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a: \(A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{2\left(\sqrt{x}-3\right)}\)
\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{2\left(\sqrt{x}-3\right)}{\sqrt{x}+1}\)
\(=\dfrac{-6}{\sqrt{x}+3}\)
b: Để A<-1/2 thì A+1/2<0
\(\Leftrightarrow-\dfrac{6}{\sqrt{x}+3}+\dfrac{1}{2}< 0\)
\(\Leftrightarrow-12+\sqrt{x}+3< 0\)
=>0<x<81 và x<>9
a: \(A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x+3}{x-9}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
\(=\dfrac{-3\sqrt{x}+3}{x-9}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)
\(=\dfrac{-3\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)
b: Khi \(x=4-2\sqrt{3}\) vào A, ta được:
\(A=\dfrac{-3\left(\sqrt{3}-1\right)+3}{\left(\sqrt{3}-1+3\right)\left(\sqrt{3}-1+1\right)}\)
\(=\dfrac{-3\sqrt{3}+6}{\sqrt{3}\cdot\left(\sqrt{3}+2\right)}=\dfrac{-3+2\sqrt{3}}{2+\sqrt{3}}\)
1. \(\left(1+\sqrt{2}+\sqrt{3}\right)\left(1+\sqrt{2}-\sqrt{3}\right)\)
\(=\left(1+\sqrt{2}\right)^2-\sqrt{3}^2\)
\(=1+2\sqrt{2}+2-3\)
\(=2\sqrt{2}\)
3. \(A=\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{\sqrt{x}+1}\right)\cdot\left(1+\dfrac{1}{\sqrt{x}}\right)\)(1)
ĐKXĐ \(x>0,x\ne1\)
pt (1) <=> \(\left(\dfrac{\sqrt{x}+1+\sqrt{x}-1}{\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\right)\cdot\left(\dfrac{\sqrt{x}+1}{\sqrt{x}}\right)\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right)\cdot\left(\sqrt{x}+1+\sqrt{x}-1\right)}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)\cdot\left(\sqrt{x}+1\right)}\)
\(\Leftrightarrow\dfrac{2\sqrt{x}}{x-\sqrt{x}}\)
\(\Leftrightarrow\dfrac{\sqrt{x}\cdot2}{\sqrt{x}\cdot\left(\sqrt{x}-1\right)}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}\)
b) Để \(\sqrt{A}>A\Leftrightarrow\sqrt{\dfrac{2}{\sqrt{x}-1}}>\dfrac{2}{\sqrt{x}-1}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}>\dfrac{4}{x-2\sqrt{x}+1}\)
\(\Leftrightarrow\dfrac{2}{\sqrt{x}-1}-\dfrac{4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\cdot\left(\sqrt{x}-1\right)-4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{2}-2-4}{x-2\sqrt{x}+1}>0\)
\(\Leftrightarrow\dfrac{2\sqrt{2}-6}{x-2\sqrt{x}+1}>0\)
Vì \(2\sqrt{2}-6< 0\Rightarrow x-2\sqrt{x}+1< 0\)
mà \(x-2\sqrt{x}+1=\left(\sqrt{x}-1\right)^2\ge0\forall x\)
Vậy không có giá trị nào của x thỏa mãn \(\sqrt{A}>A\)
(P/s Đề câu b bị sai hay sao vậy, chả có số nào mà \(\sqrt{A}>A\) cả, check lại đề giùm với nhé)
a, Rút gọn P
\(\dfrac{3}{\sqrt{x}+3}-\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2-\sqrt{x}}{\sqrt{x}+3}\right)\)
\(\Leftrightarrow\left(1-\dfrac{\sqrt{x}\left(\sqrt{x-3}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+3\right)}\right):\left(\dfrac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{x+3\sqrt{x}-2\sqrt{x}-6}-\dfrac{\sqrt{x}-3}{-\left(\sqrt{x}-2\right)\sqrt{x}+3}\right)\)
\(\Leftrightarrow\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+3}\right):\left(\dfrac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{\sqrt{x}\left(\sqrt{x}+3\right)}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2-\sqrt{x}}{\sqrt{x}+3}\right)\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}+3}:\left(\dfrac{3-\sqrt{x}}{\sqrt{x}-2}+\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2-\sqrt{x}}{\sqrt{x}+3}\right)\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}+3}:\dfrac{\left(\sqrt{x}+3\right).\left(3-\sqrt{x}\right).\left(x+\sqrt{3}\right).\left(\sqrt{x}-3\right)-\left(\sqrt{x}-2\right).\left(2-\sqrt{x}\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}+3}:\dfrac{\left(3+\sqrt{x}\right).\left(3-\sqrt{x}\right)+x-9-\left(2\sqrt{x}-x-4+2\sqrt{x}\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}+3}:\dfrac{9-x+x-9-\left(4\sqrt{x}-x-4\right)}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}+3}:\dfrac{-4\sqrt{x}+x+4}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}+3}:\dfrac{x-4\sqrt{x}+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}+3}:\dfrac{\left(\sqrt[]{x}-2\right)^2}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+3\right)}\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}+3}:\dfrac{\sqrt{x}-2}{\sqrt{x}+3}\)
\(\Leftrightarrow\dfrac{3}{\sqrt{x}+3}.\dfrac{\sqrt{x}+3}{\sqrt{x}-2}\)
\(\Leftrightarrow3.\dfrac{1}{\sqrt{x}-2}\)
\(\Leftrightarrow\)\(\dfrac{3}{\sqrt{x}-2}\)
ĐK : \(x\ge0;x\ne9\)
a) P = \(\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\)
= \(\dfrac{3-3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)
= \(\dfrac{-3}{\sqrt{x}+3}\)
a: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-\dfrac{x+9}{x-9}\right):\left(\dfrac{3\sqrt{x}+1-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\right)\)
\(=\dfrac{x-3\sqrt{x}-x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{2\sqrt{x}+4}\)
\(=\dfrac{-3\left(\sqrt{x}+3\right)}{\sqrt{x}+3}\cdot\dfrac{1}{2\sqrt{x}+4}=\dfrac{-3}{2\sqrt{x}+4}\)
b: Để A<-1 thì A+1<0
\(\Leftrightarrow\dfrac{-3+2\sqrt{x}+4}{2\sqrt{x}+4}< 0\)
\(\Leftrightarrow\dfrac{2\sqrt{x}+1}{2\sqrt{x}+4}< 0\)(vô lý)
Vậy: \(x\in\varnothing\)
a: \(P=\dfrac{2x-18-2x-6\sqrt{x}+5\sqrt{x}+20}{x-9}:\dfrac{\sqrt{x}+3-5}{\sqrt{x}+3}\)
\(=\dfrac{-\sqrt{x}+2}{x-9}\cdot\dfrac{\sqrt{x}+3}{\sqrt{x}-2}=\dfrac{-1}{\sqrt{x}-3}\)
b: Để P<-1/2 thì P+1/2<0
=>\(\dfrac{-1}{\sqrt{x}-3}+\dfrac{1}{2}< 0\)
=>\(\dfrac{-2+\sqrt{x}-3}{2\left(\sqrt{x}-3\right)}< 0\)
=>\(\dfrac{\sqrt{x}-5}{2\left(\sqrt{x}-3\right)}< 0\)
=>3<căn x<5
=>9<x<25
c: \(Q=\dfrac{-1}{\sqrt{x}-3}\cdot\sqrt{x}\left(\sqrt{x}-3\right)\left(\sqrt{x}-5\right)=-x+5\sqrt{x}\)
\(=-\left(x-5\sqrt{x}+\dfrac{25}{4}-\dfrac{25}{4}\right)=-\left(\sqrt{x}-\dfrac{5}{2}\right)^2+\dfrac{25}{4}< =\dfrac{25}{4}\)
Dấu = xảy ra khi x=25/4
a: \(=\dfrac{\sqrt{x}+1-\sqrt{x}+1-2\sqrt{x}}{x-1}=\dfrac{-2\left(\sqrt{x}-1\right)}{x-1}=\dfrac{-2}{\sqrt{x}+1}\)
b: \(=\dfrac{\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}+\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}}{1-xy}:\left(\dfrac{x+y+2xy+1-xy}{1-xy}\right)\)
\(=\dfrac{2\sqrt{x}+2y\sqrt{x}}{1-xy}\cdot\dfrac{1-xy}{x+y+xy+1}\)
\(=\dfrac{2\sqrt{x}\left(y+1\right)}{\left(y+1\right)\left(x+1\right)}=\dfrac{2\sqrt{x}}{x+1}\)
c: \(=\dfrac{3x+3\sqrt{x}-9+x+2\sqrt{x}-3-x+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{3x+5\sqrt{x}-8}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}=\dfrac{3\sqrt{x}+8}{\sqrt{x}+2}\)
a: ĐKXĐ: x>=0; \(x\notin\left\{4;9\right\}\)
b: \(A=\left(\dfrac{\sqrt{x}}{\sqrt{x}+3}-1\right):\left(\dfrac{9-x+x-9-x+4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}\right)\)
\(=\dfrac{-3}{\sqrt{x}+3}\cdot\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}{-\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{3}{\sqrt{x}+2}\)
Thay \(x=3-2\sqrt{2}\) vào A, ta được:
\(A=\dfrac{3}{\sqrt{2}-1+2}=\dfrac{3}{\sqrt{2}+1}=3\sqrt{2}-3\)
c: Để A<1 thì A-1<0
\(\Leftrightarrow\dfrac{3-\sqrt{x}-2}{\sqrt{x}+2}< 0\)
\(\Leftrightarrow1-\sqrt{x}< 0\)
hay 0<x<1