chung minh 3/1nhan4 +3/4nhan 7+...+3/91 nhan94<1
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\(\left(5+5^2+5^3+...+5^{10}\right)+4x-1=\frac{1}{4}5^{11}+\frac{1}{2}x+3\)
\(\Leftrightarrow\left(1+5+5^2+5^3+...+5^{10}\right)+4x-2=\frac{1}{4}5^{11}+\frac{1}{2}x+3\)(1)
1./ Trước tiên, ta tính:
\(S=1+5+5^2+5^3+...+5^{10}\)
\(\left(5-1\right)\cdot S=\left(5-1\right)\left(1+5+5^2+5^3+...+5^{10}\right)\)
\(\Leftrightarrow4S=5^{11}-5^{10}+5^{10}-5^9+...+5-1=5^{11}-1\)
\(\Leftrightarrow S=\frac{5^{11}-1}{4}=\frac{1}{4}5^{11}-\frac{1}{4}\)
2./ (1) trở thành:
\(\Leftrightarrow\frac{1}{4}5^{11}-\frac{1}{4}+4x-2=\frac{1}{4}5^{11}+\frac{1}{2}x+3\)
\(\Leftrightarrow4x-\frac{1}{2}x=5+\frac{1}{4}\Leftrightarrow\frac{7}{2}x=\frac{21}{4}\)
\(\Leftrightarrow x=\frac{3}{2}\).
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4. (x-3) - x = 3
4x - 12 - x = 3
3x - 12 = 3
3x = 3+12
3x = 15
x= 15:3
x = 5
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a) A=21+22+23+...+22010
A=(21+22)+(23+24)+.....+(22009+22010)
A=(21x3)+(23x3)+.....+(22009x3)
A=3x(21+23+.......+22009)
Vậy A chia hết cho 3.
NHỮNG CÂU CÒN LẠI BẠN LÀM TƯƠNG TỰ !
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A = (3^5.1 + 3^5.3) + (3^7.1 + 3^7.3) + (3^9.1 + 3^9.3)
= 3^5.(1+3) + 3^7.(1+3) + 3^9.(1+3)
= 3^5.4 + 3^7.4 + 3^9.4
=4.(3^5 +3^7 +3^9)
=4.(3^5.1 +3^5.3^2 + 3^5.3^4)
=4.3^5.(1+3^2+3^4)
=4.243.(1+9+81)
=972.91 chia hết cho 91
Vậy A chia hết cho 91
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TH1 : \(91-3x< 7+x\Rightarrow3x+x>91-7\Rightarrow4x>84\Rightarrow x>21\left(1\right)\)
TH2 : \(7+x\ge64\Rightarrow x\ge57\left(2\right)\)
\(\left(1\right),\left(2\right)\Rightarrow x\ge57\)
91 - 3\(x\) < 7 + \(x\) ≥ 64
⇒ \(\left\{{}\begin{matrix}91-3x< 7+x\\7+x\ge64\end{matrix}\right.\)
\(\left\{{}\begin{matrix}7+x+3x>91\\x\ge64-7\end{matrix}\right.\)
\(\left\{{}\begin{matrix}4x>91-7\\x\ge64-7\end{matrix}\right.\)
\(\left\{{}\begin{matrix}4x>84\\x\ge57\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x>84:4\\x\ge57\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x>21\\x\ge57\end{matrix}\right.\)
\(x\ge\) 57