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\(6-\sqrt{17}=\sqrt{36}-\sqrt{17}\)
Với :
\(\sqrt{36}-\sqrt{17}>\sqrt{31}-\sqrt{17}\)
Mặt khác :
\(\sqrt{31}-\sqrt{17}>\sqrt{31}-\sqrt{19}\)
Nên :
\(6-\sqrt{17}>\sqrt{31}-\sqrt{19}\)
Cách khác:
Ta có: \(\left(\sqrt{31}-\sqrt{19}\right)^2=50-2\sqrt{589}\)
\(\left(6-\sqrt{17}\right)^2=53-12\sqrt{17}=50+3-12\sqrt{17}\)
mà \(-2\sqrt{589}< 3-12\sqrt{17}\)
nên \(\sqrt{31}-\sqrt{19}>6-\sqrt{17}\)
\(a,10^{30}=2^{30}.5^{30}\)
\(2^{100}=\left(2^{50}\right)^2\)
\(\Rightarrow10^{30}< 2^{100}\)
tt
( 3x-1) ( x2+ 9) = (3x-1) (7x-10)
⇒( 3x-1) ( x2+ 9) - (3x-1) (7x-10) = 0
⇒( 3x-1) (( x2+ 9)-(7x-10)) = 0
⇒( 3x-1)(x2+9-7x+10)=0
⇒( 3x-1)(x2-7x+19)=0
⇒\(\left[{}\begin{matrix}3x-1=0\\x^2-7x+19=0\end{matrix}\right.\)
3x-1=0
⇒x=\(\dfrac{1}{3}\)
x2-7x+19=0
⇒ \(x^2-\dfrac{7}{2}x-\dfrac{7}{2}x+\left(\dfrac{7}{2}\right)^2+\dfrac{27}{4}=0\)
⇒ \(\left(x-\dfrac{7}{2}\right)^2+\dfrac{27}{4}=0\)
vì \(\left(x-\dfrac{7}{2}\right)^2\ge0\); \(\dfrac{27}{4}>0\)
⇒ \(\left(x-\dfrac{7}{2}\right)^2+\dfrac{27}{4}>0\)
⇒ x vô nghiệm
Vậy x= \(\dfrac{1}{3}\)
\(\left(3x-1\right)\left(x^2+9\right)=\left(3x-1\right)\left(7x-10\right)\\ \Leftrightarrow\left(3x-1\right)\left(x^2+9\right)-\left(3x-1\right)\left(7x-10\right)\\ \Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\\ \Leftrightarrow\left(3x-1\right)\left(x^2-4x-3x+12\right)=0\\ \Leftrightarrow\left(3x-1\right)\left[x\left(x-4\right)-3\left(x-4\right)\right]=0\\ \Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)
`A=4(3^2+1)(3^4+1)...(3^64+1)`
`=>2A=(3^2-1)(3^2+1)(3^4+1)...(3^64+1)`
- Ta có:
`(3^2-1)(3^2+1)=3^4-1`
`(3^4-1)(3^4+1)=3^16-1`
`....`
`(3^64-1)(3^64+1)=3^128-1`
Suy ra `2A=3^128-1=B`
`=>A<B`
đơn giản
nhưng trả lời câu hỏi của tớ đã
a, \(\sqrt{15}+\sqrt{8}< \sqrt{16}+\sqrt{9}=4+3=7\)
\(\Rightarrow\sqrt{15}+\sqrt{8}< 7\)
b, \(\sqrt{10}+\sqrt{17}+1>\sqrt{9}+\sqrt{16}+1=3+4+1=8\)
\(\sqrt{61}< \sqrt{64}=8\)
\(\Rightarrow\sqrt{10}+\sqrt{17}+1>\sqrt{61}\)
c, \(\sqrt{10}+\sqrt{5}+1>\sqrt{9}+\sqrt{4}+1=3+2+1=6\)
\(\sqrt{35}< \sqrt{36}=6\)
\(\Rightarrow\sqrt{10}+\sqrt{5}+1>\sqrt{35}\)