A= $\sqrt{3 \left(2\sqrt{3}-2\right)\sqrt{2 \sqrt{3}}}$B= $\left(1-\sqrt{3}\right)\sqrt{2 \sqrt{3}}$C= $\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}$D= \(\left(\sqrt{3}-2\right)\sqrt{\frac{2 \s...

0

QN

Quân Nguyễn

7 tháng 7 2023

a) A=$\sqrt{\left(4-\sqrt{15}\right)^2+\sqrt{15}}$

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

c) C=$\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}$

d)D=$\sqrt{29+12\sqrt{5}-\sqrt{29-12\sqrt{5}}}$

2

NL

Nguyễn Lê Phước Thịnh

7 tháng 7 2023

a: Sửa đề: $A=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}$

$=4-\sqrt{15}+\sqrt{15}=4$

b: $A=2-\sqrt{3}+\sqrt{3}-1=1$

c: $C=3\sqrt{5}-2-3\sqrt{5}-2=-4$

d: Sửa đề: $D=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}$

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

=6

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H9

HT.Phong (9A5)

CTVHS

7 tháng 7 2023

a) $A=\sqrt{\left(4-\sqrt{15}\right)^2}+\sqrt{15}$

$A=\left|4-\sqrt{15}\right|+\sqrt{15}$

$A=4-\sqrt{15}+\sqrt{15}$

$A=4$

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

$B=\left|2-\sqrt{3}\right|+\left|1-\sqrt{3}\right|$

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

$B=1$

c) $C=\sqrt{49-12\sqrt{5}}-\sqrt{49+12\sqrt{5}}$

$C=\sqrt{\left(3\sqrt{5}\right)^2-2\cdot3\sqrt{15}\cdot2+2^2}-\sqrt{\left(3\sqrt{5}\right)^2+2\cdot3\sqrt{5}\cdot2+2^2}$

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

$C=\left|3\sqrt{5}-2\right|-\left|3\sqrt{5}+2\right|$

$C=3\sqrt{5}-2-3\sqrt{5}-2$

$C=-4$

d) $D=\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}$

$D=\sqrt{\left(2\sqrt{5}\right)^2+2\cdot2\sqrt{5}\cdot3+3^2}-\sqrt{\left(2\sqrt{5}\right)^2-2\cdot2\sqrt{5}\cdot3+3^3}$

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

$D=\left|2\sqrt{5}+3\right|-\left|2\sqrt{5}-3\right|$

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

$D=6$

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B

bbiooo

10 tháng 1 2021

rút gọn các biểu thức sau:

a,$\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}$

b,$\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}$

c,$\sqrt{2+\sqrt{5-\sqrt{13-\sqrt{48}}}}$

d,$\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}$

1

NL

Nguyễn Lê Phước Thịnh

10 tháng 1 2021

a) Ta có: $\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}$

$=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2\cdot\sqrt{20}\cdot3+9}}}$

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

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

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

$=\sqrt{\sqrt{5}-\sqrt{5-2\cdot\sqrt{5}\cdot1+1}}$

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

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

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

$=\sqrt{1}=1$

b) Ta có: $\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}$

$=\sqrt{6+2\sqrt{5}-\sqrt{20-2\cdot2\sqrt{5}\cdot3+9}}$

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

$=\sqrt{6+2\sqrt{5}-\left(2\sqrt{5}-3\right)}$

$=\sqrt{6+3}=3$

c) Sửa đề: $\sqrt{2+\sqrt{5+\sqrt{13-\sqrt{48}}}}$

Ta có: $\sqrt{2+\sqrt{5+\sqrt{13-\sqrt{48}}}}$

$=\sqrt{2+\sqrt{5+\sqrt{12-2\cdot2\sqrt{3}\cdot1+1}}}$

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

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

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

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

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

d) Ta có: $\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3-\sqrt{5}}$

$=\dfrac{\left(6-2\sqrt{5}\right)\sqrt{6+2\sqrt{5}}+\left(6+2\sqrt{5}\right)\sqrt{6-2\sqrt{5}}}{2\sqrt{2}}$

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

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

$=\dfrac{4\cdot2\sqrt{5}}{2\sqrt{2}}$

$=\dfrac{8\sqrt{5}}{2\sqrt{2}}=\dfrac{4\sqrt{5}}{\sqrt{2}}=2\sqrt{10}$

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QE

Quynh Existn't

26 tháng 6 2021

...

Bài 1: Rút gọn biểu thức a) $A=\sqrt{26+15\sqrt{3}}$ b) $B=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}-\sqrt{2}$ c) $C=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}$ d) $D=\left(\sqrt{6}-2\right)\left(5+\sqrt{24}\right)\sqrt{5-\sqrt{24}}$ e) $E=\left(\sqrt{10}-\sqrt{2}\right)\left(\sqrt{3+\sqrt{5}}\right)$ f) $F=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}$ g) $G=\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}$ h)...

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)

TT

Trần Thị Hảo

16 tháng 8 2019

3

AH

Akai Haruma

Giáo viên

16 tháng 8 2019

a)

$A=\sqrt{26+15\sqrt{3}}=\sqrt{\frac{52+30\sqrt{3}}{2}}=\sqrt{\frac{27+25+2\sqrt{27.25}}{2}}$

$=\sqrt{\frac{(\sqrt{27}+\sqrt{25})^2}{2}}=\frac{\sqrt{27}+\sqrt{25}}{\sqrt{2}}=\frac{3\sqrt{3}+5}{\sqrt{2}}=\frac{3\sqrt{6}+5\sqrt{2}}{2}$

b)

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

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

$=\sqrt{(\sqrt{7}+1)^2}-\sqrt{(\sqrt{7}-1)^2}-2=\sqrt{7}+1-(\sqrt{7}-1)-2=0$

$\Rightarrow B=0$

c)

$C=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}=\sqrt{3+5-2\sqrt{3.5}}-\sqrt{3+5+2\sqrt{3.5}}$

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

AH

Akai Haruma

Giáo viên

16 tháng 8 2019

d)

$D=(\sqrt{6}-2)(5+2\sqrt{6})\sqrt{5-2\sqrt{6}}$

$=\sqrt{2}(\sqrt{3}-\sqrt{2})(2+3+2\sqrt{2.3})\sqrt{2+3-2\sqrt{2.3}}$

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

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

$=\sqrt{2}.1^2=\sqrt{2}$

e)

$E=(\sqrt{10}-\sqrt{2})\sqrt{3+\sqrt{5}}=(\sqrt{5}-1).\sqrt{2}.\sqrt{3+\sqrt{5}}$

$=(\sqrt{5}-1)\sqrt{6+2\sqrt{5}}=(\sqrt{5}-1)\sqrt{5+1+2\sqrt{5.1}}$

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

f)

$F=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20+9-2\sqrt{20.9}}}}$

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

$=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}=\sqrt{\sqrt{5}-\sqrt{5+1-2\sqrt{5}}}$

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

Tính: a) $A=\sqrt{8-2\sqrt{15}}\left(\sqrt{3}+\sqrt{5}\right)-\left(\sqrt{45}-\sqrt{20}\right)$ b) $B=\left(\frac{\sqrt{21}-\sqrt{3}}{\sqrt{7}-1}-\frac{\sqrt{15}-\sqrt{3}}{1-\sqrt{5}}\right)\left(\frac{1}{2}\sqrt{6}-\sqrt{\frac{3}{2}}+3\sqrt{\frac{2}{3}}\right)$ c) $C=2\sqrt{3}+\sqrt{7-4\sqrt{3}}+\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}+}\sqrt{3}\right):\sqrt{3}$ d)...

NC

Nguyễn Châu Mỹ Linh

12 tháng 8 2020

1

NL

Nguyễn Lê Phước Thịnh

12 tháng 8 2020

a) Ta có: $A=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{3}+\sqrt{5}\right)-\left(\sqrt{45}-\sqrt{20}\right)$

$=\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}\cdot\left(\sqrt{5}+\sqrt{3}\right)-\sqrt{5}\left(\sqrt{9}-\sqrt{4}\right)$

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

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

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

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

$=2-\sqrt{5}$

b) Ta có: $B=\left(\frac{\sqrt{21}-\sqrt{3}}{\sqrt{7}-1}-\frac{\sqrt{15}-\sqrt{3}}{1-\sqrt{5}}\right)\left(\frac{1}{2}\sqrt{6}-\sqrt{\frac{3}{2}}+3\sqrt{\frac{2}{3}}\right)$

$=\left(\frac{\sqrt{3}\left(\sqrt{7}-1\right)}{\sqrt{7}-1}+\frac{\sqrt{3}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(\sqrt{\frac{3}{2}}-\sqrt{\frac{3}{2}}+\sqrt{6}\right)$

$=\sqrt{3}+\sqrt{3}+\sqrt{6}$

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

c) Ta có: $C=2\sqrt{3}+\sqrt{7-4\sqrt{3}}+\left(\sqrt{\frac{1}{3}}-\sqrt{\frac{4}{3}}+\sqrt{3}\right):\sqrt{3}$

$=2\sqrt{3}+\sqrt{4-2\cdot2\cdot\sqrt{3}+3}+\sqrt{\frac{1}{3}:3}-\sqrt{\frac{4}{3}:3}+\sqrt{3:3}$

$=2\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\frac{1}{9}}-\sqrt{\frac{4}{9}}+\sqrt{1}$

$=2\sqrt{3}+\left|2-\sqrt{3}\right|+\frac{1}{3}-\frac{2}{3}+1$

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

$=\sqrt{3}+\frac{8}{3}$

d) Ta có: $D=\left(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}\right):\frac{1}{\sqrt{7-4\sqrt{3}}}$

$=\left(\frac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\right)\cdot\sqrt{4-2\cdot2\cdot\sqrt{3}+3}$

$=\frac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\cdot\sqrt{\left(2-\sqrt{3}\right)^2}$

$=\frac{60}{20}\cdot\left|2-\sqrt{3}\right|$

$=3\cdot\left(2-\sqrt{3}\right)$(Vì $2>\sqrt{3}$)

$=6-3\sqrt{3}$

LR

Linda Ryna Daring

16 tháng 10 2016

Bài 1: Tính

a, $4\sqrt{8}+\sqrt{18}-6\sqrt{\frac{1}{2}}-\sqrt{200}$

b, $\left(\sqrt{27}-2\sqrt{3}+\sqrt{12}\right).\sqrt{3}+\sqrt{75}$

c,$\left(\frac{5+2\sqrt{6}}{\sqrt{3}+\sqrt{2}}\right)^2-\left(\frac{5-2\sqrt{6}}{\sqrt{3}-\sqrt{2}}\right)^2$

d, $\left(2-\sqrt{2}\right).\left(-5\sqrt{2}\right)-\left(3\sqrt{2}-5\right)^2$

0

PC

Park Chanyeol

15 tháng 7 2016

thực hiện phép tính:...

1

NV

Ngọc Vĩ

16 tháng 7 2016

a/ Bạn ghi nhầm đề rồi

c/ $2\sqrt{18\sqrt{3}}-2\sqrt{5\sqrt{3}}-3\sqrt{5\sqrt{48}}$

$=2\sqrt{18}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-3\sqrt{5}.\sqrt{\sqrt{48}}$

$=2.3\sqrt{2}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-3\sqrt{5}.\sqrt{4\sqrt{3}}$

$=2.3\sqrt{2}.\sqrt{\sqrt{3}}-2\sqrt{5}.\sqrt{\sqrt{3}}-6\sqrt{5}.\sqrt{\sqrt{3}}$

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

$=2\sqrt{\sqrt{3}}\left(3\sqrt{2}-4\sqrt{5}\right)$$=2\sqrt{2\sqrt{3}}\left(3-2\sqrt{10}\right)$

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

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

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

g/ $\sqrt{5-2\sqrt{6}}+\sqrt{5+2\sqrt{6}}-2\sqrt{3}+2007$

$=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}-2\sqrt{3}+2007$

$=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}-2\sqrt{3}+2007$

$=2007$

1 . a)$A=\frac{2}{2\sqrt[3]{2}+2+\sqrt[3]{4}}$ b)$B=\frac{6}{2\sqrt[3]{2}-2+\sqrt[3]{4}}$ c)C=$\frac{2}{\sqrt[3]{4}+\sqrt[3]{2}+2}$ 2 ....

DH

Đỗ Hoàng Thanh Nga

19 tháng 8 2020

1.Tính: a, $\sqrt{\left(5-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-2\right)^2}$ b, B=$\left(2-\sqrt{3}\right).\sqrt{26+15\sqrt{3}}-\left(2+\sqrt{3}\right).\sqrt{26-15\sqrt{3}}$ c, $\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}$ d, A=$\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}$ 2.Giải pt: a,$\sqrt{x^2-3x-2}=x-2$ b,$5\sqrt{x-1}-\sqrt{36x-36}+\sqrt{9x-9}=\sqrt{8x+12}$ c,$\sqrt{x}+\sqrt{1-x}+\sqrt{x\left(1-x\right)}=1$ ...

0

ND

Ngân Đại Boss

15 tháng 9 2017

1

NL

Nguyễn Lê Phước Thịnh

27 tháng 5 2022

Bài 1:

a: $=\left|5-\sqrt{3}\right|-\left|\sqrt{3}-2\right|$

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

b; $B=\dfrac{\left(2-\sqrt{3}\right)\cdot\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\cdot\sqrt{52-30\sqrt{3}}}{\sqrt{2}}$

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

$=\dfrac{6\sqrt{3}+10-9-5\sqrt{3}-6\sqrt{3}+10-9+5\sqrt{3}}{\sqrt{2}}$

$=\dfrac{20-18}{\sqrt{2}}=\sqrt{2}$

c: $C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}$

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

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

d: $A=\left(\sqrt{5}-1\right)\cdot\sqrt{6+2\sqrt{5}}$

$=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=5-1=4$