ab = a : b

a+m/b+m = (a + m) : (b + m)

a+m/b+m >a /b

\(A=\left(-\infty;+\infty\right);B=(1;7]\)

\(A\cap B=(1;7]\)

\(A\cup B=\left(-\infty;+\infty\right)\)

\(A\backslash B=(-\infty;1]\cup\left(7;+\infty\right)\)

\(B\text{A}=\varnothing\)

a) ĐK: x ≤ 2

PT <=> 2 - x = 9

,=> x = -7

b) PT <=> \(\sqrt{\left(2-x\right)^2}=3\)

<=> 2 - x = 3

<=> x = -1

c) PT <=> \(\sqrt{4+x^2}=3-x\)

\(\Leftrightarrow4+x^2=9-6x+x^2\)

\(\Leftrightarrow-6x+5=0\)

\(\Leftrightarrow x=\dfrac{5}{6}\)

d) PT <=> \(\dfrac{1}{2}\sqrt{16\left(x-2\right)}-2\sqrt{4\left(x-2\right)}+\sqrt{9\left(x-2\right)}=5\)

\(\Leftrightarrow\dfrac{1}{2}.4.\sqrt{x-2}-2.2.\sqrt{x-2}+3.\sqrt{x-2}=5\)

\(\Leftrightarrow\sqrt{x-2}\left(\dfrac{1}{2}.4-2.2+3\right)=5\)

\(\Leftrightarrow\sqrt{x-2}=5\)

\(\Leftrightarrow x-2=25\)

<=> x = 27

   a) \(\sqrt[]{2-x}\) = 3

⇔ 2 - x = 9

⇔ x      = -7

vậy....

 b)\(\sqrt{4-4x+x^2}\) =3

⇔ \(\sqrt{\left(2-x\right)^2}\) =3

⇔ I2 - xI=3

⇔ x - 2 = 3 (vì x>2)

⇔ x = 5

vậy.....

a)\(6\sqrt[3]{81}-4\sqrt[3]{375}+3\sqrt[3]{24}\)

\(=6\sqrt[3]{3^4}-4\sqrt[3]{3.5^3}+3\sqrt[3]{3.2^3}\)

\(=6.3\sqrt[3]{3}-4.5\sqrt[3]{3}+3.2\sqrt[3]{3}=4\sqrt[3]{3}\)

b)\(\sqrt[3]{3}.\sqrt[3]{144}-\dfrac{\sqrt[3]{384}}{\sqrt[3]{3}}+2\sqrt[3]{-128}\)

\(=\sqrt[3]{432}-\sqrt[3]{\dfrac{384}{3}}+2\sqrt[3]{-2.4^3}\)

\(=\sqrt[3]{6^3.2}-\sqrt[3]{2.4^3}+2.\sqrt[3]{-2.4^3}\)

\(=6\sqrt[3]{2}-4\sqrt[3]{2}+2.-4\sqrt[3]{2}==-6\sqrt[3]{2}\)

a) \(\sqrt[3]{3x-2}=4\Rightarrow3x-2=64\Rightarrow3x=66\Rightarrow x=22\)

b) \(\sqrt[3]{x^3+7x^2}=x+4\Rightarrow x^3+7x^2=\left(x+4\right)^3\)

\(\Rightarrow x^3+7x^2=x^3+12x^2+48x+64\Rightarrow5x^2+48x+64=0\)

\(\Rightarrow\left(x+8\right)\left(5x+8\right)=0\Rightarrow\left[{}\begin{matrix}x=-8\\x=-\dfrac{8}{5}\end{matrix}\right.\)

a) Ta có: \(\sqrt[3]{3x-2}=4\)

\(\Leftrightarrow3x-2=64\)

\(\Leftrightarrow3x=66\)

hay x=22

b) Ta có: \(\sqrt[3]{x^3+7x^2}=x+4\)

\(\Leftrightarrow x^3+7x^2=\left(x+4\right)^3\)

\(\Leftrightarrow x^3+7x^2-x^3-12x^2-48x-64=0\)

\(\Leftrightarrow-5x^2-48x-64=0\)

\(\Leftrightarrow5x^2+48x+64=0\)

\(\text{Δ}=48^2-4\cdot5\cdot64=1024\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-48-32}{10}=\dfrac{-80}{10}=-8\\x_2=\dfrac{-48+32}{10}=\dfrac{-16}{10}=\dfrac{-8}{5}\end{matrix}\right.\)

\(\left[a,b\right]+\left(a,b\right)=23\\ a,b=23:2\\ a,b=11,5\\ \Rightarrow a=11\\ b=5\)

Gọi giao điểm AE và BP là F;

Gọi giao điểm QD và AB là H; 

Gọi kéo dài AD cắt BF tại P'     

Dễ cm M là trung điểm AC

Xét \(\Delta OMC\) có QD//CM\(\Rightarrow\dfrac{OD}{OM}=\dfrac{QD}{CM}\)(hệ quả tales)

Tương tự với \(\Delta OAM\) có \(\dfrac{OD}{OM}=\dfrac{DH}{AM}\) 

\(\Rightarrow\dfrac{QD}{CM}=\dfrac{DH}{AM}\)

Mà CM=AM (vì M là tđ AC)

\(\Rightarrow QD=DH\)

Dễ cm P là trung điểm BF

Xét \(\Delta ABP'\) có DH//BP'

\(\Rightarrow\dfrac{DH}{BP'}=\dfrac{AD}{AP'}\)(tales)

Tương tự với \(\Delta AFP'\) có \(\dfrac{QD}{FP'}=\dfrac{AD}{AP'}\)

\(\Rightarrow\dfrac{DH}{BP'}=\dfrac{QD}{FP'}\)

Mà DH=QD (cmt) 

\(\Rightarrow BP'=FP'\)

\(\Rightarrow\)P' là trung điểm BF

\(\Rightarrow P\equiv P'\)

\(\Rightarrow A,D,P\) thẳng hàng