a) $\sqrt{\left(5-2\sqrt{6^2}\right)}$_ $\sqrt{\left(5 2\sqrt{6^2}\right)}$b) $\sqrt{\left(2-\sqrt{3^2}\right)}$ $\sqrt{\left(1-\sqrt{3^2}\right)}$c) $\sqrt{3 \sqrt{2^2}}$

DT

Đặng Trang

28 tháng 9 2016 - olm

a) $\sqrt{\left(5-2\sqrt{6^2}\right)}$_ $\sqrt{\left(5+2\sqrt{6^2}\right)}$b) $\sqrt{\left(2-\sqrt{3^2}\right)}$+ $\sqrt{\left(1-\sqrt{3^2}\right)}$c) $\sqrt{3+\sqrt{2^2}}$_ $\sqrt{\left(1-\sqrt{2^2}\right)}$d)$\sqrt{\left(\sqrt{2+1^2}\right)}$_ $\sqrt{\left(\sqrt{2}-5^2\right)}$Mọi người giúp vs...

Đọc tiếp

#Toán lớp 9

0

QE

Quynh Existn't

26 tháng 6 2021

...

Đọc tiếp

#Toán lớp 9

2

NN

Nguyễn Ngọc Lộc

26 tháng 6 2021

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

$=54+8-32=30$

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

$=5-2\sqrt{2}$

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

$=2-2\sqrt{3}$

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

$=2\sqrt{6}$

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

Đúng(2)

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)

NT

Nguyễn Thị Thu Phương

27 tháng 7 2021

Bài 2: a) $\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}$ b) $\sqrt{\left(5-2\sqrt{6}\right)^2}-\sqrt{\left(5+2\sqrt{6}\right)^2}$c) $\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}$ d) $\sqrt{\left(\sqrt{2}+1\right)^2}-\sqrt{\left(\sqrt{2}-5\right)^2}$e) $\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)^2}$ ...

Đọc tiếp

#Toán lớp 9

3

TC

Trên con đường thành công không có....

27 tháng 7 2021

undefined

Đúng(2)

NT

Nguyễn Thị Thu Phương

27 tháng 7 2021

Em cảm ơn ạ :33

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

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

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

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

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

$=6-121$

$=-115$

Đúng(1)

NP

Nguyễn Phạm Thy Vân

13 tháng 8 2020

Tính: 1.$\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}$ 4.$\sqrt{\left(\sqrt{3}\right)^2+2.\left(\sqrt{3}\right).\left(1\right)+\left(1\right)^2}$ 2.$\sqrt{\left(\sqrt{5}-\sqrt{6}\right)^2}$ 5.$\sqrt{\left(\sqrt{5}\right)^2+2.\left(\sqrt{5}\right).\left(\sqrt{3}\right)+\left(\sqrt{3}\right)^2}$ 3.$\sqrt{\left(2\sqrt{2}+\sqrt{3}\right)^2}$...

Đọc tiếp

#Toán lớp 9

0

TH

Trịnh Hữu Hoàng

19 tháng 7 2017

Thực hiện phép tính: a) $\sqrt{\left(3-2\sqrt{2}\right)^2}+\sqrt{\left(3+2\sqrt{2}\right)^2}$ b) $\sqrt{\left(5-2\sqrt{6}\right)^2}-\sqrt{\left(5+2\sqrt{6}\right)^2}$ c) $\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1-\sqrt{3}\right)^2}$ d) $\sqrt{\left(3+\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}$ e) $\sqrt{\left(\sqrt{5-\sqrt{2}}\right)^2}+\sqrt{\left(\sqrt{5+\sqrt{2}}\right)^2}$ f)...

Đọc tiếp

#Toán lớp 9

1

QP

Quỳnh Phạm

20 tháng 7 2017

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

b$\sqrt{(5-2\sqrt{6})^2}+\sqrt{(5+2\sqrt{6})^2}=5-2\sqrt{6}+5+2\sqrt{6}=10 $

các ý còn lại làm tương tự

Đúng(0)

TH

Trịnh Hữu Hoàng

20 tháng 7 2017

hình như ở câu a) chỗ sau dấu bằng đầu tiên bạn bị sai dấu trừ cuối cùng

Đúng(0)

MT

minh thu

1 tháng 10 2021

1. Tính ( rút...

Đọc tiếp

#Toán lớp 9

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

Đúng(2)

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.

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

Đúng(0)

AQ

Anh Quynh

4 tháng 8 2021

Tính:$A=\sqrt{27}-2\sqrt{48}+3\sqrt{75}$$B=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}-3\right)^2}$$C=\sqrt{\left(2\sqrt{3}+1\right)^2}+\sqrt{\left(2\sqrt{3}-5\right)^2}$$D=\sqrt{9-4\sqrt{5}}-\sqrt{14+6\sqrt{5}}$$E=\dfrac{4}{\sqrt{5}-2}-\dfrac{32}{\sqrt{5}+1}$$M=\dfrac{10}{3\sqrt{2}-4}+\dfrac{28}{3\sqrt{2}+2}$ please help...

Đọc tiếp

#Toán lớp 9

0

NT

Nguyễn Trà Giang

24 tháng 8 2020

Rút gọn : a) $\left(\sqrt{6}+\sqrt{2}\right).\left(\sqrt{3}-2\right)\left(\sqrt{2+\sqrt{3}}\right)$ b) $\sqrt{2}.\left(\sqrt{2-\sqrt{3}}\right).\left(\sqrt{3}+1\right)$ c) $\left(\sqrt{10}-\sqrt{6}\right).\left(\sqrt{4-\sqrt{15}}\right)$ d)$\left(\sqrt{3}-\sqrt{12}\right).\left(\sqrt{5+2\sqrt{6}}\right)$ e) $\sqrt{2-\sqrt{3}}.\left(\sqrt{6}-\sqrt{2}\right).\left(2+\sqrt{3}\right)$ f)...

Đọc tiếp

#Toán lớp 9

1

NL

Nguyễn Lê Phước Thịnh

24 tháng 8 2020

a) Ta có: $\left(\sqrt{6}+\sqrt{2}\right)\cdot\left(\sqrt{3}-2\right)\cdot\left(\sqrt{2+\sqrt{3}}\right)$

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

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

$=\sqrt{3+2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)$

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

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

$=\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}+1\right)\cdot\left(\sqrt{3}-2\right)$(Vì $\sqrt{3}>1>0$)

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

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

$=2\cdot\left(3-4\right)$

$=-2$

b) Ta có: $\sqrt{2}\cdot\left(\sqrt{2-\sqrt{3}}\right)\cdot\left(\sqrt{3}+1\right)$

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

$=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}+1\right)$

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

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

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

$=3-1=2$

c) Ta có: $\left(\sqrt{10}-\sqrt{6}\right)\cdot\left(\sqrt{4-\sqrt{15}}\right)$

$=\sqrt{2}\cdot\sqrt{4-\sqrt{15}}\cdot\left(\sqrt{5}-\sqrt{3}\right)$

$=\sqrt{8-2\sqrt{15}}\cdot\left(\sqrt{5}-\sqrt{3}\right)$

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

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

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

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

$=8-2\sqrt{15}$

d) Ta có: $\left(\sqrt{3}-\sqrt{12}\right)\cdot\left(\sqrt{5+2\sqrt{6}}\right)$

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

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

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

$=-\sqrt{3}\cdot\left(\sqrt{3}+\sqrt{2}\right)$(Vì $\sqrt{3}>\sqrt{2}>0$)

$=-3-\sqrt{6}$

e) Ta có: $\sqrt{2-\sqrt{3}}\cdot\left(\sqrt{6}-\sqrt{2}\right)\cdot\left(2+\sqrt{3}\right)$

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

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

$=\sqrt{3-2\cdot\sqrt{3}\cdot1+1}\cdot\left(\sqrt{3}-1\right)\cdot\left(\sqrt{3}+2\right)$

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

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

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

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

$=\frac{16-12}{2}=\frac{4}{2}=2$

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

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

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

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

$=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\left(2+\sqrt{3}\right)}}}$(Vì $2>\sqrt{3}>0$)

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

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

$=\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{25-2\cdot5\cdot\sqrt{3}+3}}}$

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

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

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

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

$=\sqrt{4+\sqrt{25}}$

$=\sqrt{4+5}=\sqrt{9}=3$

Đúng(0)

P

prayforme

25 tháng 8 2017

Rút gọn : $\dfrac{\sqrt{x+\sqrt{4\left(x-1\right)}}-\sqrt{x-\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}.\left(\sqrt{x-1}-\dfrac{1}{\sqrt{x-1}}\right)$ b)$\left(\sqrt{2}+1\right)\left(\sqrt{3}+1\right)\left(\sqrt{6}+1\right)\left(5-2\sqrt{2}-\sqrt{3}\right)$ c)$\left(\sqrt{5}+1\right)\left(\sqrt{7}+1\right)\left(\sqrt{35}+1\right)\left(34-4\sqrt{7}-6\sqrt{5}\right)$ d)...

Đọc tiếp

#Toán lớp 9

0

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

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

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

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

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

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

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

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

$=\sqrt{29}$

Đúng(2)

Chúng tôi đề xuất

Tài nguyên

Ứng dụng mobile

Bảng xếp hạng

Các khóa học có thể bạn quan tâm

Yêu cầu VIP