Question

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

2) \(\sqrt[4]{x-1}+\sqrt[4]{5-x}=2\)

3) \(\sqrt[3]{1-2x}+\sqrt{x+3}=1\)

4) \(\dfrac{\sqrt{x}}{1+\sqrt{1-x}}=x^2-2x+2\)

Answer 1

Câu 1:

ĐKXĐ: \(x\geq \frac{1}{2}\)

Ta có: \(2\sqrt{x+3}=x-1+4\sqrt{2x-1}\)

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

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

\(\Leftrightarrow x-1+2.\frac{4(2x-1)-(x+3)}{2\sqrt{2x-1}+\sqrt{x+3}}=0\) (liên hợp)

\(\Leftrightarrow (x-1)+\frac{14(x-1)}{2\sqrt{2x-1}+\sqrt{x+3}}=0\)

\(\Leftrightarrow (x-1)\left(1+\frac{14}{2\sqrt{2x-1}+\sqrt{x+3}}\right)=0\)

Với mọi \(x\geq \frac{1}{2}\) ta luôn có \(1+\frac{14}{2\sqrt{2x-1}+\sqrt{x+3}}>0\). Do đó \(x-1=0\rightarrow x=1\) là nghiệm duy nhất

Answer 2

Câu 2:

ĐKXĐ: \(1\leq x\leq 5\)

Đặt \(\sqrt[4]{x-1}=a; \sqrt[4]{5-x}=b(a,b\geq 0)\). Khi đó ta có:

\(\left\{\begin{matrix} a+b=2\\ a^4+b^4=4\end{matrix}\right.\) \(\Rightarrow a^4+(2-a)^4=4\)

Đặt \(1-a=m\) thì pt trở thành:

\((1-m)^4+(m+1)^4=4\)

\(\Leftrightarrow 2m^4+12m^2+2=4\)

\(\Leftrightarrow m^4+6m^2-1=0\)

\(\Leftrightarrow (m^2+3)^2=10\Rightarrow m^2=\sqrt{10}-3\Rightarrow m=\pm \sqrt{\sqrt{10}-3}\)

\(\Rightarrow a=1\pm \sqrt{\sqrt{10}-3}\)

\(\Rightarrow x=(1\pm \sqrt{\sqrt{10}-3})^4+1\)

Answer 3

Câu 3:

ĐK: \(x\geq -3\). Đặt \(\sqrt[3]{1-2x}=a; \sqrt{x+3}=b\). Khi đó ta có:
\(\left\{\begin{matrix} a+b=1\\ a^3+2b^2=7\end{matrix}\right.\Rightarrow a^3+2(1-a)^2=7\)

\(\Leftrightarrow a^3+2a^2-4a-5=0\)

\(\Leftrightarrow (a+1)(a^2+a-5)=0\)

\(\Rightarrow \left[\begin{matrix} a=-1\\ a=\frac{-1\pm \sqrt{21}}{2}\end{matrix}\right.\). Vì \(b\geq 0\Rightarrow a=1-b\leq 1\). Do đó \(a=-1; a=\frac{-1-\sqrt{21}}{2}\)

\(\Rightarrow x=\frac{1-a^3}{2}=\left[\begin{matrix} 1\\ \frac{9+3\sqrt{21}}{2}\end{matrix}\right.\)

Answer 4

Câu 4: ĐK: \(0\leq x\leq 1\)

Thử ta dễ thấy $x=0$ không phải nghiệm của pt. Do đó $x\neq 0$

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

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

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

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

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

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

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

Vì với mọi \(0< x\leq 1\) thì \((\sqrt{1-x})^3+\frac{\sqrt{1-x}}{1+\sqrt{x}}+\sqrt{x}>0\), do đó \(\sqrt{1-x}=0\Rightarrow x=1\) là nghiệm duy nhất của pt.