a) Phân thức  xác định

⇔ x3 – 8 ≠ 0

⇔ x3 – 23 ≠ 0

⇔ (x – 2)(x2 + 2x + 4) ≠ 0

⇔ x – 2 ≠ 0 (Vì x2 + 2x + 4 = (x + 1)2 + 3 > 0)

⇔ x ≠ 2

Vậy phân thức xác định với mọi x ≠ 2.

b) Với x ≠ 2 ta có:

c) Tại x = 4001/2000, giá trị biểu thức bằng: 

Vậy trên 1cm2 bề mặt da có đến 6000 con vi khuẩn. Thật đáng sợ!!!!

a,ĐKXĐ \(x^3-8\ne0\Leftrightarrow x^3\ne8\Leftrightarrow x\ne2\)

b,\(\Leftrightarrow3x^2+6x+12=0\)

    \(\Leftrightarrow3\left(x^2+2x+1\right)+9=0\)

   \(\Leftrightarrow3\left(x+1\right)^2+9=0\)(VÔ LÝ VÌ 3(x+1)²>=0 =>3(x+1)²+9>0)

vì vây ko có giá trị x để F =0

C, VỚI ĐKXĐ trên ,ta có 

\(F=\frac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}\)

    \(=\frac{3}{x-2}\)

a. \(x\ne5\) là ĐKXĐ của biểu thức P

b. P =\(\dfrac{\left(x-5\right)^2}{x-5}\)=\(x-5\)

c. P = -1 <=> x-5 =-1 <=> x=4

\(P=\dfrac{3x^2+6x+3}{x+1}\)

\(a,\) Điều kiện xác định: \(x+1\ne0\Leftrightarrow x\ne-1\)

\(b,P=\dfrac{3x^2+6x+3}{x+1}=\dfrac{3\left(x^2+2x+1\right)}{x+1}=\dfrac{3\left(x+1\right)^2}{x+1}=3\left(x+1\right)=3x+3\)

\(c,x=1\Rightarrow P=3.1+3=6\)

`a,`

\(x^2-3x\ne0\)

`<=>x(x-3)`\(\ne0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x-3\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\x\ne3\end{matrix}\right.\)

`b,`

đặt `A=(x^2-6x+9)/(x^2-3x)`

`A= ((x-3)^2)/(x(x-3))`

`A= (x-3)/x`

`c, `

để `x=5`

`=> A= (x -3)/x=(5-3)/5= 2/5`

a/ ĐKXĐ: \(x^2-3x\ne0\) \(\Leftrightarrow\) x\(\ne\)0,x\(\ne\)3

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

c/ x= 5 => \(\dfrac{x-3}{x}=\dfrac{5-3}{5}=\dfrac{2}{5}\)

a) \(\frac{2x^2-4x+8}{x^3+8}\Rightarrow\) ĐKXĐ: \(x^3+8\ne0 \Leftrightarrow x^3\ne-8 \Leftrightarrow x\ne-2 \)

b) \(\frac{2x^2-4x+8}{x^3+8}=\frac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}=\frac{2}{x+2}\)

c) \(\frac{2}{x+2}\Rightarrow f\left(2\right)=\frac{2}{2+2}=\frac{1}{2}\) 

d) \(\frac{2}{x+2}=2\)
\(\Leftrightarrow x+2=1\)
\(\Leftrightarrow x=-1\)

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:

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

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c) tự làm, đkxđ: x≠1;x≠−1

ê k bn với mk ik

😘 😘 😘 😘