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a) \(=\left(a^3.a^5\right).\left(b^2.b\right)=a^8.b^3\)
b) Tương tự
c)
1. (a-b+c)-(a+c)
= a - b + c - a - c
= -b
2. (a+b)-(b+a)
= a +b - b - a
= 0
3. -(a+b-c)+(a-b-c)
= -a - b + c + a - b - c
= -2b
4. a(b+c)-a(b+d)
= ab + ac - ab - ad
= ac - ad
= a.(c-d)
5. a(b-c)-a.(b+d)
= ab - ac - ab - ad
= -ac - ad
= -a.(c+d)
6. Giống câu 5
Hok tốt !
1/.(a-b+c)-(a+c)
= a - b + c - a - c
= -b
2/(a+b)-(b+a)
= a + b - b -a
=0
3/-(a+b-c)+(a-b-c)
= -a - b + c + a - b - c
= -2b
4/a(b+c)-a(b+d)
= ab + ac - ab + ad
= ac + ad
= a( c + d)
5/a(b-c)-a.(b+d)
= ab - ac - ab + ad
= -ac + ad
6/a(b-c)-a(b+d)
= ab - ac - ab + ad
= a(b - c - b + d)
= a( -c + d)
hok tốt!
a, -( -a + c - d) - ( c - d + d) = a - c + d - c + d - d = a + d
b, - ( a+b-c+d) + (a-b-c-d) = -a -b+c-d + a-b-c-d = -2b + (-2c)= -2(b+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)-(-a+c-d)-(c-a+d)=a-c+d-c+a-d=(a+a)-(c+c)+(d-d)=2a-2c=2(a-c)
b)-(a+b-c+d)+(a-b-c-d)=-a-b+c-d+a-b-c-d=(-a+a)-(b+b)+(c-c)-(d+d)=0-2b+0-2d=-2(b-d)
c)a(b-c-d)-a(b+c-d)=ab-ac-ad-ab-ac+ad=(ab-ac)-(ac+ac)-(ad-ad)=2ac
d)đề sai
e)(a+b)(c-d)-(a-b)(c+d)=ac+b-ad+b-(ac-b+ad-b)=ac+b-ad+b-ac+b-ad+b=(ac-ac)+(b+b+b+b)-(ad+ad)=4b-2ad=2(2b-ad)
f)(a+b)2-(a-b)2=a2+2ab+b2-(a2-2ab+b2)=a2+2ab+b2-a2+2ab-b2=(a2-a2)+(2ab+2ab)+(b2-b2)=4ab
mk k chắc đâu
b) F=3.(b-c)-(a+c)+5.(a-b-c)
F=3b-3c-a-c+5a-5b-5c
F=(-a+5a)+(3b-5b)+(-3c-c-5c)
F= 4a+(-2b)+(-9c)
F=4a-2b-9c
a) E=2.(a+b)+3.(a-c)-4(b+c)
E=2a+2b+3a-3c-4b-4c
E=(2a+3a)+(2b-4b)+(-3c-4c)
E=5a+(-2b)+(-7c)=5a-2b-7c