bằng nhau

sai r bạn 

\(a,\sqrt{42}=\sqrt{3\cdot14}>\sqrt{3\cdot12}=6\\ \sqrt[3]{51}=\sqrt[3]{17}< \sqrt[3]{3\cdot72}=6\\ \Rightarrow\sqrt{42}>\sqrt[3]{51}\\ b,16^{\sqrt{3}}=4^{2\sqrt{3}}\\ 18>12\Rightarrow3\sqrt{2}>2\sqrt{3}\Rightarrow4^{3\sqrt{2}}>4^{2\sqrt{3}}\\ \Rightarrow4^{3\sqrt{2}}>16^{\sqrt{3}}\)

\(c,\left(\sqrt{16}\right)^6=16^3=4^6=4^2\cdot4^4=4^2\cdot16^2\\ \left(\sqrt[3]{60}\right)^6=60^2=4^2\cdot15^2\\ 4^2\cdot16^2>4^2\cdot15^2\Rightarrow\sqrt{16}>\sqrt[3]{60}\Rightarrow0,2^{\sqrt{16}}< 0,2^{\sqrt[3]{60}}\)

a: \(1< \sqrt{2}\)

nên \(2< \sqrt{2}+1\)

b: \(2\sqrt{31}=\sqrt{124}\)

\(10=\sqrt{100}\)

mà 124>100

nên \(2\sqrt{31}>10\)

c: \(-3\sqrt{11}=-\sqrt{99}\)

\(-\sqrt{12}=-\sqrt{12}\)

mà 99>12

nên \(-3\sqrt{11}< -\sqrt{12}\)

a) 1,2+3.1,3=5,1

b) 0,2+2.0,5=1,2

a) \(2\sqrt{31}=\sqrt{4.31}=\sqrt{124}>\sqrt{100}=10\\\Rightarrow2\sqrt{31}>10\)

Bài 1: 

a) \(\sqrt{1.44}+3\sqrt{1.69}=1.2+3\cdot1.3=1.2+3.9=5.1\)

b) \(\sqrt{0.04}+2\cdot\sqrt{0.25}=0.2+2\cdot0.5=1.2\)

đề bài là không dùng máy tính ; hoặc là không khai căn chứ

\(A^2=100.51\)

\(B^2=70^2+2+2.70.\sqrt{2}\)

\(B^2-A^2=70^2-\left(10.7\right)^2+\left(2-2.100\right)+2.70\sqrt{2}\)

\(B^2-A^2=2.70\sqrt{2}-2.99=2\left(70\sqrt{2}-99\right)\)

\(C=70.\sqrt{2};D=99\)

\(C^2=2.70^2\)

\(D^2=99^2=\left(70+29\right)^2\)

\(C^2-D^2=2.70^2-\left(70^2+2.70.29+29^2\right)=70^2-2.70.29-29^2=\left(70-29\right)^2-2.29^2=41^2-2.29^2\)\(C^2-D^2=\left(29+12\right)^2-2.29^2=29^2+12^2+2.29.12=12^2+2.29.12-29^2\)\(C^2-D^2=12^2+2.29.12-12^2-17^2-2.12.17\)\(C^2-D^2=2.12\left(29-17\right)-17^2=2.12^2-17^2\)

\(C^2-D^2=2.12^2-12^2-5^2-2.5.12=12^2-2.5.12-5^2\)

\(C^2-D^2=\left(12-5\right)^2-2.5^2=7^2-2.5^2\)

\(C^2-D^2=5^2+2.2.5+2^2-2.5^2=4.5-5^2+2^2\)

\(C^2-D^2=5\left(4-5\right)+4=4+5.\left(-1\right)=4-5=-1\)

........

=> C^2 -D^2 <0

=>C,D >0

=> C<D => C-D<0

=> B^2 -A^2 <0

A,B >0

=> B<A

kết luận

B<A

\(A=10\sqrt{51}\); \(B=70+\sqrt{2}\)

Ta có: \(A^2=5100\)

\(B^2=4900+140\sqrt{2}+2\)

So sánh \(198\) và \(140\sqrt{2}\) vì vì trừ 2 vế cho 4902.

Ta có: \(198^2=39204\)

\(\left(140\sqrt{2}\right)^2=39200\)

Vậy A > B (đpcm)

ta có :

\(A^2=\left(10\sqrt{51}\right)^2=100\cdot51=5100\)

\(B^2=\left(70+\sqrt{2}\right)^2=4900+140\sqrt{2}+2=4902+140\sqrt{2}\)

Cùng trừ 2 vế cho 4902 ta dc:

\(A^2=5100-4902=198\)

\(B^2=4902+140\sqrt{2}-4902=140\sqrt{2}\)

ta có :

A= \(198^2=39204\)

B= \(\left(140\sqrt{2}\right)^2=39200\)

vì 39204>39200 nên 10\(\sqrt{51}\)>70+\(\sqrt{2}\)

vậy A>B

a)

Có: \(2>1>0\)

\(\Rightarrow\sqrt{2}>1\Rightarrow1+\sqrt{2}>1+1\\ \Leftrightarrow1+\sqrt{2}>2\)

b) Có: \(0< \sqrt{3}< 3\)

\(\Rightarrow3+1>\sqrt{3}+1\\ \Rightarrow4>\sqrt{3}+1\)

c) Có: \(0< \sqrt{11}< \sqrt{25}\left(0< 11< 25\right)\)

\(\Rightarrow\sqrt{11}< 5\\ \Rightarrow-2\sqrt{11}>-2.5=-10\left(-2< 0\right)\)

d) Có: \(0< \sqrt{11}< \sqrt{16}=4\left(do.0< 11< 16\right)\)

\(\Rightarrow3\sqrt{11}< 3.4\\ \Leftrightarrow3\sqrt{11}< 12\)

a: 2=1+1<1+căn 2

b: 4=1+3>1+căn 3

c: -2căn 11=-căn 44

-10=-căn 100

mà 44<100

nên -2 căn 11>-10

d: 12=3*4=3*căn 16>3*căn 11