Đáp án:

 Đk: `0lexle5`

    `sqrt{x+3}+4sqrtx-2x=6-sqrt{5-x}`

    `<=>sqrt{x+3}+sqrt{5-x}=2(sqrtx-1)^2+4(1)`

   Vế trái của `(1)` bé hơn hoặc bằng `4;` vế phải lớn hơn hoặc bằng `4`

   Dấu bằng xảy ra khi và chỉ khi `<=>`$\left\{\begin{matrix} \sqrt{x+3}=\sqrt{5-x}\\   \sqrt{x}-1=0 \end{matrix}\right.$`<=>x=1`$(t/mđk)$

    Vậy phương trình có nghiệm duy nhất là `x=1` 

Đáp án: $x=1$

Giải thích các bước giải:

DKXD: $0\le x\le 5$
Ta có:
$\sqrt{x+3}+4\sqrt{x}-2x=6-\sqrt{5-x}$
$\to 2x-4\sqrt{x}-\sqrt{x+3}-\sqrt{5-x}+6=0$
$\to \left(2x-4\sqrt{x}+2\right)+\left(2-\sqrt{x+3}\right)+\left(2-\sqrt{5-x}\right)=0$
$\to 2\left(x-2\sqrt{x}+1\right)+\dfrac{2^2-\left(x+3\right)}{2+\sqrt{x+3}}+\dfrac{2^2-\left(5-x\right)}{2+\sqrt{5-x}}=0$

$\to 2\left(\sqrt{x}-1\right)^2+\dfrac{1-x}{2+\sqrt{x+3}}+\dfrac{x-1}{2+\sqrt{5-x}}=0$

$\to 2\left(\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\right)^2-\dfrac{x-1}{2+\sqrt{x+3}}+\dfrac{x-1}{2+\sqrt{5-x}}=0$

$\to 2\left(\dfrac{x-1}{\sqrt{x}+1}\right)^2-\dfrac{x-1}{2+\sqrt{x+3}}+\dfrac{x-1}{2+\sqrt{5-x}}=0$

$\to 2\cdot \dfrac{\left(x-1\right)^2}{\left(\sqrt{x}+1\right)^2}-\dfrac{x-1}{2+\sqrt{x+3}}+\dfrac{x-1}{2+\sqrt{5-x}}=0$

$\to \left(x-1\right)\left(2\cdot \dfrac{x-1}{\left(\sqrt{x}+1\right)^2}-\dfrac{1}{2+\sqrt{x+3}}+\dfrac{1}{2+\sqrt{5-x}}\right)=0$

$\to x-1=0\to x=1$

Hoặc $ 2\cdot \dfrac{x-1}{\left(\sqrt{x}+1\right)^2}-\dfrac{1}{2+\sqrt{x+3}}+\dfrac{1}{2+\sqrt{5-x}}=0$

$\to 2\cdot \dfrac{x-1}{\left(\sqrt{x}+1\right)^2}+\left(\dfrac{1}{2+\sqrt{5-x}}-\dfrac{1}{2+\sqrt{x+3}}\right)=0$

$\to 2\cdot \dfrac{x-1}{\left(\sqrt{x}+1\right)^2}+\dfrac{\sqrt{x+3}-\sqrt{5-x}}{\left(2+\sqrt{5-x}\right)\left(2+\sqrt{x+3}\right)}=0$

$\to 2\cdot \dfrac{x-1}{\left(\sqrt{x}+1\right)^2}+\dfrac{\dfrac{x+3-\left(5-x\right)}{\sqrt{x+3}+\sqrt{5-x}}}{\left(2+\sqrt{5-x}\right)\left(2+\sqrt{x+3}\right)}=0$

$\to 2\cdot \dfrac{x-1}{\left(\sqrt{x}+1\right)^2}+\dfrac{\dfrac{2\left(x-1\right)}{\sqrt{x+3}+\sqrt{5-x}}}{\left(2+\sqrt{5-x}\right)\left(2+\sqrt{x+3}\right)}=0$

$\to 2\cdot \dfrac{x-1}{\left(\sqrt{x}+1\right)^2}+\dfrac{\dfrac{2\left(x-1\right)}{\sqrt{x+3}+\sqrt{5-x}}}{\left(2+\sqrt{5-x}\right)\left(2+\sqrt{x+3}\right)}=0$

$\to\left(x-1\right)\left( 2\cdot \dfrac{1}{\left(\sqrt{x}+1\right)^2}+\dfrac{\dfrac{2}{\sqrt{x+3}+\sqrt{5-x}}}{\left(2+\sqrt{5-x}\right)\left(2+\sqrt{x+3}\right)}\right)=0$

Vì $2\cdot \dfrac{1}{\left(\sqrt{x}+1\right)^2}+\dfrac{\dfrac{2}{\sqrt{x+3}+\sqrt{5-x}}}{\left(2+\sqrt{5-x}\right)\left(2+\sqrt{x+3}\right)}>0$

$\to x-1=0\to x=1$