HOC24
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Chủ đề / Chương
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1.
a) Ta có :
\(8^{30}+8^{31}+8^{32}=8^{30}\left(8+8^2+8^3\right)\)
\(=8^{30}.584\)
* Do \(584⋮146\Rightarrow584.8^{30}⋮146\)
Vậy \(8^{30}+8^{31}+8^{32}⋮146\)
b) \(4^{25}+4^{26}+4^{27}+4^{28}+4^{29}+4^{30}\)
\(=\left(4^{25}+4^{26}\right)+\left(4^{27}+4^{28}\right)+\left(4^{29}+4^{30}\right)\)\(=4^{25}\left(1+4^2\right)+4^{27}\left(1+4^2\right)+4^{29}\left(1+4^2\right)\)\(=\left(1+4^2\right)\left(4^{25}+4^{27}+4^{29}\right)\)
\(=17\left(4^{25}+4^{27}+4^{29}\right)⋮17\)
\(\RightarrowĐpcm\)
tik mik nha !!!
a) \(\left|\dfrac{1}{2}x\right|=3-2x\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=3-2x\left(x\ge0\right)\\\dfrac{1}{2}x=2x-3\left(x< 0\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x+2x=3\\\dfrac{1}{2}x-2x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{5}{2}x=3\\-\dfrac{3}{2}x=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1,2\\x=2\end{matrix}\right.\)
\(\Rightarrow x=1,2\)
các bài khác tương tự nhé !!
\(B=12+12^2+12^3+...+12^{99}\)
\(12B=12^2+12^3+12^4+...+12^{100}\)
\(11B=12B-B=12^{100}-12\)
\(\Rightarrow B=\dfrac{12^{100}-12}{11}\)
Vậy \(B=\dfrac{12^{100}-12}{11}\)
Tik mik nha!!!
Sửa đề : \(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}< 1\)
Giải.
Ta có :
\(D=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{10^2}\)
\(D< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
\(D< 1-\dfrac{1}{10}< 1\Rightarrow D< 1\)
Vậy...
bạn bổ sung cho mình nha :
665 = 19ab
=> ab = 665 : 19
ab = 35
chưa đủ dữ kiện bạn ơi
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{100}}\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)
\(A=2A-A=1+\dfrac{1}{2}+...+\dfrac{1}{2^{99}}-\dfrac{1}{2}-\dfrac{1}{2^2}-...-\dfrac{1}{2^{100}}\)\(\Rightarrow A=1-\dfrac{1}{2^{100}}\)
\(\dfrac{x+1}{65}+\dfrac{x+3}{63}=\dfrac{x+5}{61}+\dfrac{x+7}{59}\)
\(\Leftrightarrow\dfrac{x+1}{65}+\dfrac{x+3}{63}-\dfrac{x+5}{61}-\dfrac{x+7}{59}=0\)
\(\Leftrightarrow\left(\dfrac{x+1}{65}+1\right)+\left(\dfrac{x+3}{63}+1\right)-\left(\dfrac{x+5}{61}+1\right)-\left(\dfrac{x+7}{59}+1\right)=0\)\(\Leftrightarrow\dfrac{x+66}{65}+\dfrac{x+66}{63}-\dfrac{x+66}{61}-\dfrac{x+66}{59}=0\)\(\Leftrightarrow\left(x+66\right)\left[\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\right]=0\)Nhận xét : Do \(\dfrac{1}{65}< \dfrac{1}{63}< \dfrac{1}{61}< \dfrac{1}{59}\)
\(\Rightarrow\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)< 0\)
Vậy để \(\left(x+66\right)\left[\left(\dfrac{1}{65}+\dfrac{1}{63}\right)-\left(\dfrac{1}{61}+\dfrac{1}{59}\right)\right]=0\)\(\Leftrightarrow x+66=0\Leftrightarrow x=-66\)
Vậy....
Ta có : \(x\div y=5\div\left(-3\right)\Rightarrow\dfrac{x}{5}=\dfrac{-y}{3}\)
và \(x-y=-16\)
theo tính chất dãy tỉ số bằng nhau, ta có :
\(\dfrac{x}{5}=\dfrac{-y}{3}=\dfrac{x+\left(-y\right)}{5+3}=\dfrac{x-y}{8}=\dfrac{-16}{8}=-2\)\(\Rightarrow x=-2.5=-10\)
\(\Rightarrow-y=-2.3=-6\Rightarrow y=6\)
\(C=\dfrac{258^2-242^2}{254^2-246^2}\)
\(C=\dfrac{\left(258+242\right)\left(258-242\right)}{\left(254+246\right)\left(254-246\right)}\)
\(C=\dfrac{500.16}{500.8}=2\)
Tik mik nha !!!