Toán Lớp 6: Bài 2 : Tìm x , biết : a) 2.x+4/5.x=7/25 b) 3-lx+3/7l=56/21 c) (2x-2,4)^2 : 3/4-1/3=1

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Toán Lớp 6:  Bài 2 : Tìm x , biết : a) 2.x+4/5.x=7/25 b) 3-lx+3/7l=56/21 c) (2x-2,4)^2   : 3/4-1/3=1

minhtuyethue · Answer

@Kem   Lời giải và giải thích chi tiết:    a) 2.x+4/5.x=7/25     x(2 + 4/5) = 7/25     x. 14/5 = 7/25     x = 7/25 &divide; 14/5     x = 1/10     b) 3-lx+3/7l=56/21     lx+3/7l = 3 - 56/21     lx+3/7l = 1/3     &rArr; TH1: x + 3/7 = 1/3 &hArr; x = -2/21     &rArr;TH2: x + 3/7 = - 1/3 &hArr; x = -16/21     c) (2x-2,4)^2 : 3/4-1/3=1     (2x - 2,4)&sup2; &divide; 3/4 = 1 + 1/3     (2x - 2,4)&sup2; &divide; 3/4 = 4/3     (2x - 2,4)&sup2; = 4/3 &times; 3/4      (2x - 2,4)&sup2; = 1     &rArr; TH1:2x - 2,4 = 1     2x = 1 + 2,4     2x = 3,4     x = 3,4 &divide; 2     x = 1,7     TH2: 2x - 2,4 = -1     2x = -1 + 2,4     2x = 1,4     x = 1,4 &divide; 2     x = 0,7

thaolanphuobg · Answer

text{a)}   2x + 4/5x = 7/25   -> ( 2+4/5)x = 7/25   ->  14/5x = 7/25   -> x = 7/25 : 14/5   -> x = 7/25 . 5/14 =1/10   Vậy x = 1/10   $  $    text{b)}   3 - |x+3/7| = 56/21   -> |x+3/7| = 3 - 56/21   -> |x+3/7| =  1/3    ->   ( left[  begin{array}{l}x+ dfrac{3}{7}=  dfrac{1}{3}  x+ dfrac{3}{7} = dfrac{-1}{3} end{array}  right. )    ->  ( left[  begin{array}{l}x= dfrac{-2}{21}  x= dfrac{-16}{21} end{array}  right. )    Vậy x  in {-2/21 ; -16/21}   $  $    text{c)}   (2x - 2,4)^2 : 3/4 - 1/3 =1   -> (2x - 2,4)^2 : 3/4 = 1 +1/3 = 4/3   -> (2x - 2,4)^2 = 4/3 . 3/4 =1   ->  ( left[  begin{array}{l}2x -2,4 =1  2x-2,4=-1 end{array}  right. )    ->   ( left[  begin{array}{l}x= dfrac{17}{10}  x= dfrac{7}{10} end{array}  right. )    Vậy x  in { 17/10 ; 7/10}