Bài 6: Biến đối đơn giản biểu thức chứa căn bậc hai

Sửa đề:

\(x=\dfrac{\left(\sqrt{5}+2\right)\sqrt[3]{17\sqrt{5}-38}}{\sqrt{5}+\sqrt{14-6\sqrt{5}}}\)

\(=\dfrac{\left(\sqrt{5}+2\right)\sqrt[3]{5\sqrt{5}-3.5.2+12\sqrt{5}-8}}{\sqrt{5}+\sqrt{9-6\sqrt{5}+5}}\)

\(=\dfrac{\left(\sqrt{5}+2\right)\sqrt[3]{\left(\sqrt{5}-2\right)^3}}{\sqrt{5}+\sqrt{\left(3-\sqrt{5}\right)^2}}\)

\(=\dfrac{\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)}{\sqrt{5}+\left(3-\sqrt{5}\right)}=\dfrac{1}{3}\)

Thế vô A ta được

\(A=\left(3.\dfrac{1}{3^3}+8.\dfrac{1}{3^2}+2\right)^{2018}=3^{2018}\)

Điều kiện: \(x\ge2\)

\(\left(\sqrt{x+1}-2\right)+\left(\sqrt{x-2}-1\right)+\left(3-\sqrt{2x+3}\right)+\left(\sqrt{5x+1}-4\right)=0\)

\(\Leftrightarrow\dfrac{x-3}{\sqrt{x+1}+2}+\dfrac{x-3}{\sqrt{x-2}+1}-\dfrac{2\left(x-3\right)}{\sqrt{2x+3}+3}+\dfrac{5\left(x-3\right)}{\sqrt{5x+1}+4}=0\)

\(\Leftrightarrow\left(x-3\right)\left(\dfrac{1}{\sqrt{x+1}+2}+\dfrac{1}{\sqrt{x-2}+1}-\dfrac{2}{\sqrt{2x+3}+3}+\dfrac{5}{\sqrt{5x+1}+4}\right)=0\)

\(\Leftrightarrow x=3\)

PS: Phần trong ngoặc tự chứng minh nha.

Điều kiện: \(x\ge\dfrac{1}{2}\)

\(\sqrt{2x-2\sqrt{2x-1}}-2\sqrt{2x+3-4\sqrt{2x-1}}+3\sqrt{2x+8-6\sqrt{2x-1}=0}\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-1}-1\right)^2}-2\sqrt{\left(\sqrt{2x-1}-2\right)^2}+3\sqrt{\left(\sqrt{2x-1}-3\right)^2}=0\)

\(\Leftrightarrow\left|\sqrt{2x-1}-1\right|-2\left|\sqrt{2x-1}-2\right|+3\left|\sqrt{2x-1}-3\right|=0\)

Với \(\dfrac{1}{2}\le x< 1\)

\(\Leftrightarrow1-\sqrt{2x-1}-2\left(2-\sqrt{2x-1}\right)+3\left(3-\sqrt{2x-1}\right)=0\)

\(\Leftrightarrow-2\sqrt{2x-1}+6=0\)

\(\Leftrightarrow x=5\left(l\right)\)

Tương tự cho các trường hợp: \(1\le x< \dfrac{5}{2};\dfrac{5}{2}\le x< 5;x\ge5\)

Tới đây thì kết luận thôi.

\(\sqrt{2x-2\sqrt{2x-1}}-2\sqrt{2x+3-4\sqrt{2x-1}}+3\sqrt{2x+8-6\sqrt{2x-1}}=0\)

ĐK:\(x\ge\dfrac{1}{2}\)

\(\Leftrightarrow\sqrt{2x-1-2\sqrt{2x-1}+1}-2\sqrt{2x-1-4\sqrt{2x-1}+4}+3\sqrt{2x-1-6\sqrt{2x-1}+9}=0\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-1}-1\right)^2}-2\sqrt{\left(\sqrt{2x-1}-2\right)^2}+3\sqrt{\left(\sqrt{2x-1}-3\right)^2}=0\)

\(\Leftrightarrow\sqrt{2x-1}-1-2\left(\sqrt{2x-1}-2\right)+3\left(\sqrt{2x-1}-3\right)=0\)

\(\Leftrightarrow\sqrt{2x-1}-1-2\sqrt{2x-1}+4+3\sqrt{2x-1}-9=0\)

\(\Leftrightarrow2\sqrt{2x-1}-6=0\)\(\Leftrightarrow\sqrt{2x-1}=3\)

\(\Leftrightarrow2x-1=9\Leftrightarrow2x=10\Rightarrow x=5\) *Thỏa*