Ta có: \(\dfrac{1}{\sqrt{\left(3-2x\right)^2}}=\dfrac{1}{3-2x}\)
Có nghĩa khi: \(3-2x>0\)
\(\Leftrightarrow2x>3\)
\(\Leftrightarrow x>\dfrac{3}{2}\)
Tìm x
a)\(\sqrt{x-1}=2\left(x\ge1\right)\)
b)\(\sqrt{3-x}=4\left(x\le3\right)\)
c)\(2.\sqrt{3-2x}=\dfrac{1}{2}\left(x\le\dfrac{3}{2}\right)\)
d)\(4-\sqrt{x-1}=\dfrac{1}{2}\left(x\ge1\right)\)
e)\(\sqrt{x-1}-3=1\)
f)\(\dfrac{1}{2}-2.\sqrt{x+2}=\dfrac{1}{4}\)
Tính DKXD của các căn bậc thức sau:
a)\(\sqrt{2x-4}\)
b)\(\sqrt{\dfrac{3}{-2x+1}}\)
c)\(\sqrt{\dfrac{-3x+5}{-4}}\)
d)\(\sqrt{-5\left(-2x+6\right)}\)
e)\(\sqrt{\left(x^2+2\right)\left(x-3\right)}\)
f)\(\sqrt{\dfrac{x^2+5}{-x+2}}\)
\(x=\sqrt{\dfrac{1}{2\sqrt{3}-2}-\dfrac{3}{2\left(\sqrt{3}+1\right)}}\)
Tính A = \(\dfrac{4\left(x+1\right).x^{2013}-2x^{2012}+2x+1}{2x^2+3x}\)
Bài 1: Rút gọn: A= \(\left(\dfrac{x}{x^2-49}-\dfrac{x-7}{x^2+7x}\right):\dfrac{2x-7}{x^2+7x}+\dfrac{x}{7-x}\)
Bài 2: Rút gọn: B=\(\left[\dfrac{3}{x+1}+\left(\dfrac{3}{x}-\dfrac{x}{x^2+2x+1}\right):\dfrac{2x^2+3x}{x+1}\right]:\dfrac{1+3x}{x^2+x}\)
Bài 3: Rút gọn D=\(\left(\sqrt{a}+\dfrac{b-\sqrt{ab}}{\sqrt{a}+b}\right):\left(\dfrac{a}{\sqrt{ab}+b}+\dfrac{b}{\sqrt{ab}-a}-\dfrac{a+b}{\sqrt{ab}}\right)\)
giúp mình câu rút gọn với ạ :3
\(B=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{1+x\sqrt{x}}{1+\sqrt{x}}-\sqrt{x}\right)+\dfrac{2-2\sqrt{x}}{\sqrt{x}}\)
bài 1: rút gọn
A= \(\left(\dfrac{\sqrt{x}-1}{3\sqrt{x}1}-\dfrac{1}{3\sqrt{x}+1}+\dfrac{8\sqrt{x}}{9x-1}\right):\left(1-\dfrac{3\sqrt{x}-2}{3\sqrt{x}+1}\right)\) X>=0; x#0
B= \(\left(\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right):\left(\dfrac{2}{x^2-2x+1}\right)\) x>=0; x\(\ne\)1
bài 2:
\(\dfrac{1}{2x-3\sqrt{x}+2}\) x>=0
Tìm GTNN của A
Tính giá trị của P tại x=\(2\sqrt{2}\)
\(P=\dfrac{\sqrt{\sqrt{\dfrac{x-1}{x+1}+\sqrt{\dfrac{x+1}{x-1}}}-2\left(2x+\sqrt{x^2+1}\right)}}{\sqrt{\left(x+1\right)^3}+\sqrt{\left(x-1\right)^3}}\)
Cho các số thực x, y, z thỏa mãn \(7\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)=6\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}\right)=2016\).
Tìm max: \(P=\dfrac{1}{\sqrt{3\left(2x^2+y^2\right)}}+\dfrac{1}{\sqrt{3\left(2y^2+z^2\right)}}+\dfrac{1}{\sqrt{3\left(2z^2+x^2\right)}}\)
Giải phương trình sau:
a) \(\sqrt{4x+20}-3\sqrt{5+x}+\dfrac{4}{3}\sqrt{9x+45}=6\)
b) \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
c) \(2x-x^2+\sqrt{6x^2-12x+7}=0\)
d) \(\left(x+1\right)\left(x+4\right)-3\sqrt{x^2+5x+2}=6\)
rút gọn
a, \(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}.\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
b,\(\left(\dfrac{\sqrt{x}-4}{x-2\sqrt{x}}-\dfrac{3}{2-\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\)\
c,\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
