\(p=\dfrac{\sqrt{x}-2}{x-1}-\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}:\left(\dfrac{2}{x^2-2x+1}\right)\)với x\(\ge0;x\ne1\)
1 .rút gọn p
2.tính giá trị của p khi x=7-\(4\sqrt{3}\)
3 . tính GTLN của p
giúp mình giải bpt vs
\(\dfrac{\left|2x-1\right|-x}{2x}>1;\dfrac{2-\left|x-2\right|}{x^2-1}\ge0;\dfrac{\sqrt{x+4}-2}{4-9x^2}\le0;\dfrac{x^2-2x-3}{\sqrt[3]{3x-1}+\sqrt[3]{4-5x}}\ge0;\)\(3x^2-10x+3\ge0;\left(\sqrt{2}-x\right)\left(x^2-2\right)\left(2x-4\right)< 0;\dfrac{1}{x+9}-\dfrac{1}{x}>\dfrac{1}{2};\dfrac{2}{1-2x}\le\dfrac{3}{x+1}\)
p= \(\left(\dfrac{x+2\sqrt{x}}{x-4}-\dfrac{2\sqrt{x}+1}{2x-3\sqrt{x}-2}-\dfrac{x}{\sqrt{x}-2}\right)\dfrac{x-1}{x\sqrt{x}+1}vớix\ge0;x\ne4\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{2x-4\sqrt{x}+\sqrt{x}-2}-\dfrac{x}{\sqrt{x}-2}\right)\cdot\dfrac{\sqrt{x}-1}{x-\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1-x}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}-1}{x-\sqrt{x}+1}=\dfrac{-\sqrt{x}+1}{\sqrt{x}-2}\)
Rút gọn các biểu thức sau:
\(C=\left(\dfrac{\sqrt{x}+1}{x-4}-\dfrac{\sqrt{x}-1}{x+4\sqrt{x}+4}\right).\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}-2}\)
(với \(x\ge0,x\ne4,x\ne9\))
\(D=\left(\dfrac{\sqrt{x}+2}{x-9}-\dfrac{\sqrt{x}-2}{x+6\sqrt{x}+9}\right).\dfrac{x\sqrt{x}+3x-9\sqrt{x}-27}{\sqrt{x}-2}\)
(với \(x\ge0,x\ne4,x\ne9\))
\(C=\left(\dfrac{\sqrt{x}+1}{x-4}-\dfrac{\sqrt{x}-1}{x+4\sqrt{x}+4}\right).\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8}{\sqrt{x}-2}\)
\(=\left[\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+2\right)^2}\right].\dfrac{x\left(\sqrt{x}+2\right)-4\left(\sqrt{x}+2\right)}{\sqrt{x}-2}\)
\(=\left[\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}-\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}\right].\dfrac{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}{\sqrt{x}-2}\)
\(=\left[\dfrac{x+3\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}-\dfrac{x-3\sqrt{x}+2}{\left(\sqrt{x}+2\right)^2\left(\sqrt{x}-2\right)}\right].\left(\sqrt{x}+2\right)^2\)
\(=\dfrac{6\sqrt{x}}{\sqrt{x}-2}\)
\(C=\left[\dfrac{\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+2\right)^2}\right].\dfrac{\sqrt{x}\left(x-4\right)+2\left(x-4\right)}{\sqrt{x}-2}\) (\(x\ge0,x\ne4,x\ne9\))
\(C=\left[\dfrac{\sqrt{x}+1-\sqrt{x}+1}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)^2}\right].\dfrac{\left(\sqrt{x}+2\right)\left(x-4\right)}{\sqrt{x}-2}\)
\(C=\dfrac{2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)^2}.\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+2\right)\left(\sqrt{x-2}\right)}{\sqrt{x}-2}\)
\(C=\dfrac{2}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)^2}.\left(\sqrt{x}+2\right)^2\)
\(C=\dfrac{2}{\sqrt{x}-2}\)
Rút gọn biểu thức:
E=\(\left(\dfrac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\dfrac{x+\sqrt{x}}{x-1}\right)\times\dfrac{x-1}{2x+\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}vớix\ge0,x\ne1\)
M=\(\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right)\div\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{2}{a-1}\right)vớia\ge0,a\ne1\)
Lm nhanh giúp mk nhé! Thank!
e) Ta có: \(E=\left(\dfrac{2x\sqrt{x}+x-\sqrt{x}}{x\sqrt{x}-1}-\dfrac{x+\sqrt{x}}{x-1}\right)\cdot\dfrac{x-1}{2x+\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\left(\dfrac{\sqrt{x}\left(2x+\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\left(\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}-1}\right)\cdot\dfrac{\sqrt{x}-1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)\left(\sqrt{x}-1\right)-\sqrt{x}\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(2x-3\sqrt{x}+1\right)-x\sqrt{x}-x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{2x\sqrt{x}-3x+\sqrt{x}-x\sqrt{x}-x-\sqrt{x}}{x+\sqrt{x}+1}\cdot\dfrac{1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{x\sqrt{x}-4x}{x+\sqrt{x}+1}\cdot\dfrac{1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{x\sqrt{x}-4x+\sqrt{x}\left(x+\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{x\sqrt{x}-4x+x\sqrt{x}+x+\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{2x\sqrt{x}-3x+\sqrt{x}}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}\)
m) Ta có: \(M=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{1}{a-\sqrt{a}}\right):\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{2}{a-1}\right)\)
\(=\left(\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{\sqrt{a}-1-2}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\right)\)
\(=\dfrac{\sqrt{a}+1}{\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\sqrt{a}-3}\)
\(=\left(\sqrt{a}-1\right)\cdot\dfrac{\left(\sqrt{a}+1\right)^2}{\sqrt{a}\left(\sqrt{a}-3\right)}\)
\(E=\left(\dfrac{\sqrt{x}\left(2\sqrt{x^2}+\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right).\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}-1\right)}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\left(\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right).\dfrac{\sqrt{x}+1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(2\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{\sqrt{x}}{\sqrt{x}+1}.\dfrac{\sqrt{x}+1}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}}{2\sqrt{x}-1}+\dfrac{\sqrt{x}}{2\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{x+\sqrt{x}+1}\)
\(M=\left(\dfrac{a-1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{\sqrt{a}-1-2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\left(\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\dfrac{\sqrt{a}-3}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)
\(=\left(\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}\left(\sqrt{a}-1\right)}\right)\left(\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}-3}\right)\)
\(=\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)^2}{\sqrt{a}\left(\sqrt{a}-3\right)}\)
1. Tìm x để bt có nghĩa
A=\(\dfrac{\sqrt{2x+3}}{\sqrt{x-3}}\)
B=\(\sqrt{\dfrac{2x+3}{x-3}}\)
C=\(\sqrt{-\dfrac{5}{x+2}}\)
D=\(\sqrt{-x}+\dfrac{1}{x+3}\)
2. Rút gọn bt
A=\(\sqrt{\dfrac{a+\sqrt{a^2-1}}{2}}-\sqrt{\dfrac{a-\sqrt{a^2-1}}{2}};\left(a>1\right)\)
B=\(\sqrt{\dfrac{a+\sqrt{a^2-1}}{2}}-\sqrt{\dfrac{a-\sqrt{a^2-b}}{2}};\left(a\ge\sqrt{b};b\ge0\right)\)
C=\(\left(1+\dfrac{a+\sqrt{a}}{a+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}+1}\right);\left(a\ge0,a\ne1\right)\)
D=\(\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}};\left(x>0\right)\)
Bài 1:
a: ĐKXĐ: 2x+3>=0 và x-3>0
=>x>3
b: ĐKXĐ:(2x+3)/(x-3)>=0
=>x>3 hoặc x<-3/2
c: ĐKXĐ: x+2<0
hay x<-2
d: ĐKXĐ: -x>=0 và x+3<>0
=>x<=0 và x<>-3
Rút gọn
\(\dfrac{6}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}-\dfrac{10}{\sqrt{5}}\)
B1. Với \(x\ge0,x\ne4.Chobiểuthức\)
\(A=\dfrac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}-\dfrac{1}{2-\sqrt{x}}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
\(B=\dfrac{1}{x\sqrt{x}+27}\)
a, tính giá trị biểu thức khi B= 1/4
b, Rút gọn A
c, Tìm giá trị của x để A>1/2
d, Với C= B : A. Tìm GTLN C
a) Ta có: \(\dfrac{6}{\sqrt{5}+1}+\sqrt{\dfrac{2}{3-\sqrt{5}}}-\dfrac{10}{\sqrt{5}}\)
\(=\dfrac{6\left(\sqrt{5}-1\right)}{4}+\sqrt{\dfrac{2\left(3+\sqrt{5}\right)}{4}}-2\sqrt{5}\)
\(=\dfrac{3}{2}\left(\sqrt{5}-1\right)+\dfrac{\sqrt{3+\sqrt{5}}}{\sqrt{2}}-2\sqrt{5}\)
\(=\dfrac{3}{2}\sqrt{5}-\dfrac{3}{2}-2\sqrt{5}+\dfrac{\sqrt{5}+1}{2}\)
\(=-\dfrac{1}{2}\sqrt{5}-\dfrac{3}{2}+\dfrac{1}{2}\sqrt{5}+\dfrac{1}{2}\)
=-1
Bài 1:
a) Thay \(x=\dfrac{1}{4}\)vào B, ta được:
\(B=1:\left(\dfrac{1}{4}\cdot\dfrac{1}{2}+27\right)=1:\left(27+\dfrac{1}{8}\right)=\dfrac{8}{217}\)
b) Ta có: \(A=\dfrac{x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}+\dfrac{1}{\sqrt{x}-2}-\dfrac{\sqrt{x}}{\sqrt{x}+3}\)
\(=\dfrac{x-9+\sqrt{x}+3-\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x+\sqrt{x}-6-x+2\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3\sqrt{x}-6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{3}{\sqrt{x}+3}\)
c) Để \(A>\dfrac{1}{2}\) thì \(A-\dfrac{1}{2}>0\)
\(\Leftrightarrow\dfrac{6-\sqrt{x}-3}{2\left(\sqrt{x}+3\right)}>0\)
\(\Leftrightarrow3-\sqrt{x}>0\)
\(\Leftrightarrow\sqrt{x}< 3\)
hay x<9
Kết hợp ĐKXĐ, ta được: \(\left\{{}\begin{matrix}0\le x< 9\\x\ne4\end{matrix}\right.\)
Bài 1: Cho \(A=\left(\dfrac{x-4}{\sqrt{x}-2}+\dfrac{x\sqrt{x}-8}{4-x}\right):\left[\dfrac{\left(\sqrt{x}-2\right)^2+2\sqrt{x}}{\sqrt{x}+2}\right]\)với \(x\ge0\); \(x\ne4\)
a, Rút gọn A
b, CMR: \(A< 1\) với \(x\ge0\); \(x\ne4\)
c, Tìm x để A nguyên
a: \(A=\left(\dfrac{\left(x-4\right)\left(\sqrt{x}+2\right)-x\sqrt{x}+8}{x-4}\right):\dfrac{x-2\sqrt{x}+4}{\sqrt{x}+2}\)
\(=\dfrac{x\sqrt{x}+2x-4\sqrt{x}-8-x\sqrt{x}+8}{x-4}\cdot\dfrac{\sqrt{x}+2}{x-2\sqrt{x}+4}\)
\(=\dfrac{2x-4\sqrt{x}}{\sqrt{x}-2}\cdot\dfrac{1}{x-2\sqrt{x}+4}=\dfrac{2\sqrt{x}}{x-2\sqrt{x}+4}\)
b: \(A-1=\dfrac{2\sqrt{x}-x+2\sqrt{x}-4}{x-2\sqrt{x}+4}\)
\(=\dfrac{-x+4\sqrt{x}-4}{x-2\sqrt{x}+4}=\dfrac{-\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}-1\right)^2+3}< 0\)
=>A<1
c: \(2\sqrt{x}>=0;x-2\sqrt{x}+4=\left(\sqrt{x}-1\right)^2+3>0\)
=>A>=0 với mọi x thỏa mãn ĐKXĐ
mà A<1
nên 0<=A<1
=>Để A nguyên thì A=0
=>x=0
Rút gọn biểu thức:
C=\(\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\right)\div\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)vớix\ge0,x\ne1\)
D=\(\left(\sqrt{x}+\dfrac{y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right)\div\left(\dfrac{x}{\sqrt{xy}+y}+\dfrac{y}{\sqrt{xy}-x}-\dfrac{x+y}{\sqrt{xy}}\right)\)
Lm nhanh giúp mk nhé!
\(C=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(x-1\right)}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\)
\(=\left(\dfrac{x-1-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(x-1\right)}\right):\left(\dfrac{\sqrt{x}+1-2}{x-1}\right)\)
\(=\left(\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(x-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2}.\left(\sqrt{x}+1\right)=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
Ta có: \(C=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\)
\(=\left(\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+1\right)^2}-\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}\right):\left(\dfrac{\sqrt{x}+1-2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{\sqrt{x}+1-2}{\left(\sqrt{x}+1\right)^2}:\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)^2}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
d) Ta có: \(D=\left(\sqrt{x}+\dfrac{y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right):\left(\dfrac{x}{\sqrt{xy}+y}+\dfrac{y}{\sqrt{xy}-x}-\dfrac{x+y}{\sqrt{xy}}\right)\)
\(=\left(\dfrac{x+\sqrt{xy}+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right):\left(\dfrac{x}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}+\dfrac{y}{\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)}-\dfrac{\left(x+y\right)\left(x-y\right)}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\)
\(=\dfrac{x+y}{\sqrt{x}+\sqrt{y}}:\left(\dfrac{x\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)-y\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)-x^2+y^2}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\)
\(=\dfrac{x+y}{\sqrt{x}+\sqrt{y}}:\dfrac{x^2-x\sqrt{xy}-y\sqrt{xy}-y^2-x^2+y^2}{\sqrt{xy}\left(\sqrt{x}-y\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{x+y}{\sqrt{x}+\sqrt{y}}:\dfrac{-\sqrt{xy}\left(x+y\right)}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{x+y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{xy}\cdot\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{-\sqrt{xy}\left(x+y\right)}\)
\(=-1\)
Chứng minh đẳng thức:
a) \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\sqrt{xy}\left(x\ge0,y\ge0,x^2+y^2\ne0\right)\)
b) \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\left(a\ge0,a\ne1\right)\)
c) \(\sqrt{x+2\sqrt{x-2}-1}\left(\sqrt{x-2}-1\right):\left(\sqrt{x}-\sqrt{3}\right)=\sqrt{x}+\sqrt{3}\left(x\ge2,x\ne3\right)\)
a: \(=x-\sqrt{xy}+y-x+2\sqrt{xy}-y=\sqrt{xy}\)
b: \(=\dfrac{1+\sqrt{a}}{a-\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
A=\(\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3\left(\sqrt{x}+3\right)}{x-9}\right)\)\(:\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)(với \(x\ge0;x\ne9\))
a) Rút gọn A
b) Tìm x để A<\(-\)1
\(a,A=\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}:\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\\ A=\dfrac{x-6\sqrt{x}-9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}=\dfrac{x-6\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}+1\right)}\)