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a) (1/7.x-2/7).(-1/5.x-2/5)=0
=> 1/7.x-2/7=0hoặc-1/5.x-2/5=0
*1/7.x-2/7=0
1/7.x=0+2/7
1/7.x=2/7
x=2/7:1/7
x=2
b)1/6.x+1/10.x-4/5.x+1=0
(1/6+1/10-4/5).x+1=0
(1/6+1/10-4/5).x=0-1
(1/6+1/10-4/5).x=-1
(-8/15).x=-1
x=-1:(-8/15) =15/8
a: \(=\dfrac{5\cdot\left(8-6\right)}{10}=\dfrac{5\cdot2}{10}=1\)
b: \(\dfrac{\left(-4\right)^2}{5}=\dfrac{16}{5}\)
\(B=\dfrac{3}{7}-\dfrac{1}{5}-\dfrac{3}{7}=-\dfrac{1}{5}\)
c: \(C=\left(6-2.8\right)\cdot\dfrac{25}{8}-\dfrac{8}{5}\cdot4\)
\(=\dfrac{16}{5}\cdot\dfrac{25}{8}-\dfrac{32}{5}\)
\(=5\cdot2-\dfrac{32}{5}=10-\dfrac{32}{5}=\dfrac{18}{5}\)
d: \(D=\left(\dfrac{-5}{24}+\dfrac{18}{24}+\dfrac{14}{24}\right):\dfrac{-17}{8}\)
\(=\dfrac{27}{24}\cdot\dfrac{-8}{17}=\dfrac{-9}{8}\cdot\dfrac{8}{17}=\dfrac{-9}{17}\)
\(a,\left(19x+2.5^2\right):14=\left(13-8\right)^2-4^2\)
\(\Leftrightarrow\left(19x+50\right):14=5^2-4^2\)
\(\Leftrightarrow\left(19x+50\right):14=9\)
\(\Leftrightarrow19x+50=126\)
\(\Leftrightarrow19x=76\Leftrightarrow x=4\)
b) x + ( x + 1 ) + ( x + 2 ) + ... + ( x + 30 ) = 1240
x + x + 1 + x + 2 + ... + x + 30 = 1240
( x + x + ... + x ) + ( 1 + 2 + ... + 30 ) = 1240
Số số hạng là : ( 30 - 1 ) : 1 + 1 = 30 ( số )
Tổng là : ( 30 + 1 ) . 30 : 2 = 465
=> 31x + 465 = 1240
=> 31x = 775
=> x = 25
Vậy........
\(a.-8:\left(4\dfrac{1}{5}x+\dfrac{3}{10}\right)=4\dfrac{4}{9}\)
\(4\dfrac{1}{5}x+\dfrac{3}{10}=\left(-8\right):4\dfrac{4}{9}\)
\(4\dfrac{1}{5}x+\dfrac{3}{10}=\dfrac{-9}{5}\)
\(4\dfrac{1}{5}x=\dfrac{-9}{5}-\dfrac{3}{10}\)
\(4\dfrac{1}{5}x=\dfrac{-21}{10}\)
\(x=\dfrac{-21}{10}:\dfrac{21}{5}\)
\(x=\dfrac{-1}{2}\)
Vay \(x=\dfrac{-1}{2}\).
\(b.4\dfrac{2}{3}-\left(\dfrac{3}{5}:x\right)=-20\%\)
\(\dfrac{14}{3}-\left(\dfrac{3}{5}:x\right)=\dfrac{-1}{5}\)
\(\dfrac{3}{5}:x=\dfrac{14}{3}-\dfrac{-1}{5}\)
\(\dfrac{3}{5}:x=\dfrac{73}{15}\)
\(x=\dfrac{3}{5}:\dfrac{73}{15}\)
\(x=\dfrac{9}{73}\)
Vay \(x=\dfrac{9}{73}\).
Câu c; d; e tương tự nhé.
Bài 2:
a) \(\left(x-3\right)^3+27=0\)
\(\Leftrightarrow\left(x-3\right)^3=0-27\)
\(\Leftrightarrow\left(x-3\right)^3=-27\)
\(\Leftrightarrow\left(x-3\right)^3=\left(-3\right)^3\)
\(\Leftrightarrow x-3=-3\)
\(\Leftrightarrow x=\left(-3\right)+3\)
\(\Leftrightarrow x=0\)
b) \(-125-\left(x+1\right)^3=0\)
\(\Leftrightarrow\left(x+1\right)^3=-125-0\)
\(\Leftrightarrow\left(x+1\right)^3=-125\)
\(\Leftrightarrow\left(x+1\right)^3=\left(-5\right)^3\)
\(\Leftrightarrow x+1=-5\)
\(\Leftrightarrow x=\left(-5\right)-1\)
\(\Leftrightarrow x=-6\)
c) \(\left(2x-\dfrac{1}{4}\right)^2-\dfrac{1}{16}=0\)
\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=0+\dfrac{1}{16}\)
\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\dfrac{1}{16}\)
\(\Leftrightarrow\left(2x-\dfrac{1}{4}\right)^2=\left(\dfrac{1}{4}\right)^2\)
\(\Leftrightarrow2x-\dfrac{1}{4}=\dfrac{1}{4}\)
\(\Leftrightarrow2x=\dfrac{1}{4}+\dfrac{1}{4}\)
\(\Leftrightarrow2x=\dfrac{1}{2}\)
\(\Leftrightarrow x=\dfrac{1}{2}:2\)
\(\Leftrightarrow x=\dfrac{1}{4}\)
d) \(2^x+2^{x+1}=24\)
\(\Leftrightarrow2^x+2^x.2=24\)
\(\Leftrightarrow2^x\left(1+2\right)=24\)
\(\Leftrightarrow2^x.3=24\)
\(\Leftrightarrow2^x=24:3\)
\(\Leftrightarrow2^x=8\)
\(\Leftrightarrow2^x=2^3\)
\(\Rightarrow x=3\)
e) \(\left|x+\dfrac{1}{5}\right|-\dfrac{1}{2}=1\)
\(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=1+\dfrac{1}{2}\)
\(\Leftrightarrow\left|x+\dfrac{1}{5}\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=-\dfrac{3}{2}\\x+\dfrac{1}{5}=\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{17}{10}\\x=\dfrac{13}{10}\end{matrix}\right.\)
g) \(\left|x-3\right|+2x=10\)
\(\Leftrightarrow\left|x-3\right|=10-2x\)
\(\Leftrightarrow\left|x-3\right|=2.5-2x\)
\(\Leftrightarrow\left|x-3\right|=2\left(5-x\right)\)
(không chắc có nên làm tiếp câu g không, thấy đề cứ là lạ, có j sai sai...)
Bài 1:
a) \(2^7+2^9⋮10\)
Ta có: \(2^7+2^9=2^{4.1}.2^3+2^{4.2}.2\)
\(\Leftrightarrow\overline{A6}.2^3+\overline{B6}.2\)
\(\Leftrightarrow\overline{A6}.8+\overline{B6}.2\)
\(\Leftrightarrow\overline{C8}+\overline{D2}\)
\(\Leftrightarrow\overline{E0}\)
Mà \(\overline{E0}⋮10\) \(\Rightarrow2^7+2^9⋮10\)
b) \(8^{24}.25^{10}⋮2^{36}.5^{20}\)
Ta có: \(8^{24}.25^{10}=\left(2^3\right)^{24}.\left(5^2\right)^{10}\)
\(\Leftrightarrow2^{72}.5^{20}\)
Do \(2^{72}⋮2^{36}\) và \(5^{20}⋮5^{20}\) \(\Rightarrow8^{24}.25^{10}⋮2^{36}.5^{20}\)
c) \(3^{10}+3^{12}⋮30\)
Ta có: \(3^{10}+3^{12}=3^{4.2}.3^2+3^{4.3}\)
\(\Leftrightarrow\overline{A1}.3^2+\overline{B1}\)
\(\Leftrightarrow\overline{A1}.9+\overline{B1}\)
\(\Leftrightarrow\overline{C9}+\overline{B1}\)
\(\Leftrightarrow\overline{D0}⋮10\)
(Chứng minh chia hết cho 10 rồi chứng minh chia hết cho 3, mình chưa tìm được cách làm, chờ chút)
Câu 2:
a: a=2007 nên a+1=2008
\(M=a^{11}-a^{10}\left(a+1\right)+a^9\left(a+1\right)-...-a^2\left(a+1\right)+a\left(a+1\right)\)
\(=a^{11}-a^{11}-a^{10}+a^{10}+a^9-...-a^3-a^2+a^2+a\)
=a=2007
b: a=2004 nên a-1=2003
\(N=a^{11}-a^{10}\left(a-1\right)-a^9\left(a-1\right)-...-a\left(a-1\right)-1004\)
\(=a^{11}-a^{11}+a^{10}-a^{10}+a^9-...-a^2+a-1004\)
=a-1004=1000