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

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

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

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

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

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

\(=\frac{a\sqrt{a}+4a+2\sqrt{a}+3}{\sqrt{a}\cdot\left(\sqrt{a}+1\right)}\)

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

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

\(=2+\dfrac{1}{\sqrt{a}+1}\cdot\dfrac{2a+2}{\sqrt{a}}\)

\(=\dfrac{2a+2\sqrt{a}+2a+2}{\sqrt{a}\left(\sqrt{a}+1\right)}=\dfrac{4a+2\sqrt{a}+2}{\sqrt{a}\left(\sqrt{a}+1\right)}\)

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

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

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

Bạn cần viết đề bằng công thức toán (biểu tượng $\sum$ bên trái khung soạn thảo) để được hỗ trợ tốt hơn.

a) \(\sqrt[]{x^2-4x+4}=x+3\)

\(\Leftrightarrow\sqrt[]{\left(x-2\right)^2}=x+3\)

\(\Leftrightarrow\left|x-2\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-\left(x+3\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0x=5\left(loại\right)\\x-2=-x-3\end{matrix}\right.\)

\(\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)

b) \(2x^2-\sqrt[]{9x^2-6x+1}=5\)

\(\Leftrightarrow2x^2-\sqrt[]{\left(3x-1\right)^2}=5\)

\(\Leftrightarrow2x^2-\left|3x-1\right|=5\)

\(\Leftrightarrow\left|3x-1\right|=2x^2-5\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2x^2-5\\3x-1=-2x^2+5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-4=0\left(1\right)\\2x^2+3x-6=0\left(2\right)\end{matrix}\right.\)

Giải pt (1)

\(\Delta=9+32=41>0\)

Pt \(\left(1\right)\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt[]{41}}{4}\)

Giải pt (2)

\(\Delta=9+48=57>0\)

Pt \(\left(2\right)\) \(\Leftrightarrow x=\dfrac{-3\pm\sqrt[]{57}}{4}\)

Vậy nghiệm pt là \(\left[{}\begin{matrix}x=\dfrac{3\pm\sqrt[]{41}}{4}\\x=\dfrac{-3\pm\sqrt[]{57}}{4}\end{matrix}\right.\)

ta có thể làm như sau: Bước 1: Rút gọn phần tử trong ngoặc đầu tiên: √a - 1 - 1 / √a = (√a * √a - √a - 1) / √a = (a - √a - 1) / √a Bước 2: Rút gọn phần tử trong ngoặc thứ hai: √a - 2 - √(a + 2) / √(a - 1) = (√a * √(a - 1) - 2 * √(a - 1) - √(a + 2)) / √(a - 1) = (a - √a - 2√(a - 1) - √(a + 2)) / √(a - 1) Bước 3: Thay các giá trị rút gọn vào biểu thức ban đầu: a = 1 / ((a - √a - 1) / √a) / (√a + 1 / ((a - √a - 2√(a - 1) - √(a + 2)) / √(a - 1))) Bước 4: Rút gọn biểu thức: a = √a * √(a - 1) / (a - √a - 1) * (√(a - 1) / (a - √a - 2√(a - 1) - √(a + 2))) Bước 5: Rút gọn thêm: a = √a * √(a - 1) / (a - √a - 1) * (√(a - 1) / (a - √a - 2√(a - 1) - √(a + 2))) * (√(a - 1) / (a - √a - 2√(a - 1) - √(a + 2))) Bước 6: Rút gọn thêm: a = (√a * √(a - 1))^2 / (a - √a - 1) * (√(a - 1))^2 / (a - √a - 2√(a - 1) - √(a + 2)) Bước 7: Rút gọn cuối cùng: a = (a(a - 1)) / ((a - √a - 1)(a - √a - 2√(a - 1) - √(a + 2)))