đkxđ: \(\left\{\begin{matrix} 2x-3>=0\\x-1>=0 \end{matrix}\right.\) <=>\(\left\{\begin{matrix} x>=3/2\\x>=1 \end{matrix}\right.\) <=> x>=3/2
hoặc \(\left\{\begin{matrix}2x-3<0\\x-1<0 \end{matrix}\right.\) <=>\(\left\{\begin{matrix} x<3/2\\x<1 \end{matrix}\right.\) <=> x<1
Ta có: \(\sqrt{\dfrac{2x-3}{x-1}}\) =2
<=> \(\dfrac{2x-3}{x-1}\) =4
<=> 2x-3=4x-4
<=> -2x=-1
<=> x=1/2 (T/M)
Tìm x, biết:
a) \(\sqrt{x^2-2x+1}=2\)
b)\(\sqrt{x^2-1}=x\)
c) \(\sqrt{4x-20}+3\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
d) \(x-5\sqrt{x-2}=-2\)
e) \(2x-3\sqrt{2x-1}-5=0\)
Tìm \(x\) thỏa mãn điều kiện
a) \(\sqrt{\dfrac{2x-3}{x-1}}=2\)
b) \(\dfrac{\sqrt{2x-3}}{\sqrt{x-1}}=2\)
c) \(\sqrt{\dfrac{4x+3}{x+1}}=3\)
d) \(\dfrac{\sqrt{4x+3}}{\sqrt{x+1}}=3\)
1. Tìm x để bt có nghĩa
A=\(\dfrac{\sqrt{2x+3}}{\sqrt{x-3}}\)
B=\(\sqrt{\dfrac{2x+3}{x-3}}\)
C=\(\sqrt{-\dfrac{5}{x+2}}\)
D=\(\sqrt{-x}+\dfrac{1}{x+3}\)
2. Rút gọn bt
A=\(\sqrt{\dfrac{a+\sqrt{a^2-1}}{2}}-\sqrt{\dfrac{a-\sqrt{a^2-1}}{2}};\left(a>1\right)\)
B=\(\sqrt{\dfrac{a+\sqrt{a^2-1}}{2}}-\sqrt{\dfrac{a-\sqrt{a^2-b}}{2}};\left(a\ge\sqrt{b};b\ge0\right)\)
C=\(\left(1+\dfrac{a+\sqrt{a}}{a+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}+1}\right);\left(a\ge0,a\ne1\right)\)
D=\(\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}};\left(x>0\right)\)
Tính giá trị của biểu thức
A=\(\dfrac{1+2x}{1+\sqrt{1+2x}}+\dfrac{1-2x}{1-\sqrt{1-2x}}\) với x=\(\dfrac{\sqrt{3}}{4}\)
B=\(\dfrac{2b\sqrt{x^2-1}}{x-\sqrt{x^2-1}}\) với x=\(\dfrac{1}{2}\left(\sqrt{\dfrac{a}{b}}+\sqrt{\dfrac{b}{a}}\right)\) và a>0,b>0
C=\(\dfrac{2a\sqrt{1+x^2}}{\sqrt{1+x^2}-x}\) với x=\(\dfrac{1}{2}\left(\sqrt{\dfrac{1-a}{a}}-\sqrt{\dfrac{a}{1-a}}\right)\) và 0<a<1
Rút gọn các biểu thức:
a) \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}\) ( a <0 ; b # 0 )
b) \(\sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\) ( x lớn hơn hoặc = 0)
c) \(\sqrt{\dfrac{\left(x-2\right)^2}{\left(3-x\right)^2}}+\dfrac{x^2-1}{x-3}\) ( x<3 tại x = 0,5)
d) \(\dfrac{x-1}{\sqrt{y}-1}.\sqrt{\dfrac{\left(y-2\sqrt{y}+1^2\right)}{\left(x-1\right)^4}}\) ( x # 1; y >= 0, y #1)
e) \(4x-\sqrt{8}+\dfrac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\) ( x > -2 tại x = -\(\sqrt{2}\))
Giải phương trình sau:
a)\(\sqrt{3}.x-\sqrt{12}=0\)
b)\(\sqrt{2}.x+\sqrt{2}=\sqrt{8}+\sqrt{18}\)
c)\(\sqrt{5}.x^2-\sqrt{20}=0\)
d)\(\sqrt{x^2+6x+9}=3x+6\)
e)\(\sqrt{x^2-4x+4}-2x+5=0\)
f)\(\sqrt{\dfrac{2x-3}{x-1}=2}\)
g) \(\dfrac{\sqrt{2x-3}}{\sqrt{x-1}}=2\\\)
rút gọn biểu thức
a, \(\sqrt{\dfrac{x^4+x^2}{x^2-2x+1}}\)
b, \(\dfrac{\sqrt{9x^3+3x^2-8x-4}}{\sqrt{x-1}}\)
Rút gọn các biểu thức sau:
A = \(\dfrac{3}{2\left(2x-1\right)}\sqrt{8\left(4x^2-2x+1\right)x^4}\)
B = \(\dfrac{a-b}{b^2}\sqrt{\dfrac{a^2b^4}{a^2-2ab+b^2}}\)
Câu 1:
Q= \(\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}\right):\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1}\right)\left(a\ge0,a\ne4,a\ne1\right)\)
a) Rút gon Q
b) Tìm giá trị của a để Q dương
Caau2:
B= \(\left(\dfrac{2x+1}{\sqrt{x^3}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+\sqrt{x^3}}{1+\sqrt{x}}-\sqrt{x}\right)\left(x\ge0,x\ne1\right)\)
a) rút gon B
Rút gọn biểu thức
E = \(\dfrac{x+2\sqrt{x}+1}{\sqrt{x}+1}+\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\)
F = \(\dfrac{2\sqrt{x}}{\sqrt{x}+3}-\dfrac{\sqrt{x}+1}{3-\sqrt{x}}-\dfrac{3-11\sqrt{x}}{x-9}\)
G = \(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{4\sqrt{x}-4}{4-x}\)
