=>căn(x-3)=1/2
=>x-3=1/4
hay x=3,25
\(\dfrac{2}{\sqrt{x-3}}=4\)
\(\Rightarrow4\sqrt{x-3}=2\)
\(\Rightarrow\sqrt{16}\sqrt{x-3}=2\)
\(\Rightarrow\sqrt{16x-48}=2\)
\(\Rightarrow16x-48=4\)
\(\Rightarrow x=\dfrac{13}{4}\)
=>căn(x-3)=1/2
=>x-3=1/4
hay x=3,25
\(\dfrac{2}{\sqrt{x-3}}=4\)
\(\Rightarrow4\sqrt{x-3}=2\)
\(\Rightarrow\sqrt{16}\sqrt{x-3}=2\)
\(\Rightarrow\sqrt{16x-48}=2\)
\(\Rightarrow16x-48=4\)
\(\Rightarrow x=\dfrac{13}{4}\)
Tính giá trị của P = \(\left(\dfrac{\sqrt{x-1}}{3+\sqrt{x-1}}+\dfrac{x+8}{10-x}\right):\left(\dfrac{3\sqrt{x-1}+1}{x-3\sqrt{x-1}-1}-\dfrac{1}{\sqrt{x-1}}\right)\)khi x=\(\sqrt[4]{\dfrac{3+2\sqrt{2}}{3-2\sqrt{2}}}-\sqrt[4]{\dfrac{3-2\sqrt{2}}{3+2\sqrt{2}}}\)
\(\sqrt{\dfrac{x+2}{4}}+\sqrt{25x+50}-2\sqrt{x+2}=14\) ; \(\sqrt{2x+3}=x\) ; \(\sqrt{25x^2+20x+4}=1\) ; \(\sqrt{\dfrac{x+1}{2x-1}}=2\) ; \(\dfrac{\sqrt{x-2}}{\sqrt{3x+1}}=6\)
Tìm x
P=\(\dfrac{2\sqrt{x}-3}{\sqrt{x}-4}-\dfrac{\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{2-3\sqrt{x}}{x-3\sqrt{x}-4}\)( với x >=0;x khác 16) .Rút gọn biểu thức P
( \(\dfrac{3\sqrt{x}+6}{x-4}\) + \(\dfrac{\sqrt{x}}{\sqrt{x}-2}\) ) : \(\dfrac{x-9}{\sqrt{x}-3}\)
rút gọn biểu thức
\(P=\dfrac{10\sqrt{x}}{x+3\sqrt{x}-4}-\dfrac{2\sqrt{x}-3}{\sqrt{x}+4}+\dfrac{\sqrt{x}+1}{1-\sqrt{x}}\) rút gọn
Tìm x biết :
a) \(\sqrt{9x}+\sqrt{x}=12\)
b) \(\dfrac{\sqrt{x}+3}{4}=\dfrac{\sqrt{x}}{3}\)
c) \(\dfrac{5\sqrt{x}-x}{\sqrt{x}}=2\)
Rút gọn : a) \(\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}\)
b)\(\dfrac{x+4y-4\sqrt{xy}}{\sqrt{x}-2\sqrt{y}}+\dfrac{y+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\left(x\ge0;y\ge0;x\ne4y\right)\)
c)\(\dfrac{x+4\sqrt{x}+4}{\sqrt{x}+2}+\dfrac{4-x}{\sqrt{x}-2}\left(x\ge0;x\ne4\right)\)
d)\(\dfrac{9-x}{\sqrt{3x}+3}-\dfrac{9-6\sqrt{x}+x}{\sqrt{x}-3}\)
e)\(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}-\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}\)
g)\(\left(2-\dfrac{a-3\sqrt{a}}{\sqrt{a}-3}\right)\left(2-\dfrac{5\sqrt{a}-\sqrt{ab}}{\sqrt{b}-5}\right)với\) a, b \(\ge\)0 , a \(\ne\)9; b\(\ne\)25
Mọi người giúp tớ với , cảm ơn nhiều nhiều ạ !!
\(\sqrt{x^2-\dfrac{1}{4}+\sqrt{x^2+x+\dfrac{1}{4}}=\dfrac{1}{2}\left(2x^3+x^2+2x+1\right)}\)
Cho x = \(5+4\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\)
Tính A = \(\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)
cho biểu thức K= \( \left(\dfrac{\sqrt{x}+2}{3\sqrt{x}}+\dfrac{2}{\sqrt{x}+1}-3\right):\dfrac{2-4\sqrt{x}}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1-x}{3\sqrt{x}}\)
1)rút gọn K với x> 0; x \(\ne\) \(\dfrac{1}{4}\)
2)tính giá trị của K tại x = \(\dfrac{1}{4}\)
3 ) tìm x để K < l
4) tìm x để K nguyên