\(C=\dfrac{1}{1 \sqrt{2}} \dfrac{1}{\sqrt{2} \sqrt{3}} \dfrac{1}{\sqrt{3} \sqrt{4}} ....\dfrac{1}{\sqrt{99} \sqrt{100}}\)\(C=\dfrac{1\left(1-\sqrt{2}\right)}{\left(1 \sqrt{2}\right)\left(1-\sqrt{2}\ri...

LL

Ling ling 2k7

19 tháng 9 2021

...

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#Toán lớp 9

0

M

minh

1 tháng 9 2023 - olm

tính giá trị biểu thức
a)\(\sqrt{3+2\sqrt{2}}+\sqrt{\left(\sqrt{2}-2\right)^2}\)
b)\(\dfrac{1}{5}\sqrt{50}-2\sqrt{96}-\dfrac{\sqrt{30}}{\sqrt{15}}+12\sqrt{\dfrac{1}{6}}\)
c)\(\left(\dfrac{5-\sqrt{5}}{\sqrt{5}}-2\right)\left(\dfrac{4}{1+\sqrt{5}}+4\right)\)

#Toán lớp 9

2

H9

HT.Phong (9A5)

CTVHS

1 tháng 9 2023

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

\(=\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}\cdot1+1^2}+\left|\sqrt{2}-2\right|\)

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

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

\(=\sqrt{2}+1-\sqrt{2}+2\)

\(=3\)

b) \(\dfrac{1}{5}\sqrt{50}-2\sqrt{96}-\dfrac{\sqrt{30}}{\sqrt{15}}+12\sqrt{\dfrac{1}{6}}\)

\(=\dfrac{1}{5}\cdot5\sqrt{2}-2\cdot4\sqrt{6}-\sqrt{\dfrac{30}{15}}+\sqrt{\dfrac{144}{6}}\)

\(=\sqrt{2}-8\sqrt{6}-\sqrt{2}+2\sqrt{6}\)

\(=-8\sqrt{6}+2\sqrt{6}\)

\(=-6\sqrt{6}\)

c) \(\left(\dfrac{5-\sqrt{5}}{\sqrt{5}}-2\right)\left(\dfrac{4}{1+\sqrt{5}}+4\right)\)

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

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

\(=\left(\sqrt{5}-3\right)\left(\sqrt{5}-1+4\right)\)

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

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

\(=-4\)

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ND

Nguyễn Đức Trí

1 tháng 9 2023

a) \(\sqrt[]{3+2\sqrt[]{2}}+\sqrt[]{\left(\sqrt[]{2}-2\right)^2}\)

\(=\sqrt[]{2+2\sqrt[]{2}.1+1}+\left|\sqrt[]{2}-2\right|\)

\(=\sqrt[]{\left(\sqrt[]{2}+1\right)^2}+\left(2-\sqrt[]{2}\right)\) \(\left(\left(\sqrt[]{2}\right)^2=2< 2^2=4\right)\)

\(=\left|\sqrt[]{2}+1\right|+2-\sqrt[]{2}\)

\(=\sqrt[]{2}+1+2-\sqrt[]{2}\)

\(=3\)

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AQ

Anh Quynh

4 tháng 10 2021

Tính
\(A=\sqrt{20}-3\sqrt{8}+5\sqrt{45}\)
\(B=\dfrac{30}{\sqrt{7}-1}+\dfrac{15}{\sqrt{7}+2}\)
\(C=\left(3-\dfrac{5-\sqrt{5}}{\sqrt{5}-1}\right)\left(3+\dfrac{5+\sqrt{5}}{\sqrt{5}+1}\right)\)
\(D=\sqrt{\left(3-\sqrt{2}\right)^2}-\sqrt{\left(1-\sqrt{2}\right)^2}\)
\(E=\sqrt{7-4\sqrt{3}}-\sqrt{3+2\sqrt{3}}\)

#Toán lớp 9

1

LL

Lấp La Lấp Lánh

4 tháng 10 2021

1) \(A=2\sqrt{5}-6\sqrt{2}+3\sqrt{5}=5\sqrt{5}-6\sqrt{2}\)

2) \(B=\dfrac{30\left(\sqrt{7}+1\right)}{7-1}+\dfrac{15\left(\sqrt{7}-2\right)}{7-4}=5\sqrt{7}+5+5\sqrt{7}-10=-5+10\sqrt{7}\)

3) \(C=\left(3-\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}-1}\right)\left(3+\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)=9-5=4\)

4) \(D=3-\sqrt{2}+1-\sqrt{2}=4-2\sqrt{2}\)

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AS

An Sơ Hạ

30 tháng 9 2017

Tính :

a) \(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+\dfrac{1}{\sqrt{3}+\sqrt{4}}+...\:+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

b) \(\sqrt{\dfrac{16}{\left(\sqrt{3}-\sqrt{2}\right)^2}}+\sqrt{\dfrac{9}{\left(\sqrt{3}+\sqrt{2}\right)^2}}\)

#Toán lớp 9

2

F

F.C

30 tháng 9 2017

Rút gọn biểu thức chứa căn bậc hai

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F

F.C

30 tháng 9 2017

Rút gọn biểu thức chứa căn bậc hai

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NH

Nguyễn Hồng Nhung

4 tháng 10 2018

Bài 1: Cho A = \(\dfrac{1}{2\sqrt{1}+1\sqrt{2}}+\dfrac{1}{3\sqrt{2}+2\sqrt{3}}+\dfrac{1}{4\sqrt{3}+3\sqrt{4}}+...+\dfrac{1}{100\sqrt{99}+99\sqrt{100}}\) So sánh A với 1 Bài 2: Tính A = \(\left(\dfrac{3}{\sqrt{2}+1}+\dfrac{14}{2\sqrt{2}-1}-\dfrac{4}{2-\sqrt{2}}\right).\left(\sqrt{8}+2\right)\) Bài 3: Tính...

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#Toán lớp 9

2

MN

Mẫn Nhi

4 tháng 10 2018

Bài 1:Với mọi n∈N*,ta có:

\(\dfrac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{\left(n+1\right)^2n-n^2\left(n+1\right)}=\dfrac{\left(n+1\right)\sqrt{n}-n\sqrt{n+1}}{n\left(n+1\right)}=\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\)

Do đó :

A=\(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{99}}-\dfrac{1}{\sqrt{100}}=1-\dfrac{1}{10}=\dfrac{9}{10}\)

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NL

Nguyễn Lê Phước Thịnh

6 tháng 10 2022

Bài 2:

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

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

=10

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B

bbiooo

10 tháng 1 2021

a, \(A=\left(\sqrt{2}+1\right)[\left(\sqrt{2}\right)^2+1][(\sqrt{2})^4+1][\left(\sqrt{2}\right)^8+1][1\left(\sqrt{2}\right)^{16}+1]\)

b, \(B=\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+1\sqrt{2020}}\)

c,\(C=^3\sqrt[]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}\)

#Toán lớp 9

1

NL

Nguyễn Lê Phước Thịnh

10 tháng 1 2021

a) Ta có: \(A=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^2+1\right]\left[\left(\sqrt{2}\right)^4+1\right]\left[\left(\sqrt{2}\right)^8+1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)

\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^2+1\right]\left[\left(\sqrt{2}\right)^2-1\right]\left[\left(\sqrt{2}\right)^4+1\right]\left[\left(\sqrt{2}\right)^8+1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)

\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^4-1\right]\left[\left(\sqrt{2}\right)^4+1\right]\left[\left(\sqrt{2}\right)^8+1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)

\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^8-1\right]\left[\left(\sqrt{2}\right)^8+1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)

\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^{16}-1\right]\left[\left(\sqrt{2}\right)^{16}+1\right]\)

\(=\left(\sqrt{2}+1\right)\left[\left(\sqrt{2}\right)^{32}-1\right]\)

\(=65535\sqrt{2}+65535\)

b) Ta có: \(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+\sqrt{3}}+...+\dfrac{1}{\sqrt{2019}+\sqrt{2020}}\)

\(=\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{2020}-\sqrt{2019}\)

\(=\sqrt{2020}-1\)

\(=2\sqrt{505}-1\)

c) Ta có: \(C^3=26+15\sqrt{3}+26-15\sqrt{3}+3\cdot\sqrt[3]{\left(26+15\sqrt{3}\right)\left(26-15\sqrt{3}\right)}\cdot\left(\sqrt[3]{26+15\sqrt{3}}+\sqrt[3]{26-15\sqrt{3}}\right)\)

\(\Leftrightarrow C^3=52+3\cdot C\)

\(\Leftrightarrow C^3-3\cdot C-52=0\)

\(\Leftrightarrow C^3-4C^2+4C^2-16C+13C-52=0\)

\(\Leftrightarrow C^2\left(C-4\right)+4C\left(C-4\right)+13\left(C-4\right)=0\)

\(\Leftrightarrow\left(C-4\right)\left(C^2+4C+13\right)=0\)

mà \(C^2+4C+13>0\)

nên C-4=0

hay C=4

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NP

Nguyên Phan

12 tháng 7 2021

a)\(\left(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{15}{3-\sqrt{3}}\right)\dfrac{1}{\sqrt{3}+5}\)

b)\(\dfrac{1}{1+\sqrt{2}}+\dfrac{1}{\sqrt{2}+3}+...+\dfrac{1}{\sqrt{99}+\sqrt{100}}\)

#Toán lớp 9

2

NL

Nguyễn Lê Phước Thịnh

12 tháng 7 2021

a) Ta có: \(\left(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{15}{3-\sqrt{3}}\right)\cdot\dfrac{1}{5+\sqrt{3}}\)

\(=\left(\dfrac{2\left(\sqrt{3}+1\right)}{2}-\dfrac{3\left(2+\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}+\dfrac{15\left(3+\sqrt{3}\right)}{6}\right)\cdot\dfrac{1}{5+\sqrt{3}}\)

\(=\left(\sqrt{3}+1-6-3\sqrt{3}+\dfrac{15}{2}+\dfrac{5}{2}\sqrt{3}\right)\cdot\dfrac{1}{5+\sqrt{3}}\)

\(=\left(\dfrac{5}{2}+\dfrac{\sqrt{3}}{2}\right)\cdot\dfrac{1}{5+\sqrt{3}}\)

\(=\dfrac{1}{2}\)

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NL

Nguyễn Lê Phước Thịnh

12 tháng 7 2021

b) Ta có: \(\dfrac{1}{\sqrt{2}+1}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{100}+\sqrt{99}}\)

\(=-1+\sqrt{2}-\sqrt{2}+\sqrt{3}-...-\sqrt{99}+\sqrt{100}\)

=-1+10=9

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QE

Quynh Existn't

5 tháng 7 2021

Tính:

\(A=\sqrt{20}-2\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
\(B=4\sqrt{\left(\sqrt{3}-1\right)^2}+2\sqrt{12}+4\sqrt{\dfrac{1}{2}}\)
\(C=\left(3+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)\left(3-\dfrac{3+\sqrt{3}}{1+\sqrt{3}}\right)\)
\(D=\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}\)

#Toán lớp 9

1

NL

Nguyễn Lê Phước Thịnh

5 tháng 7 2021

a) Ta có: \(A=\sqrt{20}-2\sqrt{45}+3\sqrt{18}+\sqrt{72}\)

\(=2\sqrt{5}-6\sqrt{5}+9\sqrt{2}+6\sqrt{2}\)

\(=-4\sqrt{5}+15\sqrt{2}\)

b) Ta có: \(B=4\sqrt{\left(\sqrt{3}-1\right)^2}+2\sqrt{12}+4\sqrt{\dfrac{1}{2}}\)

\(=4\left(\sqrt{3}-1\right)+2\cdot2\sqrt{3}+\dfrac{4}{\sqrt{2}}\)

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

\(=8\sqrt{3}+2\sqrt{2}-4\)

c) Ta có: \(C=\left(3+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)\left(3-\dfrac{3+\sqrt{3}}{1+\sqrt{3}}\right)\)

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

=9-3

=6

d) Ta có: \(D=\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}\)

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

=4

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MD

Mạnh Dũng

22 tháng 11 2021

Chứng minh rằng:

\(\dfrac{1}{3\left(\sqrt{2}+1\right)}+\dfrac{1}{5\left(\sqrt{3}+\sqrt{2}\right)}+\dfrac{1}{7\left(\sqrt{4}+\sqrt{3}\right)}+...+\dfrac{1}{4021\left(\sqrt{2011}+\sqrt{2010}\right)}< \dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{2011}}\right)\)

#Toán lớp 9

1

NH

Nguyễn Hoàng Minh

22 tháng 11 2021

\(\dfrac{1}{\sqrt{k}+\sqrt{k+1}}=\dfrac{\sqrt{k}-\sqrt{k+1}}{k-k-1}=\sqrt{k+1}-\sqrt{k}\\ \Leftrightarrow\text{Đặt}\text{ }A=\dfrac{1}{3\left(\sqrt{2}+\sqrt{1}\right)}+\dfrac{1}{5\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{1}{4021\left(\sqrt{2011}+\sqrt{2010}\right)}< \dfrac{1}{2\left(\sqrt{2}+\sqrt{1}\right)}+\dfrac{1}{2\left(\sqrt{3}+\sqrt{2}\right)}+...+\dfrac{1}{2\left(\sqrt{2011}+\sqrt{2010}\right)}\\ \Leftrightarrow A< \dfrac{1}{2}\left(\dfrac{1}{\sqrt{2}+\sqrt{1}}+\dfrac{1}{\sqrt{3}+\sqrt{2}}+...+\dfrac{1}{\sqrt{2011}+\sqrt{2010}}\right)\)

\(\Leftrightarrow A< \dfrac{1}{2}\left(\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+...+\sqrt{2011}-\sqrt{2010}\right)\\ \Leftrightarrow A< \dfrac{1}{2}\left(\sqrt{2011}-1\right)< \dfrac{1}{2}\cdot\dfrac{\sqrt{2011}-1}{\sqrt{2011}}=\dfrac{1}{2}\left(1-\dfrac{1}{\sqrt{2011}}\right)\)

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PT

Phương Thảo

7 tháng 7 2017

Rút gọn: A = \(\dfrac{4+\sqrt{7}}{3\sqrt{2}+\sqrt{4+\sqrt{7}}}+\dfrac{4-\sqrt{7}}{3\sqrt{2}-\sqrt{4-\sqrt{7}}}\) B = \(\dfrac{3\sqrt{2}+\sqrt{11}}{\sqrt{2}+\sqrt{6+\sqrt{11}}}+\dfrac{3\sqrt{2}-\sqrt{11}}{\sqrt{2}-\sqrt{6-\sqrt{11}}}+18\) C = \(\dfrac{1}{\sqrt{3}+\sqrt{5}}+\dfrac{1}{\sqrt{5}+\sqrt{7}}+...+\dfrac{1}{\sqrt{2n+1}+\sqrt{2n+3}}\)với n thuộc N* D =...

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#Toán lớp 9

0

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