Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a/ Chứng minh:
\(\left(x+a\right)\left(x+b\right)\)
\(=x^2+bx+ax+ab\)
\(=x^2+\left(ax+bx\right)+ab\)
\(=x^2+x\left(a+b\right)+ab=VP\) (đpcm)
b/ Chứng minh:
\(\left(x+a\right)\left(x+b\right)\left(x+c\right)\)
\(=\left(x^2+ax+bx+ab\right)\left(x+c\right)\)
\(=x^3+cx^2+ax^2+acx+bx^2+bcx+abx+abc\)
\(=x^3+\left(ax^2+bx^2+cx^2\right)+\left(abx+bcx+acx\right)+abc\)
\(=x^3+x^2\left(a+b+c\right)+x\left(ab+bc+ac\right)+abc=VP\) (đpcm)
Câu 4:
\(=\dfrac{a\left(a-b\right)-c\left(a-b\right)}{a\left(a+b\right)-c\left(a+b\right)}=\dfrac{a-b}{a+b}\)
a: =(x+y)^3+z^3-3xy(x+y)-3xyz
\(=\left(x+y+z\right)\left[\left(x+y\right)^2-z\left(x+y\right)+z^2\right]-3xy\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2-3xy\right)\)
\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)
b: \(=\left(x+y+y-z\right)^3-3\left(x+y\right)\left(y-z\right)\left(x+y+y-z\right)+\left(z-x\right)^3\)
\(=\left(x-z\right)^3+\left(z-x\right)^3-3\left(x+y\right)\left(y-z\right)\left(x-z\right)\)
\(=-3\left(x+y\right)\left(y-z\right)\left(x-z\right)\)
c: \(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)
=(x^2+x+5)(x^2+x-2)
=(x^2+x+5)(x+2)(x-1)
d: =b^2c+bc^2+ac^2-a^2c-a^2b-ab^2
=b^2c-b^2a+bc^2-a^2b+ac^2-a^2c
=b^2(c-a)+b(c^2-a^2)+ac(c-a)
=(c-a)(b^2+ac)+b(c-a)(c+a)
=(c-a)(b^2+ac+bc+ba)
=(c-a)[b^2+bc+ac+ab]
=(c-a)[b(b+c)+a(b+c)]
=(c-a)(b+c)(b+a)
\(A=6x^2+23x+21-\left(6x^2+23x-55\right)=76\\ B=x^4+x^3-x^2-2x^2-2x+2-x^4-x^3+3x^2+2x\\ =2\\ C=x^4+x^3-3x^2-2x-\left(x^4+x^3-x^2-2x^2-2x+2\right)\\ =-2\)
a. \(VT=\left(x+a\right)\left(x+b\right)=x^2+ã+bx+ab=x^2+\left(a+b\right)x+ab=VP\)
B. \(VT=\left(x+a\right)\left(x+b\right)\left(x+c\right)=\left[\left(x+a\right)\left(x+b\right)\right].\left(x+c\right)\)
\(=\left[\left(x^2+\left(a+b\right)x\right)+ab\right].\left(x+c\right)=x^3+x^2c+\left(a+b\right)x^2+c\left(a+b\right)x+abx+abc\)
\(=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc=VP\)
Lời giải:
Đặt $\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=t$
$\Rightarrow a=xt; b=yt; c=zt$. Ta có:
$a+b+c=xt+yt+zt=t(x+y+z)=t$
$a^2+b^2+c^2=t^2(x^2+y^2+z^2)=t^2$
$ab+bc+ac=\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}=\frac{t^2-t^2}{2}=0$
Ta có đpcm.
P/s : Phần b ) : \(\left(x+a\right)\left(x+b\right)\left(x+c\right)\)
a ) \(\left(x+a\right)\left(x+b\right)=x^2+ax+bx+ab=x^2+\left(a+b\right)x+ab\)
b ) \(\left(x+a\right)\left(x+b\right)\left(x+c\right)\)
\(=\left[x^2+\left(a+b\right)x+ab\right]\left(x+c\right)\)
\(=x^2\left(x+c\right)+\left(a+b\right)x\left(x+c\right)+ab\left(x+c\right)\)
\(=x^3+x^2c+\left(ax+bx\right)\left(x+c\right)+abx+abc\)
\(=x^3+x^2c+ax^2+bx^2+axc+bxc+abx+abc\)
\(=x^3+\left(x^2a+x^2b+x^2c\right)+\left(abx+bcx+axc\right)+abc\)
\(=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc\)