ĐK: $x\ge 0$
$1=\dfrac{4}{\sqrt x-2}\\\Leftrightarrow \dfrac{\sqrt x-2}{\sqrt x-2}=\dfrac{4}{\sqrt x-2}\\\Rightarrow \sqrt x-2=4\\\Rightarrow \sqrt x=6\\\Leftrightarrow x=36(TM)$
Vậy x=36
ĐK: $x\ge 0$
$1=\dfrac{4}{\sqrt x-2}\\\Leftrightarrow \dfrac{\sqrt x-2}{\sqrt x-2}=\dfrac{4}{\sqrt x-2}\\\Rightarrow \sqrt x-2=4\\\Rightarrow \sqrt x=6\\\Leftrightarrow x=36(TM)$
Vậy x=36
1.cho biểu thức A=\(\dfrac{\sqrt{x}+2}{\sqrt{x}+3}-\dfrac{5}{x+\sqrt{x}-6}-\dfrac{1}{\sqrt{x}-2}\)với(x≥0;x≠4)
a)rút gọn A
b)tính A khi x=6+4\(\sqrt{2}\)
2.cho biểu thức P=\(\left(\dfrac{4\sqrt{x}}{\sqrt{x}+2}-\dfrac{8x}{x-4}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-2}+3\right)\)với x≥0;x≠1;x≠4
a)rút gọn P
b)tìm x để P=-4
rút gọn
C=\(\left(\dfrac{1}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}\right)\div\dfrac{\sqrt{x}}{x-4}vớix>0,x\ne4\)
D=\(\dfrac{8+x\left(1+\sqrt{x-2\sqrt{x+1}}\right)}{\left(x-4\right)\left(x-2\sqrt{x}+4\right)}+\dfrac{x-3\sqrt{x}}{2\left(x-\sqrt{x}-6\right)}vớix>1,x\ne4,x\ne9\)
lm nhanhgiups mk nhé!Mk đang cần gấp!
Bài 1: Cho biểu thức A= \(\dfrac{x-4}{\sqrt{x}+2}\) B= \(\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\) - \(\dfrac{\sqrt{x}-1}{2-\sqrt{x}}\)-\(\dfrac{9-x}{4-x}\) (x ≥ 0, x ≠ 4 )
a) Tính A khi x = \(\dfrac{1}{4}\)
b) Rút gọn B
c) Tìm các giá trị x nguyên sao cho A.B ≤ 2
(mink đag cần gấp)
\(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{3}{\sqrt{x}+2}+\dfrac{12}{x-4}\)
\(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{\sqrt{x}-21}{9-x}\dfrac{1}{\sqrt{x}+3}\)
\(C=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\)
\(D=\dfrac{1}{\sqrt{x}+3}-\dfrac{\sqrt{x}}{3-\sqrt{x}}+\dfrac{2\sqrt{x}+12}{x-9}\)
\(N=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{6}{x-1}\)
\(M=\dfrac{3}{\sqrt{x}-3}+\dfrac{2}{\sqrt{x}+3}+\dfrac{x-5\sqrt{x}-3}{x-9}\)
\(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+2}-\dfrac{x+4}{x-4}\right):\left(\dfrac{2\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
1) \(\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}\)
2)\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}+\dfrac{2+5\sqrt{x}}{4-x}\)
\(\left(\dfrac{x-4}{\sqrt{x}+2}+\dfrac{9}{\sqrt{x}+4}\right).\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}-1\right)\)
RÚT GỌN
a.\(\dfrac{1}{\sqrt{x+2\sqrt{x-1}}}+\dfrac{1}{\sqrt{x-2\sqrt{x-1}}}\)
b. \(\dfrac{1}{2\sqrt{3}+\sqrt{9+4\sqrt{5}}}-\dfrac{1}{2\sqrt{3}-\sqrt{9-4\sqrt{5}}}\)
Thu gọn biểu thức
A=\(\sqrt{\dfrac{3\sqrt{3}}{2\sqrt{3}+1}}-\sqrt{\dfrac{\sqrt{3}+4}{5-2\sqrt{3}}}\)
B=\(\dfrac{x\sqrt{x}-2x+28}{x-3\sqrt{x}-4}-\dfrac{\sqrt{x}-4}{\sqrt{x}+1}+\dfrac{\sqrt{x}+8}{4-\sqrt{x}}\left(x\ge0,x\ne16\right)\)
Cho \(P=\dfrac{3x-2\sqrt{x}-4}{x+\sqrt{x}-2}-\dfrac{\sqrt{x}+1}{\sqrt{x}+2}-\dfrac{2\sqrt{x}+2}{\sqrt{x}-1}\)
a, Rút gọn P.
b, Tính P khi \(x=4+2\sqrt{3}\)
c, Tìm xϵZ để PϵZ