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a: Trên tia Ax, ta có: AB<AC
nên B nằm giữa A và C
=>AB+BC=AC
=>BC=7-5=2(cm)
b: Ta có: AB và AD là hai tia đối nhau
=>A nằm giữa B và D
=>BD=AB+AD=5+2,5=7,5(cm)
c: CB và CE là hai tia đối nhau
=>C nằm giữa B và E
=>BC+CE=BE
=>BE=2+3=5(cm)
Ta có: B nằm giữa A và C
C nằm giữa B và E
Do đó: B nằm giữa A và E
mà BA=BE(=5cm)
nên B là trung điểm của AE
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{x-2}{4}=\dfrac{5+x}{3}\)
⇒3.(x+2)=4.(5+x)
3x+6=20+x
3x+6-x=20
2x+6=20
2x=14
x=7
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 :
\(a,6^3.6^7.6^5=6^{3+7+5}=6^{15}\)
\(b,17^9:17^5:17^2=17^{9-5-2}=17^2\)
\(c,=\left(3^3\right)^3.3^3=3^9.3^3=3^{9+3}=3^{12}\)
\(d,=\left(2^4\right)^3.\left(2^6\right)^5=2^{12}.2^{30}=2^{12+30}=2^{42}\)
Bài 2 :
\(a,11^{60}:11^{58}=11^{60-58}=11^2=121\)
\(b,8^{10}:8^5:8^4=8^{10-5-4}=8^1=8\)
\(c,=\left(5^2\right)^9:\left(5^3\right)^5=5^{18}:5^{15}=5^{18-15}=5^3=125\)
\(d,=\left(2^4\right)^5:\left(2^2\right)^6:\left(2^3\right)^2=2^{20}:2^{12}:2^6=2^{20-12-6}=2^2=4\)
\(e,=10^5.\left(10^2\right)^5.\left(10^3\right)^2=10^5.10^{10}.10^6=10^{5+10+6}=10^{21}\)
Bài 3:
a)\(58.75+58.50-58.25\)
=\(58.\left(75+50-25\right)\)
=\(58.100\)
=\(5800\)
b)\(27.39+27.63-2.27\)
=\(27.\left(39+63-2\right)\)
=\(27.100\)
=\(2700\)
c)\(156.25+5.156+156.14+36.156\)
=\(156.\left(25+5+14+36\right)\)
=\(156.80\)
=\(12480\)
d)\(12.35+35.182-35.94\)
=\(35.\left(12+182-94\right)\)
=\(35.100\)
=\(3500\)
e)\(48.19+48.115+67.104\)
=\(48.\left(19+115\right)+67.104\)
=\(48.134+67.104\)
=\(48.67+48.67+67.104\)
=\(67.\left(48+48+104\right)\)
=\(67.200\)
=\(13400\)
f)\(128.72+128.67+128.72+11.72\)
=\(128.\left(72+67\right)+72.\left(128+11\right)\)
=\(128.139+72.139\)
=\(139.\left(72+128\right)\)
=\(139.200\)
=\(27800\)
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1) Ta có: \(x+\dfrac{1}{5}=0.25\)
\(\Leftrightarrow x=\dfrac{1}{4}-\dfrac{1}{5}\)
hay \(x=\dfrac{1}{20}\)
Vậy: \(x=\dfrac{1}{20}\)
2) Ta có: \(\dfrac{2}{7}-\dfrac{3}{7}:x=\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{3}{7}:x=\dfrac{2}{7}-\dfrac{1}{5}=\dfrac{3}{35}\)
hay x=5
Vậy: x=5
3) Ta có: \(\left(3x-1.5\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1.5=0\\-\dfrac{1}{2}x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=10\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{1}{2};10\right\}\)
4) Ta có: \(\left(x+3\right)^2-\dfrac{1}{3}=\dfrac{1}{9}\)
\(\Leftrightarrow\left(x+3\right)^2=\dfrac{4}{9}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=\dfrac{2}{3}\\x+3=-\dfrac{2}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{3}\\x=-\dfrac{9}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{-\dfrac{7}{3};-\dfrac{9}{3}\right\}\)
5) Ta có: \(60\%x-x+\dfrac{1}{3}=2\)
\(\Leftrightarrow\dfrac{-2}{5}x=\dfrac{5}{3}\)
hay \(x=-\dfrac{25}{6}\)
Vậy: \(x=-\dfrac{25}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
`1/4 + 3/4 : x = -2`
`3/4 : x = -2 - 1/4`
`3/4 : x = -9/4`
`x = 3/4 : (-9/4)`
`x = -1/3`
Vậy `x = -1/3`
=7/6-4/7=49/42-24/42=25/42