Rút gọn
A= a . (b - c) - b . (a + c)
B= (a + b) . (c - d) - (a - d) . (b + c)
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LP
3
NN
1
3 tháng 12 2023
a: \(\left(a-b-c\right)-\left(-c+b+a\right)-\left(a-b\right)\)
\(=a-b-c+c-b-a-a+b\)
\(=-a-b\)
b: \(a\left(b+c\right)-a\left(b+d\right)-\left(1+ac-ad\right)\)
\(=ab+ac-ab-ad-1-ac+ad\)
=-1
15 tháng 11 2021
Bài 1:
Ta có: \(a^3+b^3+c^3=3abc\)
\(\Leftrightarrow\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-3a^2b-3ab^2-3abc=0\)
\(\Leftrightarrow\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)=0\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0\left(do.a+b+c\ne0\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(a-b\right)^2=0\\\left(b-c\right)^2=0\\\left(a-c\right)^2=0\end{matrix}\right.\)\(\Leftrightarrow a=b=c\)
\(M=\dfrac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\dfrac{3a^2}{\left(3a\right)^2}=\dfrac{3a^2}{9a^2}=\dfrac{1}{3}\)
15 tháng 11 2021
Bài 2:
a) \(=\dfrac{x\left(x^2+x-6\right)}{x\left(x^2-4\right)}=\dfrac{x\left(x-2\right)\left(x+3\right)}{x\left(x-2\right)\left(x+2\right)}=\dfrac{x+3}{x+2}\)
b) \(=\dfrac{x\left(x+1\right)+7\left(x+1\right)}{x\left(x^2+2x+1\right)}=\dfrac{\left(x+1\right)\left(x+7\right)}{x\left(x+1\right)^2}=\dfrac{x+7}{x\left(x+1\right)}=\dfrac{x+7}{x^2+x}\)
PT
7 tháng 6 2021
a) - ( - a + c – d ) – ( c – a + d )
= a - c - d - c + a + d
= (a + a) + (-c - c) + (-d + d)
= 2a - 2c
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)
= -2b - 2d
7 tháng 6 2021
a) - ( - a + c - d) - ( c - a + d )
= a - c + d - c + a - d
= 2a
b) - ( a+ b - c + d ) + ( a -b -c -d )
= - a-b+c-d+a-b-c-d
=-2d -2b
c) a(b-c-d) - a(b+c-d)
= a(b-c-d-b-c+d)
= ab-ac-ad-ab-ac+ad
= -2ab-2ac
d) (a+b)(c+d)-(a+d)(b+c)
= ac+ad+bc+bd - (ab+ac+bd+cd)
= ac+ad+bc+bd-ab-ac-bd-cd
=ad+bc-ab-cd
12 tháng 1 2021
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)
10 tháng 2 2021
a, \(A=\frac{\left(x+2\right)^2}{x}\left(1-\frac{x^2}{x+2}\right)=\frac{\left(x+2\right)^2}{x}\left(\frac{x+2-x^2}{x+2}\right)\)
\(=\frac{-\left(x+2\right)^2\left(x-2\right)\left(x+1\right)}{x\left(x+2\right)}=\frac{-\left(x\pm2\right)\left(x+1\right)}{x}\)
c, Theo bài ra ta có : \(C=\frac{A}{B}\)hay \(\frac{\frac{-\left(x\pm2\right)\left(x+1\right)}{x}}{\frac{4}{\left(x-2\right)^2}}=\frac{\frac{-\left(x+2\right)\left(x+1\right)}{x}}{\frac{4}{x-2}}\)
d, Theo bài ra ta có :
\(C>0\)hay \(\frac{\frac{-\left(x+2\right)\left(x+1\right)}{x}}{\frac{4}{x-2}}>0\)
\(\Leftrightarrow\frac{-\left(x+2\right)\left(x+1\right)}{x}.\frac{x-2}{4}>0\)
\(\Leftrightarrow-\left(x+2\right)\left(x+1\right)>0\Leftrightarrow\left(x+2\right)\left(x+1\right)>0\)
\(\Leftrightarrow x>-2;x>-1\Rightarrow x>-1\)
9 tháng 8 2019
a) -(-a + c - d) - (c - a + d) = a - c + d - c + a - d = (a + a) - (c + c) + (d - d) = 2a - 2c
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) = -2b - 2d
c) (a + b - c) - (b - c + d) = a + b - c - b + c - d = a + (b - b) - (c - c) - d = a - d
d) (b + a) + (c - d) - (c + a) - (b - d) = b + a + c - d - c - a - b + d = (b - b) + (a - a) + (c - c) - (d - d) = 0
e) (a - b) - (d + a) - (c - d) + (c + b) = a - b - d - a - c + d + c + b = (a - a) - (b - b) - (d - d) - (c - c) = 0
f) -a + (c - b) - (c + a - b) = - a + c - b - c - a + b = (-a - a) + (c - c) - (b - b) = -2a
9 tháng 8 2019
a ) a - c + d - c + a - d = 2a - 2c
b ) -a - b + c - d + a - b - c - d = -2b - 2d
CÁC CÂU CÒN LẠI LÀM TƯƠNG TỰ NHÉ!!
20 tháng 7 2015
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=a.(b+c)-b.(a+c)
=ab+ac-ba+bc
=ac+bc
=c.(a+b)