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\(|x-7|+13=25\)
\(|x-7|=12\)
\(\Rightarrow\orbr{\begin{cases}x-7=12\\x-7=-12\end{cases}\Leftrightarrow\orbr{\begin{cases}x=19\\x=-5\end{cases}}}\)
\(|x-3|-16=-4\\ \Rightarrow|x-3|=14\)
\(\Rightarrow\orbr{\begin{cases}x-3=14\\x-3=-14\end{cases}\Leftrightarrow\orbr{\begin{cases}x=17\\x=-11\end{cases}}}\)
\(26-Ix+9I=-13\)
\(\Rightarrow|x+9|=39\)
\(\Rightarrow\orbr{\begin{cases}x+9=39\\x+9=-39\end{cases}\Leftrightarrow\orbr{\begin{cases}x=30\\x=-48\end{cases}}}\)
\(|x-5|=3\)
\(\Rightarrow\orbr{\begin{cases}x-5=3\\x-5=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=8\\x=2\end{cases}}}\)
\(|1-x|=7\)
\(\Rightarrow\orbr{\begin{cases}1-x=7\\1-x=-7\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-6\\x=8\end{cases}}}\)
\(|2x+5|=1\)
\(\Rightarrow\orbr{\begin{cases}2x+5=1\\2x+5=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=-3\end{cases}}}\)
\(\text{11 _ 12 +13 _ 14 + 15 - 16 + 17 _ 18 + 19 _20}\)
\(=\left(11-12\right)+\left(13-14\right)+\left(15-16\right)+\left(17-18\right)+\left(19-20\right)\)
\(=-1+\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)\)
\(=-5\)
\(\text{101 _ 102 _ ( - 103 ) _ 104 _ ( - 105 ) _ 106 _ ( -107 ) _ 108 _ ( - 109 ) _ 110}\)
\(=\left(101-102\right)+\left(-\left(-103\right)-104\right)+\left(-\left(-105\right)-106\right)\)\(+\left(-\left(-107\right)-108\right)+\left(-\left(-109\right)-110\right)\)
\(=-1+\left(-1\right)+\left(-1\right)+\left(-1\right)+\left(-1\right)\)
\(=-5\)
\(\text{1 + ( - 3 ) +5 + ( -7 ) + ... + ( - 1999 ) + 2001}\)
\(=\left(1+\left(-3\right)\right)+\left(5+\left(-7\right)\right)+...+\left(\left(-1999\right)+2001\right)\)
\(=-2+\left(-2\right)+...+\left(-2\right)\)Có 500 thừa số (-2)
\(=\left(-2\right).500\)
\(=-1000\)
\(\text{1 + ( -2 ) + 3 + ( -4) + ... + 2001 + ( - 2002)}\)
\(=\left(1+\left(-2\right)\right)+\left(3+\left(-4\right)\right)+...+\left(2001+\left(-2002\right)\right)\)
\(=-1+\left(-1\right)+...+\left(-1\right)\)Có 1000 thừa số (-1)
\(=\left(-1\right).1000\)
\(=-1000\)
chúc bạn học tốt
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`@` `\text {Ans}`
`\downarrow`
`a,`
\(x \div 0,1 + x \div 0,25 - 35 = 35\)
`\Rightarrow x * 10 + x * 4 - 35 = 35`
`\Rightarrow x*(10+4) - 35 = 35`
`\Rightarrow 14x - 35 = 35`
`\Rightarrow 14x = 70`
`\Rightarrow x=70/14 =5`
Vậy, `x=5`
`b,`
\(x \div 0,125 - x \div 0,5 = 108\)
`\Rightarrow x * 4/5 - x*2=108`
`\Rightarrow x*(4/5 - 2) = 108`
`\Rightarrow -6/5x = 108`
`\Rightarrow x=108 \div -6/5`
`\Rightarrow x=-90`
Vậy, `x=-90.`
a) x/0,1 + x/0,25 - 35 = 35
x . 10 + x . 4 = 35 + 35
14x = 70
x = 70 : 14
x = 5
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Tìm x
\(\frac{x}{24}=-\frac{11}{40}+\frac{2}{5}\)
\(\frac{x}{24}=\frac{1}{8}\)
\(=>\frac{3}{24}=\frac{1}{8}\)
\(=>x=3\)
Vậy \(x=3\)
\(\frac{x}{24}=\frac{-11}{40}+\frac{2}{5}\)
=>\(\frac{x}{24}=\frac{1}{8}\)
=>x=3
Ai tích mk mk sẽ tích lại
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(A=\frac{3^2\cdot4^2\cdot2^{32}}{11\cdot2^{13}\cdot4^{11}-16^9}\)
\(A=\frac{3^2\cdot\left(2^2\right)^2\cdot2^{32}}{11\cdot2^{13}\cdot\left(2^2\right)^{11}-16^9}\)
\(A=\frac{3^2\cdot2^4\cdot2^{32}}{11\cdot2^{13}\cdot2^{22}-16^9}=\frac{3^2\cdot2^{36}}{11\cdot2^{35}-\left(2^4\right)^9}\)
\(A=\frac{3^2\cdot2^{36}}{11\cdot2^{35}-2^{36}}\)
\(A=\frac{3^2\cdot2^{36}}{11\cdot2^{35}-2\cdot2^{35}}\)
\(A=\frac{3^2\cdot2^{36}}{\left(11-2\right)\cdot2^{35}}=\frac{9\cdot2^{36}}{9\cdot2^{35}}=\frac{2^{36}}{2^{35}}=2\)
Bài làm:
\(\frac{3^2.4^2.2^{32}}{11.2^{13}.4^{11}-16^9}\)
\(=\frac{3^2.\left(2^2\right)^2.2^{32}}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)
\(=\frac{3^2.2^4.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3^2.2^{36}}{11.2^{35}-2^{36}}\)
\(=\frac{3^2.2^{36}}{\left(11-2\right).2^{35}}\)
\(=\frac{9.2^{36}}{9.2^{35}}\)
\(=2\)
Học tốt
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a)
\(\overline{ab}\times101=\overline{ab}\times\left(100+1\right)=\overline{ab00}+\overline{ab}=\overline{abab}\)
b)
\(\overline{ab}\times10101=\overline{ab}\times\left(10000+101\right)=\overline{ab0000}+\overline{abab}=\overline{ababab}\)
c)
\(\overline{abc}\times1001=\overline{abc}\times\left(1000+1\right)=\overline{abc000}+\overline{abc}=\overline{abcabc}\)
d)
\(\overline{ab}\times1001=\overline{ab}\times\left(1000+1\right)=\overline{ab000}+\overline{ab}=\overline{ab0ab}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
HOÀNG TÚ UYÊN ƠI CHO MÌNH HỎI TÍ :
Ở CÂU 2 TẠI SAO x CÓ THỂ LÀ 0 HOẶC 5 BẠN GIẢI THÍCH TÍ CHO MÌNH ĐƯỢC KO
![](https://rs.olm.vn/images/avt/0.png?1311)
Để x thoả mãn \(-3< x\le3\) thì \(x\in\left\{-2;-1;0;1;2;3\right\}\)