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a, 3(x-1)(2x-1)=5(x+8)(x-1) b, 9x^2-1(3x-1)(4x+1) c, (x+
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Thanh Hà
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a, 3(x-1)(2x-1)=5(x+8)(x-1)
⇔3(x-1)(2x-1)-5(x+8)(x-1)=0
⇔(x-1)[3(2x-1)-5(x+8)]=0
⇔(x-1)(6x-3-5x-40)=0
⇔(x-1)(x-43)=0
⇔\(\left[ \begin{array}{l}x-1=0\\x-43=0\end{array} \right.\) <=>\(\left[ \begin{array}{l}x=1\\x=43\end{array} \right.\)
Vậy S={1;43}
b, 9x²-1=(3x-1)(4x+1)
⇔(3x-1)(3x+1)-(3x-1)(4x+1)=0
⇔(3x-1)(3x+1-4x-1)=0
⇔(3x-1).(-x)=0
⇔\(\left[ \begin{array}{l}3x-1=0\\-x=0\end{array} \right.\) <=>\(\left[ \begin{array}{l}x=1/3\\x=0\end{array} \right.\)
Vậy S={1/3;0}
c, (x+7)(3x-1)=49-x²
⇔(x+7)(3x-1)=(7-x)(7+x)
⇔(x+7)(3x-1)-(7-x)(x+7)=0
⇔(x+7)(3x-1-7+x)=0
⇔(x+7)(4x-8)=0
⇔(x+7).4(x-2)=0
⇔\(\left[ \begin{array}{l}x+7=0\\x-2=0\end{array} \right.\) <=>\(\left[ \begin{array}{l}x=-7\\x=2\end{array} \right.\)
Vậy S={-7;2}
d, x³-5x²+6x=0
⇔x(x²-5x+6)=0
⇔x(x²-3x-2x+6)=0
⇔x.[x(x-3)-2(x-3)]=0
⇔x(x-3)(x-2)=0
⇔x=0; x-3=0 hoặc x-2=0
⇔x=0; x=3 hoặc x=2
Vậy S={0;3;2}
e, $\frac{2x+1}{2x-1}-$ $\frac{2x-1}{2x+1}=$ $\frac{8}{4x^2-1}$ ⇔$\frac{(2x+1)(2x+1)-(2x-1)(2x-1)-8}{(2x+1)(2x-1)}=0$
⇔(2x+1)²-(2x-1)²-8=0
⇔(2x+1)²-(2x-1)²-8=0
⇔8(x-1)=0
⇔x-1=0
⇔x=1
Vậy S={1}
a, 3( x-1)( 2x-1)=5( x+8)( x-1)
⇔ 3.( 2x²-x-2x+1)= 5.( x²-x+8x-8)
⇔ 6x²-9x+3= 5x²+35x-40
⇔ x²-44x+43= 0
⇔ x²-43x-x+43= 0
⇔ ( x-1).( x-43)= 0
⇔ x-1= 0⇔ x= 1
hoặc x-43= 0⇔ x= 43
b, 9x²-1= ( 3x-1)( 4x+1)
⇔ 9x²-1= 12x²+3x-4x-1
⇔ 9x²= 12x²-x
⇔ 3x²-x= 0
⇔ x.( 3x-1)= 0
⇔ x= 0
hoặc 3x-1= 0⇔ x= $\frac{1}{3}$
c, ( x+7)( 3x-1)= 49-x²
⇔ 3x²-x+21x-7= 49-x²
⇔ 4x²+20x-56= 0
⇔ x²+5x-14= 0
⇔ x²-2x+7x-14= 0
⇔ ( x-2).( x+7)= 0
⇔ x-2= 0⇔ x= 2
hoặc x+7= 0⇔ x= -7
d, x³-5x²+6x= 0
⇔ x.( x²-5x+6)= 0
⇔ x.( x²-3x-2x+6)= 0
⇔ x.( x-3).( x-2)= 0
⇔ x= 0
hoặc x-3= 0⇔ x= 3
hoặc x-2= 0⇔ x= 2
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