Toán Lớp 8: tìm x 1)4x^3=36x 2)x^3-6x^2+12x-8=0 3)(5x-4)^2-81x^2=0 4)(x-4)^2-(x-4)=0 5)1-7x-(7x-1)^2 6)(2x-5)^2-(5+2x)^2=0 cảm ơn mn nhiều hứa vo

Question

Toán Lớp 8:  tìm x  1)4x^3=36x 2)x^3-6x^2+12x-8=0 3)(5x-4)^2-81x^2=0 4)(x-4)^2-(x-4)=0 5)1-7x-(7x-1)^2 6)(2x-5)^2-(5+2x)^2=0  cảm ơn mn nhiều hứa vote 5 sao

tuanh · Answer

1) 4x^3=36x <=>4x^3-36x=0 <=>4x(x^2-9)=0 Trường hợp 1: 4x=0 <=>x=0 Trường hợp 1: x^2-9=0 <=>x=3 hoặc x=-3 Vậy: S={0;±3} 2) x^3-6x^2+12x-8=0 <=>(x-2)^3=0 <=>x-2=0 <=>x=2 Vậy: S={2} 3) (5x-4)^2-81x^2=0 <=>(5x-4-9x)(5x-4+9x)=0 <=>(-4x-4)(14x-4)=0 Trường hợp 1: -4x-4=0 <=>-4x=4 <=>x=-1 Trường hợp 2: 14x-4=0 <=>14x=4 <=>x=2/7 Vậy: S={-1;2/7} 4) (x-4)^2-(x-4)=0 <=>(x-4)(x-4-1)=0 <=>(x-4)(x-5)=0 Trường hợp 1: x-4=0 <=>x=4 Trường hợp 2: x-5=0 <=>x=5 Vậy: S={4;5} 5) 1-7x-(7x-1)^2=0 <=>(7x-1)(-1-7x+1)=0 <=>(7x-1)(-7x)=0 Trường hợp 1: 7x-1=0 <=>7x=1 <=>x=1/7 Trường hợp 2: -7x=0 <=>x=0 Vậy: S={1/7;0} 6) (2x-5)^2-(5+2x)^2=0 <=>(2x-5)^2-(2x+5)^2=0 <=>[(2x-5)-(2x+5)][(2x-5)+(2x+5)]=0 <=>(2x-5-2x-5)(2x-5+2x+5)=0 <=>-10.4x=0 <=>-40x=0 <=>x=0 Vậy: S={0}

maianhtuanh · Answer

Giải đáp:   1.   ( left[  begin{array}{l}x=  x=&plusmn;3 end{array}  right. )    2. $x = 2$   3.  ( left[  begin{array}{l}x=-1  x= frac{2}{7} end{array}  right. )    4.  ( left[  begin{array}{l}x=4  x=5 end{array}  right. )    5.  ( left[  begin{array}{l}x= frac{1}{7}  x=0 end{array}  right. )    6. $x = 0$   Lời giải và giải thích chi tiết:    1. $4x^{3} = 36x$   &hArr; $4x( x^{2} - 9 ) = 0$   &hArr;  ( left[  begin{array}{l}x=0  x^{2}=9 end{array}  right. )    &hArr;  ( left[  begin{array}{l}x=  x=&plusmn;3 end{array}  right. )    2. $x^{3} - 6x^{2} + 12x - 8 = 0$   &hArr; $x^{2}( x - 2 ) - 4x( x - 2 ) + 4( x - 2 ) = 0$   &hArr; $( x - 2 )( x^{2} - 4x + 4 ) = 0$   &hArr; $( x - 2 )( x - 2 )^{2} = 0$   &hArr; $( x - 2 )^{3} = 0$   &hArr; $x = 2$   3. $( 5x - 4 )^{2} - 81x^{2} = 0$   &hArr; $( 5x - 4 )^{2} = 81x^{2}$   &hArr;  ( left[  begin{array}{l}5x-4=9x  5x-4=-9x end{array}  right. )    &hArr;  ( left[  begin{array}{l}x=-1  x= frac{2}{7} end{array}  right. )    4. $( x - 4 )^{2} - ( x - 4 ) = 0$   &hArr; $( x - 4 )( x - 4 - 1 ) = 0$   &hArr; $( x - 4 )( x - 5 ) = 0$   &hArr;  ( left[  begin{array}{l}x=4  x=5 end{array}  right. )    5. $1 - 7x - ( 7x - 1 )^{2} = 0$   &hArr; $( 7x - 1 )( -1 - 7x + 1 ) = 0$   &hArr; $( 7x - 1 )( -7x ) = 0$   &hArr;  ( left[  begin{array}{l}x= frac{1}{7}  x=0 end{array}  right. )    6. $( 2x - 5 )^{2} - ( 5 + 2x )^{2} = 0$   &hArr; $( 2x - 5 )^{2} = ( 2x + 5 )^{2}$   &hArr;  ( left[  begin{array}{l}2x-5=2x+5  2x-5=-2x-5 end{array}  right. )    &hArr;  ( left[  begin{array}{l}0x=10 ( v&ocirc; l&iacute; )  x=0 end{array}  right. )    &hArr; $x = 0$