a+b=p; a-b=q. Tính theo q và q giá trị của a.b
Phân tích đa thức thành nhân tử:
A)(a+b)(b+c)(c-a)+(b+c)(c+a)(a-b)+(c+a)(a+b)(b-c)
B)(b+c)(c+a)(b-a)+(b+c)(a+b)(a-c)+(a-b)(b-c)(a-c)
C)(a-b)(b-c)(a-c)+(a+b)(c+a)(c-b)+(b+c)(c+a)(b-a)
D)(a-b)(b-c)(a-c)+(a+b)(b+c)(a-c)+(a+b)(a+c)(c-b)
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A=(a+b)(b+c)(c+a)+abcA=(a+b)(b+c)(c+a)+abc
=a2b+ab2+a2c+ac2+b2c+bc2+2abc+abc=a2b+ab2+a2c+ac2+b2c+bc2+2abc+abc
=ab(a+b+c)+bc(a+b+c)+ca(a+b+c)=ab(a+b+c)+bc(a+b+c)+ca(a+b+c)
=(a+b+c)(ab+bc+ca)=(a+b+c)(ab+bc+ca)
Vậy....
\((a+b)(b+c)(c-a)+(b+c)(c+a)(a-b)+(c+a)(a+b)(b-c)\)
\(=(b+c)(ac-a^2+bc-ab)+(b+c)(ac-bc+a^2-ab)+(c+a)(a+b)(b-c)\)
\(=(b+c)(ac-a^2+bc-ab+ac-bc+a^2-ab)+(c+a)(a+b)(b-c)\)
\(=(b+c)(2ac-2ab)+(c+a)(a+b)(b-c)\)
\(=(2ab+2ac)(c-b)-(c+a)(a+b)(c-b)\)
\(=(c-b)(2ab+2ac-a^2-ab-ac-bc)\)
\(=(c-b)(b-a)(a-c)\)
B1: Tính
1) (a+b)(a+b)
2) (a-b)(a-b)
3) (a-b)(a+b)
4) (a+b)(a+b)(a+b)
5) (a-b)(a-b)(a-b)
6) ( a+b)(a^2-ab+b^2)
7) (a-b)(a^2+ab+b^2)
\(1)\left(a+b\right)\left(a+b\right)\)
\(=a\left(a+b\right)+b\left(a+b\right)\)
\(=a^2+ab+ba+b^2\)
\(2)\left(a-b\right)\left(a-b\right)\)
\(=a\left(a-b\right)-b\left(a-b\right)\)
\(=a^2-ab-ba-b^2\)
\(3)\left(a-b\right)\left(a+b\right)\)
\(=a\left(a+b\right)-b\left(a+b\right)\)
\(=a^2+ab-ba+b^2\)
1, (a+b)(a+b) = (a + b)2
2, (a-b)(a-b) = (a - b)2
3, (a-b)(a+b) = a2 - b2
4, (a+b)(a+b)(a+b) = (a +b)3
5, (a-b)(a-b)(a-b) = (a - b)3
6) ( a+b)(a2 - ab + b2) = a3 + b3
7) (a-b)(a^2+ab+b^2) = a3 - b3
Bài 1: Find the word which has a different sound in the part underlined
1, A. stand B. apple C. hand D. father
2, A. dark B. cancel C. action D. travel
3, A. fantastic B. part C. fact D. camera
4, A. automatic B. had C. activity D. yard
5, A. calm B. bag C. cat D. bad
6, A. animal B. card C. heart D. cart
7, A. thank B. cancer C. channel D. smart
8, A. land B. start C. stand D. plastic
9, A. bag B. that C. can D. star
10, A. plant B. hat C. far D. plan
1. A stand B apple C hand D father
2. A dark B cancel C action D travel
3. A fantastic B part C fact D camera
4 A automatic B had C activity D yard
5 A calm B bag C cat D bad
6 A animal B card C heart D cart
7 A thanks B caner C channel D smart
8 A land B start C stand D plastic
9 A bag B that C can D star
10 A plant B hat C far D plan
1.D
2.A
3.B
4.D
5.A
6.A
7.D
8.B
9.D
10.C
1. Chứng minh rằng
a) (a+b-c) + (a-b) - (a-b-c) = - (a-b+c)
b) - (a-b-c) + (-a+b-c ) - (-a+b+c) + (c-a)
2. Cho M = (-a+b) - (b+c+a) + (c-a)
Chứng minh rằng nếu a < 0 thì M > 0
3. Chứng minh A và B là 2 số đối nhau
a) A = a-b , B = b-a
b) A = a-b+c , B = -a+b-c
4. Cho a-b=1 . Tinh S , biết :
S= -(a-b-c) + (-c+b+a) - ( a+b )
Voiwsa;b;cEN*(a=b+c),ta có:
a=b+c
=>a*(a-b)=(b+c)*(a-b)
=>a^2-a*b=a*b+a*c-b^2-b*c.
=>a^2-a*b-a*c=a*b-b^2-b*c
=>a*(a-b-c)=b*(a-b-c)
=>a=b.
Chỗ sai ở đâu???
chỗ sai là ở người làm giốt bạn ạ hì hì !
À mk nghĩ ra rồi ko cần các bn giúp đâu.
Chỉ cần nhận xét thôi.
Vì a=b+c.
=>a-b-c=0.
=>Ko thể rút gọn đc như trên.
Đúng ko thế mấy bn.
rút gọn:
a) -a-(b-a-c)=
d)-(a-b+c)-(a+b+c)=
e) (a+b)-(a-b)+(a-c)-(a+c)=
b) -(a-c)-(a-b+c)=
c)b-(b+a-c)=
f) (a+b-c)+(a-b+c)-(b+c-a)-(a-b-c)=
a)-b+c
d)-2a-2c
e)2b-2c
b)-2a+b
c)-a+c
f)a
-a-b+a+c=-b+c
-a+b-c-a-b-c=-2a-2c
a+b-a-b+a-c-a-c=-2c
-a-c+a-b-c=-2c+b
b-b-a+c=-a+c
a+b-c+a-b+c-b+c-a-a+b+c=2c
hãy xác định dấu của a,b thuộc Z
a) a+b= - (/a/ +/b/)
b) a+b= (/a/+/b/)
c) a+b=- (/a/-/b/)
d) a+b= (/b/-/a/)
e) a x b= /a/ x /b/ và a+b<0
f) a x b= -/a/ x /b/ và a<b
giúp mk nha!!!
rút gọn:
a) -a-(b-a-c)=
d)-(a-b+c)-(a+b+c)=
e) (a+b)-(a-b)+(a-c)-(a+c)=
b) -(a-c)-(a-b+c)=
c)b-(b+a-c)=
f) (a+b-c)+(a-b+c)-(b+c-a)-(a-b-c)=
a. \(-a-\left(b-a-c\right)=-a-b+a+c=c-b\)
b. \(-\left(a-c\right)-\left(a-b+c\right)=-a-c-a+b-c=b-2a-2c\)
c. \(b-\left(b+a-c\right)=b-b-a+c=c-a\)
d.\(-\left(a-b+c\right)-\left(a+b+c\right)=-a+b-c-a-b-c=-2a-2c=-2\left(a+c\right)\)e. \(\left(a+b\right)-\left(a-b\right)+\left(a-c\right)-\left(a+c\right)=a+b-a+b+a-c-a-c=2b-2c=2\left(b-c\right)\)
f. \(\left(a+b-c\right)+\left(a-b+c\right)-\left(b+c-a\right)-\left(a-b-c\right)=a+b-c+a-b+c-b-c+a-a+b+c=2a\)
a) -a - (b - a - c)
= -a - b + a + c
=[ -a + a] - (b + c)
= 0 - (b + c)
= -(b + c)
d) -(a - b + c) - (a + b + c)
= -a + b - c - a - b - c
= (-a - a) + (b - b) + (c - c)
= (-a - a) + 0 + 0
= -a - a
e) (a + b) - (a - b) + (a - c) - (a + c)
= a + b - a + b + a - c - a - c
= (a - a) - (a + a) + (b + b) - (c - c)
= 0 + 2a + 2b - 0
= 2a + 2b
b) -(a - c) - (a - b + c)
= -a + c - a + b - c
= (-a - a) + (c - c) + b
= [(-a) + (-a)] + 0 + b
= 2(-a) + b
c) b - (b + a - c)
= b - b - a + c
= 0 - a - c
= -a - c
f) (a + b - c) + (a - b + c) - (b + c - a) - (a - b - c)
= 0 - (b + c - a) - (a - b - c)
= 0 - b - c + a - a + b - c
= -b - c + a - a + b - c
= (-b + b) - (c - c) + (a - a)
= 0 - 0 + 0
= 0
Biến đổi vế trái thành phải
a) a (b-c) + c (a-b) = b (a-c)
b) a(b-c) -b (a+c) = (a+b) -(-c)
c) a (b+c) - b (a-c) = (a+b) c
d) a (b-c) - a(b+d) = a (c+d)
e) (a-b) (c+d) - (a+d) (b+c)= (a-c) (d-b)
a) a(b-c)+c(a-b)=ab-ac+ca-cb=ab-cb=b(a-c)
b) a(b-c)-b(a+c)=ab-ac-ab-bc=-ac-bc=-c(a+b)
c) a(b+c)-b(a-c)=ab+ac-ab+bc=ac+bc=c(a+b)
d) a(b-c)-a(b+d)=ab-ac-ab-ad=-ac-ad=-a(a+d)
a) a(b - c) + c(a - b) = ab - ac + ac - bc = ab - bc = b(a - c)
b) a(b - c) - b(a + c) = ab - ac - ab - bc = -ac - bc = (a + b). (-c)
c) a(b + c) - b(a - c) = ab + ac - ab + bc = ac + bc = (a + b)c
d) a(b - c) - a(b + d) = ab - ac - ab - ad = -ac - ad = -a(c + d)
a) a(b-c)+c (a-b)=ab-ac+ca-cb=cb=b(a-c)
b) a(b-c)-b(a+c)=ab-ac-ab-bc=ac-bc=-c(a+b)
c) a(b+c)-a(b(a-c)=ab+ac-ab+bc=ac+bc=c(a+b)
d) a(b-c)-a(b+d)=ab-ac-ab-ad=ac-ab-ad=ac-ad=a(a+d)
rút gọn biểu thức
a) ( a + b ) - ( a - b ) + ( a - c ) - ( a + c )
b) (a + b - c ) + ( a - b + c ) - ( b + c - a ) - ( a - b - c )
c) - { - ( a + b ) - [ - (a - b ) - (a + b) ] -}
a)\(\left(a+b\right)-\left(a-b\right)+\left(a-c\right)-\left(a+c\right)=a+b-a+b+a-c-a-c\)
\(=2a-2a+2b-2c=0+2\left(b-c\right)=2\left(b-c\right)\)
b)\(\left(a+b-c\right)+\left(a-b+c\right)-\left(b+c-a\right)-\left(a-b-c\right)\)
\(=a+b-c+a-b+c-b-c+a-a+b+c\)
\(=3a-a+2b-2b+2c-2c=2a+0+0=2a\)
c)\(-\left\{-\left(a+b\right)-\left[-\left(a-b\right)-\left(a+b\right)\right]\right\}=\left\{-\left(a+b\right)-\left[b-a-a-b\right]\right\}\)
\(=-\left\{-\left(a+b\right)-b+a+a+b\right\}\)
\(=-\left\{b-a-b+2a+b\right\}\)
\(=-\left\{b-b+b+2a-a\right\}\)
\(=-\left\{b+a\right\}=-a-b\)