Question

Giải phương trình:

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

b)\(\sqrt{3+x}+\sqrt{6-x}=3\)

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

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

e)\(\sqrt{x+9}=5-\sqrt{2x+4}\)

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

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

Answer 1

a) ĐK: \(x\ge -1\)

Ta có: \(x^2+\sqrt{x+1}=1\)

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

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

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

\(\Rightarrow \left[\begin{matrix} \sqrt{x+1}=0(1)\\ (x-1)\sqrt{x+1}+1=0(2)\end{matrix}\right.\)

Với \((1)\Rightarrow x+1=0\Rightarrow x=-1\) (thỏa mãn)

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

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

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

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

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

\(\Rightarrow \left[\begin{matrix} x=0\\ x+\sqrt{x+1}=0\end{matrix}\right.\)

Với \(x+\sqrt{x+1}=0\Rightarrow x=-\sqrt{x+1}\Rightarrow \left\{\begin{matrix} x\leq 0\\ x^2=x+1\end{matrix}\right.\Rightarrow x=\frac{1-\sqrt{5}}{2}\)

Vậy \(x=\left\{-1; \frac{1-\sqrt{5}}{2}; 0\right\}\)

Answer 2

b) ĐK: \(-3\leq x\leq 6\)

Ta có: \((\sqrt{3+x}+\sqrt{6-x})^2=3+x+6-x+2\sqrt{(3+x)(6-x)}\)

\(=9+2\sqrt{(3+x)(6-x)}\geq 9\)

\(\Rightarrow \sqrt{3+x}+\sqrt{6-x}\geq 3\) do \(\sqrt{3+x}+\sqrt{6-x}\) không âm.

Dấu "=" xảy ra khi \(\sqrt{(3+x)(6-x)}=0\Leftrightarrow x=-3; x=6\)

Vậy \(x=-3\) or $x=6$

Answer 3

c) ĐK: \(x\geq 1\)

Ta có: \(\sqrt{3x-2}+\sqrt{x-1}=3\)

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

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

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

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

Dễ thấy \(\frac{3}{\sqrt{3x-2}+2}+\frac{1}{\sqrt{x-1}+1}>0\), do đó \(x-2=0\Rightarrow x=2\) (thỏa mãn)

Vậy...........

Answer 4

d) ĐK: \(-3\leq x\leq 2\)

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

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

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

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

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

Dễ thấy \(\frac{1}{\sqrt{3+x}+2}+\frac{1}{\sqrt{2-x}+1}>0\), do đó \(x-1=0\Rightarrow x=1\) (thỏa mãn)

Vậy pt có nghiệm $x=1$

Answer 5

e) ĐK: \(x\geq -2\)

Dùng phương pháp liên hợp.

Ta có: \(\sqrt{x+9}=5-\sqrt{2x+4}\)

\(\Leftrightarrow \sqrt{x+9}+\sqrt{2x+4}-5=0\)

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

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

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

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

Dễ thấy \(\frac{1}{\sqrt{x+9}+3}+\frac{2}{\sqrt{2x+4}+2}>0\), do đó \(x=0\) là nghiệm của pt.

Answer 6

f) ĐK: \(x\geq \frac{1}{2}\)

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

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

Bình phương 2 vế:

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

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

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

\(\Rightarrow (x+3)(2x-1)=1\)

\(\Rightarrow 2x^2+5x-4=0\)

\(\Rightarrow x=\frac{-5\pm \sqrt{57}}{4}\)

Kết hợp với ĐKXĐ suy ra \(x=\frac{-5+\sqrt{57}}{4}\)

Answer 7

g) ĐK: \(x\geq \frac{3}{4}\)

Ta có:

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

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

\(\Rightarrow \left\{\begin{matrix} x-2\geq 0\\ 4x-3=(x-2)^2\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} x\geq 2\\ 4x-3=x^2-4x+4\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} x\geq 2\\ x^2-8x+7=0\end{matrix}\right.\)

\(\Rightarrow \left\{\begin{matrix} x\geq 2\\ (x-1)(x-7)=0\end{matrix}\right.\)

\(\Rightarrow x=7\)

Vậy $x=7$ là nghiệm của phương trình.