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\(2\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(=3^{32}-1< 3^{32}\)
Gợi ý: Sử dụng liên tục tính chất \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
2(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)
= (3 - 1)(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)
= (32 - 1)(32 + 1)(34 + 1)(38 + 1)(316 + 1)
= (34 - 1)(34 + 1)(38 + 1)(316 + 1)
= (38 - 1)(38 + 1)(316 + 1)
= (316 - 1)(316 + 1)
= 332 - 1 < 332
- Bài 1.
a) \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)
\(\Leftrightarrow\left(25x^2+10x+1\right)-25x^2+9=30\)
\(\Leftrightarrow10x=20\Leftrightarrow x=2\)
b) \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow x^3-1-x^3+4x-5=0\)
\(\Leftrightarrow4x=6\Leftrightarrow x=\frac{3}{2}\)
- Ta có : \(A=1997.1999=\left(1998-1\right)\left(1998+1\right)=1998^2-1< 1998^2\)
\(\Rightarrow A< B\)
- Từ a+b+c=2p => \(p=\frac{a+b+c}{2}\)
Ta có : \(4p\left(p-a\right)=2\left(a+b+c\right)\left(\frac{a+b+c}{2}-a\right)=2.\left(a+b+c\right).\frac{b+c-a}{2}\)
\(=\left(a+b+c\right)\left(b+c-a\right)=\left[\left(b+c\right)+a\right]\left[\left(b+c\right)-a\right]=\left(b+c\right)^2-a^2\)
\(=b^2+c^2-a^2+2bc\)
Bài cuối bạn sửa 2ab thành 2bc nhé ^^
Đặt các cặp 1+1/3+1/5+..+1/4025 của A ra so sánh (1/2+1/4+..+1/4026)/B với 2013/2014
thấy A/B<1+2013/2014
A) Với \(x>y>0\),ta có: \(x^2+y^2< x^2+y^2+2xy=\left(x+y\right)^2\Rightarrow\frac{1}{x^2+y^2}>\frac{1}{\left(x+y\right)^2}\)
Xét: \(\frac{x^2-y^2}{x^2+y^2}>\frac{x^2-y^2}{\left(x+y\right)^2}=\frac{\left(x-y\right)\left(x+y\right)}{\left(x+y\right)^2}=\frac{x-y}{x+y}\)--->ĐPCM
B) \(3^{16}+1=\left(3^{16}-1\right)+2=\left(3^8+1\right)\left(3^8-1\right)+2\)
\(=\left(3^8+1\right)\left(3^4+1\right)\left(3^4-1\right)+2\)
\(=\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3^2-1\right)+2\)
\(=\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3+1\right)\left(3-1\right)+2\)
\(>\left(3^8+1\right)\left(3^4+1\right)\left(3^2+1\right)\left(3+1\right)\)--->ĐPCM
Tìm x
a) ( x - 1 )^3 + 1 + 3x( x - 4 ) = 0
b) x^3 - 6x^2 + 9x = 0
giúp mình với mình cần gấp
mình cảm ơn
b) \(x^3-6x^2+9x=0\)
\(\Leftrightarrow x.\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow x.\left(x-3\right)^2=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=3\end{cases}}\)
Vậy \(x=0\)hoặc \(x=3\)
a. ( x - 1 )3 + 1 + 3x ( x - 4 ) = 0
<=> x3 - 3x2 + 3x - 1 + 1 + 3x2 - 12x = 0
<=> x3 - 9x = 0
<=> x ( x2 - 9 ) = 0
<=> \(\orbr{\begin{cases}x=0\\x^2-9=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}\)
b. x3 - 6x2 + 9x = 0
<=> x ( x2 - 6x + 9 ) = 0
<=> x ( x - 3 )2 = 0
<=> \(\orbr{\begin{cases}x=0\\\left(x-3\right)^2=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=0\\x=3\end{cases}}\)