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) Ta có: \(\sqrt{3+\sqrt{5}}+\sqrt{3-\sqrt{5}}\)

\(=\dfrac{\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}+1+\sqrt{5}-1}{\sqrt{2}}=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)

b) Ta có: \(2\sqrt{2}\left(\sqrt{3}-2\right)+\left(9+4\sqrt{2}\right)\)

\(=2\sqrt{6}-4\sqrt{2}+9+4\sqrt{2}\)

\(=9+2\sqrt{6}\)

c) Ta có: \(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}\)

\(=\sqrt{4+\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}\)

\(=\sqrt{4+\sqrt{25}}=\sqrt{4+5}=3\)(đpcm)

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:

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

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