Toán Lớp 7: Tìm nghiệm. `a) 5x^2 – x` `b) x^2 –2x.` `c) x^2 –3x.` `d) 3x^2 – 4x`

Question

Toán Lớp 7:  Tìm nghiệm. a) 5x^2 – x  b) x^2 –2x.  c) x^2 –3x.  d) 3x^2 – 4x

hauthanhyen98 · Answer

Giải đáp+Lời giải và giải thích chi tiết:      a)5x^2 - x   Ta c&oacute;: 5x^2 - x = 0   => x(5x - 1) = 0   =>  ( left[  begin{array}{l}x=0  5x-1=0 end{array}  right. )    =>  ( left[  begin{array}{l}x=0  5x = 1 end{array}  right. )  =>  ( left[  begin{array}{l}x=0  x= frac{1}{5} end{array}  right. )    b) x^2 - 2x   => x(x-2)=0   =>  ( left[  begin{array}{l}x=0  x-2=0 end{array}  right. )  =>  ( left[  begin{array}{l}x=0  x=2 end{array}  right. )    c)x^2 - 3x   => x(x-3)0   =>  ( left[  begin{array}{l}x=0  x-3=0 end{array}  right. )  =>  ( left[  begin{array}{l}x=0  x=3 end{array}  right. )    d) 3x^2 -4x    => x(3x-4)=0   =>  ( left[  begin{array}{l}x=0  3x-4=0 end{array}  right. )    =>  ( left[  begin{array}{l}x=0  x=4 end{array}  right. )  =>  ( left[  begin{array}{l}x=0  x= frac{4}{3} end{array}  right. )    $@Pipimm~$

minhtuyethue · Answer

Giải đáp + giải th&iacute;ch c&aacute;c bước giải:     a) 5x^2-x=0   ->x(5x-1)=0   -> ( left[  begin{array}{l}x=0  5x-1=0 end{array}  right. )    -> ( left[  begin{array}{l}x=0  x= dfrac{1}{5} end{array}  right. )    Vậy x=0;1/5   b) x^2-2x=0   ->x(x-2)=0   -> ( left[  begin{array}{l}x=0  x-2=0 end{array}  right. )    -> ( left[  begin{array}{l}x=0  x=2 end{array}  right. )    Vậy x=0;2   c) x^2-3x=0   ->x(x-3)=0   -> ( left[  begin{array}{l}x=0  x-3=0 end{array}  right. )    -> ( left[  begin{array}{l}x=0  x=3 end{array}  right. )    Vậy x=0;3   d) 3x^2-4x=0   ->x(3x-4)=0   -> ( left[  begin{array}{l}x=0  3x-4=0 end{array}  right. )    -> ( left[  begin{array}{l}x=0  x= dfrac{4}{3} end{array}  right. )    Vậy x=0;4/3