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\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(\Rightarrow2A=8.\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
.....
\(=\left(3^{64}-1\right)\left(3^{64}+1\right)\)
\(=3^{128}-1\)
\(\Rightarrow A=\frac{3^{128}-1}{2}\)
2A = (3+1)(3-1)(3^2+1)(3^4+1)...(3^64+1)
2A= (3^2-1)(3^2+1)(3^4+1)...(3^64+1)
Cứ tiếp tục như thế ta dc
2A= 3^128 -1
A = (3^128-1)/2
\(x^2-x+1=k^2\left(k\in Z\right)\)
\(\Leftrightarrow4x^2-4x+4=4k^2\)
\(\Leftrightarrow\left(2x-1\right)^2+3=\left(2k\right)^2\)
\(\Leftrightarrow\left(2x-1\right)^2-\left(2k\right)^2=-3\)
\(\Leftrightarrow\left(2x-2k-1\right)\left(2x+2k-1\right)=-3\)
Ta có cảc trường hợp:
TH1: \(\hept{\begin{cases}2x-2k-1=1\\2x+2k-1=-3\end{cases}}\Leftrightarrow\hept{\begin{cases}x-k=1\\x+k=-1\end{cases}\Leftrightarrow}x=0\) (loại)
TH2: \(\hept{\begin{cases}2x-2k-1=-1\\2x+2k-1=3\end{cases}}\Leftrightarrow\hept{\begin{cases}x-k=0\\x+k=2\end{cases}}\Leftrightarrow x=1\) (thỏa mãn)
TH3: \(\hept{\begin{cases}2x-2k-1=3\\2x+2k-1=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x-k=2\\x+k=0\end{cases}\Leftrightarrow}x=1\) (TM)
TH4: \(\hept{\begin{cases}2x-2k-1=-3\\2x+2k-1=1\end{cases}}\Leftrightarrow\hept{\begin{cases}x-k=-1\\x+k=1\end{cases}}\Leftrightarrow x=0\) (loại)
Vậy x = 1
1. F = x2 + 2y2 + 2xy - 4x - 10y + 15
F = (x2 + 2xy + y2) - 4(x + y) + 4 + (y2 - 6y + 9) + 2
F = (x + y)2 - 4(x + y) + 4 + (y - 3)2 + 2
F = (x + y - 2)2 + (y - 3)2 + 2\(\ge\)2 với mọi x;y
Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+y-2=0\\y-3=0\end{cases}}\) <=> \(\hept{\begin{cases}x=2-y\\y=3\end{cases}}\) <=> \(\hept{\begin{cases}x=-1\\y=3\end{cases}}\)
Vậy Mìn = 2 khi x = -1 và y = 3
\(\left(2x^2+x\right)^2-4\left(2x^2+x\right)+3=0\)
\(\Leftrightarrow\left(2x^2+x\right)^2-\left(2x^2+x\right)-3\left(2x^2+x\right)+3=0\)
\(\Leftrightarrow\left(2x^2+x\right)\left(2x^2+x-1\right)-3\left(2x^2+x-1\right)=0\)
\(\Leftrightarrow\left(2x^2+x-3\right)\left(2x^2+x-1\right)=0\)
\(\Leftrightarrow\left(2x^2-2x+3x-3\right)\left(2x^2-x+2x-1\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(x-1\right)\left(2x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\)\(x=-\frac{3}{2}\) hoặc \(x=1\) hoặc \(x=\frac{1}{2}\) hoặc \(x=-1\)
\(\left(2x^2+x\right)^2-4\left(2x^2+x\right)+3=0\)
\(\Leftrightarrow\left(2x^2+x\right)^2-\left(2x^2+x\right)-3\left(2x^2+x\right)+3=0\)
\(\Leftrightarrow\left(2x^2+x\right)\left(2x^2+x-1\right)-3\left(2x^2+x-1\right)=0\)
\(\Leftrightarrow\left(2x^2+x-1\right)\left(2x^2+x-3\right)=0\)
\(\Leftrightarrow\left(2x^2+2x-x-1\right)\left(2x^2-2x+3x-3\right)=0\)
\(\Leftrightarrow\left[2x\left(x+1\right)-\left(x+1\right)\right]\left[2x\left(x-1\right)+3\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x-1\right)\left(2x+3\right)\left(x-1\right)=0\)
Nếu x + 1 = 0 thì x = -1
Hoặc 2x - 1= 0 thì x = 1/2
Hoặc 2x + 3 = 0 thì x = -3/2
Hoặc x - 1 = 0 thì x = 1
Vậy ....