Bài 2:

a: =>2x^2-4x+1=x^2+x+5

=>x^2-5x-4=0

=>\(x=\dfrac{5\pm\sqrt{41}}{2}\)

b: =>11x^2-14x-12=3x^2+4x-7

=>8x^2-18x-5=0

=>x=5/2 hoặc x=-1/4

Ghi đề rõ ra b nhé

a.

\(3\sqrt{-x^2+x+6}\ge2\left(1-2x\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-x^2+x+6\ge0\\1-2x< 0\end{matrix}\right.\\\left\{{}\begin{matrix}1-2x\ge0\\9\left(-x^2+x+6\right)\ge4\left(1-2x\right)^2\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}-2\le x\le3\\x>\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\25\left(x^2-x-2\right)\le0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}< x\le3\\\left\{{}\begin{matrix}x\le\dfrac{1}{2}\\-1\le x\le2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-1\le x\le3\)

b.

ĐKXĐ: \(x\ge0\)

\(\Leftrightarrow\sqrt{2x^2+8x+5}-4\sqrt{x}+\sqrt{2x^2-4x+5}-2\sqrt{x}=0\)

\(\Leftrightarrow\dfrac{2x^2+8x+5-16x}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-4x+5-4x}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\dfrac{2x^2-8x+5}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{2x^2-8x+5}{\sqrt{2x^2-4x+5}+2\sqrt{x}}=0\)

\(\Leftrightarrow\left(2x^2-8x+5\right)\left(\dfrac{1}{\sqrt{2x^2+8x+5}+4\sqrt{x}}+\dfrac{1}{\sqrt{2x^2-4x+5}+2\sqrt{x}}\right)=0\)

\(\Leftrightarrow2x^2-8x+5=0\)

\(\Leftrightarrow x=\dfrac{4\pm\sqrt{6}}{2}\)

Câu b còn 1 cách giải nữa:

Với \(x=0\) không phải nghiệm

Với \(x>0\) , chia 2 vế cho \(\sqrt{x}\) ta được:

\(\sqrt{2x+8+\dfrac{5}{x}}+\sqrt{2x-4+\dfrac{5}{x}}=6\)

Đặt \(\sqrt{2x-4+\dfrac{5}{x}}=t>0\Leftrightarrow2x+8+\dfrac{5}{x}=t^2+12\)

Phương trình trở thành:

\(\sqrt{t^2+12}+t=6\)

\(\Leftrightarrow\sqrt{t^2+12}=6-t\)

\(\Leftrightarrow\left\{{}\begin{matrix}6-t\ge0\\t^2+12=\left(6-t\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\le6\\12t=24\end{matrix}\right.\)

\(\Rightarrow t=2\)

\(\Rightarrow\sqrt{2x-4+\dfrac{5}{x}}=2\)

\(\Leftrightarrow2x-4+\dfrac{5}{x}=4\)

\(\Rightarrow2x^2-8x+5=0\)

\(\Leftrightarrow...\)

ĐKXĐ : \(\left\{{}\begin{matrix}x\ne2\\x\ne\dfrac{3}{4}\end{matrix}\right.\)

PT \(\Leftrightarrow\dfrac{5\left(3-4x\right)+6\left(x-2\right)}{\left(x-2\right)\left(3-4x\right)}=0\)

\(\Leftrightarrow5\left(3-4x\right)+6\left(x-2\right)=0\)

\(\Leftrightarrow15-20x+6x-12=0\)

\(\Leftrightarrow-14x+3=0\)

\(\Leftrightarrow x=\dfrac{3}{14}\)

Vậy ...

\(\frac{5-x}{3}+\frac{3x-2}{5}=\frac{4x+3}{6-2x}\)

\(\frac{5\left(5-x\right)}{15}+\frac{3\left(3x-2\right)}{15}=\frac{2\left(2x+2\right)-1}{2\left(3-x\right)}\)

\(25-x+9x-6=\frac{2x+1}{3-x}\)

\(19+8x=\frac{2x+1}{3-x}\)

\(19+8x=\left(2x+1\right).\frac{1}{3-x}\)

\(19+8x=\frac{2x}{3-x}+\frac{1}{3-x}\)

\(19+8x=2x+1\)

\(19+8x-2x-1=0\)

\(18+6x=0\Leftrightarrow6x=-18\Leftrightarrow x=-3\)

ai mà biết "-"

a: Ta có: \(3x+5\le4x-9\)

\(\Leftrightarrow-x\le-14\)

\(\Leftrightarrow x\ge14\)

b: Ta có: \(6-2x< 6-x\)

\(\Leftrightarrow-x< 0\)

hay x>0

c: Ta có: \(7\left(x-1\right)+5>-3x\)

\(\Leftrightarrow7x-7+5+3x>0\)

\(\Leftrightarrow10x>2\)

hay \(x>\dfrac{1}{5}\)