Toán Lớp 8: Tìm `x` `a)` `3x(x+1)-2x(x+2)= -1-x` `b)` `4x(x-2019)-x+2019=0` `c)` `(x-4)^2-36=0`

Question

Toán Lớp 8:  Tìm x a) 3x(x+1)-2x(x+2)= -1-x b) 4x(x-2019)-x+2019=0 c) (x-4)^2-36=0

kymmailien · Answer

Giải đáp:     a) x &isin; &empty;   b) x &isin; { 1/4 ; 2019 }   c) x &isin; { 10 ; - 2 }     Lời giải và giải thích chi tiết:     a) 3x . ( x + 1 ) - 2x . ( x + 2 ) = - 1 - x   => 3x^2 + 3x - 2x^2 - 4x + 1 + x = 0   => ( 3x^2 - 2x^2 ) + ( 3x - 4x + x ) = - 1   => x^2 = - 1   Do x^2 &ge; 0 ( $ forall$ x )  ; - 1 < 0   => x &isin; &empty;   Vậy x &isin; &empty;   b) 4x . ( x - 2019 ) - x + 2019 = 0   => 4x . ( x - 2019 ) - ( x - 2019 ) . 1 = 0   => ( 4x - 1 )( x - 2019 ) = 0   =>  ( left[  begin{array}{l}4x - 1 =0  x-2019=0 end{array}  right. )    =>  ( left[  begin{array}{l}4x = 1  x= 2019 end{array}  right. )    => x &isin; { 1/4 ; 2019 }   Vậy x &isin; { 1/4 ; 2019 }   c) ( x - 4 )^2 - 36 = 0   => ( x - 4 )^2 - 6^2 = 0   &Aacute;p dụng hằng đẳng thức a^2 - b^2 = ( a - b )( a + b ) , ta được :   ( x - 4 )^2 - 6^2 = 0   => ( x - 4 - 6 )( x - 4 + 6 ) = 0   => ( x - 10 )( x + 2 ) = 0   =>  ( left[  begin{array}{l}x - 10=0  x+2=0 end{array}  right. )    =>  ( left[  begin{array}{l}x =10  x= -2 end{array}  right. )    Vậy x &isin; { 10 ; - 2 }

hiennhi98 · Answer

a,   3x.(x+1)-2x.(x+2)=-1-x   =>3x.x+3x.1-2x.x-2x.2=-1-x   =>3x^2+3x-2x^2-4x+1+x=0   =>(3x^2-2x^2)+(3x+x-4x)+1=0   =>x^2+1=0   =>x^2=-1   => Kh&ocirc;ng thoả m&atilde;n.   $  $   b,   4x.(x-2019)-x+2019=0   =>4x.(x-2019)-(x-2019)=0   =>(x-2019).(4x-1)=0   => ( left[  begin{array}{l}x-2019=0  4x-1=0 end{array}  right. )    => ( left[  begin{array}{l}x=2019  4x=1 end{array}  right. )    => ( left[  begin{array}{l}x=2019  x= dfrac{1}{4} end{array}  right. )    Vậy x=2019; x= frac{1}{4}   $  $   c,   (x-4)^2-36=0   =>(x-4)^2-6^2=0   =>(x-4-6).(x-4+6)=0   =>(x-10).(x+2)=0   => ( left[  begin{array}{l}x-10=0  x+2=0 end{array}  right. )    => ( left[  begin{array}{l}x=10  x=-2 end{array}  right. )    Vậy x=10; x=-2