a: Trường hợp 1: m=0

Pt sẽ là -x+1=0

hay x=1(nhận)

Trườg hợp 2: m<>0

\(\text{Δ}=\left(m-1\right)^2-4\cdot2m\cdot\left(-3m+1\right)\)

\(=\left(m-1\right)^2+8m\left(3m-1\right)\)

\(=m^2-2m+1+24m^2-8m\)

\(=25m^2-10m+1=\left(5m-1\right)^2>=0\)

Do đó: Phương trình luôn có nghiệm

b: Trường hợp 1: m=1

Pt sẽ là 3x-2=0

hay x=2/3(nhận)

Trường hợp 2: m<>1

\(\text{Δ}=\left(2m+1\right)^2-4\left(m-1\right)\left(m-3\right)\)

\(=4m^2+4m+1-4\left(m^2-4m+3\right)\)

\(=4m^2+4m+1-4m^2+16m-12\)

=20m-11

Để phương trình có nghiệm thì 20m-11>=0

hay m>=11/20

\(a,\) \(\Delta=b^2-4ac=\left(-4\right)^2-4.\left(-3m-1\right)=16+12m+4=12m+20\)

Để pt trên có nghiệm thì \(\Delta\ge0\Rightarrow12m+20\ge0\Rightarrow m\ge-\dfrac{5}{3}\)

\(b,\Delta=b^2-4ac=\left(-2m\right)^2-4\left(m-2\right)\left(m+3\right)\)

                          \(=4m^2-4\left(m^2+3m-2m-6\right)\)

                          \(=4m^2-4m^2-4m+24\)

                          \(=-4m+24\)

Để pt trên có nghiệm thì \(\Delta\ge0\Rightarrow-4m+24\ge0\Rightarrow m\le6\)

a) 3sinx= 1-m => \(-3\le1-m\le3\) \(\Leftrightarrow-2\le m\le4\)

b, \(4cos^2x=m+3\)

\(\Leftrightarrow4cos^2x-2=m+1\)

\(\Leftrightarrow2cos2x=m+1\)

\(\Leftrightarrow cos2x=\dfrac{m+1}{2}\)

Phương trình có nghiệm khi:

\(-1\le\dfrac{m+1}{2}\le1\)

\(\Leftrightarrow-2\le m+1\le2\)

\(\Leftrightarrow-3\le m\le1\)

a, \(3sinx+m-1=0\)

\(\Leftrightarrow sinx=\dfrac{1-m}{3}\)

Phương trình có nghiệm khi:

\(-1\le\dfrac{1-m}{3}\le1\)

\(\Leftrightarrow-3\le1-m\le3\)

\(\Leftrightarrow-2\le m\le4\)

\(a,\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x-6=0\Leftrightarrow x=6\\2x-5=0\Leftrightarrow x=\dfrac{5}{2}\\3x+9=0\Leftrightarrow x=-3\end{matrix}\right.\)

\(b,2x\left(x-3\right)+5\left(x-3\right)=0\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\2x+5=0\Leftrightarrow x=-\dfrac{5}{2}\end{matrix}\right.\)

\(c,x^2-4-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(x=-7\left(2m-5\right)x-2m^2+8\Leftrightarrow x+7\left(2m-5\right)=8-2m^2\Leftrightarrow x\left(14m-34\right)=8-2m^2\)

\(ycđb\Leftrightarrow14m-34\ne0\Leftrightarrow m\ne\dfrac{34}{14}\)\(\Rightarrow x=\dfrac{8-2m^2}{14m-34}\)

\(3.17\Leftrightarrow4x^2-4x+1-2x-1=0\Leftrightarrow4x^2-6x=0\Leftrightarrow x\left(4x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

3.15:

a, \(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\2x-5=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=\dfrac{5}{2}\\x=-\dfrac{9}{3}=-3\end{matrix}\right.\)

b, \(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)

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

c, \(\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

3.16

\(\Leftrightarrow\left(2m-5\right).-7-2m^2+8=0\)

\(\Leftrightarrow-14m+35-2m^2+8=0\)

\(\Leftrightarrow-14m-2m^2+43=0\)

\(\Leftrightarrow-2\left(7m+m^2\right)=-43\)

\(\Leftrightarrow m\left(7-m\right)=\dfrac{43}{2}\)

\(\Leftrightarrow\dfrac{m\left(7-m\right)}{1}-\dfrac{43}{2}=0\)

\(\Leftrightarrow\dfrac{14m-2m^2}{2}-\dfrac{43}{2}=0\)

pt vô nghiệm

a: \(\Delta=\left(2m-6\right)^2-4\cdot1\cdot\left(m-3\right)\)

\(=4m^2-24m+36-4m+12\)

\(=4m^2-28m+48\)

\(=4\left(m-3\right)\left(m-4\right)\)

Để phương trình có nghiệm kép thì (m-3)(m-4)=0

=>m=3 hoặc m=4

b: Trường hợp 1: m=7/2

Phương trình sẽ là \(2\cdot\left(2\cdot\dfrac{7}{2}+5\right)x-14\cdot\dfrac{7}{2}+1=0\)

\(\Leftrightarrow24x-48=0\)

hay x=2

=>Nhận

Trường hợp 2: m<>7/2

\(\Delta=\left(4m+10\right)^2-4\cdot\left(2m-7\right)\left(-14m+1\right)\)

\(=16m^2+80m+100-4\left(-28m^2+2m+98m-7\right)\)

\(=16m^2+80m+100+112m^2-400m+28\)

\(=128m^2-320m+128\)

\(=64\left(2m^2-5m+2\right)\)

Để phương trình có hai nghiệm phân biệt thì (2m-1)(m-1)=0

=>m=1 hoặc m=1/2