2.

a, \(P=\left(\frac{a\sqrt{a}+1}{a-1}-\frac{a-1}{\sqrt{a}-1}\right):\left(\sqrt{a}-\frac{\sqrt{a}}{\sqrt{a}-1}\right)\)

\(=\left[\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}-\frac{a-1}{\sqrt{a}-1}\right]:\frac{a-\sqrt{a}-\sqrt{a}}{\sqrt{a}-1}\)

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

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

b, \(a=3-2\sqrt{2}=\left(\sqrt{2}-1\right)^2\Rightarrow\sqrt{a}=\sqrt{2}-1\)

Khi đó \(P=-\frac{1}{\sqrt{a}}=-\frac{1}{\sqrt{2}-1}=-\sqrt{2}-1\)

1.

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

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

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

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

b, \(A=a-\sqrt{a}=\left(\sqrt{a}-\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)

\(\Rightarrow MinA=-\frac{1}{4}\Leftrightarrow x=\frac{1}{4}\)

a/ \(\frac{2}{a}.\frac{4\left|a\right|}{3}=\frac{-8a}{3a}=-\frac{8}{3}\)

b/ \(\frac{3}{a-1}\sqrt{\frac{4\left(a-1\right)^2}{25}}=\frac{3}{\left(a-1\right)}.\frac{2\left|a-1\right|}{5}=\frac{6\left(a-1\right)}{5\left(a-1\right)}=\frac{6}{5}\)

c/ \(\frac{3\sqrt{9a^2b^4}}{\sqrt{a^2b^2}}=\frac{9.\left|a\right|.b^2}{\left|a\right|\left|b\right|}=9\left|b\right|\)

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

a/ \(=\frac{2}{a}.\frac{4\left|a\right|}{3}=\frac{2}{a}.\frac{-4a}{3}=\frac{-8}{3}\)

b/ \(=\frac{3}{a-1}.\frac{\left|2a-2\right|}{5}=\frac{3}{a-1}.\frac{2\left(a-1\right)}{5}=\frac{6}{5}\)

c/ \(=\sqrt{\frac{162a^2b^4}{2a^2b^2}}=\sqrt{81b^2}=9\left|b\right|\)

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

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

a)

\(=\frac{\left(\sqrt{a}+1\right)^2-\left(\sqrt{a}-1\right)^2+4\sqrt{a}\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}.\frac{1}{2a\sqrt{a}}\)

\(=\frac{a+2\sqrt{a}+1-a+2\sqrt{a}-1+4a\sqrt{a}-4\sqrt{a}}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}.\frac{1}{2a\sqrt{a}}\)

\(=\frac{4a\sqrt{a}}{a-1}.\frac{1}{2a\sqrt{a}}=\frac{2}{a-1}\)

b) \(\frac{2}{a-1}=a\Rightarrow a^2-a-2=0\)

Ta có: 1+1+(-2)=0, nên pt có 2 nghiệm a₁=-1<0 (không thỏa mãn đk)=> loại

a₂=2(thỏa mãn đk)=> chọn

Vậy a=2 thì P=a

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

\(A="\frac{1a+\sqrt{a}-1}{1-a}-\frac{1a\sqrt{a}-\sqrt{a}+a}{1-a\sqrt{a}}"\times\frac{a-\sqrt{a}}{1\sqrt{a}-1}\)

P/s: Ko chắc đâu nhé 

Đọc tiếp

.......

Điều kiện: x \(\ne\) 1;  1/4 ; x \(\ge\) 0

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

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

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

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

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

Các bài tập dạng này hoàn toàn làm tương tự!!!

a) ĐKXĐ : \(a\ge0;a\ne1\)

Ta có: \(P=\left(\frac{2a+1}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\right)\cdot\left(\frac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\left(\sqrt{a}+1\right)}-\sqrt{a}\right)\) \(=\frac{2a+1-a+\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+\sqrt{a}+1\right)}\cdot\left(a-\sqrt{a}+1-\sqrt{a}\right)\)

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