Rút gọn:
a) P= \(\frac{a+\sqrt{ab}}{b+\sqrt{ab}}\) (với a>0, b>0)
b) \(\frac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{y}}\) (với x>0, y>0)
Giúp mik với ạ. Cảm ơn mn nhiều!
CHỨNG MINH
a) \(\frac{\left(\sqrt{a}+1\right)^2-4\sqrt{a}}{\sqrt{a}-1}+\frac{a+\sqrt{a}}{\sqrt{a}}=2\sqrt{a}\) \(\left(a>0;a\ne1\right)\)
b) \(\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2=\sqrt{xy}\) \(\left(x\ge0;y\ge0\right)\)
c) \(\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}:\frac{a-b}{\sqrt{a}-\sqrt{b}}=1\) \(\left(a>0;b>0;a\ne b\right)\)
d) \(\left[\frac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\sqrt{a}+\sqrt{b}}-\frac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}\right]:\sqrt{b}=2\) \(\left(a>0;b>0\right)\)
Giúp mình với, cảm ơn mn <3
cau c í mk thấy bn chép sai đề nên mk sửa lại đề rồi bạn xem lại đề rồi so với bài làm của mk nha có j ko hiểu thì ib mk nha
\(a)VT = \dfrac{{{{\left( {\sqrt a + 1} \right)}^2} - 4\sqrt a }}{{\sqrt a - 1}} + \dfrac{{a + \sqrt a }}{{\sqrt a }}\\ = \dfrac{{a + 2\sqrt a + 1 - 4\sqrt a }}{{\sqrt a - 1}} + \dfrac{{\sqrt a \left( {\sqrt a + 1} \right)}}{{\sqrt a }}\\ = \dfrac{{a - 2\sqrt a + 1}}{{\left( {\sqrt a - 1} \right)}} + \sqrt a + 1\\ = \dfrac{{{{\left( {\sqrt a - 1} \right)}^2}}}{{\sqrt a - 1}} + \sqrt a + 1\\ = \sqrt a - 1 + \sqrt a + 1\\ = 2\sqrt a = VP (đpcm) \)
\(b)VT = \dfrac{{x\sqrt x + y\sqrt y }}{{\sqrt x + \sqrt y }} - {\left( {\sqrt x - \sqrt y } \right)^2}\\ = \dfrac{{\left( {\sqrt x + \sqrt y } \right)\left( {x - \sqrt {xy} + y} \right)}}{{\sqrt x + \sqrt y }} - \left( {x - 2\sqrt {xy} + y} \right)\\ = x - \sqrt {xy} + y - x + 2\sqrt {xy} - y\\ = \sqrt {xy} (đpcm)\\ c)VT = \dfrac{{a\sqrt b - b\sqrt a }}{{\sqrt {ab} }}:\dfrac{{a - b}}{{\sqrt a + \sqrt b }}\\ = \dfrac{{\sqrt {ab} \left( {\sqrt a - \sqrt b } \right)}}{{\sqrt {ab} }}.\dfrac{{\sqrt a + \sqrt b }}{{a - b}}\\ = \sqrt a - \sqrt b .\dfrac{{\sqrt a + \sqrt b }}{{a - b}}\\ = \dfrac{{\left( {\sqrt a - \sqrt b } \right)\left( {\sqrt a + \sqrt b } \right)}}{{a - b}}\\ = \dfrac{{a - b}}{{a - b}} = 1 (đpcm)\\ d)VT = \left[ {\dfrac{{{{\left( {\sqrt a - \sqrt b } \right)}^2} + 4\sqrt {ab} }}{{\sqrt a + \sqrt b }} - \dfrac{{a\sqrt b - b\sqrt a }}{{\sqrt {ab} }}} \right]:\sqrt b \\ = \dfrac{{a - 2\sqrt {ab} + b + 4\sqrt {ab} }}{{\sqrt a + \sqrt b }} - \dfrac{{\sqrt {ab} \left( {\sqrt a - \sqrt b } \right)}}{{\sqrt {ab} }}:\sqrt b \\ = \dfrac{{{{\left( {\sqrt a + \sqrt b } \right)}^2}}}{{\sqrt a + \sqrt b }} - \left( {\sqrt a - \sqrt b } \right):\sqrt b \\ = \sqrt a + \sqrt b - \sqrt a + \sqrt b :\sqrt b \\ = \dfrac{{2\sqrt b }}{{\sqrt b }} = 2 (đpcm) \)
Câu c đề sai (đã sửa)
Rút gọn
1) \(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}-\sqrt{ab}}.\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\) với a,b lớn hơn hoặc bằng 0,a khác b.
2) \(\left(2-\frac{7+3\sqrt{7}}{\sqrt{7}+3}\right).\left(2-\frac{5\sqrt{7}-\sqrt{14}}{\sqrt{2}-5}\right)\)
3) \(\left(\frac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\frac{\sqrt{x}-\sqrt{y}}{1-\sqrt{xy}}\right):\left(\frac{x+xy}{1-xy}\right)\)với x,y lớn hơn 0,x,y khác 1
3)\(...=\left[\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(1+\sqrt{xy}\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}\right].\frac{1-xy}{x+xy}\)
= \(\frac{\sqrt{x}+x\sqrt{y}+\sqrt{y}+y\sqrt{x}+\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}}{1-xy}.\frac{1-xy}{x\left(1+y\right)}\)= \(\frac{2\sqrt{x}+2y\sqrt{x}}{x\left(1+y\right)}=\frac{2\sqrt{x}\left(1+y\right)}{x\left(1+y\right)}=\frac{2}{\sqrt{x}}\)
rút gọn:
\(a,\frac{\sqrt{4mn^2}}{\sqrt{20m}}\left(m>0,n>0\right)\)
\(b,\frac{\sqrt{16a^4b^6}}{\sqrt{12a^6b^6}}\left(a< 0,b\ne0\right)\)
\(c,\frac{y-\sqrt{xy}}{x-\sqrt{xy}}\)với\(xy>0,y\ne1\)
\(d,\frac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)với \(x>0,y>0,y\ne1\)
\(e,\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}\)với \(x>0\)
a) \(\frac{\sqrt{4mn^2}}{\sqrt{20m}}=\sqrt{\frac{4mn^2}{20m}}=\sqrt{\frac{n^2}{5}}=\frac{n}{\sqrt{5}}\)
b) \(\frac{\sqrt{16a^4b^6}}{\sqrt{12a^6b^6}}=\sqrt{\frac{16a^4b^6}{12a^6b^6}}=\sqrt{\frac{4}{3a^2}}=\frac{2}{\sqrt{3}.\left|a\right|}=-\frac{2}{a\sqrt{3}}\)
d) \(\frac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)
e) \(\sqrt{\frac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}}=\sqrt{\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}+1\right)^2}}=\frac{\left|\sqrt{x}-1\right|}{\sqrt{x}+1}\)
Thu gọn
a) \(\frac{y}{x}.\sqrt{\frac{x^2}{y^4}}\)với x >0; y khác 0
b) \(5xy.\sqrt{\frac{x^2}{y^6}}\)với x < 0
c) \(0.2x^3y^3.\sqrt{\frac{16}{x^4y^8}}\)Với x và y khác 0
Giải từng bước giúp mình với nhé! Cảm ơn nhiều ạ!
a, Ta có : \(\frac{y}{x}.\sqrt{\frac{x^2}{y^4}}=\frac{y}{x}.\frac{x}{y^2}=\frac{1}{y}\)
b , Ta có : \(5xy\sqrt{\frac{x^2}{y^6}}=5xy\frac{x}{y^3}=\frac{5x^2}{y^2}\)
c, Ta có : \(0,2x^3y^3\sqrt{\frac{16}{x^4y^8}}=0,2x^3y^3.\frac{4}{x^2y^4}=\frac{0,8x}{y}\)
Giúp mình với <3
B = \(\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}+\frac{x\sqrt{x}-y\sqrt{y}}{y-x}\right):\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
a, Rút gọn B
b, Chứng minh : B > hoặc = 0
c, So sánh B với \(\sqrt{B}\)
Rút gọn các biểu thức:
a)\(\frac{1}{\sqrt{x}-1}+\frac{1}{1+\sqrt{x}}+1\)với x>0 và \(x\ne1\)
b)\(\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{4-x}\)với x>0 và \(x\ne4\)
c)\(5\sqrt{\frac{x}{y}}-4\sqrt{\frac{y}{x}}+\sqrt{\frac{1}{xy}}\)với x>0, y>0
a) \(\frac{1}{\sqrt{x}-1}+\frac{1}{1+\sqrt{x}}=\frac{1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}=\frac{2\sqrt{x}}{x-1}\)( x > 0 ; x ≠ 1 )
b) \(\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{4-x}=\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}+\frac{\sqrt{x}}{x-4}\)
\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2-2\sqrt{x}-4+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-6}{x-4}\)( x > 0 ; x ≠ 4 )
a) Với \(x>0\)và \(x\ne1\)ta có:
\(\frac{1}{\sqrt{x}-1}+\frac{1}{1+\sqrt{x}}+1\)
\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}+1+\sqrt{x}-1+x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b) Với \(x>0\)và \(x\ne4\)ta có:
\(\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{4-x}=\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{x-4}\)
\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)-2\left(\sqrt{x}+2\right)+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2-2\sqrt{x}-4+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-6}{x-4}\)
Bài 1: Rút gọn biểu thức:
\(A=\left(\frac{2\sqrt{xy}}{x-y}+\frac{\sqrt{x}-\sqrt{y}}{2\sqrt{x}+2\sqrt{y}}\right).\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{y}-\sqrt{x}}\) Với x>0, y>0, x#y
Ta có \(A=\left(\frac{2\sqrt{xy}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}+\frac{\sqrt{x}-\sqrt{y}}{2\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}+\frac{\sqrt{y}}{\sqrt{y}-\sqrt{x}}\)
\(=\left(\frac{4\sqrt{xy}+\left(\sqrt{x}-\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\) (Quy đồng biểu thức đầu và đổi dấu số hạng cuối)
\(=\left(\frac{4\sqrt{xy}+x-2\sqrt{xy}+y}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
\(=\frac{\sqrt{x}+\sqrt{y}}{2\left(\sqrt{x}-\sqrt{y}\right)}\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}}{\sqrt{x}-\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}=1.\)
Vậy giá trị biểu thức \(A=1.\)
\(=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{2\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\cdot\frac{2\sqrt{x}}{\sqrt{x}+\sqrt{y}}-\frac{\sqrt{y}}{\sqrt{x}-\sqrt{y}}\)
1. Rút gọn:
\(\sqrt{8-2\sqrt{7}}-\sqrt{9-2\sqrt{14}}\)
2. Tính:
2 + \(\sqrt{17-4\sqrt{9}+4\sqrt{5}}\)
3. CM:
a. \(\frac{x+y}{2}\) >= \(\sqrt{xy}\) với x, y >= 0
b. \(\frac{x}{y}+\frac{y}{x}\) >= 2 với x,y >= 0
c. a + b + 1 >= \(\sqrt{ab}\) + \(\sqrt{a}+\sqrt{b}\) với a,b >= 0.
3/a) \(BĐT\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\)(đúng với mọi x, y không âm)
Đẳng thức xảy ra khi x = y
b) \(BĐT\Leftrightarrow\frac{\left(x-y\right)^2}{xy}\ge0\) (đúng với mọi x, y không âm)
"=" <=> x = y
c) BĐT \(\Leftrightarrow2a+2b+2\ge2\sqrt{ab}+2\sqrt{a}+2\sqrt{b}\)
\(\Leftrightarrow\left(a-2\sqrt{ab}+b\right)+\left(a-2\sqrt{a}+1\right)+\left(b-2\sqrt{b}+1\right)\ge0\)
\(\Leftrightarrow\left(\sqrt{a}-\sqrt{b}\right)^2+\left(\sqrt{a}-1\right)^2+\left(\sqrt{b}-1\right)^2\ge0\) (đúng)
"=" <=> a = b = 1
1/ \(A=\sqrt{7-2\sqrt{7}.1+1}-\sqrt{7-2\sqrt{7}.\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}\)
\(=\left|\sqrt{7}-1\right|-\left|\sqrt{7}-\sqrt{2}\right|\) (thực ra em nghĩ ko cần thêm trị tuyệt đối đâu nhưng thêm cho chắc:D)
\(=\sqrt{7}-1-\sqrt{7}+\sqrt{2}=\sqrt{2}-1\)
2/Em thấy nó sai sai nên thôi:(
GIÚP MIK GIẢI 2 CÂU NÀY VỚI CẦN GẤP...........................
làm ơn .................
CÂU 1
\(A=\left(\frac{x-y}{\sqrt{x}-\sqrt{y}}\right)+\left(\frac{x\sqrt{x}-y\sqrt{y}}{y-x}\right):\frac{\left(\sqrt{x}-\sqrt{y}\right)^2+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
a,RÚT GỌN
b,CHỨNG MINH A>=0
c,SO SÁNH A VỚI \(\sqrt{A}\)
CÂU 2
\(P=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-x}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
a, Rút gọn
b, x=? để P = 1/3
c,Tìm GTLN khi Q= P-9x
NHANH LÊN NHÉ