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a) Đặt A = (2 + 1)(22 + 1)(24 + 1 )(28 +1)( 216 +1 )
=> A = ( 22 - 1 ) (22 + 1)(24 + 1 )(28 +1)( 216 +1 )
=> A = (24 - 1)(24 + 1 )(28 +1)( 216 +1 )
=> A = (28 - 1)(28 +1)( 216 +1 )
=> A= (216 -1 ) (216 + 1) = 232 - 1 => đpcm
b) 1002 + 1032 + 1052 + 942 = 1012 + 982 + 962 + 1072
<=> \(\left(100^2-98^2\right)+\left(103^2-101^2\right)+\left(105^2-107^2\right)+\left(94^2-96^2\right)\) = 0
<=> \(\left(100-98\right)\left(100+98\right)+\left(103-101\right)\left(103+101\right)\)+ (105 -107)(105+107) + (94 - 96)(96 + 94) = 0
<=> \(2.198+2.204-2.212-2.190\) = 0
<=> \(2\left(198+204-212-190\right)=0\)
<=> \(\left(198-190\right)+\left(204-212\right)=0\)
<=> \(-8+8=0\) (luôn đúng) => đpcm
P/s: đây ko phải bài lớp 10 đâu!
Đề bài trên sai. Đề đúng: CM: \(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{97}{98}.\dfrac{99}{100}>\dfrac{\sqrt{2}}{20}\).
Câu 1:
a) = \(\dfrac{-7}{2}\) x \(\dfrac{45}{32}\) = \(\dfrac{-315}{64}\)
b) = \(\dfrac{18}{7}\) : \(\dfrac{-27}{14}\) = \(\dfrac{18}{7}\) x \(\dfrac{14}{-27}\) = \(\dfrac{-4}{3}\)
c) = \(\dfrac{-3}{8}\) x ( \(\dfrac{5}{11}\) + \(\dfrac{6}{11}\) + 2 ) = \(\dfrac{-3}{8}\) x 3 = \(\dfrac{-9}{8}\)
Câu 2:
\(\dfrac{-3}{5}\) . x + \(\dfrac{7}{6}\) = \(\dfrac{5}{4}\)
\(\Leftrightarrow\) \(\dfrac{-3}{5}\) . x = \(\dfrac{5}{4}\) - \(\dfrac{7}{6}\)
\(\Leftrightarrow\) \(\dfrac{-3}{5}\) . x = \(\dfrac{1}{12}\)
\(\Leftrightarrow\) x = \(\dfrac{1}{12}\) : \(\dfrac{-3}{5}\)
\(\Leftrightarrow\) x = \(\dfrac{-5}{36}\)
Bài 3: a) Xét A=(1+1/2+1/3+....+1/98).2.3.4.5.....98
=(1+1/2+1/3+....+1/98).(9.11).2.3.4.....98
=(1+1/2+1/3+....+1/98).99.2.3.4....98⋮99
(đpcm)
a) \(x\cdot3\dfrac{1}{4}+\left(-\dfrac{7}{6}\right)\cdot x-1\dfrac{2}{3}=\dfrac{5}{12}\)
\(\Rightarrow\dfrac{3}{4}x-\dfrac{7}{6}x-\dfrac{2}{3}=\dfrac{5}{12}\)
\(\Leftrightarrow9x-14x-8=5\)
\(\Leftrightarrow-5x-8=5\)
\(\Leftrightarrow-5x=5+8\)
\(\Leftrightarrow-5x=13\)
\(\Rightarrow x=-\dfrac{13}{5}\)
Vậy \(x=-\dfrac{13}{5}\)
b) \(5\dfrac{8}{17}:x+\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)
\(\Rightarrow5\dfrac{8}{17}:x+\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\left(đk:x\ne0\right)\)
\(\Leftrightarrow\dfrac{93}{17}\cdot\dfrac{1}{x}+\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{93}{17x}+\left|2x-\dfrac{3}{4}\right|=-\dfrac{7}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{93}{17x}+2x-\dfrac{3}{4}=-\dfrac{7}{4}\left(đk:2x-\dfrac{3}{4}\ge0\right)\\\dfrac{93}{17x}-\left(2x-\dfrac{3}{4}\right)=-\dfrac{7}{4}\left(đk:2x-\dfrac{3}{4}< 0\right)\end{matrix}\right.\)
đến đây bạn giải tiếp nhé
c) \(\left(x+\dfrac{1}{2}\right)\cdot\left(\dfrac{2}{3}-2x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0-\dfrac{1}{2}\\2x=0+\dfrac{2}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{2}{3}:2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{1}{2};x_2=\dfrac{1}{3}\)
ta có : A= 22 + 42 + 62 + ....+ 982 + 1002
A= 22( 12 + 22 + 32 + ....+ 492 + 502)
A= 22. \(\dfrac{50\left(50+1\right)\left(2.50+1\right)}{6}\)
A= 171700
Hí. Mơn bn nha ≥ω≤ ♥♥♥