Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a^2+ac-b^2-bc=\left(a^2-b^2\right)+\left(ac-bc\right)=\left(a+b\right)\left(a-b\right)+c\left(a-b\right)=\)\(\left(a-b\right)\left(a+b+c\right)\)
Tương tự:
\(b^2+ab-c^2-ac=\left(b-c\right)\left(a+b+c\right)\)
\(c^2+bc-a^2-ab=\left(c-a\right)\left(a+b+c\right)\)
\(Q=\frac{1}{\left(b-c\right)\left(a-b\right)\left(a+b+c\right)}+\frac{1}{\left(c-a\right)\left(b-c\right)\left(a+b+c\right)}+\frac{1}{\left(a-b\right)\left(c-a\right)\left(a+b+c\right)}\)
\(=\frac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+b+c\right)}=0\)
với ab+bc+ca=1
=>\(a^2+1=a^2+ab+bc+ca=\left(a+b\right)\left(a+c\right)\)
tương tự mấy cái kia rồi thay vào, ta có
A=\(\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}=1\)
b),ta có \(a^2+2bc-1=a^2+bc-ab-ac=\left(a-b\right)\left(a-c\right)\)
tương tự mấy cái kia, rồi thay váo, ta có
\(B=\frac{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}=1\)
^_^
Ta có: MS = (1+a2).(1+b2).(1+c2)
= (ab + ac + bc + a2).(ab + ac + bc + b2).(ab + bc + ac + c2)
= [ (a2 + ac) + (ab + bc) ] . [ (ab + b2) + (ac + bc) ] . [ (ab + bc) + (ac + c2) ]
= [ a(a + c) + b(a + c) ] . [ b(a + b) + c(a + b) ] . [ b(a + c) + c(a + c) ]
= (a + b)(a + c)(b + c)(a + b)(b + c)(a + c)
= (a + b)2(b + c)2(a + c)2 = TS
Vậy A = 1
làm cái đề ra ấy, ngại viết lại đề :P
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=4\left(a^2+b^2+c^2\right)-4\left(ab+bc+ca\right)\)
\(\Leftrightarrow2\left(a^2+b^2+c^2\right)-2\left(ab+bc+ca\right)=0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}a=b\\b=c\\c=a\end{cases}}\)
\(\Rightarrow M=1^{2018}+1^{2019}+1^{2020}=1+1+1=3\)
1a)
Đặt \(a^2+a+1=t\Rightarrow a^2+a+2=t+1\)
\(\Rightarrow A=t\left(t+1\right)-12=t^2+t-12=t^2-3t+4t-12=\left(t-3\right)\left(t+4\right)\)
\(=\left(a^2+a-2\right)\left(a^2+a+5\right)\)
Mà \(a>1\Rightarrow\hept{\begin{cases}a^2+a-2>0\\a^2+a+5>0\end{cases}}\forall a>1\)
Vậy A là hợp số
1b)
Ta có :
\(B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\cdot...\cdot\left(2^{1006}+1\right)+1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\cdot...\cdot\left(2^{1006}+1\right)+1=....=\left(2^{1006}-1\right)\left(2^{1006}+1\right)+1\)
\(=2^{2012}-1+1=2^{2012}\)
a) Ta có : \(a^2+1=a^2+ab+bc+ac=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)
Tương tự : \(b^2+1=\left(b+a\right)\left(b+c\right)\) ; \(c^2+1=\left(c+a\right)\left(c+b\right)\)
Suy ra \(\left(1+a^2\right)\left(1+b^2\right)\left(1+c^2\right)=\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2\)
Vậy \(A=\frac{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}{\left(a+b\right)^2\left(b+c\right)^2\left(c+a\right)^2}=1\)
b) Ta có ; \(a^2+2bc-1=a^2+2bc-\left(ab+bc+ac\right)=a^2-ab+bc-ac=a\left(a-b\right)-c\left(a-b\right)\)
\(=\left(a-b\right)\left(a-c\right)\)
Tương tự : \(b^2+2ac-1=\left(a-b\right)\left(c-b\right)\) ; \(c^2+2ab-1=\left(a-c\right)\left(b-c\right)\)
Suy ra \(\left(a^2+2bc-1\right)\left(b^2+2ac-1\right)\left(c^2+2ab-1\right)=\left(a-b\right)^2.\left(c-a\right)^2.\left[-\left(b-c\right)^2\right]\)
Vậy : \(B=\frac{-\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2}{\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)}=-1\)
ta có : \(P=\left(a+1\right)^2+\left(b+1\right)^2+\left(c+1\right)^2+2\left(ab+bc+ca\right)\)
\(P=a^2+2a+1+b^2+2b+1+c^2+2c+1+2ab+2bc+2ca\)
\(P=\left(a^2+b^2+c^2+2ab+2bc+2ca\right)+2a+2b+2c+3\)
\(P=\left(a+b+c\right)^2+2\left(a+b+c\right)+3\)
ta có : \(Q=\left(a+b+c+1\right)^2=\left(\left(a+b+c\right)+1\right)^2\)
\(Q=\left(a+b+c\right)^2+2\left(a+b+c\right)+1\)
\(\Leftrightarrow P-Q=\left(a+b+c\right)^2+2\left(a+b+c\right)+3-\left(a+b+c\right)^2-2\left(a+b+c\right)-1=2\)
vậy \(P-Q=2\)