-1 có căn k bạn?

39 < 103

1 + căn 15 ......căn 24

4.872983346...\(\approx\) 4.872    và 4.898979486...\(\approx\)4.898

=> 4.872 < 4.898

=> 1 + căn 15 < căn 24

Tham khảo nhé ~.~ 

Ta có : 

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

\(\left(\sqrt{24}\right)^2=24=16+8=16+2.4=16+2\sqrt{16}\)

Ta thấy \(16+2\sqrt{15}< 16+2\sqrt{16}\) nên \(\left(1+\sqrt{15}\right)^2< \left(\sqrt{24}\right)^2\)

\(\Rightarrow\)\(1+\sqrt{15}< \sqrt{24}\)

Vậy \(1+\sqrt{15}< \sqrt{24}\)

Chúc bạn học tốt ~ 

* \(1+\sqrt{3}< 2+\sqrt{2}\)

* \(\sqrt{5}+\sqrt{3}>3\)

(Đúng thì k cho mình nhá!)

\(8=\sqrt{64}\)

vì 64>63

8>căn 63

\(13=\sqrt{169}\)

vì 170>169

căn 170 > 13

\(15=\sqrt{225}\)

vì 225<227

15 < căn 227

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

\(\sqrt{235}\)=15,32970972

=>15<\(\sqrt{235}\)

Ta có : 

\(\sqrt{7}< \sqrt{9}\)

\(\sqrt{15}< \sqrt{16}\)

\(\Rightarrow\)\(\sqrt{7}+\sqrt{15}< \sqrt{9}+\sqrt{16}=3+4=7\)

Vậy \(\sqrt{7}+\sqrt{15}< 7\)

Chúc bạn học tốt ~ 

Ta có: \(\sqrt{7}+\sqrt{15}< \sqrt{9}+\sqrt{16}=3+4=7\)

\(\Rightarrow\sqrt{7}+\sqrt{15}< 7\)

Chúc bn hc tốt!