Toán Lớp 8: Cho e hỏi bài này với ạ Bài 3 Tìm x A)25x^2-9=0 B)3(x-1)^2-3x(x-5)=1 C)(x-3)^2-4=0

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Toán Lớp 8:  Cho e hỏi bài này với ạ Bài 3 Tìm x A)25x^2-9=0 B)3(x-1)^2-3x(x-5)=1 C)(x-3)^2-4=0

minhhaanh · Answer

a) 25x&sup2; - 9 = 0   &rArr; (5x)&sup2; - 3&sup2; = 0   &rArr; (5x - 3) (5x + 3) = 0   &rArr; ( left[  begin{array}{l}5x - 3=0  5x + 3= 0 end{array}  right. )    &rArr;  ( left[  begin{array}{l}5x=3  5x=-3 end{array}  right. )    &rArr;  ( left[  begin{array}{l}x=3/5  x=-3/5 end{array}  right. )    b) 3(x - 1)&sup2; - 3x(x - 5) = 1   &rArr; 3(x&sup2; - 2x + 1) - 3x&sup2; + 15x = 1   &rArr; 3x&sup2; - 6x + 3 - 3x&sup2; + 15x = 1   &rArr; (3x&sup2; - 3x&sup2;) + (-6x + 15x) + 3 = 1   &rArr; 9x + 3 = 1   &rArr; 9x = 1 - 3   &rArr; 9x = -2   &rArr; x = $ frac{-2}{9}$    c) (x - 3)&sup2; - 4 = 0   &rArr; (x - 3)&sup2; - 2&sup2; = 0   &rArr; (x - 3 - 2) (x - 3 + 2) = 0   &rArr;  ( left[  begin{array}{l}x - 3 - 2=0  x - 3 + 2=0 end{array}  right. )    &rArr;  ( left[  begin{array}{l}x=5  x=1 end{array}  right. )     Ch&uacute;c bn học tốt!!!!     vote 5*, ctlhn nha bn

thaolanphuobg · Answer

Giải đáp:      (a) ,S= left {{ dfrac{3}{5};- dfrac{3}{5}} right }  b) ,S= left {{- dfrac{2}{9}} right }  c) ,S= left {{1;5} right } )     Lời giải và giải thích chi tiết:      (a) ,25x^2-9=0   Leftrightarrow25x^2=9   Leftrightarrow x^2= dfrac{9}{25}   Leftrightarrow x= pm dfrac{3}{5} )Vậy  ( rm S= left {{ dfrac{3}{5};- dfrac{3}{5}} right } ) (  b) ,3(x-1)^2-3x(x-5)=1   Leftrightarrow3(x^2-2x+1)-3x^2-15x-1=0   Leftrightarrow3x^2-6x+3-3x^2+15x-1=0   Leftrightarrow9x+2=0   Leftrightarrow9x=-2   Leftrightarrow x=- dfrac{2}{9} )Vậy  ( rm S= left {{- dfrac{2}{9}} right } ) (  c) ,(x-3)^2-4=0   Leftrightarrow x^2-6x+9-4=0   Leftrightarrow x^2-6x+5=0   Leftrightarrow x^2-x-5x+5=0   Leftrightarrow x(x-1)-5(x-1)=0   Leftrightarrow (x-1)(x-5)=0   Leftrightarrow left[  begin{array}{l}x-1=0  x-5=0 end{array}  right.  Leftrightarrow  left[  begin{array}{l}x=1  x=5 end{array}  right. )Vậy  ( rm S= left {{1;5} right } )