Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 4:
Ta có:
\(a^2-2a+b^2+4b+4c^2-4c+6=0\)
\(\Leftrightarrow a^2-2a+1+b^2+4b+4+4c^2-4c+1\)
\(\Leftrightarrow\left(a^2-2b+1\right)+\left(b^2+4b+4\right)+\left(4c^2-4c+1\right)\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2\)
Mà \(\hept{\begin{cases}\left(a-1\right)^2\ge0\\\left(b+2\right)^2\ge0\\\left(2c-1\right)^2\ge0\end{cases}}\)
\(\Rightarrow\left(a-1\right)^2+\left(b+2\right)^2+\left(2c-1\right)^2\ge0\)
Dấu "=" xảy ra khi \(\hept{\begin{cases}\left(a-1\right)^2=0\\\left(b+2\right)^2=0\\\left(2c-1\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a=1\\b=-2\\c=\frac{1}{2}\end{cases}}}\)
Vậy \(\left(a,b,c\right)=\left(1;-2;\frac{1}{2}\right)\)
a)310*210-67(63-1)=610-610-67=0-67=-67
b)2xn(3xn+1-1)-3xn+1(2xn-1)=2xn*3xn+1-2xn-3xn+1*2xn-3xn+1=(2xn*3xn+1-2xn*3xn+1)-(2xn-3xn+1)=0-(2xn-3xn+1)=-2xn+3xn+1
Bài làm:
a) \(M=90.10^n-10^{n+2}+10^{n+1}\)
\(M=9.10.10^n-10^{n+2}+10^{n+1}\)
\(M=10^{n+1}\left(9-10+1\right)\)
\(M=10^{n+1}.0=0\)
b) \(N=x\left(x+y\right)-y\left(x+y\right)\)
\(N=\left(x-y\right)\left(x+y\right)\)
\(N=x^2-y^2\)
c) \(P=y\left(x^{n-1}+y^{n-1}\right)-x^{n-1}\left(x+y\right)\)
\(P=x^{n-1}y+y^n-x^n-x^{n-1}y\)
\(P=y^n-x^n\)
Học tốt!!!!
a, 5^6 -10^4=5^2. 5^4 -5^4. 2^4
=5^4(5^2 -2^4)
=5^4. 9 \(⋮\) 9
b, (n+3)2- (n -1)2=(n+3- n+1)(n+3+ n- 1)
=4(2n+2)
=8n+ 8\(⋮8\)
a)10n+1-6.10n
=10n.10-6.10n
=10n(10-6)
=10n.4
b)90.10n-10n+2+10n+1
=90.10n-10n.100+10n+10
=10n(90-100+10)
=10n.0
=0
a, \(10^{n+1}-6.10^n\)
= \(10^n.10-6.10^n\)
=\(10^n.\left(10-6\right)\)
=\(10^n.4\)
b, \(90.10^n-10^{n+2}-10^{n+1}\)
= \(90.10^n-10^n.10^2-10^n.10\)
= \(10^n.\left(90-10^2-10\right)\)
= \(10^n.\left(-20\right)\)
nhớ k cho mik nha!!!!!!!!!!!!!