Question

[Hoàng Thùy Linh]

Giải phương trình:

a, \(\sqrt{2x^2-9x+4}+3\sqrt{2x-1}=\sqrt{2x^2+21x-11}\)

b, \(\sqrt{1-x}+\sqrt{x^2-3x+2}+\left(x-2\right)\sqrt{\frac{x-1}{x-2}=3}\)

Answer 1

[Nguyễn Ngọc Lộc]

a, ĐKXĐ : Tự tìm hộ hen :)

Ta có : \(\sqrt{2x^2-9x+4}+3\sqrt{2x-1}=\sqrt{2x^2+21x-11}\)

=> \(\sqrt{2x^2-9x+4}+3\sqrt{2x-1}-\sqrt{2x^2+21x-11}=0\)

=> \(\sqrt{\left(x-4\right)\left(2x-1\right)}+3\sqrt{2x-1}-\sqrt{\left(2x-1\right)\left(x+11\right)}=0\)

=> \(\sqrt{2x-1}\left(\sqrt{x-4}+3-\sqrt{x+11}\right)=0\)

=> \(\left[{}\begin{matrix}\sqrt{2x-1}=0\\\sqrt{x-4}+3=\sqrt{x+11}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}2x-1=0\\x-4+6\sqrt{x-4}+9=x+11\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}2x-1=0\\6\sqrt{x-4}=6\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}2x-1=0\\x-4=1\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{1}{2}\\x=5\end{matrix}\right.\) ( TM )

Vậy ...

b, ĐKXĐ : Tiếp tục tìm hộ nha :)

Ta có : \(\sqrt{1-x}+\sqrt{x^2-3x+2}+\left(x-2\right)\sqrt{\frac{x-1}{x-2}}=3\)

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

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

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

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

=> \(\sqrt{1-x}\left(1+\sqrt{2-x}-\sqrt{2-x}\right)=3\)

=> \(\sqrt{1-x}=3\)

=> \(1-x=9\)

=> \(x=-8\left(TM\right)\)

Vậy ...