Question

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

2)\(\sqrt{25-x^2}=x-1\)

3)(x+4).\(\sqrt{10-x^2}=x^2+2x-8\)

4)(x-3).\(\sqrt{x^2-3x+2}=x^2-8x+15\)

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

6)\(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\)

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

8)x-4\(\sqrt{2x+4}-2\sqrt{1-x}+10=0\)

Answer 1

1.

ĐK: $-x^2+2x+4\geq 0$

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

\(\Leftrightarrow \left\{\begin{matrix} x\geq 2\\ 6x=2x^2\end{matrix}\right.\Rightarrow x=3\) (thỏa mãn)

Vậy...........

Answer 2

2)

ĐK: $-5\leq x\leq 5$

PT \(\Rightarrow \left\{\begin{matrix} x-1\geq 0\\ 25-x^2=(x-1)^2=x^2-2x+1\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ 2x^2-2x-24=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ x^2-x-12=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 1\\ (x+3)(x-4)=0\end{matrix}\right.\)

\(\Rightarrow x=4\) (thỏa mãn)

Answer 3

3)

ĐK: $x^2\leq 10$

PT $\Leftrightarrow (x+4)\sqrt{10-x^2}=(x+4)(x-2)$

$\Leftrightarrow (x+4)[\sqrt{10-x^2}-(x-2)]=0$

Nếu $x+4=0\Rightarrow x=-4$ (không thỏa mãn ĐKXĐ)

Nếu $\sqrt{10-x^2}-(x-2)=0$

$\Leftrightarrow \sqrt{10-x^2}=x-2$

\(\Rightarrow \left\{\begin{matrix} x-2\geq 0\\ 10-x^2=(x-2)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 2\\ 2x^2-4x-6=0\end{matrix}\right.\)

\(\Leftrightarrow \left\{\begin{matrix} x\geq 0\\ 2(x-3)(x+1)=0\end{matrix}\right.\Rightarrow x=3\)

Answer 4

4)

ĐKXĐ: $x\geq 2$ hoặc $x\leq 1$
PT $\Leftrightarrow (x-3)\sqrt{x^2-3x+2}=(x-3)(x-5)$

$\Leftrightarrow (x-3)[\sqrt{x^2-3x+2}-(x-5)]=0$

Nếu $x-3=0\Rightarrow x=3$ (thỏa mãn)

Nếu $\sqrt{x^2-3x+2}-(x-5)=0$

$\Leftrightarrow \sqrt{x^2-3x+2}=x-5$

\(\Rightarrow \left\{\begin{matrix} x-5\geq 0\\ x^2-3x+2=(x-5)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 5\\ 7x-23=0\end{matrix}\right.\) (không có giá trị nào thỏa mãn)

Vậy $x=3$

Answer 5

5)

ĐKXĐ: $x\geq 1$

PT $\Leftrightarrow \sqrt{(x-1)-4\sqrt{x-1}+4}+\sqrt{(x-1)-6\sqrt{x-1}+9}=1$

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

$\Leftrightarrow |\sqrt{x-1}-2|+|\sqrt{x-1}-3|=1$

Ta thấy:

$|\sqrt{x-1}-2|+|\sqrt{x-1}-3|=|\sqrt{x-1}-2|+|3-\sqrt{x-1}|\geq |\sqrt{x-1}-2+3-\sqrt{x-1}|=1$

Dấu "=" xảy ra khi $(\sqrt{x-1}-2)(3-\sqrt{x-1})\geq 0$

$\Leftrightarrow 3\geq \sqrt{x-1}\geq 2$

$\Leftrightarrow 10\geq x\geq 5$

Kết hợp đkxđ ta suy ra $x\in [5;10]$

Answer 6

6)

ĐKXĐ: $x\geq \frac{5}{2}$

PT $\Leftrightarrow \sqrt{2x+4+6\sqrt{2x-5}}+\sqrt{2x-4-2\sqrt{2x-5}}=4$

$\Leftrightarrow \sqrt{(2x-5)+6\sqrt{2x-5}+9}+\sqrt{(2x-5)-2\sqrt{2x-5}+1}=4$

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

$\Leftrightarrow |\sqrt{2x-5}+3|+|1-\sqrt{2x-5}|=4$

Ta thấy: $ |\sqrt{2x-5}+3|+|1-\sqrt{2x-5}|\geq |\sqrt{2x-5}+3+1-\sqrt{2x-5}|=4$

Dấu "=" xảy ra khi $(\sqrt{2x-5}+3)(1-\sqrt{2x-5})\geq 0$

$\Leftrightarrow 1-\sqrt{2x-5}\geq 0$

$\Leftrightarrow \sqrt{2x-5}\leq 1$

$\Leftrightarrow \frac{5}{2}\leq x\leq 3$

Answer 7

7)

ĐKXĐ: $x\geq -1$

PT $\Leftrightarrow x^2+[(x+1)-2\sqrt{x+1}+1]=0$

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

Ta thấy:

$x^2\geq 0; (\sqrt{x+1}-1)^2\geq 0$ với mọi $x\geq -1$

Do đó để tổng của chúng bằng $0$ thì $x^2=(\sqrt{x+1}-1)^2=0$

$\Leftrightarrow x=0$ (thỏa mãn)

Vậy.......

Answer 8

8) ĐKXĐ: $-2\leq x\leq 1$

PT $\Leftrightarrow (2x+4)-4\sqrt{2x+4}+4+[(1-x)-2\sqrt{1-x}+1]=0$

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

Dễ thấy: $(\sqrt{2x+4}-2)^2; (\sqrt{1-x}-1)^2\geq 0$ với mọi $x\in [-2;1]$ nên để tổng của chúng bằng $0$ thì:

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

$\Leftrightarrow \sqrt{2x+4}=2; \sqrt{1-x}-1=0$

$\Leftrightarrow x=0$ (thỏa mãn)

Vậy.....