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

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

P/s: Ko chắc đâu nhé 

Đọc tiếp

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

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

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

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

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

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

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

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

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

\(=\frac{1+\sqrt{a}+\sqrt{a}}{1+\sqrt{a}}\)

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

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!!!

xin lỗi,giờ mình mới học lớp 6 thôi

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

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

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

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

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

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

\(\Leftrightarrow a^2-2\sqrt{a}.a+2a-\sqrt{a}-2a=2a-2\sqrt{a}+2\)

\(\Leftrightarrow a^2-2\sqrt{a}.a+2a-\sqrt{a}-2a=-2\sqrt{a}+2\)

\(\Leftrightarrow-2\sqrt{a}.a+2a-\sqrt{a}-2a=-2\sqrt{a}+2-a^2\)

\(\Leftrightarrow-2\sqrt{a}.a-\sqrt{a}=-2\sqrt{a}+2-a^2\)

\(\Leftrightarrow-2a\sqrt{a}+\sqrt{a}=2-a^2\)

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

\(\Leftrightarrow\left[\sqrt{a}.\left(2a+1\right)\right]^2=\left(2-a^2\right)^2\)

\(\Leftrightarrow4a^3-4a^2+a=4-4a^2+a^4\)

\(\Leftrightarrow\orbr{\begin{cases}a=4\left(\text{thỏa mãn}\right)\\a=1\left(\text{loại}\right)\end{cases}}\)

=> a = 4

Cách ngắn hơn :

\(đkxđ\Leftrightarrow x\ge0\)

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

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

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

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

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

\(b,A=2\Rightarrow a-\sqrt{a}=2\)

\(\Rightarrow a-\sqrt{a}-2=0\)

\(\Rightarrow a+\sqrt{a}-2\sqrt{a}-2=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}\sqrt{a}=2\\\sqrt{a}=-1\end{cases}\Rightarrow\orbr{\begin{cases}a=4\\a\in\varnothing\end{cases}}}\)

\(\Rightarrow a=4\)

\(c,A=a-\sqrt{a}=\sqrt{a}^2-2.\sqrt{a}.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}\)

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

\(\Rightarrow A_{min}=-\frac{1}{4}\Leftrightarrow\left(\sqrt{a}-\frac{1}{2}\right)^2=0\)

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

Vậy với \(a=\frac{1}{4}\)thì A có giá trị nhỏ nhất là \(-\frac{1}{4}\)

Làm màu quá :))

ĐKXĐ :

\(a-\sqrt{a}+1\ne0\)

\(\sqrt{a}>0\)

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

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

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

\(=-\sqrt{a}\)