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Ta có:
\(\overline{xxyy}=x.1000+x.100+y.10+y=x.1100+y.11=11\left(x.100+y\right)\)
\(\overline{\left(x+1\right)\left(x+1\right)}.\overline{\left(y+1\right)\left(y+1\right)}=\overline{x+1}.11.\overline{y+1}.11\)
=> \(\overline{xxyy}=\overline{\left(x+1\right)\left(x+1\right)}.\overline{\left(y+1\right)\left(y+1\right)}\)
\(\Leftrightarrow11\left(x.100+y\right)=\overline{\left(x+1\right)}.11.\overline{\left(y+1\right)}.11\)
\(\Leftrightarrow x.100+y=11.\overline{x+1}.\overline{y+1}\)
\(\Leftrightarrow\overline{x0y}=11.\overline{x+1}.\overline{y+1}\)(1)
=> \(\overline{x0y}⋮11\)=> \(x-0+y⋮11\Rightarrow x+y⋮11\)=> x+y=11
và \(\overline{x0y}⋮x+1;\overline{x0y}⋮y+1\)
Em thay các giá trị x, y vào thử nhé
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126 chia hết x; 210 chia hết x
=>x thuộc ƯC(126;210)
Ta có:
\(\text{126=2.3^2.7}\)
\(210=2.3.5.7\)
\(\Rightarrow\)UCLN(126;210)=2*3*7=42
\(\Rightarrow\)x thuộc Ư(42).
\(\Rightarrow x\in\left\{1;2;3;6;7;14;21;42\right\}\)
\(\Rightarrow x=\left\{{}\begin{matrix}14\\21\end{matrix}\right.\) (thỏa mãn)
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x có thể\(\in\){ 16 ; 17 ; 18 ; 19 ; 20 ; 21 ; 22 ; 23 ; 24 ; 25 ; 26 ; 27 ; 28 ; 29 }
Để 126 chia hết cho x ; 210 chia hết cho x mà 15 < x < 30 -> Thì x = 21
( Vì 126 và 210 chia hết cho 21 ; 15 < 21 < 30 nên x = 21 )
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Vì \(126⋮x;210⋮x\Rightarrow x\inƯC\left(126;210\right)\)
Mà ƯCLN(126; 210) = 42
\(\Rightarrow x\inƯ\left(42\right)\)
Mặt khác, theo đề bài: 15 < x < 30
=> x = 21
Vậy x = 21
Vì : \(126⋮x\) và \(210⋮x\)
\(\Rightarrow x\inƯCLN\left(126;210\right)\)
Ta có :
\(126=3^2.2.7\)
\(210=3.2.5.7\)
\(\RightarrowƯCLN\left(126;210\right)=2.3.7=42\)
\(\RightarrowƯ\left(42\right)=\left\{1;2;3;6;7;14;21;42\right\}\)
Mà : \(15< x< 30\Rightarrow x=21\)
Vậy số tự nhiên x là 21
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\(\frac{9}{11}< \frac{x}{15}< \frac{10}{11}\)
\(\frac{135}{165}< \frac{11x}{165}< \frac{150}{165}\)
135 < 11x < 150
Chỉ có 11x = 143 là thích hợp.
Vậy x = 143 : 11 = 13.
Ta có: \(\frac{9}{11}< \frac{x}{15}< \frac{10}{11}\)
=> \(\frac{135}{165}< \frac{x.11}{165}< \frac{150}{165}\)
=> 135 < x.11 < 150
=> x.11 E (thuộc) {136; 137;...;149}
Và x.11 phải chia hết cho 11 nên x = 143
Vậy: x = 143
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a: \(\dfrac{7}{11}< x-\dfrac{1}{7}< \dfrac{10}{13}\)
\(\Leftrightarrow\dfrac{7}{11}+\dfrac{1}{7}< x< \dfrac{10}{13}+\dfrac{1}{7}\)
hay 60/77<x<83/91
b: \(\dfrac{7}{9}< \dfrac{13}{11}-x< \dfrac{15}{16}\)
\(\Leftrightarrow\dfrac{-7}{9}>x-\dfrac{13}{11}>-\dfrac{15}{16}\)
\(\Leftrightarrow-\dfrac{7}{9}+\dfrac{13}{11}>x>\dfrac{-15}{16}+\dfrac{13}{11}\)
\(\Leftrightarrow\dfrac{40}{99}>x>\dfrac{43}{176}\)
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a) \(\frac{1}{3}+\frac{3}{35}< \frac{x}{210}< \frac{7}{7}+\frac{3}{5}+\frac{1}{3}\)
\(\Rightarrow\frac{44}{105}< \frac{x}{210}< \frac{29}{15}\)
\(\Rightarrow\frac{88}{210}< \frac{x}{210}< \frac{406}{210}\)
\(\Rightarrow x\in\left\{89;90;91;...;405\right\}\)
b) \(\frac{5}{3}+-\frac{14}{3}< x< \frac{8}{5}+\frac{8}{10}\)
\(\Rightarrow-3< x< 2\frac{2}{5}\)
=> x thuộc {-2;-1;0;1;2} ( nếu x là số nguyên)
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a: \(\Leftrightarrow70+18< x< 120+126+70\)
=>88<x<316
hay \(x\in\left\{89;90;...;315\right\}\)
b: \(\Leftrightarrow-\dfrac{9}{3}< x< \dfrac{8}{5}+\dfrac{9}{5}=\dfrac{17}{5}\)
=>-3<x<3,4
hay \(x\in\left\{-2;-1;0;1;2;3\right\}\)
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a ) Vì 126 chia hết cho x nên x là ước của 126
ư (126 ) = { 1,2,3,6,7,14,18,21,42,63,126 } mà 15<x<30 nên a = 18,21
Vậy a = 18 hoặc 21
b ) vì 210 chia hết cho x nên x là ước của 210
ư (210 ) = { 1,2,3,5,6,7,10,14,15,21,30,35,42,
tôi chưa viết xong để toi viết tiếp
ư ( 210 ) = { 1,2,3,5,6,7,10,14,15,21,30,35,42,105,210 }
Vậy X = 1,2,3,5,6,7,10,14,15,21,30,35,42,105,210
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\(A=1+5+5^2+5^3+...+5^{2011}\)
\(5A=5+5^2+5^3+...+5^{2012}\)
=>\(5A-A=5^{2012}-1\Rightarrow A=\frac{5^{2012}-1}{4}\)
Phương trình ban đầu tương đương với: \(\frac{5^{2012}-1}{4}\left|x-1\right|=5^{2012}-1\)
\(\Leftrightarrow\left|x-1\right|=4\Leftrightarrow\orbr{\begin{cases}x-1=4\\x-1=-4\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}\)
\(514+11< \overline{53x}-10< 737-210\)
⇒\(525< 52x< 527\)
⇒\(x=6\)
\(514+11< \overline{53x}-10< 737-210\)
\(\Leftrightarrow525< \overline{52x}< 527\)
\(\Rightarrow25< \overline{2x}< 27\Rightarrow x=6\)
Đs....