1. Rút gọn: $\sqrt{8-2\sqrt{7}}-\sqrt{9-2\sqrt{14}}$ $\sqrt{5-2\sqrt{6}}-\sqrt{11-4\sqrt{6}}$ 2. Tính: a. 2 $\sqrt{17-4\sqrt{9} 4\sqrt{5}}$ b. \(\sqrt{5\sqrt{3} 5\sqrt{48-10\sqrt{7} 4\sqr...

NH

Nguyễn Hàn Băng

6 tháng 7 2019

1. Rút gọn: $\sqrt{8-2\sqrt{7}}-\sqrt{9-2\sqrt{14}}$ $\sqrt{5-2\sqrt{6}}-\sqrt{11-4\sqrt{6}}$ 2. Tính: a. 2 + $\sqrt{17-4\sqrt{9}+4\sqrt{5}}$ b. $\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7}+4\sqrt{3}}}$ c. $\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}$ 3. CM: a. $\frac{x+y}{2}$ >= $\sqrt{xy}$ với x, y >= 0 b. $\frac{x}{y}+\frac{y}{x}$ >= 2 với x,y >= 0 c. a + b + 1 >= $\sqrt{ab}$ + $\sqrt{a}+\sqrt{b}$ với a,b >=...

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#Toán lớp 9

0

BG

Bao Gia

12 tháng 7 2021

$\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}$

$\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}$

$\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}$

$\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2^6\right)}$

rút gọn:giải chi tiết hộ mình nha

#Toán lớp 9

1

NL

Nguyễn Lê Phước Thịnh

12 tháng 7 2021

a) Ta có: $\sqrt{3-2\sqrt{2}}-\sqrt{11+6\sqrt{2}}$

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

=-4

b) Ta có: $\sqrt{4-2\sqrt{3}}-\sqrt{7-4\sqrt{3}}+\sqrt{19+8\sqrt{3}}$

$=\sqrt{3}-1-2+\sqrt{3}+4+\sqrt{3}$

$=3\sqrt{3}+1$

c) Ta có: $\sqrt{6-2\sqrt{5}}+\sqrt{9+4\sqrt{5}}-\sqrt{14-6\sqrt{5}}$

$=\sqrt{5}-1+\sqrt{5}-2-3+\sqrt{5}$

$=3\sqrt{5}-6$

d) Ta có: $\sqrt{11-4\sqrt{7}}+\sqrt{23-8\sqrt{7}}+\sqrt{\left(-2\right)^6}$

$=\sqrt{7}-2+4-\sqrt{7}+8$

=10

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AD

Alice dono

18 tháng 7 2020

Rút gọn biểu thức a) $\sqrt{11-2\sqrt{10}}$ b) $\sqrt{9-2\sqrt{14}}$ c) $\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}$ d) $\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}$ e) $\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}$ g) $\sqrt{3}+\sqrt{11+6\sqrt{2}}+\sqrt{5+2\sqrt{6}}$ h) $\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}$ k) $\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}$ i)...

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#Toán lớp 9

1

NL

Nguyễn Lê Phước Thịnh

18 tháng 7 2020

a) Ta có: $\sqrt{11-2\sqrt{10}}$

$=\sqrt{10-2\cdot\sqrt{10}\cdot1+1}$

$=\sqrt{\left(\sqrt{10}-1\right)^2}$

$=\left|\sqrt{10}-1\right|=\sqrt{10}-1$

b) Ta có: $\sqrt{9-2\sqrt{14}}$

$=\sqrt{7-2\cdot\sqrt{7}\cdot\sqrt{2}+2}$

$=\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}$

$=\left|\sqrt{7}-\sqrt{2}\right|$

$=\sqrt{7}-\sqrt{2}$

c) Ta có: $\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}$

$=\sqrt{3+2\cdot\sqrt{3}\cdot1+1}+\sqrt{3-2\cdot\sqrt{3}\cdot1+1}$

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

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

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

$=2\sqrt{3}$

d) Ta có: $\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}$

$=\sqrt{5-2\cdot\sqrt{5}\cdot2+4}-\sqrt{5+2\cdot\sqrt{5}\cdot2+4}$

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

$=\left|\sqrt{5}-2\right|-\left|\sqrt{5}+2\right|$

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

$=\sqrt{5}-2-\sqrt{5}-2$

$=-4$

e) Ta có: $\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}$

$=\frac{\sqrt{2}\cdot\left(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\right)}{\sqrt{2}}$

$=\frac{\sqrt{2}\cdot\left(\sqrt{4-\sqrt{7}}\right)-\sqrt{2}\cdot\left(\sqrt{4+\sqrt{7}}\right)}{\sqrt{2}}$

$=\frac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}$

$=\frac{\sqrt{7-2\cdot\sqrt{7}\cdot1+1}-\sqrt{7+2\cdot\sqrt{7}\cdot1+1}}{\sqrt{2}}$

$=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}$

$=\frac{\left|\sqrt{7}-1\right|-\left|\sqrt{7}+1\right|}{\sqrt{2}}$

$=\frac{\sqrt{7}-1-\left(\sqrt{7}+1\right)}{\sqrt{2}}$

$=\frac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}$

$=\frac{-2}{\sqrt{2}}=-\sqrt{2}$

g) Ta có: $\sqrt{3}+\sqrt{11+6\sqrt{2}}+\sqrt{5+2\sqrt{6}}$

$=\sqrt{3}+\sqrt{9+2\cdot3\cdot\sqrt{2}+2}+\sqrt{2+2\cdot\sqrt{2}\cdot\sqrt{3}+3}$

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

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

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

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

h) Ta có: $\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}$

$=\sqrt{5\sqrt{3}+5\sqrt{48-10\cdot\sqrt{3+2\cdot\sqrt{3}\cdot2+4}}}$

$=\sqrt{5\sqrt{3}+5\sqrt{48-10\cdot\sqrt{\left(\sqrt{3}+2\right)^2}}}$

$=\sqrt{5\sqrt{3}+5\cdot\sqrt{48-10\cdot\left(\sqrt{3}+2\right)}}$

$=\sqrt{5\sqrt{3}+5\cdot\sqrt{48-10\sqrt{3}-20}}$

$=\sqrt{5\sqrt{3}+5\cdot\sqrt{28-10\sqrt{3}}}$

$=\sqrt{5\sqrt{3}+5\cdot\sqrt{25-2\cdot5\cdot\sqrt{3}+3}}$

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

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

$=\sqrt{5\sqrt{3}+25-5\sqrt{3}}$

$=\sqrt{25}=5$

k) Ta có: $\sqrt{94-42\sqrt{5}}-\sqrt{94+42\sqrt{5}}$

$=\sqrt{49-2\cdot7\cdot\sqrt{45}+45}-\sqrt{49+2\cdot7\cdot\sqrt{45}+45}$

$=\sqrt{\left(7-\sqrt{45}\right)^2}-\sqrt{\left(7+\sqrt{45}\right)^2}$

$=\left|7-\sqrt{45}\right|-\left|7+\sqrt{45}\right|$

$=7-\sqrt{45}-\left(7+\sqrt{45}\right)$

$=7-\sqrt{45}-7-\sqrt{45}$

$=-2\sqrt{45}=-6\sqrt{5}$

i) Đặt $A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}$

$\Leftrightarrow A^2=\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)^2$

$=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\cdot\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\cdot\left(4-\sqrt{10+2\sqrt{5}}\right)}$

$=8+2\cdot\sqrt{16-\left(10+2\sqrt{5}\right)}$

$=8+2\cdot\sqrt{6-2\sqrt{5}}$

$=8+2\cdot\sqrt{\left(\sqrt{5}-1\right)^2}$

$=8+2\cdot\left(\sqrt{5}-1\right)$

$=8+2\sqrt{5}-2$

$=6+2\sqrt{5}$

$=\left(\sqrt{5}+1\right)^2$

$\Leftrightarrow A=\sqrt{5}+1$

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NU

Nguyễn Uyên Minh

14 tháng 5 2022

B 5. Rút gọn các biểu thức sau:

a)$\sqrt{7+4\sqrt{3}}$ b)$\sqrt{9-4\sqrt{5}}$

c)$\sqrt{14+6\sqrt{5}}$ d)$\sqrt{17-12\sqrt{2}}$

#Toán lớp 9

1

NN

Nguyễn Ngọc Huy Toàn

14 tháng 5 2022

a.$\sqrt{7+4\sqrt{3}}=\sqrt{\left(\sqrt{3}+2\right)^2}=\left|\sqrt{3}+2\right|=\sqrt{3}+2$

b.$\sqrt{9-4\sqrt{5}}=\sqrt{\left(\sqrt{5}-2\right)^2}=\left|\sqrt{5}-2\right|=\sqrt{5}-2$

c.$\sqrt{14+6\sqrt{5}}=\sqrt{\left(\sqrt{5}+3\right)^2}=\left|\sqrt{5}+3\right|=\sqrt{5}+3$

d.$\sqrt{17-12\sqrt{2}}=\sqrt{\left(2\sqrt{2}-3\right)^2}=\left|2\sqrt{2}-3\right|=3-2\sqrt{2}$

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DM

Đỗ Minh Anh

12 tháng 7 2019 - olm

Rút gọn biểu thức sau:

A = $\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}$

B = $\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}$

C = $\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}$

D = $\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}$

E = $\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}$

#Toán lớp 9

2

B

💋Bevis💋

12 tháng 7 2019

$A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}$

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

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

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

$=2\sqrt{3}$

$B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}$

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

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

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

$=2\sqrt{2}$

$C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}$

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

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

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

$=6$

$D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}$

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

$=|2+\sqrt{5}|-|2-\sqrt{5}|$

$=2+\sqrt{5}-\sqrt{5}+2$

$=4$

$E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}$

$=\sqrt{\left(1+\sqrt{5}\right)^2}-\sqrt{\left(1-\sqrt{5}\right)^2}$

$=|1+\sqrt{5}|-|1-\sqrt{5}|$

$=1+\sqrt{5}-\sqrt{5}+1$

$=2$

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K

KAl(SO4)2·12H2O

12 tháng 7 2019

$A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}$

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

A = 2 + 2

A = 4

$B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}$

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

B = 3 + 3

B = 6

$C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}$

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

C = 3 + 3

C = 6

$D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}$

$D=\sqrt{5}+2-\sqrt{5}+2$

D = 2 + 2

D = 4

$E=\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}$

$E=\sqrt{5}+1-\sqrt{5}+1$

E = 1 + 1

E = 2

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TT

Trần Thị Quỳnh Mai

12 tháng 8 2017 - olm

Câu hỏi hay

Rút gọn biểu...

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#Toán lớp 9

4

LT

lê thị thu huyền

23 tháng 9 2017

a) đặt $A=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}$

nhân cả hai vế với $\sqrt{2}$, ta được:

$\sqrt{2}A=\sqrt{2}\sqrt{4-\sqrt{7}}-\sqrt{2}\sqrt{4+\sqrt{7}}$

$=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}$

$=\sqrt{\left(1-\sqrt{7}\right)^2}-\sqrt{\left(1+ \sqrt{7}\right)^2}$

$=\left|1-\sqrt{7}\right|-\left|1+\sqrt{7}\right|$

$=\sqrt{7}-1-\sqrt{7}-1$

$=-2$

$\Rightarrow A=-\frac{2}{\sqrt{2}}=-\sqrt{2}$

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M

Mafia

12 tháng 5 2018

a) $\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}$

$=\frac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}}{\sqrt{2}}$

$=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}$

$=\frac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}$

$=\frac{-2}{\sqrt{2}}=-\sqrt{2}$

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A4

AK-47

26 tháng 8 2023

Thực hiện phép tính (rút gọn biểu thức)

a) $\sqrt{9+4\sqrt{5}}$ - $\sqrt{9-4\sqrt{5}}$

b) $\sqrt{12-6\sqrt{3}}$ + $\sqrt{12+6\sqrt{3}}$

c) $\sqrt{6\sqrt{2}+11}$ - $\sqrt{11-6\sqrt{2}}$

#Toán lớp 9

1

AH

Akai Haruma

Giáo viên

26 tháng 8 2023

Lời giải:
a.

$=\sqrt{5+2.2\sqrt{5}+2^2}-\sqrt{5-2.2\sqrt{5}+2^2}$

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

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

b.

$=\sqrt{3-2.3\sqrt{3}+3^2}+\sqrt{3+2.3.\sqrt{3}+3^2}$

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

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

$=(3-\sqrt{3})+(\sqrt{3}+3)=6$

c.

$=\sqrt{2+2.3\sqrt{2}+3^2}-\sqrt{2-2.3\sqrt{2}+3^2}$

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

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

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TB

Trang Be

2 tháng 10 2015 - olm

Rút gọn:

$A=\sqrt{11-2\sqrt{10}}+\sqrt{9-2\sqrt{4}}-\sqrt{10}-\sqrt{7}$

$B=\frac{\sqrt{3}+\sqrt{11+2\sqrt{6}}-\sqrt{5+2\sqrt{6}}}{\sqrt{2}+\sqrt{6+2\sqrt{5}}-\sqrt{7+2\sqrt{10}}}$

#Toán lớp 9

2

MT

Minh Triều

2 tháng 10 2015

câu a coi lai có sai sót j ko

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MT

Minh Triều

2 tháng 10 2015

$A=\sqrt{11-2\sqrt{10}}+\sqrt{9-2\sqrt{4}}-\sqrt{10}-\sqrt{7}$

$=\sqrt{\left(\sqrt{10}-1\right)^2}+\sqrt{5}-\sqrt{10}-\sqrt{7}=\sqrt{10}-1+\sqrt{5}-\sqrt{10}-\sqrt{7}$

$=\sqrt{5}-\sqrt{7}-1$

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....

24 tháng 6 2021

rút gọn các biểu thức sau:

a $\sqrt[3]{8\sqrt{5}-16}.\sqrt[3]{8\sqrt{5}+16}$

b $\sqrt[3]{7-5\sqrt{2}}-\sqrt[6]{8}$

c $\sqrt[3]{4}.\sqrt[3]{1-\sqrt{3}}.\sqrt[6]{4+2\sqrt{3}}$

d $\dfrac{2}{\sqrt[3]{3}-1}-\dfrac{4}{\sqrt[3]{9}-\sqrt[3]{3}+1}$

#Toán lớp 9

2

Y

Yeutoanhoc

24 tháng 6 2021

`c)root{3}{4}.root{3}{1-sqrt3}.root{6}{(sqrt3+1)^2}`

`=root{3}{4(1-sqrt3)}.root{3}{1+sqrt3}`

`=root{3}{4(1-sqrt3)(1+sqrt3)}`

`=root{3}{4(1-3)}=-2`

`d)2/(root{3}{3}-1)-4/(root{9}-root{3}{3}+1)`

`=(2(root{3}{9}+root{3}{3}+1))/(3-1)-(4(root{3}{3}+1))/(3+1)`

`=root{3}{9}+root{3}{3}+1-root{3}{3}-1`

`=root{3}{9}`

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Y

Yeutoanhoc

24 tháng 6 2021

`a)root{3}{8sqrt5-16}.root{3}{8sqrt5+16}`

`=root{3}{(8sqrt5-16)(8sqrt5+16)}`

`=root{3}{320-256}`

`=root{3}{64}=4`

`b)root{3}{7-5sqrt2}-root{6}{8}`

`=root{3}{1-3.sqrt{2}+3.2.1-2sqrt2}-root{6}{(2)^3}`

`=root{3}{(1-sqrt2)^3}-sqrt2`

`=1-sqrt2-sqrt2=1-2sqrt2`

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AS

Akashi Seijuro

24 tháng 6 2019

Bài 1: Tính

$\sqrt{3+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}\\ \sqrt{12+6\sqrt{3}+\sqrt{12-6\sqrt{3}}}\\ \sqrt{9-4\sqrt{2}+\sqrt{9+4\sqrt{2}}}$

$\sqrt{\sqrt{2}+2+\sqrt{4+\sqrt{9-\sqrt{32}}}}\\ \sqrt{6+2\sqrt{5}-\sqrt{29+12\sqrt{5}}}\\ \sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{\sqrt{49}+\sqrt{40}}\\ \sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}$

#Toán lớp 9

4

AH

Akai Haruma

Giáo viên

20 tháng 7 2020

7.

$\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{4+3+2\sqrt{4.3}}}}}$

$=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{(\sqrt{4}+\sqrt{3})^2}}}}$

$=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10(2+\sqrt{3})}}}$

$=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}$

$=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{25+3-2.5\sqrt{3}}}}$

$=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{(5-\sqrt{3})^2}}}$

$=\sqrt{4+\sqrt{5\sqrt{3}+5(5-\sqrt{3})}}=\sqrt{4+\sqrt{25}}=\sqrt{4+5}=3$

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AH

Akai Haruma

Giáo viên

20 tháng 7 2020

5.

$\sqrt{6+2\sqrt{5}-\sqrt{29+12\sqrt{5}}}=\sqrt{6+2\sqrt{5}-\sqrt{20+9+2\sqrt{20.9}}}$

$=\sqrt{6+2\sqrt{5}-\sqrt{(\sqrt{20}+3)^2}}=\sqrt{6+2\sqrt{5}-(\sqrt{20}+3)}=\sqrt{3}$

6.

$\sqrt{8+\sqrt{8}+\sqrt{20}+\sqrt{40}}-\sqrt{\sqrt{49}+\sqrt{40}}$

$=\sqrt{8+2\sqrt{2}+2\sqrt{5}+2\sqrt{10}}-\sqrt{7+2\sqrt{10}}$

$=\sqrt{(2+5+2\sqrt{2.5})+2(\sqrt{2}+\sqrt{5})+1}-\sqrt{2+5+2\sqrt{2.5}}$

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

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

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P

phamthiminhanh

26 tháng 6 2021

Tính:A=$\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}$B=$\sqrt{9-4\sqrt{5}}+\sqrt{9+4\sqrt{5}}$C=$\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}$D=$\sqrt{5\sqrt{3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}$E=$\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}$(2...

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#Toán lớp 9

4

AH

Akai Haruma

Giáo viên

26 tháng 6 2021

$A=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{3+1+2\sqrt{3.1}}-\sqrt{3+1-2\sqrt{3.1}}$

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

$B=\sqrt{4+5-2\sqrt{4.5}}+\sqrt{4+5+2\sqrt{4.5}}=\sqrt{(\sqrt{4}-\sqrt{5})^2}+\sqrt{(\sqrt{4}+\sqrt{5})^2}$

$=|\sqrt{4}-\sqrt{5}|+|\sqrt{4}+\sqrt{5}|=2\sqrt{5}$

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AH

Akai Haruma

Giáo viên

26 tháng 6 2021

$C\sqrt{2}=\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}=\sqrt{7+1-2\sqrt{7.1}}-\sqrt{7+1+2\sqrt{7.1}}$

$=\sqrt{(\sqrt{7}-1)^2}-\sqrt{(\sqrt{7}+1)^2}$

$=|\sqrt{7}-1|-|\sqrt{7}+1|=-2\Rightarrow C=-\sqrt{2}$

----------------------------

$7+4\sqrt{3}=(2+\sqrt{3})^2\Rightarrow 10\sqrt{7+4\sqrt{3}}=10(2+\sqrt{3})$

$\Rightarrow \sqrt{48-10\sqrt{7+4\sqrt{3}}}=\sqrt{28-10\sqrt{3}}=\sqrt{(5-\sqrt{3})^2}=5-\sqrt{3}$

$\Rightarrow 3+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}=3+5(5-\sqrt{3})=28-5\sqrt{3}$

$\Rightarrow D=\sqrt{5\sqrt{28-5\sqrt{3}}}$

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