bạn viết có thánh đọc ra á :v

Bạn viết như vậy vẫn nhìn đc nhưng nhìn hơi khó

Thì các bạn vít ra giấy là hỉu nk mong giải giúp mk cái

`+)axx2+bxx1=cxx2+axx1<=>2a+b=2c+a<=>2c-a=b`

`+)cxx3+axx1=bxx2+axx1<=>3c+a=2b+a<=>3c=2b<=>c=2/3b`

mà `2c-a=b` nên `a=2c-b=4/3b-b=1/3b`

Khi đó: `cxx2+axx2=2(a+c)=2(1/3b+2/3b)=2b`

Vậy dấu hỏi chấm cần điền là `2`

Bài 1: 

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

b: Để A=3 thì 3x-9=x+1

=>2x=10

hay x=5

Bài 2: 

a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)

\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)

b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{3;1;5;-1\right\}\)

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

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

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

b: |x-3|=2

=>x-3=2 hoặc x-3=-2

=>x=5(nhận) hoặc x=1(loại)

Khi x=5 thì \(E=\dfrac{5^2}{5-1}=\dfrac{25}{4}\)

c: Để E=1/2 thì \(\dfrac{x^2}{x-1}=\dfrac{1}{2}\)

\(\Leftrightarrow2x^2-x+1=0\)

hay \(x\in\varnothing\)

f) \(A=\dfrac{x^2}{x-1}=\dfrac{x^2-x+x-1+1}{x-1}=\dfrac{x\left(x-1\right)+x-1+1}{x-1}=x+1+\dfrac{1}{x-1}=x-1+\dfrac{1}{x-1}+2\ge2\sqrt{\left(x-1\right).\dfrac{1}{x-1}}+2=4\)\(A=4\Leftrightarrow x=2\)

-Vậy \(A_{min}=4\)

a, ĐK : \(x\ne0;1\)

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

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

b, Thay x = 3 vào A ta được : \(\dfrac{9}{2}\)

c, \(A=4\Rightarrow\dfrac{x^2}{x-1}=4\Rightarrow x^2=4x-4\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x=2\)

d, \(A< 2\Rightarrow\dfrac{x^2}{x-1}-2< 0\Leftrightarrow\dfrac{x^2-2x+1}{x-1}< 0\Rightarrow x-1< 0\Leftrightarrow x>1\)

a,\(\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}-\dfrac{1}{1-x}+\dfrac{2-x^2}{x^2-x}\right)\)

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

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

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

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

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

b, Thay x=3 vào bt A ta có :

\(\dfrac{x^2}{x-1}=\dfrac{3^2}{3-1}=4,5\)

Vậy giá trị của biểu thức A bằng 4,5 khi x=3

giúp mình vớiiii

\(a,A=\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}+\dfrac{x^2-1}{9-x^2}\right):\left(2-\dfrac{x+5}{3+x}\right)\\ =\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}-\dfrac{x^2-1}{x^2-9}\right):\left(\dfrac{2\left(3+x\right)}{3+x}-\dfrac{x+5}{3+x}\right)\\ =\left(\dfrac{x}{x+3}-\dfrac{2}{x-3}-\dfrac{x^2-1}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{2\left(3+x\right)-\left(x+5\right)}{3+x}\\ =\left(\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{x^2-1}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{6+2x-x-5}{3+x}\)

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

\(b,A=x^2-x-2=0\\ \Leftrightarrow x^2+x-2x-2=0\\ \Leftrightarrow x\left(x+1\right)-2\left(x+1\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)

\(c,\dfrac{-5}{x-3}=\dfrac{1}{2}\\ \Leftrightarrow-10=x-3\\ \Leftrightarrow-x+3=10\\ \Leftrightarrow-x=7\\ \Leftrightarrow x=7\)

Để `A=1/2` thì `x=7`

Lời giải:
$A(x)+B(x)=(x^3-3x^2+3x-1)+(2x^3+x^2-x+5)$

$=3x^3-2x^2+2x+4$

b.

$A(x)C(x)=(x^3-3x^2+3x-1)(x-2)=x(x^3-3x^2+3x-1)-2(x^3-3x^2+3x-1)$

$=(x^4-3x^3+3x^2-x)-(2x^3-6x^2+6x-2)$

$=x^4-5x^3+9x^2-7x+2$