Tìm x nguyên, biết: (x-1)^2=/x^2-1/ x+(x+1)+(x+2)+(x+3)+…+(x+2010)=2029099 1+5+9+13+…+x=501501 4(x-1)-(3x-7)=0 Tìm số nguyên n biết : 3n+5 chia h

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1. a, (x-1)²=|x²-1|

   ⇒(x-1)²=x²-1

   ⇒x²-2x+1=x²-1

   ⇒x²-x²-2x=-2

   ⇒-2x=-2

   ⇒x=1

   b, x+(x+1)+(x+2)+(x+3)+…+(x+2010)=2029099

   ⇒2011x+(1+2+3+…+2010)=2029099

   ⇒2011x+2021055=2029099

   ⇒2011x=8044

   ⇒x=4

   c, 1+5+9+13+…+x=501501

   ⇒(x+1)×$(\frac{x-1}{4}+1):2$=501501

   ⇒(x+1)×$(\frac{x-1}{4}+1)$=1003002

   ⇒$\frac{(x+1).(x-3)}{4}$=1003002

   ⇒(x+1)(x+3)=4012008

   ⇒(x+1).(x+3)=2002.2004

   ⇒x+1=2002

   ⇒x=2001

   d, 4(x-1)-(3x-7)=0

   ⇒4x-4-3x+7=0

   ⇒x=-3

   Bài 2:

   Ta có: 3n+5$\vdots$n-2

   ⇒3(n-2)+11$\vdots$n-2

   ⇒n-2∈Ư(11)={±1;±11}

   Ta có: 3n+2$\vdots$2n-1

   ⇒2.(3n+2)$\vdots$2n-1

   ⇒6n+4$\vdots$2n-1

   ⇒3(2n-1)+7$\vdots$2n-1

   ⇒2n-1∈Ư(7)={±1;±7}

   Ta có: n²+3n+9$\vdots$n+1

   ⇒(n²-1)+3(n+1)+7$\vdots$n+1

   ⇒(n-1)(n+1)+3(n+1)+7$\vdots$n+1

   ⇒n+1∈Ư(7)={±1;±7}

   Ta có: n+2$\vdots$n²-2

   ⇒(n+2)(n-2)$\vdots$n²-2

   ⇒n²-4$\vdots$n²-2

   ⇒(n²-2)-2$\vdots$n²-2

   ⇒n²-2∈Ư(2)={±1;±2}

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2. Câu 1:

   a, ( x-1)²= | x²-1|

   ⇔ | x-1|²= | x-1|.| x+1|

   ⇔ | x-1|.( | x-1|-| x+1|)= 0

   b, x+( x+1)+( x+2)+…+( x+2010)= 2029099

   ⇔ x+x+…+x+( 0+1+2+….+2010)= 2029099

   ⇔ 2011.x+( 2010+0).[( 2010-0):1+1]:2= 2029099

   ⇔ 2011.x+2021055= 2029099

   ⇔ 2011.x= 8044

   ⇔ x= 4

   c, 1+5+9+13+…+x= 501501

   ⇔ ( x+1).[( x-1):4+1]:2= 501501

   ⇔ ( x+1).( 0,25.x+0,75)= 1003002

   ⇔ 0,25.x²+0,75.x+0,25.x+0,75= 1003002

   ⇔ x²+4x+3= 4012008

   ⇔ x²+4x-4012008= 0

   Giải nghiệm ta có x∉ Z

   ⇒ x∈ ∅

   d, 4.( x-1)-( 3x-7)= 0

   ⇔ 4x-4-3x+7= 0

   ⇔ x+3= 0

   ⇔ x= -3

   Bài 2:

   a, 3n+5⋮ n-2

   ⇔ 3.( n-2)+11⋮ n-2

   ⇒ 11⋮ n-2

   ⇒ n-2∈ Ư( 11)= { -11; -1; 1; 11}

   ⇔ n∈ { -9; 1; 3; 13}

   b, 3n+2⋮ 2n-1

   ⇒ 2.( 3n+2)⋮ 2n-1

   ⇔ 6n+4⋮ 2n-1

   ⇔ 3.( 2n-1)+7⋮ 2n-1

   ⇒ 2n-1∈ Ư( 7)= { -7; -1; 1; 7}

   ⇔ n∈ { -3; 0; 1; 4}

   c, n²+3n+9⋮ n+1

   ⇔ n²+2n+1+n+1+7⋮ n+1

   ⇔ ( n+1)²+n+1+7⋮ n+1

   ⇒ 7⋮ n+1

   ⇒ n+1∈ Ư( 7)= { -7; -1; 1; 7}

   ⇔ n∈ { -8; -2; 0; 6}

   d, n+2⋮ n²-2

   ⇔ n.( n+2)⋮ n²-2

   ⇔ n²-2+2n+2⋮ n²-2

   ⇒ 2n+2⋮ n²-2

   Mà n+2⋮ n²-2

   ⇒ 2n+4⋮ n²-2

   ⇒ 2n+4-2n-2⋮ n²-2

   ⇒ 2⋮ n²-2

   ⇒ n²-2∈ U( 2)= { -2; -1; 1; 2}

   ⇒ n²∈ { 0; 1; 3; 4}

   Vì n∈ Z⇒ n∈ { 0; ±1; ±2}

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