Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\left(x+a\right)\left(x+b\right)\left(x+c\right)\)
\(=\left[x^2+\left(a+b\right)x+ab\right]\left(x+c\right)\)
\(=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc\)
b) \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ca-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
c) \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)+b^2c-ab^2+c^2a-bc^2\)
\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b-c\right)\left(b+c\right)\)
\(=\left(b-c\right)\left(a^2+bc-ab-ca\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
Nhầm đoạn cuối là \(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
\(A=x^2-y^2-x+y\)
\(=\left(x^2-y^2\right)-\left(x-y\right)\)
\(=\left(x+y\right)\left(x-y\right)-\left(x-y\right)\)
\(=\left(x+y-1\right)\left(x-y\right)\)
\(B=ax-ab+b-x\)
\(=\left(ax-ab\right)-\left(x-b\right)\)
\(=a\left(x-b\right)-\left(x-b\right)\)
\(=\left(a-1\right)\left(x-b\right)\)
\(D=x^2-2xy+y^2-m^2+2mn-n^2\)
\(=\left(x^2+y^2-2xy\right)-\left(m^2+n^2-2mn\right)\)
\(=\left(x-y\right)^2-\left(m-n\right)^2\)
\(=\left(x-y-m+n\right)\left(x-y+m-n\right)\)
\(E=x^2-y^2-2yz-z^2\)
\(=x^2-\left(y^2+z^2+2yz\right)\)
\(=x^2-\left(y-z\right)^2\)
\(=\left(x+y-z\right)\left(z-y+z\right)\)
\(=>A=\left(x-y\right)\left(x+y\right)-\left(x-y\right)\\ =>A=\left(x-y\right)\left(x+y-1\right)\) ( dấu phía sau bị lỗi nha )
\(=>B=a\left(x-b\right)-\left(x-b\right)\\ =>B=\left(x-b\right)\left(a-1\right)\)
\(=>C=\left(a+b+c\right)\left(3x^2+36xy+108y^2\right)\)
\(=>C=3\left(a+b+c\right)\left(x^2+12xy+36y^2\right)\\ =>C=3\left(a+b+c\right)\left(x+6y\right)^2\)
\(\Rightarrow D=\left(x-y\right)^2-\left(m^2-2mn+n^2\right)\\ =>D=\left(x-y\right)^2-\left(m-n\right)^2\)
\(=>D=\left(x-y+m-n\right)\left(x-y-m+n\right)\)
\(=>E=x^2-\left(y^2+2yz+z^2\right)\\ =>E=x^2-\left(y+z\right)^2\)
\(=>E=\left(x-y-z\right)\left(x+y+z\right)\)
T I C K ủng hộ nha
CHÚC BẠN HỌC TỐT
1) \(a^2+b^2+1\ge ab+a+b\)
\(\Leftrightarrow a^2+b^2+1-ab+a+b\ge0\)
\(\Leftrightarrow2a^2+2b^2+2-2ab+2a+2b\ge0\)
\(\Leftrightarrow\left(a^2+2ab+b^2\right)+\left(a^2+2a+1\right)+\left(b^2+2b+1\right)\ge0\)
\(\Leftrightarrow\left(a+b\right)^2+\left(a+1\right)^2+\left(b+1\right)^2\ge0\) (luôn đúng)
Dấu "=" xảy ra \(\Leftrightarrow a=b=-1\)
2/ \(\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge4\)
Áp dụng bđt cosi : \(\left(a+b\right)\left(\frac{1}{a}+\frac{1}{b}\right)\ge2\sqrt{ab}\cdot2\sqrt{\frac{1}{a}.\frac{1}{b}}=4\)(ĐPCM)
Dấu "=" xảy ra \(\Leftrightarrow a=b\)
3/ \(\frac{a^2+a+1}{a^2-a+1}>0\)
Vì \(\hept{\begin{cases}a^2+a+1=\left(a+\frac{1}{2}\right)^2+\frac{3}{4}>0\\a^2-a+1=\left(a-\frac{1}{2}\right)^2+\frac{3}{4}>0\end{cases}}\Leftrightarrow\frac{a^2+a+1}{a^2-a+1}>0\)(ĐPCM)
a) giả sử a^2-ab+b^2>/ab
<=> a^2-ab+b^2-ab>/0
<=> a^2-2ab+b^2>/0
<=> (a-b)^2>/0 (đúng với mọi a,b)
vậy a^2-ab+b^2>/ab
b) giả sử (a+b)^2.(a-b)^2>/4ab(a-b)^2
<=> (a+b)^2(a-b)^2-4ab(a-b)^2>/0
<=> (a-b)^2(a^2+2ab+b^2-4ab)>/0
<=> (a-b)^2(a-b)^2>/0
<=> (a-b)^4>/0 (đúng với mọi a,b)
vậy (a+b)^2(a-b)^2>/4ab(a-b)^2