a)
=(x-2)³
b)\(\left(2-x\right)^3\)
c)\(\left(x+\dfrac{1}{3}\right)^3\)
d)\(\left(\dfrac{x}{2}+y\right)^3\)
e)
\(=\left(x-1\right)^2\left(x-1-15\right)+25\left[3\left(x-1\right)-5\right]\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-3-5\right)\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-8\right)\)
 

a: ĐKXĐ: \(x\ne\dfrac{3}{2}\)

b: ĐKXĐ: \(x\in R\)

Ta có : \(B\text{=}4x^2-12x+9\)

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

Với \(x\text{=}\dfrac{1}{2}\)

\(\Rightarrow B\text{=}\left(2.\dfrac{1}{2}-3\right)^2\)

\(B\text{=}\left(-2\right)^2\text{=}4\)

Ta có : \(A\text{=}5\left(x+3\right)\left(x-3\right)+\left(2x+3\right)^2+\left(x-6\right)^2\)

\(A\text{=}5\left(x^2-9\right)+\left(2x+3\right)^2+\left(x-6\right)^2\)

\(A\text{=}5x^2-45+4x^2+12x+9+x^2-12x+36\)

\(A\text{=}10x^2\)

Với \(x\text{=}-\dfrac{1}{5}\)

\(\Rightarrow A\text{=}10.\left(-\dfrac{1}{5}\right)^2\text{=}\dfrac{2}{5}\)

B = 4x² - 12x + 9

= (2x - 3)²

Tại x = 1/2 ta có:

B = (2.1/2 - 3)²

= (-2)²

= 4

-------------------

A = 5(x + 3)(x - 3) + (2x + 3)² + (x - 6)²

= 5x² - 45 + 4x² + 12x + 9 + x² - 12x + 36

= 10x²

Tại x = 1/5 ta có:

A = 10.(1/5)²

= 2/5

\(B=4x^2-12x+9\)

\(B=\left(2x\right)^2-2\cdot2x\cdot3+3^2\)

\(B=\left(2x-3\right)^2\)

Thay \(x=\dfrac{1}{2}\) vào B ta có: 

\(B=\left(2\cdot\dfrac{1}{2}-3\right)^2=4\)

_______________________

\(A=5\left(x+3\right)\left(x-3\right)+\left(2x+3\right)^2+\left(x-6\right)^2\)

\(A=5\left(x^2-9\right)+4x^2+12x+9+x^2-12x+36\)

\(A=5x^2-45+5x^2+45\)

\(A=10x^2\)

Thay \(x=\dfrac{1}{5}\)vào A ta có:

\(A=10\cdot\left(-\dfrac{1}{5}\right)^2=\dfrac{2}{5}\)

a) \(\sqrt{\left|x-1\right|-3}\) 

Với \(x\ge1\) thì

\(\sqrt{x-1-3}=\sqrt{x-4}\) được xác định khi:
\(x\ge4\)

Với \(x< 1\) thì

\(\sqrt{-\left(x-1\right)-3}=\sqrt{-x+1-3}=\sqrt{-x-2}\) được xác đinh khi:

\(x\le-2\)

\(a,\sqrt{\left|x-1\right|-3}\) xác định \(\Leftrightarrow\left|x-1\right|-3\ge0\Leftrightarrow\left|x-1\right|\ge3\)

\(TH_1:x\ge1\\ x-1\ge3\Leftrightarrow x\ge4\left(tm\right)\\ TH_2:x< 1\\ x-1\ge-3\\ \Leftrightarrow x\ge-2\left(tm\right)\)

Vậy căn thức trên xác định \(\Leftrightarrow x\ge4\)

\(b,\sqrt{x-2\sqrt{x-1}}\) xác định \(\Leftrightarrow\left[{}\begin{matrix}x-2\sqrt{x-1}\ge0\\x-1\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}\le\dfrac{x}{2}\\x\ge1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x-1\le\dfrac{x^2}{4}\\x\ge1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}4x-4-x^2\le0\\x\ge1\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}-\left(x^2-4x+4\right)\le0\\x\ge1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-2\right)^2\ge0\left(LD\right)\\x\ge1\end{matrix}\right.\)\(\Leftrightarrow x\ge1\)

Vậy căn thức trên xác định \(\Leftrightarrow x\ge1\)

\(c,\dfrac{1}{\sqrt{9-12x+4x^2}}=\dfrac{1}{\sqrt{\left(3-2x\right)^2}}=\dfrac{1}{3-2x}\) xác định \(\Leftrightarrow3-2x\ne0\Leftrightarrow x\ne\dfrac{3}{2}\)

Vậy căn thức trên xác định \(\Leftrightarrow x\ne\dfrac{3}{2}\)

c) \(\dfrac{1}{\sqrt{9-12x+4x^2}}=\dfrac{1}{\sqrt{\left(3-2x\right)^2}}=\dfrac{1}{3-2x}\) 

Xác định khi:

\(3-2x\ne0\)

\(\Leftrightarrow x\ne\dfrac{3}{2}\)

1)ĐK:`4x^2-12x+9>0`

`<=>(2n-3)^2>0`

`<=>2n-3 ne 0`

`<=>n ne 3/2`

`d)x^2-x+1`

`=(x-1/2)^2+3/4>0AAx`

`=>` bt xd `AAx in RR`

e)ĐK:`x^2-8x+15>0`

`<=>x^2-3x-5x+15>0`

`<=>x(x-3)-5(x-3)>0`

`<=>(x-3)(x-5)>0`

`TH1:` \(\begin{cases}x-3>0\\x-5>0\\\end{cases}\)

`<=>` \(\begin{cases}x>3\\x>5\\\end{cases}\)

`<=>x>5`

`TH2:` \(\begin{cases}x-3<0\\x-5<0\\\end{cases}\)

`<=>` \(\begin{cases}x<3\\x<5\\\end{cases}\)

`<=>x<3`

f)ĐK:`3x^2-7x+20>0`

`<=>x^2-2x+1+2x^2-5x+19>0`

`<=>(x-1)^2+2(x-5/2)^2+13/2>0` luôn đúng

c) Để biểu thức \(\dfrac{1}{\sqrt{4x^2-12x+9}}\) có nghĩa thì \(4x^2-12x+9>0\)

\(\Leftrightarrow\left(2x-3\right)^2>0\)

\(\Leftrightarrow2x-3\ne0\)

\(\Leftrightarrow2x\ne3\)

hay \(x\ne\dfrac{3}{2}\)

d) Để biểu thức \(\dfrac{1}{\sqrt{x^2-x+1}}\) có nghĩa thì \(x^2-x+1>0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}>0\)

\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)(luôn đúng)

e) Để biểu thức \(\dfrac{1}{\sqrt{x^2-8x+15}}\) có nghĩa thì \(x^2-8x+15>0\)

\(\Leftrightarrow\left(x-4\right)^2>1\)

\(\Leftrightarrow\left[{}\begin{matrix}x-4>1\\x-4< -1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>5\\x< 3\end{matrix}\right.\)

f) Để biểu thức \(\dfrac{1}{\sqrt{3x^2-7x+20}}\) có nghĩa thì \(3x^2-7x+20>0\)

\(\Leftrightarrow x^2-\dfrac{7}{3}x+\dfrac{20}{3}>0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{191}{36}>0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2+\dfrac{191}{36}>0\)(luôn đúng)