Mít cứ bình phương lên là ok

(2\(\sqrt{7}\))² =28 (1)

(3\(\sqrt{3}\))² =27 (2)

vậy (1) > (2)

cứ thế mà làm là hết mít

a) có \(\sqrt{2}\) <\(\sqrt{3}\)

5= \(\sqrt{25}\) >\(\sqrt{11}\)

=>\(\sqrt{2}+\sqrt{11}< \sqrt{3}+5\)

b)có \(\sqrt{21}>\sqrt{20}\)

-\(\sqrt{5}\) >-\(\sqrt{6}\)

=>\(\sqrt{21}-\sqrt{5}>\sqrt{20}-\sqrt{6}\)

1)\(x>y\)

2)\(x< y\)

3)\(x< y\)

a) Ta có 290>289

<=>  \(\sqrt{290}\)   >       \(\sqrt{289}\)

<=>  \(\sqrt{290}\)   >        17

Vậy ..........

\(a,290>289\)

\(\Rightarrow\sqrt{290}>\sqrt{289}\)

\(\Rightarrow\sqrt{290}>17\)

\(b,\sqrt{7}+\sqrt{15}< \sqrt{9}+\sqrt{16}\)

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

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

a: \(\left(\sqrt{7}+\sqrt{15}\right)^2=22+2\sqrt{105}=7+15+2\sqrt{105}\)

\(7^2=49=7+42\)

mà \(15+2\sqrt{105}< 42\)

nên \(\sqrt{7}+\sqrt{15}< 7\)

b: \(\left(\sqrt{2}+\sqrt{11}\right)^2=13+2\sqrt{22}\)

\(\left(5+\sqrt{3}\right)^2=28+10\sqrt{3}=13+15+10\sqrt{3}\)

mà \(2\sqrt{22}< 15+10\sqrt{3}\)

nên \(\sqrt{2}+\sqrt{11}< 5+\sqrt{3}\)

a,x>y

b,x<y

c,ko biết

a) \(x>y\)

b)\(x< y\)

c)\(x< y\)

4 8/3 và 3\(\sqrt2\)

4\(\frac83\) = \(\frac{20}{3}\) = \(\sqrt{\frac{400}{3}}\)

3\(\sqrt2\) = \(\sqrt{9.2}\) = \(\sqrt{18}\) = \(\sqrt{\frac{54}{3}}\)

Vì 400/3 > 54/3 nên \(\sqrt{\frac{400}{3}}\) > \(\sqrt{\frac{54}{3}}\)

Vậy: 4\(\frac83\) > 3\(\sqrt2\)

So sánh:

5\(\sqrt{\left(-10\right)^2}\) và 10\(\sqrt{\left(-5\right)^2}\)

5\(\sqrt{\left(-10\right)^2}\) = \(\sqrt{100.5^2}\) = \(\sqrt{2500}\)

10\(\sqrt{\left(-5\right)^2}\) = \(\sqrt{10^2.25}\) = \(\sqrt{2500}\)

Vậy 5\(\sqrt{\left(-10\right)^2}\) = 10\(\sqrt{\left(-5\right)^2}\)

a)>

b)<

c)>

a, >

b, <

c, >