1.Tính và thu gọn: a) $\sqrt{\left(\sqrt{5} 1\right)}-\sqrt{\left(\sqrt{5}-2\right)^2}$ b) $\sqrt{5 2\sqrt{6}} \sqrt{5-2\sqrt{6}}-2\sqrt{4-2\sqrt{3}}$ c) Chứng minh: \(\sqrt{6 \sqrt{6 \sqrt{6 \...

NJ

Nozomi Judo

21 tháng 12 2018

1.Tính và thu gọn: a) $\sqrt{\left(\sqrt{5}+1\right)}-\sqrt{\left(\sqrt{5}-2\right)^2}$ b) $\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}-2\sqrt{4-2\sqrt{3}}$ c) Chứng minh: $\sqrt{6+\sqrt{6+\sqrt{6+\sqrt{6+...+\sqrt{6}}}}} 3$ ( Có 2009 dấu căn ) d) Chứng minh: $\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{2...\sqrt{2}}}}}} 2$ ( Có 2010 dấu căn ) 2. Tìm x biết: * $\sqrt{x^2+8x+16}+\sqrt{y^2-y+\dfrac{1}{4}}=0$ 3.Tính: a)...

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#Toán lớp 9

1

NL

Nguyễn Lê Phước Thịnh

10 tháng 12 2022

Câu 2:

=>|x+4|+|y-1/2|=0

=>x+4=0 và y-1/2=0

=>x=-4 và y=1/2

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NK

ngan kim

1 tháng 10 2023

RÚT GỌN BIỂU THỨC

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

B= $\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)$$\left(\sqrt{6}+11\right)$

#Toán lớp 9

1

H9

HT.Phong (9A5)

CTVHS

1 tháng 10 2023

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

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

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

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

$A=4-5$

$A=-1$

____

$B=\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)$

$B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3+\sqrt{6}\right)\left(3-\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)$

$B=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{2}-\dfrac{12\left(3+\sqrt{6}\right)}{3}\right]\left(\sqrt{6}+11\right)$

$B=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)$

$B=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)$

$B=6-121$

$B=-115$

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ND

nguyễn đăng khôi

30 tháng 5 2023

rút gọn A)$\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5-3}\right)^2}}$
B) $\sqrt{6+2\sqrt{5-\sqrt{\left(2\sqrt{3+1}\right)^2}}}$
C) $\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}$

#Toán lớp 9

1

Y

YangSu

30 tháng 5 2023

$c,\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{\left(2+\sqrt{3}\right)^2}}}}$

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

$=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-20-10\sqrt{3}}}}$

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

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

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

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

$=\sqrt{29}$

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HA

Hoài An

24 tháng 9 2021

a) $\sqrt{\left(2\sqrt{6}-4\right)^2}+\sqrt{15-6\sqrt{6}}$

b) $\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{19+2\sqrt{18}}$

c) $\sqrt{9+4\sqrt{5}}-\sqrt{\left(1-\sqrt{5}^2\right)}$

#Toán lớp 9

3

NH

Nguyễn Hoàng Minh

24 tháng 9 2021

$a,=2\sqrt{6}-4+\sqrt{\left(3-\sqrt{6}\right)^2}=2\sqrt{6}-4+3-\sqrt{6}=\sqrt{6}-1\\ b,=3-2\sqrt{2}+\sqrt{\left(3\sqrt{2}+1\right)^2}=3-2\sqrt{2}+3\sqrt{2}+1=4+\sqrt{2}\\ c,=\sqrt{\left(\sqrt{5}+2\right)^2}-\left(\sqrt{5}-1\right)=\sqrt{5}+2-\sqrt{5}+1=3$

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LL

Lấp La Lấp Lánh

24 tháng 9 2021

a) $=2\sqrt{6}-4+\sqrt{\left(3-\sqrt{6}\right)^2}=2\sqrt{6}-4+3-\sqrt{6}=-1+\sqrt{6}$

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

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

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HN

Hương Nguyễn

28 tháng 7 2018 - olm

Bài 1: Rút gọna. $\left(5-2\sqrt{3}\right)^2+\left(5+2\sqrt{3}\right)^2$b. $\left(\sqrt{5}+\sqrt{2}\right)^2-\left(2\sqrt{5}+1\right)\left(2\sqrt{5}-1\right)-\sqrt{40}$c. $\left(\sqrt{2}-1\right)^2-\frac{2}{3}\sqrt{4}+\frac{4\sqrt{2}}{5}+\sqrt{1\frac{11}{15}}-\sqrt{2}$d. $\left(\sqrt{6}-\sqrt{18}+5\sqrt{2}-\frac{1}{2}\sqrt{8}\right)2\sqrt{6}+2\sqrt{3}$e. $\left(2\sqrt{3}-3\sqrt{2}\right)^2+6\sqrt{6}+3\sqrt{24}$Bài 2: Rút gọnA...

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#Toán lớp 9

0

M

manh

29 tháng 9 2023

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

b : $\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}-\sqrt{2}$

c : $\left(2+\dfrac{5-2\sqrt{5}}{2-\sqrt{5}}\right).\left(2+\dfrac{5-3\sqrt{5}}{3-\sqrt{5}}\right)$

d : $\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right).\left(\sqrt{6}+11\right)$

#Toán lớp 9

1

H9

HT.Phong (9A5)

CTVHS

30 tháng 9 2023

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

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

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

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

$=2\sqrt{3}$

b) $\dfrac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}-\sqrt{2}$

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

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

$=0$

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

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

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

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

$=9-4\sqrt{5}$

d) $\left(\dfrac{15}{\sqrt{6}+1}+\dfrac{4}{\sqrt{6}-2}-\dfrac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)$

$=\left[\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}+\dfrac{4\left(\sqrt{6}+2\right)}{\left(\sqrt{6}-2\right)\left(\sqrt{6}+2\right)}-\dfrac{12\left(3+\sqrt{6}\right)}{\left(3-\sqrt{6}\right)\left(3+\sqrt{6}\right)}\right]\left(\sqrt{6}+11\right)$

$=\left[\dfrac{15\left(\sqrt{6}-1\right)}{5}+\dfrac{4\left(\sqrt{6}+2\right)}{6-4}-\dfrac{12\left(3+\sqrt{6}\right)}{9-6}\right]\left(\sqrt{6}+11\right)$

$=\left(3\sqrt{6}-3+2\sqrt{6}+4-12-4\sqrt{6}\right)\left(\sqrt{6}+11\right)$

$=\left(\sqrt{6}-11\right)\left(\sqrt{6}+11\right)$

$=6-121$

$=-115$

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NN

Nguyễn Như Nguyệt

27 tháng 6 2017

Tính : a) $\dfrac{5+2\sqrt{5}}{\sqrt{5}}+\dfrac{3+\sqrt{3}}{\sqrt{3}}-\left(\sqrt{5}+\sqrt{3}\right)$ b) $\left(\dfrac{1}{2-\sqrt{5}}+\dfrac{2}{\sqrt{5}+\sqrt{3}}\right):\dfrac{1}{\sqrt{21+12\sqrt{3}}}$ c) $\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}$ d) $\sqrt{21-6\sqrt{6}}+\sqrt{9+2\sqrt{18}}-2\sqrt{6+3\sqrt{3}}$ e) $\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}$ f)...

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#Toán lớp 9

1

NL

Nguyễn Lê Phước Thịnh

19 tháng 5 2022

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

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

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

$=-\sqrt{3}\left(7+4\sqrt{3}\right)=-7\sqrt{3}-12$

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

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QE

Quynh Existn't

26 tháng 6 2021

...

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#Toán lớp 9

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`

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M

manh

30 tháng 9 2023

a : $\sqrt{\left(2\sqrt{2}-1\right)^2}-\sqrt{17+12\sqrt{2}}$

b : $\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}$

c : $\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{19+6\sqrt{2}}$

#Toán lớp 9

1

H

⭐Hannie⭐

30 tháng 9 2023

$\sqrt{\left(2\sqrt{2-1}\right)^2}-\sqrt{17+12\sqrt{2}}\\ =\left|2\sqrt{2}-1\right|-\sqrt{9+2\cdot3\cdot2\sqrt{2}+\left(2\sqrt{2}\right)^2}\\ =2\sqrt{2}-1-\sqrt{\left(3+2\sqrt{2}\right)^2}\\=2\sqrt{2}-1-\left(3+2\sqrt{2}\right)\\ =2\sqrt{2}-1-3-2\sqrt{2}\\ =-4$

__

$\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{14-6\sqrt{5}}\\ =\left|2-\sqrt{5}\right|+\sqrt{9-2\cdot3\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\\ =2-\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}\\ =2-\sqrt{5}+3-\sqrt{5}\\ =5-2\sqrt{5}$

__

$\sqrt{\left(4-3\sqrt{2}\right)^2}-\sqrt{19+6\sqrt{2}}\\ =\left|4-3\sqrt{2}\right|-\sqrt{18+2\cdot3\cdot\sqrt{2}+1}\\ =4-3\sqrt{2}-\sqrt{\left(3\sqrt{2}+1\right)^2}\\ =4-3\sqrt{2}-3\sqrt{2}-1\\ =3-6\sqrt{2}$

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BB

Byun Baekhyun

10 tháng 7 2017

$A=\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}+2\sqrt{2}\\ B=\left(5+2\sqrt{6}\right)\cdot\left(49-20\sqrt{6}\right)\cdot\sqrt{5-2\sqrt{6}}$

$C=\dfrac{1}{2}\left(\sqrt{6}+\sqrt{5}\right)^2-\dfrac{1}{4}\sqrt{120}-\sqrt{\dfrac{15}{2}}$

$D=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}+\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}$

#Toán lớp 9

3

BL

Bình Lê

10 tháng 7 2017

$A=\sqrt{8}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =2\sqrt{2}-\sqrt{7}+5\sqrt{7}+2\sqrt{2}\\ =4\sqrt{2}+4\sqrt{7}$

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BL

Bình Lê

10 tháng 7 2017

$B=\left(3+2\sqrt{6}+2\right)\left(25-20\sqrt{6}+24\right)\sqrt{3-2\sqrt{6}+2}\\ =\left(\sqrt{3}+\sqrt{2}\right)^2\left(5-2\sqrt{6}\right)^2\left(\sqrt{3}-\sqrt{2}\right)\\ =\left(\sqrt{3}+\sqrt{2}\right)\left(3-2\sqrt{6}+2\right)^2\\ =\left(\sqrt{3}-\sqrt{2}\right)^3\\ =9\sqrt{3}-11\sqrt{2}$

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PH

Phạm Hà Linh

7 tháng 9 2023

Thực hiện phép tính và thu gọn biểu thức:

B= $\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)$

Thực hiện phép tính:

$\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}$

#Toán lớp 9

1

H9

HT.Phong (9A5)

CTVHS

8 tháng 9 2023

$B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)$

$B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)$

$B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)$

$B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)$

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

$B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)$

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

$B=-\left(5-36\right)$

$B=-\left(-31\right)$

$B=31$

_____________________________

$\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}$

$=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}$

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

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

$=2\sqrt{3}+1$

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