Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a ) \(A=\frac{ax^2\left(a-x\right)-a^2x\left(x-a\right)}{3a^2-3x^2}=\frac{ax\left(a-x\right)\left(a+x\right)}{3\left(a-x\right)\left(a+x\right)}=\frac{ax}{3}\)
Thay \(a=\frac{1}{2};x=-3\), ta có :
\(A=\frac{\frac{1}{2}.-3}{3}=-\frac{1}{2}\)
b ) \(B=\frac{\left(ab+bc+cd+da\right)abcd}{\left(c+d\right)\left(a+b\right)+\left(b-c\right)\left(a-d\right)}=\frac{\left[\left(ab+ad\right)+\left(bc+cd\right)\right]abcd}{ca+cb+da+db+ba-bd-ca+cd}\)
\(=\frac{\left[a\left(b+d\right)+c\left(b+d\right)\right]abcd}{ba+da+cb+cd}=\frac{\left(b+d\right)\left(a+c\right)abcd}{\left(b+d\right)\left(a+c\right)}=abcd\)
Thay \(a=-3;b=-4;c=2;d=3\), ta có :
\(B=\left(-3\right).\left(-4\right).2.3=72\)
\(a,\left(3x+5\right)^2+\left(3x-5\right)^2-\left(3x+2\right)\left(3x-2\right)=9x^2+30x+25+9x^2-30x+25-9x^2+4=9x^2+54\)
\(b,BT=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x=x^3-16x^2+25x\)
\(c,BT=\left(x+y-z\right)^2-2\left(x+y-z\right)\left(x+y\right)+\left(x+y\right)^2=\left(x+y-z-x-y\right)^2=z^2\)
a: =x^2-4-(x^2-2x-3)
=x^2-4-x^2+2x+3
=2x-1
b: \(=2x^2+3x-10x-15-2x^2+6x+x+7\)
=-8
c: \(=a^2+2ab+b^2-a^2+2ab-b^2=4ab\)
a) \(\left(x-5\right)\left(x+8\right)-\left(x+4\right)\left(x-1\right)\)
\(=\left(x^2+3x-40\right)-\left(x^2+3x-4\right)\)
\(=x^2+3x-40-x^2-3x+4\)
\(=-36\)
b)\(x^4\left(x^2-1\right)\left(x^2+1\right)\)
\(=x^4\left(x^4-1\right)\)
\(=x^8-x^4\)
a,=0 hoac =2a
b,=0 hoac =-(2a)
c,=a2 hoac =-(a2)
d,=1 hoac = -a/a
còn lại hông bít làm
e: TH1: x<-3
A=3x-3-2(-x-3)=3x-3+2x+6=5x+3
TH2: x>=-3
A=3x-3-2(x+3)=3x-3-2x-6=x-9
g: TH1: x<1/4
B=2(3-x)-(1-4x)
=6-2x-1+4x=2x+5
TH2: 1/4<=x<3
B=2(3-x)-4x+1=6-2x-4x+1=-6x+7
TH3: x>=3
B=2(x-3)-4x+1=2x-6-4x+1=-2x-5
A=-x+2.5+x-1.7
A=0.8
B=-x+1.5+x-2/5
B=1.5-2/5
B=11/10hayB=1.1