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18 tháng 12 2019

a, \(\left(\sqrt{3}-1\right).\sqrt{4+2\sqrt{3}}\)

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

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

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

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

b, Với x không âm ⇔ \(x\ge0\) ta có:

\(5\sqrt{2x}-3\sqrt{8x}+\sqrt{50x}-7\)

\(=5\sqrt{2x}-3\sqrt{2^2.2x}+\sqrt{5^2.2x}-7\)

\(=5\sqrt{2x}-6\sqrt{2x}+5\sqrt{2x}-7\)

\(=\left(5-6+5\right).\sqrt{2x}-7\)

\(=4\sqrt{2x}-7\)

Vậy với \(x\ge0\) thì biểu thức có giá trị \(=4\sqrt{2x}-7\).

20 tháng 12 2019

không có j
cùng khối 9 gọi bạn ik hihi

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

12 tháng 7 2019

\(A=\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\)

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

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

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

\(=2\sqrt{3}\)

\(B=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}\)

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

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

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

\(=2\sqrt{2}\)

\(C=\sqrt{17+12\sqrt{2}}+\sqrt{17-12\sqrt{2}}\)

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

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

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

\(=6\)

\(D=\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

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

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

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

\(=4\)

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

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

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

\(=1+\sqrt{5}-\sqrt{5}+1\)

\(=2\)

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

Bài 2: 

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

Ta có: \(P=x^2-2x+2020\)

\(=4+2\sqrt{3}-2\left(\sqrt{3}-1\right)+2020\)

\(=4+2\sqrt{3}-2\sqrt{3}+2+2020\)

=2026

Bài 1: 

\(A=-\dfrac{3}{4}\cdot\sqrt{9-4\sqrt{5}}\cdot\sqrt{\left(-8\right)^2\cdot\left(2+\sqrt{5}\right)^2}\)

\(=\dfrac{-3}{4}\cdot8\cdot\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)\)

=-6

24 tháng 9 2021

1)\(=\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{26^2}=\sqrt{5}-2+26=24-\sqrt{5}\)

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

3) \(=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}\)\(=\left[{}\begin{matrix}1\left(x>1\right)\\-1\left(x< 1\right)\end{matrix}\right.\)

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

24 tháng 9 2021

2. \(\dfrac{x^2-5}{x+\sqrt{5}}=\dfrac{x^2-\left(\sqrt{5}\right)^2}{x+\sqrt{5}}=\dfrac{\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)}{x+\sqrt{5}}=x-\sqrt{5}\)

3. \(\dfrac{\sqrt{x^2-2x+1}}{x-1}=\dfrac{\sqrt{x^2-2.x.1+1^2}}{x-1}=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{|x-1|}{x-1}=\left[{}\begin{matrix}x-1>0\left(x>1\right)\\x-1< 0\left(x< 1\right)\end{matrix}\right.=\left[{}\begin{matrix}=1\\=\dfrac{x+1}{x-1}\end{matrix}\right.\)

7 tháng 6 2023

\(1,P=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{2x}{9-x}\right):\left(\dfrac{\sqrt{x}-1}{x-3\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\left(dkxd:x\ge0,x\ne9\right)\)

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

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

\(=\dfrac{x-3\sqrt{x}-2x}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\sqrt{x}-1-2\sqrt{x}+6}\)

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

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

\(=-\dfrac{x}{5-\sqrt{x}}\)

\(2,x=\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}\)

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

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

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

\(x=4\Rightarrow P=-\dfrac{4}{5-\sqrt{4}}=\dfrac{-4}{5-2}=-\dfrac{4}{3}\)

7 tháng 6 2023

cảm ơn bạn nha!

9 tháng 11 2021

\(a,=27-5\sqrt{3x}\\ b,=3\sqrt{2x}-10\sqrt{2x}+21\sqrt{2x}+28=14\sqrt{2x}+28\)

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.

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

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

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