Câu hỏi của Julian Edward

Question

giải pt

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

b) $\sqrt{x-2}-\sqrt{x+2}=2\sqrt{x^2-4}-2x+2$

c) $\sqrt{2x+3}+\sqrt{x+1}=3x+2\sqrt{2x^2+5x+3}-16$

d) \(2\sqrt{x}\left(\sqrt{x+1}-2\sqrt{x}...

Nguyễn Việt Lâm · Accepted Answer

a/ ĐKXĐ: $x\ge0$ $\Leftrightarrow\sqrt{x+1}+\sqrt{x}+2x+1+2\sqrt{x^2+x}-2=0$ Đặt $\sqrt{x+1}+\sqrt{x}=a>0\Rightarrow a^2=2x+1+2\sqrt{x^2+x}$ $\Rightarrow a+a^2-2=0\Rightarrow\left[{}\begin{matrix}a=1\a=-2\left(l\right)\end{matrix}\right.$ $\Rightarrow\sqrt{x+1}+\sqrt{x}=1$ Mà $x\ge0\Rightarrow\left\{{}\begin{matrix}\sqrt{x}\ge0\\sqrt{x+1}\ge1\end{matrix}\right.$ $\Rightarrow\sqrt{x+1}+\sqrt{x}\ge1$ Dấu "=" xảy ra khi và chỉ khi $x=0$ b/ ĐKXĐ: $x\ge2$ $\Leftrightarrow\sqrt{x-2}-\sqrt{x+2}+2x-2\sqrt{x^2-4}-2=0$ Đặt $\sqrt{x-2}-\sqrt{x+2}=a< 0$ $\Rightarrow a^2=2x-2\sqrt{x^2-4}$ , pt trở thành: $a+a^2-2=0\Rightarrow\left[{}\begin{matrix}a=1\left(l\right)\a=-2\end{matrix}\right.$ $\Rightarrow\sqrt{x-2}-\sqrt{x+2}=-2$ $\Leftrightarrow\sqrt{x-2}+2=\sqrt{x+2}$ $\Leftrightarrow x+2+4\sqrt{x-2}=x+2$ $\Leftrightarrow4\sqrt{x-2}=0\Rightarrow x=2$Câu trả lời: a/ ĐKXĐ: $x\ge0$ $\Leftrightarrow\sqrt{x+1}+\sqrt{x}+2x+1+2\sqrt{x^2+x}-2=0$ Đặt $\sqrt{x+1}+\sqrt{x}=a>0\Rightarrow a^2=2x+1+2\sqrt{x^2+x}$ $\Rightarrow a+a^2-2=0\Rightarrow\left[{}\begin{matrix}a=1\a=-2\left(l\right)\end{matrix}\right.$ $\Rightarrow\sqrt{x+1}+\sqrt{x}=1$ Mà $x\ge0\Rightarrow\left\{{}\begin{matrix}\sqrt{x}\ge0\\sqrt{x+1}\ge1\end{matrix}\right.$ $\Rightarrow\sqrt{x+1}+\sqrt{x}\ge1$ Dấu "=" xảy ra khi và chỉ khi $x=0$ b/ ĐKXĐ: $x\ge2$ $\Leftrightarrow\sqrt{x-2}-\sqrt{x+2}+2x-2\sqrt{x^2-4}-2=0$ Đặt $\sqrt{x-2}-\sqrt{x+2}=a< 0$ $\Rightarrow a^2=2x-2\sqrt{x^2-4}$ , pt trở thành: $a+a^2-2=0\Rightarrow\left[{}\begin{matrix}a=1\left(l\right)\a=-2\end{matrix}\right.$ $\Rightarrow\sqrt{x-2}-\sqrt{x+2}=-2$ $\Leftrightarrow\sqrt{x-2}+2=\sqrt{x+2}$ $\Leftrightarrow x+2+4\sqrt{x-2}=x+2$ $\Leftrightarrow4\sqrt{x-2}=0\Rightarrow x=2$

Nguyễn Việt Lâm · Answer

c/ĐKXĐ: $x\ge-1$

$\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}-\left(\sqrt{2x+3}+\sqrt{x+1}\right)-20=0$

Đặt $\sqrt{2x+3}+\sqrt{x+1}=a>0$

$\Rightarrow a^2=3x+4+2\sqrt{2x^2+5x+3}$, ta được:

$a^2-a-20=0\Rightarrow\left[{}\begin{matrix}a=5\a=-4\left(l\right)\end{matrix}\right.$

$\Rightarrow\sqrt{2x+3}+\sqrt{x+1}=5$

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

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

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

$\Rightarrow x=3$Câu trả lời: c/ĐKXĐ: $x\ge-1$

$\Leftrightarrow3x+4+2\sqrt{2x^2+5x+3}-\left(\sqrt{2x+3}+\sqrt{x+1}\right)-20=0$

Đặt $\sqrt{2x+3}+\sqrt{x+1}=a>0$

$\Rightarrow a^2=3x+4+2\sqrt{2x^2+5x+3}$, ta được:

$a^2-a-20=0\Rightarrow\left[{}\begin{matrix}a=5\a=-4\left(l\right)\end{matrix}\right.$

$\Rightarrow\sqrt{2x+3}+\sqrt{x+1}=5$

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

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

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

$\Rightarrow x=3$

Nguyễn Việt Lâm · Answer

d/ ĐKXĐ: $x\ge0$

$\Leftrightarrow2\sqrt{x^2+x}-4x+\sqrt{x+1}+\sqrt{x}+6x-1=0$

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

Đến đây thì nó giống hệt câu a không khác 1 chữ nào

e/ ĐKXĐ: $x\ge-3$

$\Leftrightarrow x^2+x+3+2x\sqrt{x+3}+x+\sqrt{x+3}-12=0$

Đặt $x+\sqrt{x+3}=a\ge-3\Rightarrow a^2=x^2+x+3+2x\sqrt{x+3}$

Phương trình trở thành:

$a^2+a-12=0\Rightarrow\left[{}\begin{matrix}a=3\a=-4\left(l\right)\end{matrix}\right.$

$\Rightarrow x+\sqrt{x+3}=3$

$\Leftrightarrow\sqrt{x+3}=3-x$ ($x\le3$)

$\Leftrightarrow x+3=\left(3-x\right)^2$

$\Leftrightarrow x^2-7x+6=0\Rightarrow\left[{}\begin{matrix}x=1\x=6\left(l\right)\end{matrix}\right.$Câu trả lời: d/ ĐKXĐ: $x\ge0$