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a: \(A=\dfrac{2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}-1}{\sqrt{x}}=\dfrac{2\sqrt{x}+1}{x+\sqrt{x}}\)
Mạn phép ko chép lại đề , mk làm luôn
a) \(D=\left[\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(1+\sqrt{ab}\right)+\left(\sqrt{a}+\sqrt{b}\right)\left(1-\sqrt{ab}\right)}{1-ab}\right]:\dfrac{a+b+2ab+1-ab}{1-ab}\)\(D=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(1+\sqrt{ab}+1-\sqrt{ab}\right)}{1-ab}.\dfrac{1-ab}{a+b+ab+1}\)
\(D=\dfrac{2\left(\sqrt{a}+\sqrt{b}\right)}{\left(b+1\right)\left(a+1\right)}\)
D=A/B
a)
B=1+(a+b+2ab)/(1-ab)=(a+b+ab)/(1-ab)
dk: a,b≥0; a.b≠1
1/B=(1-ab)/(a+b+ab)
A=√a+√b)[(1+√ab)+(1-√ab)]/(1-ab)=2(√a+√b)/(1-ab)
D=2(√a+√b)/[(a+1)(b+1)]
b)
a=2/(√3+2)=2(2-√3)/[(2+√3)(2-√3)]=2(2-√3)=(√3-1)^2
Bài này chính xác là của lớp 9 nè!!
Đề bạn ghi sai hay sao ý, pn xem lại xem, mk sửa đề như dưới, pn tham khảo:
Ta có: \(D=\left(\sqrt{a}+\dfrac{b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\dfrac{a}{\sqrt{ab}+b}+\dfrac{b}{\sqrt{ab}-a}-\dfrac{a+b}{\sqrt{ab}}\right)\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}:\left(\dfrac{a}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}+\dfrac{b}{\sqrt{a}\left(\sqrt{b}-\sqrt{a}\right)}-\dfrac{a+b}{\sqrt{ab}}\right)\)
\(=\dfrac{a+\sqrt{ab}+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}:\left(\dfrac{a.\sqrt{a}.\left(\sqrt{b}-\sqrt{a}\right)+b.\sqrt{b}.\left(\sqrt{a}+\sqrt{b}\right)-\left(a+b\right)\left(b-a\right)}{\sqrt{ab}\left(b-a\right)}\right)\)
\(=\dfrac{a+b}{\sqrt{a}+\sqrt{b}}:\left(\dfrac{a\sqrt{ab}-a^2+b\sqrt{ab}+b^2-b^2+a^2}{\sqrt{ab}\left(b-a\right)}\right)\)
\(=\dfrac{a+b}{\sqrt{a}+\sqrt{b}}:\dfrac{a\sqrt{ab}+b\sqrt{ab}}{\sqrt{ab}\left(b-a\right)}\)
\(=\dfrac{a+b}{\sqrt{a}+\sqrt{b}}:\dfrac{\sqrt{ab}\left(a+b\right)}{\sqrt{ab}\left(b-a\right)}=\dfrac{a+b}{\sqrt{a}+\sqrt{b}}:\dfrac{a+b}{b-a}\)
\(=\dfrac{a+b}{\sqrt{a}+\sqrt{b}}.\dfrac{b-a}{a+b}=\dfrac{b-a}{\sqrt{a}+\sqrt{b}}=\dfrac{\left(\sqrt{b}-\sqrt{a}\right)\left(\sqrt{b}+\sqrt{a}\right)}{\sqrt{a}+\sqrt{b}}\)
\(=\sqrt{b}-\sqrt{a}\)
\(D=\left(\dfrac{a+b\sqrt{a}+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\left(\dfrac{a}{\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)}-\dfrac{b}{\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)}-\dfrac{a+b}{\sqrt{ab}}\right)\)
\(=\left(\dfrac{a+b\sqrt{a}+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\dfrac{a\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-b\sqrt{b}\left(\sqrt{a}+\sqrt{b}\right)-\left(a^2-b^2\right)}{\sqrt{ab}\left(a-b\right)}\)
\(=\left(\dfrac{a+b\sqrt{a}+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\dfrac{a^2-a\sqrt{ab}-b\sqrt{ab}-b^2-a^2+b^2}{\sqrt{ab}\left(a-b\right)}\)
\(=\left(\dfrac{a+b\sqrt{a}+b-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\right):\dfrac{-\sqrt{ab}\left(a+b\right)}{\sqrt{ab}\left(a-b\right)}\)
\(=\dfrac{a+b+b\sqrt{a}-\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\cdot\dfrac{-\left(a-b\right)}{a+b}\)
\(=\dfrac{-\left(a+b+b\sqrt{a}-\sqrt{ab}\right)\left(\sqrt{a}-\sqrt{b}\right)}{a+b}\)
Cần điều kiện a;b;c dương
\(\dfrac{bc}{\sqrt{a.1+bc}}=\dfrac{bc}{\sqrt{a\left(a+b+c\right)+bc}}=\dfrac{bc}{\sqrt{\left(a+b\right)\left(a+c\right)}}\le\dfrac{1}{2}\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+c}\right)\)
Tương tự: \(\dfrac{ca}{\sqrt{b+ca}}\le\dfrac{1}{2}\left(\dfrac{ca}{a+b}+\dfrac{ca}{b+c}\right)\) ; \(\dfrac{ab}{\sqrt{c+ab}}\le\dfrac{1}{2}\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)\)
Cộng vế với vế:
\(A\le\dfrac{1}{2}\left(\dfrac{bc+ca}{a+b}+\dfrac{bc+ab}{a+c}+\dfrac{ca+ab}{b+c}\right)=\dfrac{1}{2}\left(a+b+c\right)=\dfrac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{1}{3}\)
a: \(P=\dfrac{x-2+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)
b:Sửa đề: 2A
2A=2căn x+5
=>(2căn x+2)/căn x=2căn x+5
=>2x+5căn x-2căn x-2=0
=>2x+3căn x-2=0
=>(căn x+2)(2căn x-1)=0
=>x=1/4
Ta có: \(P=\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right)\cdot\left(1-\dfrac{1}{\sqrt{a}}\right)\)
\(=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{-\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\cdot\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
\(=\dfrac{-2\sqrt{a}}{\sqrt{a}\left(\sqrt{a}+1\right)}=\dfrac{-2}{\sqrt{a}+1}\)
