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Ta có:(x+1)(x+2)=x(x+2)+(x+2)=x^2+2x+x+2
=x(x+2+1)+2=x(x+3)+2
(x-3)(x-2)=x(x-2)-3(x-2)=x^2-2x-3x+6
=x(x-2-3)+6=x(x-5)+6
(a-b)(a-b)=a(a-b)-b(a-b)=a^2-ab-ba+2b
=a(a-b-b)+2b=a(a-2b)+2b
Bài 1:
\(a,\left(a+b-c\right)-\left(b-c-d\right)\)
\(=a+b-c-b+c+d\)
\(=a+d\)
\(b,-\left(a-b+c\right)+\left(a-b+d\right)\)
\(=-a+b-c+a-b+d\)
\(=-c+d\)
\(c,\left(a+b\right)-\left(-a+b-c\right)\)
\(=a+b+a-b+c\)
\(=2a+c\)
\(d,-\left(a+b\right)+\left(a+b+c\right)\)
\(=-a-b+a+b+c\)
\(=c\)
Bài 3 :
\(a,15-\left(4-x\right)=6\)
\(4-x=15-6\)
\(4-x=9\)
\(x=4-9\)
\(x=-5\)
\(b,-30+\left(25-x\right)=-1\)
\(25-x=-1+30\)
\(25-x=29\)
\(x=25-29\)
\(x=-4\)
\(c,x-5=-1\)
\(x=-1+5\)
\(x=4\)
\(d,x-4=-10\)
\(x=-10+4\)
\(x=-6\)
\(e,x+3=-8\)
\(x=-8-3\)
\(x=-11\)
\(g,x+6=0\)
\(x=-6\)
Câu 1:
A, (a+b-c)-(b-c-d)
= a+b-c-b+c+d
= a+(b-b)+(c-c)
= a
B, -(a-b+c)+(a-b+d)
= -a+b-c+a+b+d
= (a-a)+(b+b)+d-c
= 2b+d-c
C, (a+b)-(-a+b-c)
= a+b+a-b+c
= (a+a)+(b-b)+c
= 2a+c
D, -(a+b)+(a+b+c)
= -a-b+a+b+c
= (-a+a)+(b-b)+c
= c
=
bài 1:
a) ta có: 3x + 5 = (3(x+1)+2)\(⋮\)(x+1)
vì (3(x+1)\(⋮\)(x+1) nên 2 \(⋮\)(x+1) => (x+1) \(\in\)Ư(2) => (x+1) \(\in\)\(\xi\)-2;-1;1;2 \(\xi\)=> x \(\in\)\(\xi\)-3; -2; 0; 1 \(\xi\)
vậy, x= -3; -2; 0; 1
a, \(\left(-a\right)\left(b-c-d\right)=-ab+ac+ad\)
b, \(x\left(y-z-2\right)=xy-xz-2x\)
c, \(\left(-3\right)\left(2x-5\right)=-6x+15\)
1.
Ix+1I + I2x+2I =3
Ix+1I và I2x+2I thuộc N
=> x thuộc N
x+1+2x+2=3
3x+1+2=3
3x = 3-1-2
3x = 0
x = 0:3
x = 0
3 x 8/15 x 12 = 3x8/15 x12 = 24/15 x 12 = 288/15
7 x 25-49/7 x 24+21= 175 - 7 x24 + 21 =4053
2 x (-13) x 9 x 10/(-3) x 4 x (-5) x 26= -26 x 9 x 10 / 1560 = -234 x 1/156 = -3/2
Bài 1:
a)
<=> 3x - 18 - 5x + 10 = 24
<=> 3x - 5x = 24 + 18 - 10
<=> -2x = 32
<=> x = 32 : (-2)
<=> x = -16
b)
<=> -4x + 20 - 8x + 16 = 48
<=> -4x - 8x = 48 - 20 - 16
<=> -12x = 12
<=> x = 12 : (-12)
<=> x = -1
Bài 2:
\(=a^2-ab+ab-b^2\)
\(=a^2-b^2\)
a) 3(x-6)-5(x-2) = 24
<=> 3x -36 -5x + 10 =24
<=> -2x = 50
<=> x = -25
b) -4(x-5) -8(x-2) = 48
<=> -4x +20 - 8x +16 = 48
<=> -12x = 12
<=> x = -1
(a+b)(a-b) = a^2 -ab +ab -b^2 = a^2 - b^2