K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

20 tháng 8 2021

\(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}\left(a,b>0;a\ne b\right)\\ =\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\\ =\dfrac{4\sqrt{ab}+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\\ =\dfrac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)

Tick plz

Ta có: \(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}\)

\(=\dfrac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\dfrac{4b+4\sqrt{ab}}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\dfrac{4\sqrt{b}\left(\sqrt{b}+\sqrt{a}\right)}{2\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{b}+\sqrt{a}\right)}\)

\(=\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)

20 tháng 6 2021

a) \(\dfrac{a^2+3}{\sqrt{a^2+2}}=\sqrt{a^2+2}+\dfrac{1}{\sqrt{a^2+2}}\ge2\sqrt{\sqrt{a^2+2}.\dfrac{1}{\sqrt{a^2+2}}}=2\)

Dấu = xảy ra khi \(\sqrt{a^2+2}=\dfrac{1}{\sqrt{a^2+2}}\Leftrightarrow a^2=-1\left(vn\right)\)

\(\Rightarrow\) Dấu "=" không xảy ra

Vậy \(\dfrac{a^2+3}{\sqrt{a^2+2}}>2\)

b)Với x,y>0,ta cm bđt phụ sau:

\(x^3+y^3\ge xy\left(x+y\right)\) (1)

Thật vậy (1)\(\Leftrightarrow\left(x+y\right)\left(x^2-xy+y^2\right)-xy\left(x+y\right)\ge0\)

\(\Leftrightarrow\cdot\left(x+y\right)\left(x^2-2xy+y^2\right)\ge0\)

\(\Leftrightarrow\left(x+y\right)\left(x-y\right)^2\ge0\) (lđ)

Áp dụng (1) có:

\(\dfrac{a}{\sqrt{b}}+\dfrac{b}{\sqrt{a}}=\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}.\sqrt{b}}\ge\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}=\sqrt{a}+\sqrt{b}\)

Dấu "=" xra khi a=b

Vậy...

a: \(\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\sqrt{ab}\right)\left(\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\right)\)

\(=\left(a-\sqrt{ab}+b-\sqrt{ab}\right)\cdot\dfrac{1}{\sqrt{a}-\sqrt{b}}\)

\(=\dfrac{\left(\sqrt{a}-\sqrt{b}\right)^2}{\sqrt{a}-\sqrt{b}}=\sqrt{a}-\sqrt{b}\)

b: \(VT=\dfrac{\sqrt{2}\left(2+\sqrt{2}\right)}{2+\sqrt{3}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{2}\right)}{2-\left(\sqrt{3}-1\right)}\)

\(=\dfrac{\sqrt{2}\left(2+\sqrt{2}\right)}{3+\sqrt{3}}+\dfrac{\sqrt{2}\left(2-\sqrt{2}\right)}{3-\sqrt{3}}\)

\(=\dfrac{2\left(\sqrt{2}+1\right)\left(\sqrt{3}-1\right)+2\left(\sqrt{2}-1\right)\left(\sqrt{3}+1\right)}{\sqrt{3}\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=\dfrac{2\left(\sqrt{6}-\sqrt{2}+\sqrt{3}-1+\sqrt{6}+\sqrt{2}-\sqrt{3}-1\right)}{\sqrt{3}\cdot2}\)

\(=\dfrac{2\left(2\sqrt{6}-2\right)}{2\sqrt{3}}=\dfrac{2\sqrt{6}-2}{\sqrt{3}}\)

13 tháng 11 2021

Câu b bạn sửa lại đề

\(a,VT=\left[1+\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}\right]\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right]\\ =\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x=VP\\ b,VT=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{ab}}+\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}\\ =\sqrt{a}-\sqrt{b}+\sqrt{a}+\sqrt{b}=2\sqrt{a}=VP\)

13 tháng 11 2021

a: \(=\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)=1-x\)

13 tháng 7 2018

b)CM: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}-\sqrt{a^2b^2+1}=0\)

\(VT=ab\sqrt{\dfrac{a^2b^2+1}{\left(ab\right)^2}}-\sqrt{a^2b^2+1}\)

\(VT=ab\dfrac{\sqrt{a^2b^2+1}}{ab}-\sqrt{a^2b^2+1}\)

\(VT=\sqrt{a^2b^2+1}-\sqrt{a^2b^2+1}\)

\(VT=0=VP\)

3 tháng 7 2021

a) 

\(P=\left(\dfrac{b-a}{\sqrt{b}-\sqrt{a}}-\dfrac{a\sqrt{a}-b\sqrt{b}}{a-b}\right):\dfrac{\left(\sqrt{b}-\sqrt{a}\right)^2+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\)

\(=\left[\sqrt{b}+\sqrt{a}-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\right]:\dfrac{b-\sqrt{ab}+a}{\sqrt{a}+\sqrt{b}}\)

\(=\left(\sqrt{b}+\sqrt{a}-\dfrac{a+\sqrt{ab}+b}{\sqrt{a}+\sqrt{b}}\right).\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\)

\(=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-a-\sqrt{ab}-b}{\sqrt{a}+\sqrt{b}}.\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\)

\(=\dfrac{\sqrt{ab}}{\sqrt{a}+\sqrt{b}}.\dfrac{\sqrt{a}+\sqrt{b}}{a-\sqrt{ab}+b}\)\(=\dfrac{\sqrt{ab}}{a-\sqrt{ab}+b}\)

b) \(P=\dfrac{\sqrt{ab}}{a-\sqrt{ab}+b}=\dfrac{\sqrt{ab}}{\left(\sqrt{a}-\dfrac{1}{2}\sqrt{b}\right)^2+\dfrac{3}{4}b}\)

Vì \(\left(\sqrt{a}-\dfrac{1}{2}\sqrt{b}\right)^2+\dfrac{3}{4}b>0;\forall a\ge0;b\ge0;a\ne b\)

\(\sqrt{ab}\ge0\)\(\forall a\ge0;b\ge0\)

\(\Rightarrow P=\dfrac{\sqrt{ab}}{\left(\sqrt{a}-\dfrac{1}{2}\sqrt{b}\right)^2+\dfrac{3}{4}b}\ge0\)

Vậy...

6 tháng 7 2021

cảm ơn tất cả mọi người

7 tháng 10 2018

b) \(\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{a-b}\)

\(=\dfrac{\sqrt{a}}{\sqrt{a}-\sqrt{b}}-\dfrac{\sqrt{b}}{\sqrt{a}+\sqrt{b}}-\dfrac{2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)-\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)-2b}{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{a}+\sqrt{b}\right)}\)

\(=\dfrac{a+\sqrt{ab}-\sqrt{ab}+b-\sqrt{ab}+b-2b}{a-b}\)

\(=\dfrac{a}{a-b}\)

7 tháng 10 2018

khúc \(\dfrac{a}{a-b}\) sai nhé

\(=\dfrac{a-b}{a-b}=1\)

b) Ta có: \(4x^2+x-5=0\)

\(\Leftrightarrow4x^2-4x+5x-5=0\)

\(\Leftrightarrow4x\left(x-1\right)+5\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\4x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{5}{4}\left(loại\right)\end{matrix}\right.\)

Thay x=1 vào biểu thức \(B=\dfrac{\sqrt{x}-1}{\sqrt{x}}\), ta được:

\(B=\dfrac{\sqrt{1}-1}{\sqrt{1}}=0\)

Vậy: Khi \(4x^2+x-5=0\) thì B=0

27 tháng 5 2017

Ôn tập Căn bậc hai. Căn bậc ba

21 tháng 9 2017

a) \(\dfrac{\sqrt{a}+\sqrt{b}}{2\sqrt{a}-2\sqrt{b}}-\dfrac{\sqrt{a}-\sqrt{b}}{2\sqrt{a}+2\sqrt{b}}-\dfrac{2b}{b-a}\)

=\(\dfrac{\left(\sqrt{a}+\sqrt{b}\right)^2-\left(\sqrt{a}-\sqrt{b}\right)^2+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

=\(\dfrac{a+2\sqrt{ab}+b-a+2\sqrt{ab}-b+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

=\(\dfrac{4\sqrt{ab}+4b}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}=\dfrac{4\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}{2\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}\)

=\(\dfrac{2\sqrt{b}}{\sqrt{a}-\sqrt{b}}\)(đpcm)