a: \(\Leftrightarrow\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)-\sqrt{x}\left(\sqrt{x}+1\right)=0\)
=>x-2 căn x+1=0
=>x=1
b: \(\Leftrightarrow9\sqrt{x}-12=4\sqrt{x}-2\)
=>5 căn x=10
=>x=4
a: \(\Leftrightarrow\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)-\sqrt{x}\left(\sqrt{x}+1\right)=0\)
=>x-2 căn x+1=0
=>x=1
b: \(\Leftrightarrow9\sqrt{x}-12=4\sqrt{x}-2\)
=>5 căn x=10
=>x=4
a : \(\dfrac{x}{x-4}+\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\)với a ≥ 0 x ≠ 4
b : \(\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right).\dfrac{\sqrt{x}}{x+\sqrt{x}}\)
c : \(\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{x-1}\)
d : \(\left[\dfrac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\dfrac{a\sqrt{a}}{a-1}\right]:\left(\dfrac{1}{\sqrt{a}-1}+\dfrac{1}{\sqrt{a}+1}\right)\)
A=\(\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right):\dfrac{x-2\sqrt{x}+1}{x-1}\) ( ĐKXĐ x>0;x≠4)
P=\(\left(\dfrac{3\sqrt{x}}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{2-\sqrt{x}}+\dfrac{8\sqrt{x}}{x-4}\right):\left(2-\dfrac{2\sqrt{x}+3}{\sqrt{x}+2}\right)\)
E=\(\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+3}+\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{x+\sqrt{x}-6}\right)\) (ĐKXĐ x≥0;x≠4)
RÚT GỌN GIÚP MÌNH VỚI A
a : \(\dfrac{3}{\sqrt{x}-5}+\dfrac{20-2\sqrt{x}}{x-25}\)với x ≥ 0 x ≠ 25
b : \(\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{2\sqrt{x}-2}{x-9}\)với x ≥ 0 x ≠ 9
c : \(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}+\dfrac{5\sqrt{x}-2}{x-4}\)với x ≥ 0 x ≠ 4
d : \(\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)với ≥ 0 x ≠ 1
Rút gọn:
A= \(\dfrac{10\sqrt{x}}{x+3\sqrt{x}-4}\) - \(\dfrac{2\sqrt{x}-3}{\sqrt{x}+4}\) + \(\dfrac{\sqrt{x}+1}{1-\sqrt{x}}\) ( với x ≥ 0; x ≠ 1)
(\(\dfrac{\sqrt{x}}{\sqrt{x}-2}\) + \(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\)-\(\dfrac{3\sqrt{x}+2}{x-4}\) ) : \(\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\) ( với x ≥ 0; x ≠ 4)
RÚT GỌN Ạ
\(choP=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{3-\sqrt{x}}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right)..a,tìmdkxd..b,rút.gọn.P..c,tính.P.khi.x=4-2\sqrt{3}..d,tìm.x.để.P>0..e,tìm.x.để.P< \dfrac{1}{2}\)
( \(\dfrac{\sqrt{x}}{\sqrt{x}-2}\) + \(\dfrac{\sqrt{x}-1}{\sqrt{x}+2}\) - \(\dfrac{3\sqrt{x}+2}{x-4}\) ) : \(\dfrac{\sqrt{x}-2}{x-4}\) ( với x ≥ 0; x ≠ 4)
RÚT GỌN Ạ
a \(\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\)
b \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}=4\)
c \(\dfrac{1}{3}\sqrt{x-2}-\dfrac{2}{3}\sqrt{9x-18}+6\sqrt{\dfrac{x-2}{81}=-4}\)
d \(\sqrt{9x+27}+4\sqrt{x+3}-\dfrac{3}{4}\sqrt{16x+48}=0\)
P=(\(\dfrac{x-2}{x+2\sqrt{x}}\)+\(\dfrac{1}{\sqrt{x+2}}\)).\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\) với x >0 và x≠1
A=(\(\dfrac{2}{\sqrt{x}-2}\)+\(\dfrac{3}{2\sqrt{x+1}}\)-\(\dfrac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\)):\(\dfrac{2\sqrt{x}+3}{5x-10\sqrt{x}}\) với x > 0 và x≠4
A=\(\dfrac{x+1-2\sqrt{x}}{\sqrt{x-1}}\)+\(\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\) với x≥0 và x≠1
V=(\(\dfrac{1}{\sqrt{x}+2}\)+\(\dfrac{1}{\sqrt{x}-2}\))\(\dfrac{\sqrt{x}+2}{\sqrt{x}}\) với x>0 và x≠4
A=(\(\dfrac{1}{x-1}\)+\(\dfrac{3\sqrt{x}+5}{x\sqrt{x}-x-\sqrt{x}+1}\))(\(\dfrac{\left(\sqrt{x}+1\right)^2}{4\sqrt{x}}\)-1)
P=\(\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}\)+\(\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}\)+\(\dfrac{2\sqrt{x}+3}{3+\sqrt{x}}\)
MỌI NGƯỜI GIÚP ĐỠ MÌNH RÚT GỌN MẤY BIỂU THỨC NÀY VỚI Ạ . EM XIN CẢM ƠN
Rút gọn biểu thức sau :
\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x}-4}\right):\dfrac{x+16}{\sqrt{x}+2}\) ( x ≥ 0 ; x ≠ 16 )
\(P=\dfrac{x}{x-1}+\dfrac{3}{x+1}-\dfrac{6x-4}{x^2-1}\)
\(A=\dfrac{1}{x+\sqrt{x}}+\dfrac{2\sqrt{x}}{x-1}-\dfrac{1}{x-\sqrt{x}}\)
\(B=\left(2-\sqrt{3}\right)\sqrt{26+15\sqrt{3}-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}}\)
\(A=\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}\)