Question

Giải các phương trình sau :

a) \(\dfrac{14x-5-8x^2}{3x-1-2x^2}+\dfrac{3-2x}{x-1}=2\)

b) \(\sqrt{x^2-3x+8}+4=x\)

c) \(\sqrt{x^2-5x-2}=8-x\)

d) \(2-\dfrac{3}{3-x}=\dfrac{3-2x}{x^2-7x+12}\)

Answer 1

mng giúp e với ạ !

Answer 2

b)\(\Leftrightarrow\left\{{}\begin{matrix}x-4\ge0\\\sqrt{x^2-3x+8}=x-4\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\x^2-3x+8=\left(x-4\right)^2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\x^2-3x+8=x^2-8x+16\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\5x=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge4\\x=\dfrac{8}{5}\left(loại\right)\end{matrix}\right.\)=> pt vô nghiệm

c)\(\left\{{}\begin{matrix}8-x\ge0\\x^2-5x-2=\left(8-x\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le8\\x^2-5x-2=x^2-16x+64\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le8\\11x=66\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le8\\x=6\left(nhận\right)\end{matrix}\right.\)

Answer 3

d: \(\Leftrightarrow2+\dfrac{3}{x-3}=\dfrac{3-2x}{\left(x-3\right)\left(x-4\right)}\)

\(\Leftrightarrow2\left(x^2-7x+12\right)+3\left(x-4\right)=3-2x\)

=>2x^2-14x+24+3x-12=3-2x

=>2x^2-11x+12-3+2x=0

=>2x^2-9x+9=0

=>2x^2-3x-6x+9=0

=>(2x-3)(x-3)=0

=>x=3(loại) hoặc x=3/2(nhận)

a: \(\Leftrightarrow\dfrac{8x^2-14x+5}{\left(2x-1\right)\left(x-1\right)}+\dfrac{3-2x}{x-1}=2\)

\(\Leftrightarrow8x^2-14x+5+\left(3-2x\right)\left(2x-1\right)=2\left(2x^2-3x+1\right)\)

\(\Leftrightarrow8x^2-14x+5+6x-3-4x^2+2x=4x^2-6x+2\)

\(\Leftrightarrow4x^2-6x+2=4x^2-6x+2\)(luôn đúng)

Vậy: S=R\{1/2;1}