mong mọi người giải giúp em vs 

      \(\left(4x-5\right)^2\left(2x-3\right)\left(x-1\right)=9\)

\(\Leftrightarrow\left(4x-5\right)^2\left(2x-3\right).2.\left(x-1\right).4=9.2.4\)

\(\Leftrightarrow\left(4x-5\right)^2\left(4x-6\right)\left(4x-4\right)=72\)(1)

Đặt \(4x-5=a\)

Khi đó (1) trở thành: 

      \(a^2\left(a-1\right)\left(a+1\right)=72\)

\(\Leftrightarrow a^2\left(a^2-1\right)=72\)    

\(\Leftrightarrow a^4-a^2-72=0\)

\(\Leftrightarrow a^4-9a^2+8a^2-72=0\)

\(\Leftrightarrow a^2\left(a^2-9\right)+8\left(a^2-9\right)=0\)

\(\Leftrightarrow\left(a^2-9\right)\left(a^2+8\right)=0\)

\(\Leftrightarrow a^2-9=0\) (vì \(a^2+8>0\forall a\) )

\(\Leftrightarrow\orbr{\begin{cases}a=3\\a=-3\end{cases}}\)

- Với \(a=3\Rightarrow4x-5=3\Rightarrow x=2\)

-Với \(a=-3\Rightarrow4x-5=-3\Rightarrow x=\frac{1}{2}\)

Vậy \(x=2,x=\frac{1}{2}\)

Chúc bạn học tốt.

a, Ta có: \(\Delta'=1-m+3=4-m\)

Phương trình có 2 nghiệm phân biệt \(\Leftrightarrow\Delta'>0\Leftrightarrow4-m>0\Leftrightarrow m< 4\)

b, ĐXXĐ: \(x\le\frac{9}{4}\)

\(pt\Leftrightarrow\sqrt{\left(9-4x\right)\left(x-3\right)^2}=\left|-2x+5\right|\sqrt{9-4x}\)

\(\Leftrightarrow\sqrt{9-4x}\left(\left|x-3\right|-\left|-2x+5\right|\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}9-4x=0\\\left|x-3\right|=\left|-2x+5\right|\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}9-4x=0\\x-3=-2x+5\\x-3=2x-5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{9}{4}\\x=\frac{8}{3}\left(l\right)\\x=2\end{matrix}\right.\)

Vậy pt đã cho có 2 nghiệm \(x=2;x=\frac{9}{4}\)

a/ \(x\ge-3\)

\(\Leftrightarrow\left(2x-1\right)^2=\left(x+3\right)^2\)

\(\Leftrightarrow3x^2-10x-8=0\Rightarrow\left[{}\begin{matrix}x=4\\x=-\frac{2}{3}\end{matrix}\right.\)

b/ \(x\ge-\frac{5}{2}\)

\(\Leftrightarrow\left(4x+7\right)^2=\left(2x+5\right)^2\)

\(\Leftrightarrow x^2+3x+2=0\Rightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)

c/ \(x\ge1\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x^2-8x=0\\2x^2+2x-10=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=4\\x=\frac{-1+\sqrt{21}}{2}\\x=\frac{-1-\sqrt{21}}{2}\left(l\right)\end{matrix}\right.\)

d/ \(x\ge\frac{17}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=4x-17\\x^2-4x-5=17-4x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-8x+12=0\\x^2=22\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=6\\x=2\left(l\right)\\x=\sqrt{22}\\x=-\sqrt{22}\left(l\right)\end{matrix}\right.\)

e/ \(\left[{}\begin{matrix}x\ge1\\x\le-\frac{2}{3}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}3x^2-2x=0\\3x^2=4\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\left(l\right)\\x=\frac{2}{3}\left(l\right)\\x=\frac{2\sqrt{3}}{3}\\x=\frac{-2\sqrt{3}}{3}\end{matrix}\right.\)

f/

- Với \(x\ge-\frac{1}{4}\) pt tương đương:

\(x^2+2x-4=4x+1\)

\(\Leftrightarrow x^2-2x-5=0\Rightarrow\left[{}\begin{matrix}x=1+\sqrt{6}\\x=1-\sqrt{6}\left(l\right)\end{matrix}\right.\)

- Với \(x< -\frac{1}{4}\) pt tương đương:

\(-4x-1=x^2+2x-4\)

\(\Leftrightarrow x^2+6x-3=0\Rightarrow\left[{}\begin{matrix}x=-3+2\sqrt{3}\left(l\right)\\x=-3-2\sqrt{3}\end{matrix}\right.\)

f/ \(\Leftrightarrow x^2+6x+9=\left(2x-1\right)^2\)

\(\Leftrightarrow3x^2-10x-8=0\Rightarrow\left[{}\begin{matrix}x=4\\x=-\frac{2}{3}\end{matrix}\right.\)

Làm cho bạn 1 con thôi dài quá trôi hết màn hình:

c) có vẻ khó nhất (con khác tương tự)

đặt 2x+2=t=> x+1=t/2

\(\left(t-1\right).\left(\frac{t}{2}\right)^{^2}.\left(t+1\right)=18\Leftrightarrow\left(t^2-1\right)t^2=4.18\)

\(t^4-t^2=4.18\Leftrightarrow y^2-2.\frac{1}{2}y+\frac{1}{4}=4.18+\frac{1}{4}=\frac{16.18+1}{4}=\left(\frac{17}{2}\right)^2\)

<=> \(\left(y-\frac{1}{2}\right)^{^2}=\left(\frac{17}{2}\right)^2\Rightarrow\left[\begin{matrix}y=\frac{1}{2}-\frac{17}{2}=-8\\y=\frac{1}{2}+\frac{17}{2}=9\end{matrix}\right.\Rightarrow\left[\begin{matrix}2x+2=-8\Rightarrow x=-5\\2x+2=9\Rightarrow x=\frac{7}{2}\end{matrix}\right.\)

bậc nhất môt ẩn đây ak

e sẽ cố gắng !!! 

\(3x-15=2x\left(x-5\right)\)

\(3x-15=2x^2-10x\)

\(3x-15-2x^2+10x=0\)

\(13x-15-2x^2=0\)

\(x\left(13-2x\right)-15=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\13-2x-15=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\-2-2x=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\2x=-2\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-1\end{cases}}}\)

\(f,x\left(2x-7\right)-4x+14=0\)

\(2x^2-7x-4x+14=0\)

\(2x^2-11x+14=0\)

\(x\left(2x-11\right)=-14\)

\(\Rightarrow\orbr{\begin{cases}x=-14\\2x-11=-14\end{cases}\Rightarrow\orbr{\begin{cases}x=-14\\2x=-3\end{cases}\Rightarrow}\orbr{\begin{cases}x=-14\\x=-\frac{3}{2}\end{cases}}}\)

\(e,\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)

\(3x+36-8x=x^2+4x+4\)

\(-5x+36-x^2-4x-4=0\)

\(-9x+32-x^2=0\)

\(x\left(-9-x\right)+32=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\23-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x=23\end{cases}}}\)

chúc cj hay a hc tốt