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

b, Thay \(x=-2\left(t/mđk\right)\) vào 

\(-2+3=1\)

Vậy tại \(x=-2\) thì biểu thức = 1 

\(A=\dfrac{x^2+6x+9}{x+3}\)

\(A=\dfrac{x^2+2.x.3+3^2}{x+3}\)

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

\(A=x+3\)

b) Thay x = -2 vào A ta được A = -2 + 3 = 1

Vậy khi x = -2 thì A = 1

cảm ơn bạn nhiều.

a. ĐKXĐ: x \(\ne\pm3\)

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

= \(\frac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{6x}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\)

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

c. M = 0 hay \(\frac{x+3}{x-3}=0\) => x + 3 = 0 <=> x = -3 (Loại)

\(A=\left(\frac{3-x}{x+3}\times\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\) \(\left(ĐKXĐ:x\ne\pm3\right)\)

\(A=\left(\frac{3-x}{x+3}\times\frac{x+3}{x-3}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)

\(A=\left(\frac{3-x}{x-3}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)

\(A=\left[\frac{\left(3-x\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right]:\frac{3x^2}{x+3}\)

\(A=\left(\frac{9-3x}{\left(x-3\right)\left(x+3\right)}\right):\frac{3x^2}{x+3}\)

\(A=\left(\frac{-3\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\frac{3x^2}{x+3}\)

\(A=\frac{-3}{x+3}\times\frac{x+3}{3x^2}\)

\(A=\frac{-1}{x^2}\)

Ta có :\(x^2+x-6=0\)

\(\Leftrightarrow\left(x^2-2x\right)+\left(3x-6\right)=0\)

\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\left(L\right)\\x=2\left(tm\right)\end{cases}}\)

\(\Rightarrow A=\frac{-1}{2^2}\)

\(A=\frac{-1}{4}\)

a) ĐKXĐ: 

\(x^2-1\ne0\Leftrightarrow x\ne\pm1\)

b) \(A=\dfrac{x^2-2x+1}{x^2-1}\)

\(A=\dfrac{x^2-2\cdot x\cdot1+1^2}{x^2-1^2}\)

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

\(A=\dfrac{x-1}{x+1}\)

c) Thay x = 3 vào A ta có:

\(A=\dfrac{3-1}{3+1}=\dfrac{2}{4}=\dfrac{1}{2}\)

a) ĐKXĐ: 

\(9x^2-y^2\ne0\Leftrightarrow\left(3x\right)^2-y^2\ne0\Leftrightarrow\left(3x-y\right)\left(3x+y\right)\ne0\)

\(\Leftrightarrow3x\ne\pm y\) 

b) \(B=\dfrac{6x-2y}{9x^2-y^2}\)

\(B=\dfrac{2\cdot3x-2y}{\left(3x\right)^2-y^2}\)

\(B=\dfrac{2\left(3x-y\right)}{\left(3x+y\right)\left(3x-y\right)}\)

\(B=\dfrac{2}{3x+y}\)

Thay x = 1 và \(y=\dfrac{1}{2}\) và B ta có:

\(B=\dfrac{2}{3\cdot1+\dfrac{1}{2}}=\dfrac{2}{3+\dfrac{1}{2}}=\dfrac{2}{\dfrac{7}{2}}=\dfrac{4}{7}\)

Bài 2:

a: Ta có: \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

hay \(x=\dfrac{2}{7}\)

b: Ta có: \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow x^3=-8\)

hay x=-2

Bài 1: 

a: Ta có: \(I=x\left(y^2-xy^2\right)+y\left(x^2y-xy+x\right)\)

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

\(=xy\)

=1

b: Ta có: \(K=x^2\left(y^2+xy^2+1\right)-\left(x^3+x^2+1\right)\cdot y^2\)

\(=x^2y^2+x^3y^2+x^2-x^3y^2-x^2y^2-y^2\)

\(=x^2-y^2\)

\(=\dfrac{1}{4}-\dfrac{1}{4}=0\)

a: ĐKXĐ: x<>-3

b: =x+3

còn c và d thôi bạn ơi

giúp mình với

a) Tại x=16 thì A = \(\dfrac{\sqrt{16}-1}{\sqrt{16}+2}=\dfrac{4-1}{4+2}=\dfrac{1}{2}\)

b) B = \(\dfrac{\sqrt{x}+1+\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\div\dfrac{\sqrt{x}}{x+\sqrt{x}}\)

        = \(\dfrac{\sqrt{x}+1+x-\sqrt{x}}{x+\sqrt{x}}\times\dfrac{x+\sqrt{x}}{\sqrt{x}}\) 

        = \(\dfrac{x+1}{\sqrt{x}}\)

B = \(\dfrac{x+1}{\sqrt{x}}\)= 2

   ⇒ x + 1 = 2\(\sqrt{x}\) 

   ⇒ x - \(2\sqrt{x}\) +1 = 0

   ⇒ \(\left(\sqrt{x}-1\right)^2\) = 0

   ⇒ \(\sqrt{x}-1=0\)

⇒  x = 1