Question

[Lê Ánh ethuachenyu]

Giải pt sau:

a. \(\sqrt{2x^2-3x-11}=\sqrt{x^2-1}\)

b. \(\sqrt{2x^2+3x-5}=x+1\)

c. \(\sqrt{\left(x+2\right)^2}=\sqrt{3x^2-5x+14}\)

d. \(\sqrt{\left(x-1\right)\cdot\left(2x-3\right)}=-x-9\)

Answer 1

[Nguyễn Việt Lâm]

a/ \(\Rightarrow2x^2-3x-11=x^2-1\)

\(\Leftrightarrow x^2-3x-10=0\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

Thay 2 nghiệm vào cả 2 căn thức thấy đều xác định

Vậy nghiệm của pt là ...

b/ \(\left\{{}\begin{matrix}x\ge-1\\2x^2+3x-5=\left(x+1\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x^2+x-6=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x\ge-1\\\left[{}\begin{matrix}x=2\\x=-3\left(l\right)\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow x=2\)

Answer 2

[Nguyễn Việt Lâm]

c/

\(\Leftrightarrow x^2+4x+4=3x^2-5x+14\)

\(\Leftrightarrow2x^2-9x+10=0\)

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

d/

\(\Leftrightarrow\left\{{}\begin{matrix}-x-9\ge0\\\left(x-1\right)\left(2x-3\right)=\left(-x-9\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le-9\\2x^2-5x+3=x^2+18x+81\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le-9\\x^2-23x-78=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=26\left(ktm\right)\\x=-3\left(ktm\right)\end{matrix}\right.\)

Vậy pt vô nghiệm