\(\left|4x-1\right|=5-x\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-1=5-x\left(x\ge\dfrac{1}{4}\right)\\4x-1=x-5\left(x< \dfrac{1}{4}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{6}{5}\left(nhận\right)\\x=-\dfrac{4}{3}\left(nhận\right)\end{matrix}\right.\)

\(\Leftrightarrow x\sqrt{2x-1}-4x+2=0\)0

\(\Leftrightarrow x\sqrt{2x-1}-2\left(2x-1\right)=0\)

\(\Leftrightarrow\sqrt{2x-1}\left(x-2\sqrt{2x-1}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{2x-1}=0\\x-2\sqrt{2x-1}=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=2\sqrt{2x-1}\left(1\right)\end{cases}}\)

+) giải phương trình (1) ta có 

\(x=2\sqrt{2x-1}\)

\(\Leftrightarrow x^2=4.\left(2x-1\right)=0\)

\(\Leftrightarrow x^2-8x+4=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=4-2\sqrt{3}\\x=4+2\sqrt{3}\end{cases}}\)

Vậy phương trình đã cho có 3 nghiệm là \(x=\frac{1}{2};x=4+2\sqrt{3};x=4-2\sqrt{3}\)

Đặt \(\sqrt{2x-1}=t\Rightarrow t^2=2x-1\Rightarrow x=\frac{t^2+1}{2}\)

Vậy pt đã cho \(\Leftrightarrow\frac{t^2+1}{2}\cdot t=2t^2\\ \Leftrightarrow t^3+t-4t^2=0\Rightarrow t\left(t^2-4t+1\right)=0\)

\(t=0\Rightarrow x=\frac{1}{2}\left(tm\right)\)

\(t^2-4t+1=0\Rightarrow\orbr{\begin{cases}t=2-\sqrt{3}\\t=2+\sqrt{3}\end{cases}}\)

\(t=2-\sqrt{3}\Rightarrow2x-1=7-4\sqrt{3}\Rightarrow2x=8-4\sqrt{3}\\ \Rightarrow x=4-2\sqrt{3}\)

\(t=2+\sqrt{3}\Rightarrow2x-1=7+4\sqrt{3}\Rightarrow2x=8+4\sqrt{3}\\ \Rightarrow x=4+2\sqrt{3}\)

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

Nếu    \(x^2-9x+14=\left(x-7\right)\left(x-2\right)\ge0\)

\(\Leftrightarrow\)\(x\ge7;\)\(x\le2\)

thì  \(\left|x^2-9x+14\right|=x^2-9x+14\)

Khi đó bpt trở thành:          \(x^2-9x+14+3x>x^2-4\)

                               \(\Leftrightarrow\)\(-6x>-18\)

                              \(\Leftrightarrow\) \(x< 3\)(thỏa mãn)

Nếu  \(x^2-9x+14=\left(x-7\right)\left(x-2\right)< 0\)

\(\Leftrightarrow\)\(2< x< 7\)

thì    \(\left|x^2-9x+14\right|=-x^2+9x-14\)

Khi đó bpt trở thành:     \(-x^2+9x-14+3x>x^2-4\)

                              \(\Leftrightarrow\)\(-2x^2+12x-10>0\)

                             \(\Leftrightarrow\) \(x^2-6x+5< 0\)

                            \(\Leftrightarrow\)\(\left(x-1\right)\left(x-5\right)< 0\)

                           \(\Leftrightarrow\) \(1< x< 5\) (thỏa mãn)

  Vậy...

Vô nghiệm.