Toán Lớp 8: Tìm x: a,8x.(x-2)+x-2=0 b,3.(x-5)+x-5=0 c,(x-4).(x-3)-(x+2).(x-1)=2021 Giúp mình với,mình đang cần gấp

Question

Toán Lớp 8:  Tìm x: a,8x.(x-2)+x-2=0 b,3.(x-5)+x-5=0 c,(x-4).(x-3)-(x+2).(x-1)=2021 Giúp mình với,mình đang cần gấp

minhtuquynh · Answer

a, 8x(x-2) + x-2=0   8x(x-2) + (x-2)=0   (x-2)(8x+1)=0   => ( left[  begin{array}{l}x-2=0  8x+1=0 end{array}  right. )    => ( left[  begin{array}{l}x=2  x=- frac{-1}{8} end{array}  right. )    text{Vậy} x &isin;{ -1/8 ;2}   b, 3(x-5) + x-5=0   3(x-5) + (x-5)=0   (x-5)(3+1)=0   (x-5).4=0   => x-5=0   => x = 5   c, (x-4)(x-3) - (x+2)(x-1)=2021   x^2 - 3x - 4x + 12 - (x^2 - x + 2x -2)=2021   x^2 - 3x - 4x + 12 - x^2 +x -2x +2=2021   (x^2 - x^2 ) + (-3x -4x +x - 2x) + (12+2)=2021   -8x + 14=2021   => - 8x =2007   =>x= -2007/8

thaolanphuobg · Answer

Giải đáp: a. 8x(x-2)+(x-2)=0 <=> (x-2)(8x+1)=0 <=> [(x-2=0),(8x+1=0):} <=> [(x=2),(x=-1/8):} Vậy S={-1/8;2} b. 3(x-5)+(x-5)=0 <=> 4(x-5)=0 <=> x-5=0 <=> x=5 Vậy S={5} c. (x-4)(x-3)-(x+2)(x-1)=2021 <=> x^2-3x-4x+12-(x^2-x+2x-2)=2021 <=> x^2-7x+12-x^2-x+2=2021 <=> -8x+14=2021 <=> -8x=2007 <=> x=-2007/8 Vậy S={-2007/8}