Rút gọn
(a+b+c+d)^2+(a+b-c-d)^2+(a+c-b-d)^2+(a+d-b-c)^2
Bài 1: bỏ dấu ngoặc rồi rút gọn biểu thức a, - ( - a + c - d ) - ( c - d + d) b, - ( a + b - c + d ) + (a - b - c - d) c, a( b - c - d ) - a( b + c -d ) d*, (a + b).(c+d) - ( a+d).(b+c) e*, (a+b).(c-d) - (a-b).(c+d) f*, (a+b)2 - (a-b)2
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)
Rút gọn biểu thức: A = (a+b+c+d)2+(a+b-c-d)2+(a+c-b-d)2+(a+d-b-c)2
\(A=\left[\left(a+b\right)+\left(c+d\right)\right]^2+\left[\left(a+b\right)-\left(c+d\right)\right]^2+\left[\left(a-b\right)+\left(c-d\right)\right]^2+\left[\left(a-b\right)-\left(c-d\right)\right]^2\)
Ta có
\(\left[\left(a+b\right)+\left(c+d\right)\right]^2=\left(a+b\right)^2+2\left(a+b\right)\left(c+d\right)+\left(c+d\right)^2\)
\(\left[\left(a+b\right)-\left(c+d\right)\right]^2=\left(a+b\right)^2-2\left(a+b\right)\left(c+d\right)+\left(c+d\right)^2\)
\(\left[\left(a-b\right)+\left(c-d\right)\right]^2=\left(a-b\right)^2+2\left(a-b\right)\left(c-d\right)+\left(c-d\right)^2\)
\(\left[\left(a-b\right)-\left(c-d\right)\right]^2=\left(a-b\right)^2-2\left(a-b\right)\left(c-d\right)+\left(c-d\right)^2\)
\(A=2\left(a+b\right)^2+2\left(a-b\right)^2+2\left(c+d\right)^2+2\left(c-d\right)^2\)
\(A=2\left(a^2+2ab+b^2+a^2-2ab+b^2+c^2+2cd+d^2+c^2-2cd+d^2\right)\)
\(A=4\left(a^2+b^2+c^2+d^2\right)\)
Rút gọn biểu thức: D = ( a + b + c + d )2 + ( a + b + c - d )2 + (a + b - c - d )2 + ( a + d - b - c )2
rút gọn
(a+b+c+d)2 +(a+b-c-d)2 +(a+c-b-d)2 +(a+d-b-c)
làm bừa thui,ai tích mình mình tích lại
số dư lớn nhất bé hơn 175 là 174
số nhỏ nhất có 4 chữ số là 1000
Mà 1000:175=5( dư 125)
số đó là:
Bỏ dấu ngoặc rồi rút gọn biểu thức.
a) (a+b)*(c+d)-(a+d)*(b+c)
b) (a+b)*(c-d)-(a-b)*(c+d)
c) (a+b)^2-(a-b)^2
Bài 1 : Rút gọn
a) (a + b + c + d) (a - b -c - d)
b) (a - b - c - d) (a + b - c - d)
c) (a - b - c) ( a2 + b2 + c2 + 2ab - ac - bc)
d) (a - b - c + d) (a + b - c - d)
e) (a - b - c) ( a2 + b2 + c2 - 2ab + ac - bc)
a) (a + b + c + d)(a - b - c - d)
= a(a + b + c + d) - b(a + b + c + d) - c(a + b + c + d) - d(a + b + c + d)
= (aa + ab + ac + ad) - (ba + bb + bc + bd) - (ca + cb + cc + cd) - (da + db + dc + dd)
= aa - bb - cc - dd
rút gọn biểu thức:
\(\left(a+b+c+d\right)^2+\left(a+c-c-d\right)^2+\left(a-b+c-d\right)^2+\left(a-b-c+d\right)^2\)
\(\left(a+b+c+d\right)^2+\left(a+b-c-d\right)^2+\left(a-b+c-d\right)^2+\left(a-b-c+d\right)^2\)(Sửa lại nha bn viết sai để)
Đặt x=a+b , y=c+d , z=a-b , t=c-d
Khi đó biểu thức bằng
\(\left(x+y\right)^2+\left(x-y\right)^2+\left(z+t\right)^2+\left(z-t\right)^2\)
\(=x^2+y^2+2xy+x^2+y^2-2xy+z^2+t^2+2zt+z^2+t^2-2zt\)
\(=2\left(x^2+y^2+z^2+t^2\right)=2\left[\left(a+b\right)^2+\left(a-b\right)^2+\left(c+d\right)^2+\left(c-d\right)^2\right]\)
\(=2(a^2+b^2-2ab+a^2+b^2-2ab+c^2+d^2+2cd+c^2+d^2-2cd)\)
\(=2\left(2a^2+2b^2+2c^2+2d^2\right)=4\left(a^2+b^2+c^2+d^2\right)\)
Bỏ dấu ngoặc rồi rút gọn biểu thức
a) -(-a+c-d)-(c-a+d)
b) -(a+b-c+d)+(a-b-c-d)
c) a(b-c-d)-a(b+c-d)
d*) (a+b)(c+d0-(a+d)(b+c)
e*) (a+b)(c-d)-(a-b)(c+d)
f*) (a+b)^2-(a-b)^2
Bỏ dấu ngoặc rồi rút gọn biểu thức
a) -(-a+c-d)-(c-a+d)
b) -(a+b-c+d)+(a-b-c-d)
c) a(b-c-d)-a(b+c-d)
d*) (a+b)(c+d0-(a+d)(b+c)
e*) (a+b)(c-d)-(a-b)(c+d)
f*) (a+b)^2-(a-b)^2
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
-(a+b-c)+(b+c-d)-(c+d-a)
Bỏ dấu ngoặc rồi rút gọn biểu thức
(a+b).(c+d)-(a+d).(b+c)
(a+b).(c-d)-(a-b).(c+d)
(a+b)2-(a-b)2
1) \(=ac+ad+bc+bd-ab-ac-db-dc=ad+bc-dc-ab=d\left(a-c\right)-b\left(a-c\right)=\left(a-c\right)\left(d-b\right)\)
2) \(=ac-ad+bc-bd-ac-ad+bc+bd=2bc-2ad=2\left(bc-ad\right)\)
3) \(\left(a+b\right)\left(a+b\right)-\left(a-b\right)\left(a-b\right)=a^2+2ab+b^2-a^2+2ab-b^2=4ab\)