Question

[Ly nguyễn gia]

giải phương trình ( đặt ẩn phụ )

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

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

3) 4x² +10x+9=5. \(\sqrt{2x^2+5x+3}\)

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

5)\(\sqrt{x+1}\)+\(\sqrt{4-x}\)+\(\sqrt{\left(x+1\right)\left(4-x\right)}\)=5

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

Answer 1

[Nguyễn Việt Lâm]

f/

ĐKXĐ: ...

Đặt \(\sqrt{2-x}+\sqrt{x+2}=a>0\)

\(\Rightarrow a^2=4+2\sqrt{4-x^2}\Rightarrow\sqrt{4-x^2}=\frac{a^2-4}{2}\)

Phương trình trở thành:

\(a+\frac{a^2-4}{2}=2\)

\(\Leftrightarrow a^2+2a-8=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-4\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{4-x^2}=\frac{a^2-4}{2}=0\)

\(\Rightarrow4-x^2=0\Rightarrow x=\pm2\)

Answer 2

[Nguyễn Việt Lâm]

e/ ĐKXĐ: ...

Đặt \(\sqrt{x+1}+\sqrt{4-x}=a>0\)

\(\Rightarrow a^2=5+2\sqrt{\left(x+1\right)\left(4-x\right)}\Rightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=\frac{a^2-5}{2}\)

Pt trở thành:

\(a+\frac{a^2-5}{2}=5\)

\(\Leftrightarrow a^2+2a-15=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-5\left(l\right)\end{matrix}\right.\)

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

\(\Leftrightarrow5+2\sqrt{\left(x+1\right)\left(4-x\right)}=9\)

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

\(\Leftrightarrow\left(x+1\right)\left(4-x\right)=4\)

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