Question

[bach nhac lam]

gpt : a) \(\sqrt{x}+\sqrt{x+\sqrt{1-x}}=1\)

b) \(\sqrt{1-\sqrt{x^2-x}}=\sqrt{x}-1\)

c) \(\sqrt{x^2+6}=x-2\sqrt{x^2-1}\)

Answer 1

[Trần Thanh Phương]

c) \(\sqrt{x^2+6}=x-2\sqrt{x^2-1}\)

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

\(\Leftrightarrow x^2+6+4\left(x^2-1\right)+4\sqrt{\left(x^2+6\right)\left(x^2-1\right)}=x^2\)

\(\Leftrightarrow6+4x^2-4+4\sqrt{\left(x^2+6\right)\left(x^2-1\right)}=0\)

\(\Leftrightarrow4x^2+2+4\sqrt{\left(x^2+6\right)\left(x^2-1\right)}=0\)

\(\Leftrightarrow2x^2+2\sqrt{\left(x^2+6\right)\left(x^2-1\right)}+1=0\)

Dễ thấy \(VT>0\forall x\)

Do đó pt vô nghiệm

Answer 2

[Akai Haruma]

Lời giải:
a)

ĐK: \(0\leq x\leq 1\)

PT \(\Leftrightarrow \sqrt{x+\sqrt{1-x}}=1-\sqrt{x}\)

\(\Rightarrow x+\sqrt{1-x}=1+x-2\sqrt{x}\) (bình phương 2 vế)

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

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

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

Ta thấy \(\sqrt{1-x}+1\geq 1\Rightarrow \frac{\sqrt{x}}{\sqrt{1-x}+1}\leq \sqrt{x}\leq 1< 2\) với mọi $0\leq x\leq 1$

\(\Rightarrow 2-\frac{\sqrt{x}}{\sqrt{1-x}+1}>0\Rightarrow 2-\frac{\sqrt{x}}{\sqrt{1-x}+1}\neq 0\)

Do đó $\sqrt{x}=0\Leftrightarrow x=0$ là nghiệm duy nhất

b)

ĐK: \(1 \leq x\leq \frac{1+\sqrt{5}}{2}\) hoặc \(0\geq x\geq \frac{1-\sqrt{5}}{2}\)

PT \(\Rightarrow \left\{\begin{matrix} \sqrt{x}-1\geq 0\\ 1-\sqrt{x^2-x}=x-2\sqrt{x}+1\end{matrix}\right.\) (bình phương 2 vế)

\(\Leftrightarrow \left\{\begin{matrix} x\geq 1(1)\\ x+\sqrt{x^2-x}-2\sqrt{x}=0(2)\end{matrix}\right.\)

(1) kết hợp với ĐKXĐ suy ra \(1\leq x\leq \frac{1+\sqrt{5}}{2}(*)\)

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

Từ $(*)$ suy ra $x\neq 0$. Do đó \(\sqrt{x}+\sqrt{x-1}-2=0\)

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

\(\Rightarrow x-1=4+x-4\sqrt{x}\) (bình phương)

\(\Leftrightarrow 4\sqrt{x}=5\Rightarrow x=\frac{25}{16}\) (thỏa mãn $(*)$)

Vậy......