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a. Đề:\(3\left(2x-1\right)^2+7\left(3y+5\right)^2=0\)
Giải :\(\Rightarrow\hept{\begin{cases}2x-1=0\\3y+5=0\end{cases}}\Rightarrow\hept{\begin{cases}2x=0+1=1\\3y=0-5=-5\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=-\frac{5}{3}\end{cases}}}\)
b. Đề : \(x^2+y^2-2x+10y+26=0\)
Giải : \(\Leftrightarrow x^2-2.1.x+1+y^2+2.5.y+25=0\)
\(\Leftrightarrow\left(x-1\right)^2+\left(y+5\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x-1=0\\y+5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0+1=1\\y=0-5=-5\end{cases}}}\)
Đây là bài 1 bài 2 đang ghi nha
t i c k nha cảm ơn
\(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1\)
\(B=2^{32}\)
=> \(A< B\)
ta có A= \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
=(2-1)(2+1)\(\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
=\(2^{32}-1\) (ấp dụng các hằng đẳng thức )
=> A=232-1
B=232
=> A<B
Ta có: \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1< 2^{32}\)
\(\Leftrightarrow A< B\)
\(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(A=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(A=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(A=\left(2^8-1\right)\left(2^8+1\right)\)
\(A=2^{16}-1< 2^{16}\)
Mình làm theo cách tính nhé !
\(A=\left(2+1\right).\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right)\)
\(A=3.\left(4+1\right).\left(16+1\right).\left(256+1\right)\)
\(A=3.5.17.257\)
\(\Rightarrow A=65535\)
\(B=2^{16}=65536\)
Từ đó \(\Rightarrow A< B\)
A = (2 - 1)(2 + 1)(2^2 + 1 )(2^4 + 1 ) (2^8 + 1)(2^16 + 1) ( nhân vói 2 - 1 = 1 Gía không thay dổi)
A = ( 2 ^2 - 1 )(2^2 + 1 )(2^4 + 1 )(2^8 + 1 )(2^16 + 1 )
A = ( 2^4 - 1 )(2^4 + 1)(2^8 + 1)(2^16 + 1)
A = (2^8 - 1)(2^8 + 1)(2^16 + 1)
A = (2^16 - 1)(2^16 + 1 )
A = 2^32 - 1 <2^32 = B
VẬy A < B
A=(2+1)x(22+1)x(24+1)x(28+1)x(216+1)
= 3.5.17.257.65537
= 42949672995 = 232 ⇒A = B mình nhé
1/
\(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^{16}+1\right)=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^{16}+1\right)=\left(2^4-1\right)\left(2^4+1\right)\left(2^{16}-1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1\)
Vì 232 > 223 => 232-1>223-1 hay A>B
2/
\(A=x^2+y^2=\left(x-y\right)^2+2xy=5^2+2.14=25+28=53\)
\(B=\left(x+y\right)^2=\left(x-y\right)^2+4xy=5^2+4.14=25+56=81\)
