1: ĐKXĐ: \(x\in R\)
2: ĐKXĐ: 12x+5>=0
=>>x=-5/12
3 ĐKXĐ: (3x+2)(x-1)>=0
=>x>=1 hoặc x<=-2/3
1: ĐKXĐ: \(x\in R\)
2: ĐKXĐ: 12x+5>=0
=>>x=-5/12
3 ĐKXĐ: (3x+2)(x-1)>=0
=>x>=1 hoặc x<=-2/3
1.\(\sqrt{-4x^2+25}=x\)
2.\(\sqrt{3x^2-4x+3}=1-2x\)
3. \(\sqrt{4\left(1-x\right)^2}-\sqrt{3}=0\)
4.\(\dfrac{3\sqrt{x+5}}{\sqrt{ }x-1}< 0\)
5. \(\dfrac{3\sqrt{x-5}}{\sqrt{x+1}}\ge0\)
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) \(\left(\sqrt{7}+1\right)\left(2\sqrt{2}-1\right)\left(2\sqrt{14}-1\right)\left(55+12\sqrt{2}-7\sqrt{7}\right)\)
e)\(\left(3\sqrt{2}+1\right)\left(2\sqrt{3}+1\right)\left(6\sqrt{6}+1\right)\left(215-34\sqrt{3}-33\sqrt{2}\right)\)
bài 1 tính
a, \(\sqrt{\left(1-\sqrt{5}\right)^2}+1\)
b, \(\sqrt{3+2\cdot\sqrt{2}}-2\)
c, \(\sqrt{b^2-b+\dfrac{1}{4}}-\left(2b-\dfrac{1}{2}\right)\left(vsb\ge\dfrac{1}{2}\right)\)
d, \(\sqrt{7+2\cdot\sqrt{10}}\)
e. \(\sqrt{11-4\cdot\sqrt{7}}\)
f, \(\sqrt{x-2\cdot\sqrt{x-1}}\)
g, \(3x+\sqrt{x^2-2x+1}\)
h, \(\sqrt{y+2\sqrt{y^2-2y+1}}\) (voi y>1)
i, \(\sqrt{17-2\sqrt{32}}+\sqrt{17+2\sqrt{32}}\)
k, \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
Bài 2: Tìm sự xác định của các biểu thức chứa căn
1> \(\sqrt{6x+1}\)
2> \(\sqrt{\dfrac{-3}{2+x}}\)
3> \(\sqrt{-8x}\)
4> \(\sqrt{4-5x}\)
5> \(\sqrt{\left(x+5\right)^2}\)
6> \(\sqrt{\dfrac{\sqrt{6}-4}{m+2}}\)
7> \(\sqrt{\left(\sqrt{3}-x\right)^2}\)
8> \(\dfrac{16x-1}{\sqrt{x}-7}\)
9> \(\sqrt{x^2+2x+1}\)
10> \(\sqrt{2x+5}\)
11> \(\sqrt{-12x+5}\)
12> \(\dfrac{3}{\sqrt{12x-1}}\)
13> 2 - \(4\sqrt{5x+8}\)
14> \(\sqrt{x^2+3}\)
15> \(\sqrt{\dfrac{5}{x^2}}\)
16> \(\sqrt{\dfrac{x+3}{7-x}}\)
17> \(\sqrt{x-x^2}\)
a:\(\dfrac{b}{\left(a-4\right)^2}.\sqrt{\dfrac{\left(a-4\right)^4}{b^2}}\left(b>0;a\ne4\right)\)
b:\(\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\left(x\ge0;y\ge0;x\ne0\right)\)
c:\(\dfrac{a}{\left(b-2\right)^2}.\sqrt{\dfrac{\left(b-2\right)^4}{a^2}\left(a>0;b\ne2\right)}\)
d:\(\dfrac{x}{\left(y-3\right)^2}.\sqrt{\dfrac{\left(y-3\right)^2}{x^2}\left(x>0;y\ne3\right)}\)
e:2x +\(\dfrac{\sqrt{1-6x+9x^2}}{3x-1}\)
Rút gọn
a.\(\dfrac{\sqrt{7}-5}{2}-\dfrac{6}{\sqrt{7}-2}+\dfrac{1}{3+\sqrt{7}}+\dfrac{3}{5+2\sqrt{7}}\)
b.\(\left(\sqrt{10}+\sqrt{2}\right).\left(6-2\sqrt{5}\right).\sqrt{3+\sqrt{5}}\)
Bài 3: Giari phương trình \(\sqrt{A}\)=B
a> \(\sqrt{3x-1}\)
b> \(\sqrt{-3x+4}\) = 12
c> \(\sqrt{x^2-8x+16}\) = 4
d> \(\sqrt{9\left(x-1\right)}\) = 21
g> \(\sqrt{2-3x}\) = 10
h> \(\sqrt{4x}\) = \(\sqrt{5}\)
i> \(\sqrt{4-5x}\) = 12
p> \(\sqrt{16x}\) = 8
q> \(\sqrt{\left(x-3\right)^2}\) = 3
v> \(\sqrt{\dfrac{-6}{1+x}}\) = 5
w> \(\sqrt{4x-20}-3\)\(\sqrt{\dfrac{x-5}{9}}\) = \(\sqrt{1-x}\)
x> \(\sqrt{4x+8}\) + 2\(\sqrt{x+2}\) - \(\sqrt{9x+18}\) = 1
a'> \(\sqrt{x^2-6x+9}\)+x=11
z> \(\sqrt{16\left(x+1\right)^2}\) - \(\sqrt{9\left(x+1\right)}\) = 4
b'> \(\sqrt{9x+9}\) + \(\sqrt{4x+4}\) = \(\sqrt{x+1}\)
Tính:
a) \(\left(\sqrt{1\dfrac{9}{16}}-\sqrt{\dfrac{9}{16}}\right):5\)
b) \(\left(\sqrt{3}-2\right)^2\left(\sqrt{3}+2\right)^2\)
c) \(\left(11-4\sqrt{3}\right)\left(11+4\sqrt{3}\right)\)
d) \(\left(\sqrt{2}-1\right)^2-\dfrac{3}{2}\sqrt{\left(-2\right)^2}+\dfrac{4\sqrt{2}}{5}+\sqrt{1\dfrac{11}{25}}.\sqrt{2}\)
e) \(\left(1+\sqrt{2021}\right)\sqrt{2022-2\sqrt{2021}}\)
1. A= \(\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{4}{x+2\sqrt{x}}\right):\left(1+\dfrac{1}{\sqrt{x}}\right)=\dfrac{\sqrt{x}-2}{\sqrt{x}+1}\)
Chứng minh: A<1
Giải phương trình:
1) \(x^2-4x-2\sqrt{2x-5}+5=0\)
2)\(x+y+4=2\sqrt{x}+4\sqrt{y-1}\)
3)\(\sqrt{x-2}+\sqrt{y-3}+\sqrt{z-5}=\dfrac{1}{2}\left(x+y+z-7\right)\)
4)\(\sqrt{x-2}+\sqrt{10-x}=x^2-12x+40\)