left(x+y\right)}=\dfrac{x+y}{3x}=VT\)

Vậy ta có đpcm 

a) Ta có: \(\dfrac{x^2+4x+3}{x^2+6x+9}\)

\(=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x+3\right)\left(x+3\right)}\)

\(=\dfrac{x+1}{x+3}\)

b: Ta có: \(\dfrac{3x\left(x+y\right)^2}{9x^2\left(x+y\right)}\)

\(=\dfrac{3x\left(x+y\right)\left(x+y\right)}{3x\cdot3x\cdot\left(x+y\right)}\)

\(=\dfrac{x+y}{3x}\)

1: \(MTC=2\left(x-y\right)\left(x+y\right)\)

\(\dfrac{x-y}{2x^2-4xy+2y^2}=\dfrac{x-y}{2\left(x-y\right)^2}=\dfrac{1}{2\left(x-y\right)}=\dfrac{1\cdot\left(x+y\right)}{2\left(x-y\right)\left(x+y\right)}=\dfrac{x+y}{2\left(x-y\right)\left(x+y\right)}\)

\(\dfrac{x+y}{2x^2+4xy+2y^2}\)

\(=\dfrac{x+y}{2\left(x^2+2xy+y^2\right)}\)

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

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

2: \(\dfrac{1}{x^2+8x+15}=\dfrac{1}{\left(x+3\right)\left(x+5\right)}=\dfrac{x+3}{\left(x+3\right)^2\cdot\left(x+5\right)}\)

\(\dfrac{1}{x^2+6x+9}=\dfrac{1}{\left(x+3\right)^2}=\dfrac{x+5}{\left(x+3\right)^2\cdot\left(x+5\right)}\)

3: \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}=\dfrac{1\cdot\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\dfrac{a-c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(\dfrac{1}{\left(c-b\right)\left(c-a\right)}=\dfrac{1}{\left(b-c\right)\left(a-c\right)}=\dfrac{a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(\dfrac{1}{\left(b-a\right)\left(a-c\right)}=\dfrac{-1}{\left(a-b\right)\left(a-c\right)}=\dfrac{-\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)