\(\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2=1\) (\(a\ge0,b\ge0,a\ne b\))
Rút gọn biểu thức
a) \(\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2+4\sqrt{ab}}{\left(\sqrt{a+\sqrt{b}}\right)^2-4\sqrt{ab}}.\dfrac{a-b}{\left(\sqrt{a}-\sqrt{b}\right)^2}\) \(\left(đkxđ:a\ne b;a\ge0;b\ge0\right)\)
b) \(\dfrac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\dfrac{a-b}{\left(\sqrt{a}+\sqrt{b}\right)^2}\)\(\left(đkxđ:a\ne b;a\ge0;b\ge0\right)\)
HELP ME PLSSSSSSSSSS
câu a ở phần mẫu của cụm đầu tiên cái \(\left(\sqrt{a+\sqrt{b}}\right)^2\rightarrow\left(\sqrt{a}+\sqrt{b}\right)^2\) giúp em với ạ ( em cảm ơn )
a
\(=\dfrac{a-2\sqrt{ab}+b+4\sqrt{ab}}{a+2\sqrt{ab}+b-4\sqrt{ab}}.\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)^2}\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}.\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2.\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)^2}\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^3}{\left(\sqrt{a}-\sqrt{b}\right)^3}\)
rút gọn:
A=\(\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{a^3}-\sqrt{b^3}}{a-b}\left(a,b\ge0,a\ne b\right)\)
B=\(\left(\dfrac{\sqrt{x^3}+\sqrt{y^3}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right)\cdot\left(\dfrac{\sqrt{x}+\sqrt{y}}{x-y}\right)\left(x,y\ge0,x\ne y\right)\)
c. rút gọn biểu thức
\(C=\left(\sqrt{a}+\dfrac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\dfrac{a}{\sqrt{ab}+b}-\dfrac{b}{\sqrt{ab}-a}-\dfrac{a+b}{\sqrt{ab}}\right)\) vs \(a\ge0,a\ne4,a\ne9\)
Rút gọn : a) \(\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}\)
b)\(\dfrac{x+4y-4\sqrt{xy}}{\sqrt{x}-2\sqrt{y}}+\dfrac{y+\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\left(x\ge0;y\ge0;x\ne4y\right)\)
c)\(\dfrac{x+4\sqrt{x}+4}{\sqrt{x}+2}+\dfrac{4-x}{\sqrt{x}-2}\left(x\ge0;x\ne4\right)\)
d)\(\dfrac{9-x}{\sqrt{3x}+3}-\dfrac{9-6\sqrt{x}+x}{\sqrt{x}-3}\)
e)\(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)^2+4\sqrt{xy}}{\sqrt{x}+\sqrt{y}}-\dfrac{x\sqrt{y}+y\sqrt{x}}{\sqrt{xy}}\)
g)\(\left(2-\dfrac{a-3\sqrt{a}}{\sqrt{a}-3}\right)\left(2-\dfrac{5\sqrt{a}-\sqrt{ab}}{\sqrt{b}-5}\right)với\) a, b \(\ge\)0 , a \(\ne\)9; b\(\ne\)25
Mọi người giúp tớ với , cảm ơn nhiều nhiều ạ !!
a: \(=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}-\sqrt{ab}=\sqrt{ab}-\sqrt{ab}=0\)
b: \(=\dfrac{\left(\sqrt{x}-2\sqrt{y}\right)^2}{\sqrt{x}-2\sqrt{y}}+\dfrac{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)
\(=\sqrt{x}-2\sqrt{y}+\sqrt{y}=\sqrt{x}-\sqrt{y}\)
c: \(=\sqrt{x}+2-\dfrac{x-4}{\sqrt{x}-2}\)
\(=\sqrt{x}+2-\sqrt{x}-2=0\)
tính giá trị của biểu thức:
a) \(\sqrt{75}+\sqrt{48}-\sqrt{300}\)
b) \(\sqrt{81a}-\sqrt{36a}+\sqrt{144a}\left(a\ge0\right)\)
c) \(\dfrac{4}{\sqrt{5}-2}-\dfrac{4}{\sqrt{5}+2}\)
d) \(\dfrac{a\sqrt{b}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}\left(a\ge0;b\ge0;a\ne b\right)\)
a, \(\sqrt{75}+\sqrt{48}-\sqrt{300}\)
\(=5\sqrt{3}+4\sqrt{3}-10\sqrt{3}\)
\(=-\sqrt{3}\)
b, \(\sqrt{81a}-\sqrt{36a}+\sqrt{144a}\)
\(=9\sqrt{a}-6\sqrt{a}+12\sqrt{a}\)
\(=15\sqrt{a}\)
c, \(\dfrac{4}{\sqrt{5}-2}-\dfrac{4}{\sqrt{5}+2}\)
\(=\dfrac{4\sqrt{5}+8-4\sqrt{5}+8}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)
\(=\dfrac{16}{5-4}=16\)
d, \(\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}=\sqrt{ab}\)
Chứng minh các đẳng thức sau:
a) \(\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right)=1-x\)
(Với \(x\ge0;x\ne1\))
b) \(\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{ab}}+\dfrac{a-b}{\sqrt{a}-b}=2\sqrt{a}\)
(Với a>0; b>0; \(a\ne b\))
Câu b bạn sửa lại đề
\(a,VT=\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\\ =\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=VP\\ b,VT=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}+\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\\ =\sqrt{a}-\sqrt{b}+\sqrt{a}+\sqrt{b}=2\sqrt{a}=VP\)
a: \(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)
tính giá trị của biểu thức\(\dfrac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}\left(a\ge0;b\ge0\right);a\ne b\)
Ta có : \(a\sqrt{a}-b\sqrt{b}=\left(\sqrt{a}\right)^3-\left(\sqrt{b}\right)^3=\left(\sqrt{a}-\sqrt{b}\right)\cdot\left(a+\sqrt{ab}+b\right)\)Suy ra : a\(\dfrac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}=a+\sqrt{ab}+b\)
rút gọn biểu thức sau
\(\dfrac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}\left(vớia\ge0;b\ge0;a\ne b\right)\)
\(\dfrac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}=\dfrac{\sqrt{a^3}-\sqrt{b^3}}{\sqrt{a}-\sqrt{b}}\)
\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\sqrt{a}-\sqrt{b}}\)
\(=a+\sqrt{ab}+b\)
Chứng minh các đẳng thức sau :
a) \(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{6}\right).\dfrac{1}{\sqrt{6}}=-1,5\)
b) \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}=-2\)
c) \(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}=a-b\) với a, b dương và \(a\ne b\)
d) \(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)=1-a\) với \(a\ge0\) và \(a\ne1\)