\(\dfrac{\sqrt{x}-2}{\sqrt{x}-1}\ge0\)
LT
Những câu hỏi liên quan
Rút gọn biểu thức
\(a.\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(b.\sqrt{41-\sqrt{160}}+\sqrt{49+\sqrt{90}}\)
\(c.\dfrac{x-y}{\sqrt{x}-\sqrt{y}}\left(x\ge0;y\ge0;x\ne y\right)\)
\(d.\dfrac{y+1-2\sqrt{y}}{\sqrt{y}-1}\left(y\ge0;y\ne1\right)\)
\(e.\sqrt{x+2+2\sqrt{x+1}}-\sqrt{x+2-2\sqrt{x+1}}\)
a: \(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
\(=\sqrt{5}+1-\sqrt{5}+1\)
=2
c: \(\dfrac{x-y}{\sqrt{x}-\sqrt{y}}=\sqrt{x}+\sqrt{y}\)
d: \(\dfrac{y-2\sqrt{y}+1}{\sqrt{y}-1}=\sqrt{y}-1\)
Đúng 0
Bình luận (0)
Chứng minh đẳng thức:
a) dfrac{xsqrt{x}+ysqrt{y}}{sqrt{x}+sqrt{y}}-left(sqrt{x}-sqrt{y}right)^2sqrt{xy}left(xge0,yge0,x^2+y^2ne0right)
b) left(dfrac{1}{a-sqrt{a}}+dfrac{1}{sqrt{a}-1}right):dfrac{sqrt{a}+1}{a-2sqrt{a}+1}left(age0,ane1right)
c) sqrt{x+2sqrt{x-2}-1}left(sqrt{x-2}-1right):left(sqrt{x}-sqrt{3}right)sqrt{x}+sqrt{3}left(xge2,xne3right)
Đọc tiếp
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: \(=\dfrac{1+\sqrt{a}}{a-\sqrt{a}}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Đúng 0
Bình luận (0)
\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}+\dfrac{2\sqrt{x}}{\sqrt{x}+2}2-\dfrac{2+5\sqrt{x}}{x-4},x\ge0,x\ne4\)
Lời giải:
Nếu yêu cầu đề là rút gọn thì giải như sau:
\(BT = \frac{(\sqrt{x}+1)(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{2\sqrt{x}(\sqrt{x}-2)}{(\sqrt{x}+2)(\sqrt{x}-2)}-\frac{2+5\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\)
\(=\frac{(x+3\sqrt{x}+2)+(2x-4\sqrt{x})-(2+5\sqrt{x})}{(\sqrt{x}-2)(\sqrt{x}+2)}=\frac{3x-6\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}=\frac{3\sqrt{x}(\sqrt{x}-2)}{(\sqrt{x}-2)(\sqrt{x}+2)}=\frac{3\sqrt{x}}{\sqrt{x}+2}\)
Đúng 1
Bình luận (0)
p= \(\left(\dfrac{x+2\sqrt{x}}{x-4}-\dfrac{2\sqrt{x}+1}{2x-3\sqrt{x}-2}-\dfrac{x}{\sqrt{x}-2}\right)\dfrac{x-1}{x\sqrt{x}+1}vớix\ge0;x\ne4\)
\(=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{2\sqrt{x}+1}{2x-4\sqrt{x}+\sqrt{x}-2}-\dfrac{x}{\sqrt{x}-2}\right)\cdot\dfrac{\sqrt{x}-1}{x-\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}-1-x}{\sqrt{x}-2}\cdot\dfrac{\sqrt{x}-1}{x-\sqrt{x}+1}=\dfrac{-\sqrt{x}+1}{\sqrt{x}-2}\)
Đúng 1
Bình luận (0)
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
Đúng 2
Bình luận (0)
dfrac{2sqrt{x}}{sqrt{x+}3}+dfrac{sqrt{x}+1}{sqrt{x}-3}+dfrac{3-11sqrt{x}}{9-x},xge0,xne9
Đọc tiếp
\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}+\dfrac{3-11\sqrt{x}}{9-x}\\ =\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}+1}{\sqrt{x}-3}-\dfrac{3-11\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{3-11\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}+\sqrt{x}+3-3+11\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ =\dfrac{3x+15\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)
Đúng 2
Bình luận (2)
rút gọn biểu thức
a) A=\(\dfrac{\sqrt{x}-3}{\sqrt{x-2}}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{x-3\sqrt{x}+2}vớix\ge0,x\ne4,x\ne1\)
b)\(\left(\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right)\div\dfrac{\sqrt{x}-1}{2}vớix>0,x\ne1\)
`a)(sqrtx-3)/(sqrtx-2)-(2sqrtx-1)/(sqrtx-1)+(x-2)/(x-3sqrtx+2)`
`=(x-4sqrtx+3-(2sqrtx-1)(sqrtx-2)+x-2)/(x-3sqrtx+2)`
`=(2x-4sqrtx+1-2x+5sqrtx-2)/(x-3sqrtx+2)`
`=(sqrtx-1)/(x-3sqrtx+2)`
`=1/(sqrtx-2)`
`b)((x+2)/(xsqrtx-1)-sqrtx/(x+sqrtx+1)+1/(1-sqrtx)):(sqrtx-1)/2`
`=((x+2-x+sqrtx-x-sqrtx-1)/(xsqrtx-1))*2/(sqrtx-1)`
`=(1-x)/(xsqrtx-1)*2/(sqrtx-1)`
`=(-2(sqrtx+1))/(x+sqrtx+1)`
Đúng 1
Bình luận (0)
a) Ta có: \(A=\dfrac{\sqrt{x}-3}{\sqrt{x}-2}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x-2}{x-3\sqrt{x}+2}\)
\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}-\dfrac{\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}+\dfrac{x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{x-4\sqrt{x}+3-2x+4\sqrt{x}+\sqrt{x}-2+x-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}\)
\(=\dfrac{1}{\sqrt{x}-2}\)
b) Ta có: \(\left(\dfrac{x+2}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\)
\(=\dfrac{x+2-\sqrt{x}\left(\sqrt{x}-1\right)-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{-\sqrt{x}+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)
\(=\dfrac{-2\sqrt{x}-2}{x\sqrt{x}-1}\)
Đúng 0
Bình luận (0)
Đề : Trục căn thức ở mẫuf) dfrac{2}{sqrt{6}-sqrt{5}} l) dfrac{3}{sqrt{10}+sqrt{7}} m) dfrac{1}{sqrt{x}-sqrt{y}} ( x0 ,y0,xne y )o) dfrac{2ab}{sqrt{a}-sqrt{b}} (age0,bge0,ane b) P) dfrac{P}{2sqrt{P}-1} (Pge0 , Pnedfrac{1}{4})
Đọc tiếp
Đề : Trục căn thức ở mẫu
f) \(\dfrac{2}{\sqrt{6}-\sqrt{5}}\) l) \(\dfrac{3}{\sqrt{10}+\sqrt{7}}\) m) \(\dfrac{1}{\sqrt{x}-\sqrt{y}}\) ( x>0 ,y>0,\(x\ne y\) )
o) \(\dfrac{2ab}{\sqrt{a}-\sqrt{b}}\) (\(a\ge0,b\ge0,a\ne b\))
P) \(\dfrac{P}{2\sqrt{P}-1}\) (\(P\ge0\) , \(P\ne\dfrac{1}{4}\))
f: \(\dfrac{2}{\sqrt{6}-\sqrt{5}}=2\sqrt{6}+2\sqrt{5}\)
l: \(\dfrac{3}{\sqrt{10}+\sqrt{7}}=\sqrt{10}-\sqrt{7}\)
m: \(\dfrac{1}{\sqrt{x}-\sqrt{y}}=\dfrac{\sqrt{x}+\sqrt{y}}{x-y}\)
Đúng 1
Bình luận (0)
rủ gọn biểu thức
\(C=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}\right)\) : \(\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\)vs \(x\ge0,x\ne1\)
\(C=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-2}{\sqrt{x}\left(x-1\right)+\left(x-1\right)}\right):\left(\dfrac{\sqrt{x}+1-2}{x-1}\right)\)
\(=\dfrac{x-1-2\sqrt{x}+2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{x-1}{\sqrt{x}-1}\)
\(=\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
Đúng 2
Bình luận (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!
\(C=\left(\dfrac{1}{\sqrt{x}+1}-\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(x-1\right)}\right):\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{2}{x-1}\right)\)
\(=\left(\dfrac{x-1-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(x-1\right)}\right):\left(\dfrac{\sqrt{x}+1-2}{x-1}\right)\)
\(=\left(\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}+1\right)\left(x-1\right)}\right):\left(\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)\left(x-1\right)}\right):\left(\dfrac{1}{\sqrt{x}+1}\right)\)
\(=\dfrac{\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2}.\left(\sqrt{x}+1\right)=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
Đúng 2
Bình luận (0)
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: \(D=\left(\sqrt{x}+\dfrac{y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right):\left(\dfrac{x}{\sqrt{xy}+y}+\dfrac{y}{\sqrt{xy}-x}-\dfrac{x+y}{\sqrt{xy}}\right)\)
\(=\left(\dfrac{x+\sqrt{xy}+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right):\left(\dfrac{x}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}+\dfrac{y}{\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)}-\dfrac{\left(x+y\right)\left(x-y\right)}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\)
\(=\dfrac{x+y}{\sqrt{x}+\sqrt{y}}:\left(\dfrac{x\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)-y\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)-x^2+y^2}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\)
\(=\dfrac{x+y}{\sqrt{x}+\sqrt{y}}:\dfrac{x^2-x\sqrt{xy}-y\sqrt{xy}-y^2-x^2+y^2}{\sqrt{xy}\left(\sqrt{x}-y\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{x+y}{\sqrt{x}+\sqrt{y}}:\dfrac{-\sqrt{xy}\left(x+y\right)}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\dfrac{x+y}{\sqrt{x}+\sqrt{y}}\cdot\dfrac{\sqrt{xy}\cdot\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{-\sqrt{xy}\left(x+y\right)}\)
\(=-1\)
Đúng 1
Bình luận (0)