Lời giải:

a.

$\sqrt{8}+\sqrt{15}+1<\sqrt{9}+\sqrt{16}+1=3+4+1=8=\sqrt{64}< \sqrt{65}$

$\Rightarrow \sqrt{8}+\sqrt{15}< \sqrt{65}-1$
b.

$(2\sqrt{3}+6\sqrt{2})^2=84+24\sqrt{6}< 84+24\sqrt{9}< 169$

$\Rightarrow 2\sqrt{3}+6\sqrt{2}< 13$

$\Rightarrow \frac{13-2\sqrt{3}}{6}> \sqrt{2}$

a,35+55=90 nên sin 35=cos 55

b,12+78=90 nên tan 12 =cot 78

tik mik nha

a: \(\sin35^0=\cos55^0\)

b: \(\tan12^0=\cot78^0\)

chơi nhau à

có cc=)))

a: \(\sqrt{12}<\sqrt{12,25}=3,5\)

\(\sqrt{20}<\sqrt{20,25}=4,5\)

\(\sqrt{30}<\sqrt{30,25}=5,5\)

\(\sqrt{42}<\sqrt{42,25}=6,5\)

Do đó: \(\sqrt{12}+\sqrt{20}+\sqrt{30}+\sqrt{42}\) <3,5+4,5+5,5+6,5

=>A<20

b: \(B=\sqrt{196}-\frac{1}{\sqrt3}-1=14-\frac{1}{\sqrt3}-1=13-\frac{1}{\sqrt3}\)

\(C=\sqrt{169}+\frac{1}{-\sqrt2}=13-\frac{1}{\sqrt2}\)

3>2

=>\(\frac{1}{\sqrt3}<\frac{1}{\sqrt2}\)

=>\(-\frac{1}{\sqrt3}>-\frac{1}{\sqrt2}\)

=>\(-\frac{1}{\sqrt3}+13>-\frac{1}{\sqrt2}+13\)

=>B>C

`a)`

  `m > n`

`<=>2m > 2n`

`<=>2m+3 > 2n+3`

Vậy `2n+3 < 2m+3`

_________________________

`b)`

   `m > n`

`<=>-m < -n`

`<=>-m-5 < -n-5`

Vậy `-n-5 > -m-5`

a)\(m>n\Rightarrow2m>2n\Rightarrow2m+3>2n+2\)

b)\(m>n\Rightarrow-m< -n\Rightarrow-m-5< -n-5\)

theo ket qua cho thay:9.4594<10

Ta có :

\(\sqrt{3}< \sqrt{4}=2\)

\(\sqrt{8}< \sqrt{9}=3\)

\(\sqrt{24}< \sqrt{25}=5\)

\(\Rightarrow\sqrt{3}+\sqrt{8}+\sqrt{24}< 2+3+5=10\)(đpcm)

Vậy ...

\(a,2^{300}=\left(2^3\right)^{100}=8^{100}\)

\(3^{200}=\left(3^2\right)^{100}=9^{100}\)

Vì \(8^{100}< 9^{100}\) nên \(2^{300}< 3^{200}\)

\(b,8^5=32768\)

\(6^6=46656\)

Vì \(32768< 46656\) nên \(8^5< 6^6\)

\(c,3^{450}=\left(3^3\right)^{150}=27^{150}\)

\(5^{300}=\left(5^2\right)^{150}=25^{150}\)

Vì \(27^{150}>25^{150}\) nên \(3^{450}>5^{300}\)

#Ayumu

a) ta thấy -59/1310 <0 còn -1/-9=1/9 nên > 0. Vì vậy phân số -1/-9> -59/1310

b)-3/7<0 còn -1/-5> 0 nên -3/7<-1/-5

c) ta có:13/17 <1 còn -23/-27=23/27> 1nen -23/-27>13/17