1/ bình phương hai vế được (căn11)^2+(căn5)^2=11+5   4^2=16 vậy căn 11+căn 5=4

2/ tương tự (3 căn3 )^2=27   (căn19)^2-(căn 2)^2=19-2=17  vậy 3 căn 3 >căn 19-căn2

* \(1+\sqrt{3}< 2+\sqrt{2}\)

* \(\sqrt{5}+\sqrt{3}>3\)

(Đúng thì k cho mình nhá!)

Bài 2:

a: ĐKXĐ: x>=0

\(\sqrt{3x}-5\sqrt{12x}+7\cdot\sqrt{27x}=12\)

=>\(\sqrt{3x}-5\cdot2\sqrt{3x}+7\cdot3\sqrt{3x}=12\)

=>\(12\sqrt{3x}=12\)

=>\(\sqrt{3x}=1\)

=>3x=1

=>x=1/3(nhận)

Bài 1:

a: \(A=\left(\sqrt{\frac23}+\sqrt{\frac{50}{3}}-\sqrt{24}\right)\cdot\sqrt6\)

\(=\left(\frac{2\sqrt6}{6}+\sqrt{\frac{100}{6}}-2\sqrt6\right)\cdot\sqrt6\)

\(=2+\sqrt{100}-2\cdot6=2+10-12=0\)

b: \(B=\left(\frac{\sqrt{14}-\sqrt7}{\sqrt2-1}+\frac{\sqrt{15}-\sqrt5}{\sqrt3-1}\right):\frac{1}{\sqrt7-\sqrt5}\)

\(=\left(\frac{\sqrt7\left(\sqrt2-1\right)}{\sqrt2-1}+\frac{\sqrt5\left(\sqrt3-1\right)}{\sqrt3-1}\right)\cdot\left(\sqrt7-\sqrt5\right)\)

\(=\left(\sqrt7+\sqrt5\right)\left(\sqrt7-\sqrt5\right)\)

=7-5

=2

\(5\sqrt{2}+\sqrt{75}=5\sqrt{2}+5\sqrt{3}\)

\(5\sqrt{3}+\sqrt{50}=5\sqrt{3}+5\sqrt{2}\)

\(\Rightarrow5\sqrt{2}+\sqrt{75}=5\sqrt{3}+\sqrt{50}\)

b: \(\sqrt{\dfrac{3}{2}}>\sqrt{\dfrac{2}{2}}=1\)

a: \(\left(2\sqrt{5}-3\sqrt{2}\right)^2=38-12\sqrt{10}=1+37-12\sqrt{10}\)

\(1^2=1\)

mà \(37-12\sqrt{10}< 0\)

nên \(2\sqrt{5}-3\sqrt{2}< 1\)

3+căn 5 < 2 căn 2 + căn 6

\(3+\sqrt{5}\approx5,23\)

\(2\sqrt{2}+\sqrt{6}\approx5,27\)

Vì 5,23 < 5,27 nên \(3+\sqrt{5}< 2\sqrt{2}+\sqrt{6}\)

1/

Ta có:  \(\left(1+\sqrt{15}\right)^2\)= 1 + 15 + \(2\sqrt{15}\)= 16 + \(2\sqrt{15}\)

              \(\sqrt{24}^2\)= 24 = 16 + 8

Vì:     \(\sqrt{15}^2\)= 15 < 16 =\(4^2\)

Nên:   \(\sqrt{15}< 4\)

=>       \(2\sqrt{15}< 8\)

=>       \(16+2\sqrt{15}< 24\)

=>      \(\left(1+\sqrt{15}\right)^2< \sqrt{24}^2\)

Vậy     \(1+\sqrt{15}< \sqrt{24}\)

2/

b/    \(3x-7\sqrt{x}=20\)\(\left(x\ge0\right)\)

<=> \(3x-7\sqrt{x}-20=0\)

<=> \(3x-12\sqrt{x}+5\sqrt{x}-20=0\)

<=> \(3\sqrt{x}\left(\sqrt{x}-4\right)+5\left(\sqrt{x}-4\right)=0\)

<=> \(\left(\sqrt{x}-4\right)\left(3\sqrt{x}+5\right)=0\)

<=> \(\sqrt{x}-4=0\)hoặc \(3\sqrt{x}+5=0\)

<=>   \(\sqrt{x}=4\)hoặc \(3\sqrt{x}=-5\)(vô nghiệm)

<=>   \(x=16\)

Vậy S=\(\left\{16\right\}\)

c/    \(1+\sqrt{3x}>3\)

<=> \(\sqrt{3x}>2\)

<=>   \(3x>4\)

<=>  \(x>\frac{4}{3}\)

d/      \(x^2-x\sqrt{x}-5x-\sqrt{x}-6=0\)(\(x\ge0\))

<=>   \(\left(x^2-5x-6\right)-\left(x\sqrt{x}+\sqrt{x}\right)=0\)

<=>   \(\left(x^2-6x+x-6\right)-\left(x\sqrt{x}+\sqrt{x}\right)=0\)

<=>    \([x\left(x-6\right)+\left(x-6\right)]-\sqrt{x}\left(x+1\right)=0\)

<=>   \(\left(x-6\right)\left(x+1\right)-\sqrt{x}\left(x+1\right)=0\)

<=>   \(\left(x+1\right)\left(x-6-\sqrt{x}\right)=0\)

<=>    \(\left(x+1\right)\left(x-3\sqrt{x}+2\sqrt{x}-6\right)=0\) 

<=>    \(\left(x+1\right)[\sqrt{x}\left(\sqrt{x}-3\right)+2\left(\sqrt{x}-3\right)]=0\)

<=>    \(\left(x+1\right)\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)=0\)

<=>     \(x+1=0\)  hoặc \(\sqrt{x}-3=0\)hoặc \(\sqrt{x}+2=0\)

<=>     \(x=-1\)(loại)  hoặc \(x=9\)hoặc \(\sqrt{x}=-2\)(vô nghiệm)

Vậy S={  9 }

Lời giải:

$\sqrt{3}+5> \sqrt{1}+5=6$

$\sqrt{2}+\sqrt{11}< \sqrt{4}+\sqrt{16}=6$

$\Rightarrow \sqrt{3}+5> \sqrt{2}+\sqrt{11}$