chắc b

(3x-1)²-5(2x+1)²+(6x-3)(2x+1)=(x-1)²

<=> (3x-1)²+2(3x-1)(2x+1)+(2x+1)²-6(2x+1)²=(x-1)²

<=> (5x)²-6(4x²+4x+1)-(x²-2x+1)=0

<=> -22x-7=0

=> x=-7/22

\(\left(3x-1\right)^2-5\left(2x+1\right)^2+\left(6x-3\right)\left(2x+1\right)=\left(x-1\right)^2\)

\(\Leftrightarrow9x^2-6x+1+\left(2x+1\right)\left[-5\left(2x+1\right)+6x-3\right]=x^2-1\)

\(\Leftrightarrow9x^2-6x+1+\left(2x+1\right)\left[-10x-5+6x-3\right]=x^2-1\)

\(\Leftrightarrow9x^2-6x+1+\left(2x+1\right)\left[-4x-8\right]=x^2-1\)

\(\Leftrightarrow9x^2-6x+1-4x\left(2x+1\right)-8\left(2x+1\right)=x^2-1\)

\(\Leftrightarrow9x^2-6x+1-8x^2-4x-16x-8=x^2-1\)

\(\Leftrightarrow\left(9x^2-8x^2-x^2\right)-\left(4x+6x+16x\right)+\left(1-8\right)=-1\)

\(\Leftrightarrow0-26x-7=-1\)

\(\Leftrightarrow-26x=-1+7\)

\(\Leftrightarrow-26x=6\)

\(\Leftrightarrow x=\frac{-3}{13}\)

Ta có: 3(x-2)=2x-9

\(\Leftrightarrow3x-6-2x+9=0\)

\(\Leftrightarrow x=-3\)

Để (1) và (2) tương đương thì \(-3\left(m-3\right)=m+1\)

\(\Leftrightarrow-3m+9-m-1=0\)

\(\Leftrightarrow-4m=-8\)

hay m=2

Vậy: Để hai phương trình tương đương thì m=2

Ta có: 3(x-2)=2x-9

Để (1) và (2) tương đương thì 

hay m=2

Vậy: Để hai phương trình tương đương thì m=2

\(a,\Leftrightarrow\left[{}\begin{matrix}x+5=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-\dfrac{1}{2}\end{matrix}\right.\\ b,\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\\ c,\Leftrightarrow2x^2-10x-3x-2x^2=26\\ \Leftrightarrow-13x=26\Leftrightarrow x=-2\\ d,\Leftrightarrow x^2-18x+16=0\\ \Leftrightarrow\left(x^2-18x+81\right)-65=0\\ \Leftrightarrow\left(x-9\right)^2-65=0\\ \Leftrightarrow\left(x-9+\sqrt{65}\right)\left(x-9-\sqrt{65}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=9-\sqrt{65}\\9+\sqrt{65}\end{matrix}\right.\)

\(e,\Leftrightarrow x^2-10x-25=0\\ \Leftrightarrow\left(x-5\right)^2-50=0\\ \Leftrightarrow\left(x-5-5\sqrt{2}\right)\left(x-5+5\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5+5\sqrt{2}\\x=5-5\sqrt{2}\end{matrix}\right.\\ f,\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\\ g,\Leftrightarrow2\left(x+5\right)-x\left(x+5\right)=0\\ \Leftrightarrow\left(2-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\\ h,\Leftrightarrow x^2+2x+3x+6=0\\ \Leftrightarrow\left(x+3\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\\ i,\Leftrightarrow4x^2-12x+9-4x^2+4=49\\ \Leftrightarrow-12x=36\Leftrightarrow x=-3\)

\(j,\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\\ k,\Leftrightarrow x^2\left(x-1\right)=4\left(x-1\right)^2\\ \Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

a: =>(3x+1)(3x-1)-(3x+1)(2x-3)=0

=>(3x+1)(3x-1-2x+3)=0

=>(3x+1)(x+2)=0

=>x=-1/3 hoặc x=-2

b: =>(3x+1)(6x+2)-(3x+1)(x-2)=0

=>(3x+1)(6x+2-x+2)=0

=>(3x+1)(5x+4)=0

=>x=-1/3 hoặc x=-4/5

x(4x - 1)²(2x - 1)= 3/2

<=>(16x² - 8x + 1)( 2x² - x)= 3/2

<=>(16x² - 8x + 1)( 16x² - 8x)= 12

Đặt 16x² - 8x= y, ta có phương trình:

(y + 1) . y= 12

<=>y² + y - 12=0

<=>y² + 4x - 3x - 12=0

<=>y(y + 4) - 3(x + 4)=0

<=>(y + 4)(y - 3)=0

Đến đây tự làm tiếp nha.

x(4x-1)^2(2x+1)=3/2

<=>8x(4x-1)^2(2x-1)=8.3/2

<=>(16x^2-8x+1)(16x^2-8x)=12     (1)

đặt 16x^2-8x=y  ta có

 (y+1)y=12

<=>y^2+y-12=0

<=>y^2-3y+4y-12=0

<=>y(y-3)+4(y-3)=0

<=>(y-3)(y+4)=0

thay y=x^2+8x rồi giải phương trình

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