Bài 2: Tính chất cơ bản của phân thức

\(\dfrac{x^4-10x^2+9}{x^4+8x^3+22x^2+24x+9}\)

\(=\dfrac{x^4-x^2-9x^2+9}{x^4+x^3+7x^3+7x^2+15x^2+15x+9x+9}\)

\(=\dfrac{x^2\left(x^2-1\right)-9\left(x^2-1\right)}{x^3\left(x+1\right)+7x^2\left(x+1\right)+15x\left(x+1\right)+9\left(x+1\right)}\)

\(=\dfrac{\left(x^2-3^2\right)\left(x^2-1\right)}{\left(x+1\right)\left(x^3+7x^2+15x+9\right)}\)

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

= \(\dfrac{\left(x+3\right)\left(x-3\right)\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left[x^2\left(x+1\right)+6x\left(x+1\right)+9\left(x+1\right)\right]}\)

= \(\dfrac{\left(x+3\right)\left(x-3\right)\left(x-1\right)\left(x+1\right)}{\left(x+1\right)\left(x+1\right)\left(x^2+2.3x+3^2\right)}\)

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

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

Các bạn ơi giúp mình với mai mình học rồi

Help me!

các bạn ơi giúp mình với

Vì mk ko biết trình bày bài 3 nên mk chỉ ghi mỗi kết quả thôi

Ơ Phân thức đâu hả bạn ???

phân thức của bn là : 12x²y

a)\(3x-y\) = \(\dfrac{7\left(3x-y\right)}{7}=\dfrac{21x-7y}{7}\)

b) \(3x-y=\dfrac{\left(3x-y\right)x}{x}=\dfrac{3x^2-xy}{x}\)

c) \(3x-y=\dfrac{\left(3x-y\right)\left(9x+y\right)}{9x+y}\)

d) \(3x-y=\dfrac{\left(3x-y\right)\left(3x-y\right)}{3x-y}=\dfrac{\left(3x-y\right)^2}{3x-y}\)

a) 3x - y = \(\dfrac{7\left(3x-y\right)}{7}=\dfrac{21x-7y}{7}\)

3x-y=\(\dfrac{7\cdot\left(3x-y\right)}{7}\) =\(\dfrac{21x-7y}{7}\)

3x-y=\(\dfrac{x\cdot\left(3x-y\right)}{x}\) =\(\dfrac{3x^2-xy}{x}\)

A, C

a)

\(\dfrac{H}{x^2+9x+14}=\dfrac{1-x}{x+2}\)

\(\Rightarrow\dfrac{H}{x^2+7x+2x+14}=\dfrac{1-x}{x+2}\)

\(\Rightarrow\dfrac{H}{\left(x+7\right)\left(x+2\right)}=\dfrac{1-x}{x+2}\)

\(\Rightarrow\left(x+2\right)\left(x+7\right)\left(1-x\right)=H.\left(x+2\right)\)

\(\Rightarrow H=\left(x+7\right)\left(1-x\right)\)

b)

\(\dfrac{2x^2-5x+2}{x^2+5x-14}=\dfrac{2x-1}{H}\)

\(\Rightarrow\dfrac{2x^2-4x-x+2}{x^2+7x-2x-14}=\dfrac{2x-1}{H}\)

\(\Rightarrow\dfrac{\left(2x-1\right)\left(x-2\right)}{\left(x+7\right)\left(x-2\right)}=\dfrac{2x-1}{H}\)

\(\Rightarrow\left(2x-1\right)\left(x-2\right).H=\left(2x-1\right)\left(x+7\right)\left(x-2\right)\)

\(\Rightarrow H=x+7\)

Ta có : 1/M=a²+2bc+b²+2ac+c²+2ab

=(a+b+c)² ➝ M=1/(a+b+c)²

mik nghĩ là thế

Có:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\)

\(\Leftrightarrow\dfrac{ab+bc+ac}{abc}=0\)

\(\Leftrightarrow ab+bc+ac=0\)

\(1\Leftrightarrow a^2+2bc=a^2+bc-ab-ac\)

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

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

\(2\Leftrightarrow b^2+2ac=b^2+ac-ab-bc\)

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

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

\(3.c^2+2ab=c^2+ab-bc-ac\)

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

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

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

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

\(\Rightarrow M=\dfrac{b-c-a+c+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(\Rightarrow M=0\)

Để \(\dfrac{x^3-x^2+2}{x-1}\)chia hết cho x-1 thì

\(x^3-x^2+2\) chia hết cho x-1

=>\(x^2.\left(x-1\right)+2\) chia hết cho x-1

=>2 chia hết cho x-1

=>x-1\(\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

Ta có bảng sau:

Vậy \(x\in\left\{2;0;-1\right\}\)