\(\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{a+\sqrt{2̣̣̣\left(a-2\right)}}}}}}}}}}}\)
Các bn xem bài này mk làm đúng không
a)\(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2=1\)
VT=\(\left(\frac{a\sqrt{a}+b\sqrt{b}-\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
=\(\left(\frac{a\sqrt{a}+b\sqrt{b}-a\sqrt{b}-b\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
=\(\left(\frac{\left(a\sqrt{a}-a\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\)
=\(\left(\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(a-b\right)}{\sqrt{a+\sqrt{b}}}\right)\frac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(a-b\right)^2}\)
= \(\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{a-b}=\frac{a-b}{a-b}=1\Rightarrow\left(=VP\right)\)
b)\(\frac{a+b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=a-b\)
VT=\(\frac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}.\sqrt{a}+\sqrt{b}=\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)\)
=\(a+\sqrt{ab}-\sqrt{ab}-b=a-b\Rightarrow\left(=VP\right)\)
Đây là đề chứng minh hả !
Phần a , b đúng r
Nhưng phần b chỗ \(\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)=\left(\sqrt{a}\right)^2-\left(\sqrt{b}\right)^2\) = a - b
Dùng hằng đẳng thức thức 3 như vậy sẽ hay hơn !
Chúc bạn học tốt!
Chứng minh các đẳng thức sau:
a) \(\left(1-a^2\right):\left[\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1
+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\right]+1=\frac{2}{1-a}\)
b) \(\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a}{\sqrt{ab}+b}
+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)=\sqrt{b}-\sqrt{a}\)
c) \(\frac{\sqrt{a}+\sqrt{b}-1}{a
+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a
+\sqrt{ab}}\right)=\frac{\sqrt{a}}{a}\)
d) \(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2=1\)
Chứng minh các đẳng thức sau:
a) \(\left(1-a^2\right):\left(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\right)+1=\frac{2}{1-a}\)
b) \(\left(\sqrt{a}+\frac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\frac{a}{\sqrt{ab}+b}+\frac{b}{\sqrt{ab}-a}-\frac{a+b}{\sqrt{ab}}\right)=\sqrt{b}-\sqrt{a}\)
c) \(\frac{\sqrt{a}+\sqrt{b}-1}{a+\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{2\sqrt{ab}}.\left(\frac{\sqrt{b}}{a-\sqrt{ab}}+\frac{\sqrt{b}}{a+\sqrt{ab}}\right)=\frac{\sqrt{a}}{a}\)
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)
C/m biểu thức
a)\(\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\frac{\sqrt{a}+\sqrt{b}}{a-b}\right)=1\)(a,b>0,a\(\ne\)0
b)\(\frac{a-b-2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}:\frac{1}{\sqrt{a}+\sqrt{b}}=a-b\left(a,b>0,a\ne b\right)\)
c)\(\left(2+\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\left(2-\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)=4-a\left(a>0,a\ne1\right)\)
d)\(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)=\left(1-a\right)^2\left(a\ge0,a\ne1\right)\)
Giải giúp mk với. THứ 3 tuần sau là phải nộp rồi
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
\(=\dfrac{a-2\sqrt{ab}+b+4\sqrt{ab}}{a+2\sqrt{ab}+b-4\sqrt{ab}}.\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)^2}\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}-\sqrt{b}\right)^2}.\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}-\sqrt{b}}\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2.\left(\sqrt{a}+\sqrt{b}\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)^2}\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^3}{\left(\sqrt{a}-\sqrt{b}\right)^3}\)
câu1 : a) A= \(\dfrac{\sqrt{15}-\sqrt{12}}{\sqrt{5}-2}-\dfrac{1}{2-\sqrt{3}}\)
b) \(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}-\dfrac{4}{\sqrt{a}}\right)\)
Câu 2 :
a) A= \(\left(2\sqrt{4+\sqrt{6-2\sqrt{5}}}\right).\left(\sqrt{10}-\sqrt{2}\right)\)
b) B= \(\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right).\left(1-\dfrac{2}{a+1}\right)^2\)
Câu 2:
a: \(=2\left(\sqrt{4+\sqrt{5}-1}\right)\left(\sqrt{10}-\sqrt{2}\right)\)
\(=\sqrt{2}\cdot\sqrt{6+2\sqrt{5}}\cdot\left(\sqrt{10}-\sqrt{2}\right)\)
\(=2\cdot\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)=8\)
b: \(=\dfrac{a-2\sqrt{a}+1+a+2\sqrt{a}+1}{a-1}\cdot\left(\dfrac{a+1-2}{a+1}\right)^2\)
\(=\dfrac{2\left(a+1\right)}{a-1}\cdot\dfrac{\left(a-1\right)^2}{\left(a+1\right)^2}=\dfrac{2\left(a-1\right)}{a+1}\)
Rút gọn:
n) N = \(\left(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right)\left(\dfrac{\sqrt{x}+\sqrt{y}}{x-y}\right)^2\)
o) O = \(\left(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}+\dfrac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}\right):\left(\dfrac{a-b}{\sqrt{a}-\sqrt{ }b}\right)^2\)
p) P = \(\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{x\sqrt{x}+1}{\sqrt{x}+1}-\sqrt{x}\right)\)
q) Q = \(\left(\dfrac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}\right):\dfrac{x+xy}{1-xy}\)
n) Ta có: \(N=\left(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right)\left(\dfrac{\sqrt{x}+\sqrt{y}}{x-y}\right)^2\)
\(=\left(\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-\sqrt{xy}\right)\left(\dfrac{1}{\sqrt{x}-\sqrt{y}}\right)^2\)
\(=\left(\sqrt{x}-\sqrt{y}\right)^2\cdot\dfrac{1}{\left(\sqrt{x}-\sqrt{y}\right)^2}\)
=1
o) Ta có: \(O=\left(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{a}+\sqrt{b}}+\dfrac{a\sqrt{a}-b\sqrt{b}}{\sqrt{a}-\sqrt{b}}\right):\left(\dfrac{a-b}{\sqrt{a}-\sqrt{b}}\right)^2\)
\(=\left(\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}+\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\sqrt{a}-\sqrt{b}}\right):\left(\sqrt{a}+\sqrt{b}\right)^2\)
\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2}{\left(\sqrt{a}+\sqrt{b}\right)^2}\)
=1
p) Ta có: \(P=\left(\dfrac{2x+1}{x\sqrt{x}-1}-\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\right)\left(\dfrac{x\sqrt{x}+1}{\sqrt{x}+1}-\sqrt{x}\right)\)
\(=\dfrac{2x+1-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}+1}-\sqrt{x}\right)\)
\(=\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)^2\)
\(=\dfrac{x+\sqrt{x}+1}{x+\sqrt{x}+1}\cdot\left(\sqrt{x}-1\right)\)
\(=\sqrt{x}-1\)
q) Ta có: \(Q=\left(\dfrac{\sqrt{x}+\sqrt{y}}{1-\sqrt{xy}}+\dfrac{\sqrt{x}-\sqrt{y}}{1+\sqrt{xy}}\right):\dfrac{x+xy}{1-xy}\)
\(=\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{xy}+1\right)+\left(\sqrt{x}-\sqrt{y}\right)\left(1-\sqrt{xy}\right)}{\left(1-\sqrt{xy}\right)\left(1+\sqrt{xy}\right)}:\dfrac{x+xy}{1-xy}\)
\(=\dfrac{x\sqrt{y}+\sqrt{x}+y\sqrt{x}+\sqrt{y}+\sqrt{x}-x\sqrt{y}-\sqrt{y}+y\sqrt{x}}{x+xy}\)
\(=\dfrac{2\sqrt{x}+2y\sqrt{x}}{x+xy}\)
\(=\dfrac{2\sqrt{x}\left(1+y\right)}{x\left(1+y\right)}\)
\(=\dfrac{2}{\sqrt{x}}\)
\(P=\left[\left(\frac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right):\left(a-b\right)+\frac{2\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right].\frac{a\sqrt{a}-8}{a-\sqrt{a}-2}\)