Question

Giải các phương trình sau :

a)\(\left|\dfrac{x+5}{-x^2+9}\right|=2\)

b)\(\dfrac{4}{\sqrt{2-x}}-\sqrt{2-x}=2\)

c)\(^{x^2-6x+9=4\sqrt{x^2-6x+6}}\)

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

e)\(\sqrt{x+1}=8-\sqrt{3x+1}\)

f')(x-2)\(\sqrt{2x+7}=x^2-4\)

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

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

i) \(\sqrt{x+4}-\sqrt{3x+1}+2\sqrt{3x^2+13x+4}=51-4x\)

k)\(\dfrac{x-2}{1-x}+\dfrac{x-3}{x+1}=\dfrac{x^2+4x+15}{x^2-1}\)

Answer 1

Câu a:

ĐKXĐ: \(x\neq \pm 3\)

\(\left|\frac{x+5}{-x^2+9}\right|=2\Rightarrow \left[\begin{matrix} \frac{x+5}{-x^2+9}=2\\ \frac{x+5}{-x^2+9}=-2\end{matrix}\right.\)

\(\Rightarrow \left[\begin{matrix} x+5=2(-x^2+9)\\ x+5=-2(-x^2+9)\end{matrix}\right.\Rightarrow \left[\begin{matrix} 2x^2+x-13=0\\ 2x^2-x-23=0\end{matrix}\right.\)

\(\Rightarrow \left[\begin{matrix} x=\frac{-1\pm \sqrt{105}}{4}\\ x=\frac{1\pm \sqrt{185}}{4}\end{matrix}\right.\) (đều thỏa mãn )

Vậy.......

Answer 2

Câu b:

ĐKXĐ: \(x< 2\)

Ta có: \(\frac{4}{\sqrt{2-x}}-\sqrt{2-x}=2\)

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

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

\(\Leftrightarrow 5=(2-x)+2\sqrt{2-x}+1=(\sqrt{2-x}+1)^2\)

\(\Rightarrow \sqrt{2-x}+1=\sqrt{5}\) (do \(\sqrt{2-x}+1>0\) )

\(\Rightarrow \sqrt{2-x}=\sqrt{5}-1\)

\(\Rightarrow 2-x=6-2\sqrt{5}\)

\(\Rightarrow x=-4+2\sqrt{5}\) (thỏa mãn)

Vậy...........

Answer 3

Câu c:

ĐKXĐ: \(x\leq 3-\sqrt{3}\) hoặc \(x\geq 3+\sqrt{3}\)

Ta có: \(x^2-6x+9=4\sqrt{x^2-6x+6}\)

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

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

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

\(\Rightarrow \left[\begin{matrix} \sqrt{x^2-6x+6}-2=1\\ \sqrt{x^2-6x+6}-2=-1\end{matrix}\right.\Rightarrow \left[\begin{matrix} \sqrt{x^2-6x+6}=3\\ \sqrt{x^2-6x+6}=1\end{matrix}\right.\)

\(\Rightarrow \left[\begin{matrix} x^2-6x-3=0\\ x^2-6x+5=0\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=3\pm 2\sqrt{3}\\ x=5\\ x=1\end{matrix}\right.\) (đều thỏa mãn)

Vậy PT có 4 nghiệm \(x\in \left\{3+2\sqrt{3}; 3-2\sqrt{3}; 1;5\right\}\)

Answer 4

Câu d:

ĐKXĐ : \(x\geq 3\)

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

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

\(\Rightarrow (x-3)(x-2)=4\)

\(\Leftrightarrow x^2-5x+6=4\)

\(\Leftrightarrow x^2-5x+2=0\)

\(\Rightarrow \left[\begin{matrix} x=\frac{5+\sqrt{17}}{2}\\ x=\frac{5-\sqrt{17}}{2}\end{matrix}\right.\)

Kết hợp với ĐKXĐ suy ra \(x=\frac{5+\sqrt{17}}{2}\)

Answer 5

Câu e:

ĐKXĐ: \(x\geq \frac{-1}{3}\)

Ta có: \(\sqrt{x+1}=8-\sqrt{3x+1}\)

\(\Leftrightarrow \sqrt{x+1}+\sqrt{3x+1}-8=0\)

\(\Leftrightarrow (\sqrt{x+1}-3)+(\sqrt{3x+1}-5)=0\)

\(\Leftrightarrow \frac{x-8}{\sqrt{x+1}+3}+\frac{3(x-8)}{\sqrt{3x+1}+5}=0\) (liên hợp)

\(\Leftrightarrow (x-8)\left(\frac{1}{\sqrt{x+1}+3}+\frac{3}{\sqrt{3x+1}+5}\right)=0\)

Dễ thấy biểu thức \(\frac{1}{\sqrt{x+1}+3}+\frac{3}{\sqrt{3x+1}+5}>0\). Do đó \(x-8=0\Rightarrow x=8\) (t.m)

Vậy PT có nghiệm duy nhất $x=8$

Answer 6

Câu f:

ĐKXĐ: \(x\geq \frac{-7}{2}\)

Ta có: \((x-2)\sqrt{2x+7}=x^2-4\)

\(\Leftrightarrow (x-2)\sqrt{2x+7}-(x^2-4)=0\)

\(\Leftrightarrow (x-2)\sqrt{2x+7}-(x-2)(x+2)=0\)

\(\Leftrightarrow (x-2)[\sqrt{2x+7}-(x+2)]=0\)

\(\Rightarrow \left[\begin{matrix} x-2=0\\ \sqrt{2x+7}=x+2\end{matrix}\right.\)

Với \(x-2=0\Rightarrow x=2\) (thỏa mãn)

Với \(\sqrt{2x+7}=x+2\Rightarrow \left\{\begin{matrix} x\geq -2\\ 2x+7=(x+2)^2\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x\geq -2\\ x^2+2x-3=0\end{matrix}\right.\)

\(\Rightarrow x=1\)

Vậy \(x\in\left\{1;2\right\}\)

Answer 7

Câu g:

ĐK: \(x\geq 1\)

Ta có:

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

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

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

Nếu \(x\geq 2\) thì:

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

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

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

\(\Rightarrow \sqrt{x-1}-2=0\) (do \(x\geq 2\Rightarrow \sqrt{x-1}\neq 0\) )

\(\Rightarrow x=5\) (thỏa mãn)

Nếu \(1\leq x< 2\) thì:

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

\(\Leftrightarrow 2=x-1\Rightarrow x=3\) (không thỏa mãn)

Vậy PT có nghiệm duy nhất \(x=5\)

Answer 8

h)

ĐKXĐ: \(-4\leq x\leq \frac{1}{2}\)

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

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

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

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

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

Dễ thấy \(\frac{1}{\sqrt{x+4}+2}+\frac{1}{\sqrt{1-x}+1}+\frac{2}{\sqrt{1-2x}+1}>0\)

Do đó $x=0$ là nghiệm duy nhất của pt.

Answer 9

Câu i:

Bạn xem lại \(51-4x\) hay \(51+4x\).

Câu k:

ĐKXĐ: \(x\neq \pm 1\)

\(\frac{x-2}{1-x}+\frac{x-3}{x+1}=\frac{x^2+4x+15}{x^2-1}\)

\(\Leftrightarrow \frac{(x-2)(x+1)+(x-3)(1-x)}{(1-x)(1+x)}=\frac{x^2+4x+15}{x^2-1}\)

\(\Leftrightarrow \frac{3x-5}{1-x^2}=\frac{x^2+4x+15}{x^2-1}\)

\(\Leftrightarrow \frac{5-3x}{x^2-1}=\frac{x^2+4x+15}{x^2-1}\)

\(\Rightarrow 5-3x=x^2+4x+15\)

\(\Leftrightarrow x^2+7x+10=0\)

\(\Leftrightarrow (x+2)(x+5)=0\Rightarrow \left[\begin{matrix} x=-2\\ x=-5\end{matrix}\right.\)(đều thỏa mãn)