\(A=\frac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\frac{2a+\sqrt{a}}{\sqrt{a}}+1\)

\(A=\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\frac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)

\(A=\sqrt{a}\left(\sqrt{a}+1\right)-\left(2\sqrt{a}+1\right)+1\)

\(A=a+\sqrt{a}-2\sqrt{a}-1+1\)

\(A=a-\sqrt{a}\)

Giúp m với

Bạn có thể đăng từng bài k như thế nhìn đã sợ ai làm

1)đặt nhân tử chung quy đồng là xong

2)phân tích x+2cănx-3=(1-cănx)(3+cănx)

3)2a+căn a đặt căn a ra r rút gọn

a) P = \(\left(\frac{3\sqrt{a}}{a+\sqrt{a}+b}-\frac{3a}{a\sqrt{a}-b\sqrt{b}}+\frac{1}{\sqrt{a}-\sqrt{b}}\right):\frac{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}{\left(2.a+2.\sqrt{ab}+2.b\right)}\)

        = \(\left(\frac{3\sqrt{a}.\left(\sqrt{a}-\sqrt{b}\right)-3.a+a+\sqrt{ab}+b}{\left(\sqrt{a}-\sqrt{b}\right).\left(a+\sqrt{ab}+b\right)}\right).\frac{2.\left(a+\sqrt{ab}+b\right)}{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}\)

        = \(\frac{a-2.\sqrt{ab}+b}{\sqrt{a}-\sqrt{b}}.\frac{2}{\left(a-1\right).\left(\sqrt{a}-\sqrt{b}\right)}\)

          = \(\frac{2}{a-1}\)

b) P nguyên <=> \(\frac{2}{a-1}\)nguyên => 2 \(⋮\)a - 1 

=> ( a- 1 ) = { \(\pm\)1 ; \(\pm\) 2} => a = { -1 ; 0 ; 2 ;3 } 

ĐK \(a\ge0\)

\(A=\frac{\sqrt{a}\left(a\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\frac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)

\(=\frac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\left(2\sqrt{a}+1\right)+1\)

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

b. \(A=2\Rightarrow a-\sqrt{a}-2=0\Rightarrow\left(\sqrt{a}-2\right)\left(\sqrt{a}+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}\sqrt{a}=2\\\sqrt{a}=-1\left(l\right)\end{cases}\Rightarrow a=4}\)

Vậy a=4 thì A=2

c. \(A=a-\sqrt{a}=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\forall a\Rightarrow A\ge-\frac{1}{4}\)

Vậy \(MinA=-\frac{1}{4}\Rightarrow\sqrt{a}=\frac{1}{2}\Rightarrow a=\frac{1}{4}\)