Toán Lớp 8: e. ( 3x – 1)^2 – ( x +5)^2 =0 f. ( 2x -1)^2 – ( x -3)^2 =0

Question

Toán Lớp 8:  e. ( 3x – 1)^2 – ( x +5)^2 =0             f. ( 2x -1)^2 – ( x -3)^2 =0

bichhhathu · Answer

e. ( 3x – 1)^2 – ( x +5)^2 =0 <=> ( 3x - 1 - x - 5 ) ( 3x - 1 + x + 5) = 0 <=> ( 2x - 6 )( 4x + 4 ) = 0 <=> 2x - 6 =0 4x + 4 = 0 <=> 2x = 6 4x = - 4 <=> x = 3 x = -1 f. ( 2x -1)^2 – ( x -3)^2 =0 <=> ( 2x - 1 - x + 3 ) ( 2x - 1 + x - 3) = 0 <=> ( x + 2)( 3x - 4) = 0 <=> x + 2 = 0 3x - 4 = 0 <=> x = - 2 3x = 4 <=> x = -2 x = 4/3

maianhtuanh · Answer

Giải đáp+Lời giải và giải thích chi tiết: e, (3x-1)^2-(x+5)^2=0 <=> (3x-1-x-5)(3x-1+x-5)=0 <=> (2x-6)(4x-6)=0 <=> ( left[ begin{array}{l}2x-6=0 4x-6=0 end{array} right. ) <=> ( left[ begin{array}{l}2x=6 4x=6 end{array} right. ) <=> ( left[ begin{array}{l}x=3 x= dfrac{6}{4} end{array} right. ) Vậy x in {3;6/4} f, (2x-1)^2-(x-3)^2=0 <=> (2x-1-x+3)(2x-1+x-3)=0 <=> (x+2)(3x-4)=0 <=> ( left[ begin{array}{l}x+2=0 3x-4=0 end{array} right. ) <=> ( left[ begin{array}{l}x=-2 3x=4 end{array} right. ) <=> ( left[ begin{array}{l}x=-2 x= dfrac{4}{3} end{array} right. ) Vậy x in {-2;4/3} $@Maruko-chan$