1.rút gọn a) $\sqrt{\left(6 2\sqrt{5}\right)^3}-\sqrt{\left(6-2\sqrt{5}\right)^3}$b) $\sqrt{\left(3-2\sqrt{2}\right)\left(4-2\sqrt{3}\right)}$2.chứng minh rằng số \(x=\sqrt{2 \sqrt{2 \sqrt{3}...

4

AN

alibaba nguyễn

29 tháng 6 2017

1/ a/ $\sqrt{\left(6+2\sqrt{5}\right)^3}-\sqrt{\left(6-2\sqrt{5}\right)^3}$

$=\sqrt{\left(\sqrt{5}+1\right)^6}-\sqrt{\left(\sqrt{5}-1\right)^6}$

$=\left(\sqrt{5}+1\right)^3-\left(\sqrt{5}-1\right)^3$

$=32$

b/ $\sqrt{\left(3-2\sqrt{2}\right)\left(4-2\sqrt{3}\right)}$

$=\sqrt{\left(\sqrt{2}-1\right)^2\left(\sqrt{3}-1\right)^2}$

$=\left(\sqrt{2}-1\right)\left(\sqrt{3}-1\right)$

$=\sqrt{6}-\sqrt{2}-\sqrt{3}+1$

AN

alibaba nguyễn

29 tháng 6 2017

Câu 3/ $A=\sqrt{2+\sqrt{2+\sqrt{2+...+\sqrt{2+\sqrt{2}}}}}$

$< \sqrt{2+\sqrt{2+\sqrt{2+...+\sqrt{2+\sqrt{4}}}}}=2$

Ta lại có:

$A=\sqrt{2+\sqrt{2+\sqrt{2+...+\sqrt{2+\sqrt{2}}}}}>\sqrt{2}>1$

$\Rightarrow1< A< 2$

Vậy $A\notin N$

MT

minh thu

1 tháng 10 2021

1. Tính ( rút...

2

NL

Nguyễn Lê Phước Thịnh

1 tháng 10 2021

a: Ta có: $\sqrt{\left(5-\sqrt{19}\right)^2}-\sqrt{\left(4-\sqrt{19}\right)^2}$

$=5-\sqrt{19}-\sqrt{19}+4$

$=9-2\sqrt{19}$

b: Ta có: $\sqrt{\left(3-2\sqrt{2}\right)^2}-\sqrt{\left(2\sqrt{2}-3\right)^2}$

$=3-2\sqrt{2}-3+2\sqrt{2}$

=0

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AH

Akai Haruma

Giáo viên

2 tháng 10 2021

c.

Căn bậc 2 không xác định do $2-\sqrt{5}< 0$

d.

$=\sqrt{(3+\sqrt{3})^2}(3+\sqrt{3})=|3+\sqrt{3}|(3+\sqrt{3})=(3+\sqrt{3})^2=12+6\sqrt{3}$

e.

$=(2-\sqrt{5})\sqrt{(2+\sqrt{5})^2}=(2-\sqrt{5})|2+\sqrt{5}|=(2-\sqrt{5})(2+\sqrt{5})=4-5=-1$

YN

Yến Nhi Nguyễn

12 tháng 12 2018

Rút gọn:

$\left(\sqrt{6}+\sqrt{2}\right).\sqrt{2-\sqrt{3}}$

$\left(\sqrt{4+\sqrt{15}}-\sqrt{16-3\sqrt{15}}\right).\left(\sqrt{3}+\sqrt{5}\right)$

1

I

Incursion_03

12 tháng 12 2018

a,$\left(\sqrt{6}+\sqrt{2}\right)\sqrt{2-\sqrt{3}}$

$=\sqrt{2}\left(\sqrt{3}+1\right)\sqrt{2-\sqrt{3}}$

$=\left(\sqrt{3}+1\right)\sqrt{4-2\sqrt{3}}$

$=\left(\sqrt{3}+1\right)\sqrt{\left(\sqrt{3}-1\right)^2}$

$=\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)$

$=3-1$

$=2$

b, $\left(\sqrt{4+\sqrt{15}}-\sqrt{16-3\sqrt{15}}\right)\left(\sqrt{3}+\sqrt{5}\right)$

$=\frac{\sqrt{8+2\sqrt{15}}-\sqrt{32-6\sqrt{15}}}{\sqrt{2}}.\left(\sqrt{3}+\sqrt{5}\right)$

$=\frac{\sqrt{3+2\sqrt{3}.\sqrt{5}+5}-\sqrt{27-2.3\sqrt{3}.\sqrt{5}+5}}{\sqrt{2}}\left(\sqrt{3}+\sqrt{5}\right)$

$=\frac{\sqrt{\left(\sqrt{3}+\sqrt{5}\right)^2}-\sqrt{\left(3\sqrt{3}-\sqrt{5}\right)^2}}{\sqrt{2}}\left(\sqrt{3}+\sqrt{5}\right)$

$=\frac{\sqrt{3}+\sqrt{5}-3\sqrt{3}+\sqrt{5}}{\sqrt{2}}\left(\sqrt{3}+\sqrt{5}\right)$

$=\frac{2\sqrt{5}-2\sqrt{3}}{\sqrt{2}}\left(\sqrt{5}+\sqrt{3}\right)$

$=\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{3}\right)$

$=\sqrt{2}\left(5-3\right)$

$=2\sqrt{2}$

Dạng 3.Chứng minh đẳng thức Bài 1: CMa)$\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=2$b)$\left(5+\sqrt{21}\right)\left(\sqrt{14}-\sqrt{6}\right)\sqrt{5-\sqrt{21}}=8$Bài 2...

2

ＴＲ ᗩ ＮＧ ²ᵏ⁶

17 tháng 6 2021

1

D

๖ۣۜDũ๖ۣۜN๖ۣۜG

17 tháng 6 2021

Bài 1

a) Đặt VT = A

<=> $2\sqrt{2}A=\left(8+2\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{8-2\sqrt{15}}$

<=> $2\sqrt{2}A=\left(\sqrt{5}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{5}-\sqrt{3}\right).\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}$

<=> $2A=\left(\sqrt{5}+\sqrt{3}\right)^2.\left(\sqrt{5}-\sqrt{3}\right)^2$

<=> 2A = $\left(5-3\right)^2=4$

<=> A = 2

b) Đặt VT = B

<=> $2\sqrt{2}B=\left(10+2\sqrt{21}\right).\left(\sqrt{14}-\sqrt{6}\right)\sqrt{10-2\sqrt{21}}$

<=> $2\sqrt{2}B=\left(\sqrt{7}+\sqrt{3}\right)^2.\sqrt{2}\left(\sqrt{7}-\sqrt{3}\right).\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}$

<=> $2B=\left(\sqrt{7}+\sqrt{3}\right)^2.\left(\sqrt{7}-\sqrt{3}\right)^2=\left(7-3\right)^2=16$

<=> B = 8

Bài 2

Đặt VT = A

<=> A² = $\dfrac{\sqrt{5}+2+\sqrt{5}-2+2\sqrt{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}}{2}$

<=> A² = $\dfrac{2\sqrt{5}+2\sqrt{5-4}}{2}=\dfrac{2\sqrt{5}+2}{2}=\sqrt{5}+1$

<=> $A=\sqrt{\sqrt{5}+1}$

QE

Quynh Existn't

26 tháng 6 2021

...

2

NN

Nguyễn Ngọc Lộc

26 tháng 6 2021

$A=2.\left|\left(-3\right)\right|^3+2.\left(-2\right)^2-4\left|\left(-2\right)^3\right|$

$=54+8-32=30$

$B=\left|\sqrt{2}-2\right|+\left|\sqrt{2}-3\right|=2-\sqrt{2}+3-\sqrt{2}$

$=5-2\sqrt{2}$

$C=\left|3-\sqrt{3}\right|-\left|1+\sqrt{3}\right|=3-\sqrt{3}-1-\sqrt{3}$

$=2-2\sqrt{3}$

$D=\left|5+\sqrt{6}\right|-\left|\sqrt{6}-5\right|=5+\sqrt{6}-5+\sqrt{6}$

$=2\sqrt{6}$

$E=\sqrt{15^2}-\sqrt{5^2}=15-5=10$

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Y

Yeutoanhoc

26 tháng 6 2021

`A=2sqrt{(-3)^6}+2sqrt{(-2)^4}-4sqrt{(-2)^6}=2|(-3)^3|+2|(-2)^2|-4|(-2)^3|=54+8-32=30` $\\$ `B=sqrt{(sqrt2-2)^2}+sqrt{(sqrt2-3)^2}=2-sqrt2+3-sqrt2=5-2sqrt2` $\\$ `C=sqrt{(3-sqrt3)^2}-sqrt{(1+sqrt3)^2}=3-sqrt3-sqrt3-1=2-2sqrt3` $\\$ `D=sqrt{(5+sqrt6)^2}-sqrt{(sqrt6-sqrt5)^2}=5+sqrt6-5+sqrt6=2sqrt6` $\\$ `E=sqrt{17^2-8^2}-sqrt{3^2+4^2}=sqrt{289-64}-sqrt{9+16}=sqrt(225)-sqrt{25}=15-5=10`

Đúng(1)

M

Minh_28_Anh_09_Lê

31 tháng 8 2015

Rút gọn biểu thức :

a) $\sqrt{5-2\sqrt{6}}-\sqrt{5+2\sqrt{6}}$

b) $\left(2-\sqrt{3}\right)\left(\sqrt{6}+\sqrt{2}\right)\sqrt{2+\sqrt{3}}$