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|5.6 - \(x\)| = 4.6
|30 - \(x\)| = 24
\(x< 30\) ⇒ 30 - \(x\) = 24 ⇒ \(x\) = 30 - 24 = -6
\(x>30\) ⇒ -(30 - \(x\)) = 24 ⇒ -30 + \(x\) = 24 ⇒ \(x\) = 24 + 30 = 54
Vậy \(x\) \(\in\) { -6; 54}
`@` `\text {Ans}`
`\downarrow`
\(\left|5\cdot6-x\right|=4\cdot6\)
`\Rightarrow`\(\left|5\cdot6-x\right|=24\)
`\Rightarrow`\(\left[{}\begin{matrix}5\cdot6-x=24\\5\cdot6-x=-24\end{matrix}\right.\)
`\Rightarrow`\(\left[{}\begin{matrix}30-x=24\\30-x=-24\end{matrix}\right.\)
`\Rightarrow`\(\left[{}\begin{matrix}x=30-24\\x=30-\left(-24\right)\end{matrix}\right.\)
`\Rightarrow`\(\left[{}\begin{matrix}x=6\\x=54\end{matrix}\right.\)
Vậy, `x={6; 54}.`
![](https://rs.olm.vn/images/avt/0.png?1311)
Gọi biểu thức trên là A
Ta có:
2A = (\(\dfrac{1}{2.4}\)+\(\dfrac{1}{4.6}\)+...+\(\dfrac{1}{x.\left(x+2\right)}\)).2
2A = \(\dfrac{2}{2.4}\)+\(\dfrac{2}{4.6}\)+...+\(\dfrac{2}{x\left(x+2\right)}\)
2A = \(\dfrac{1}{2}\)-\(\dfrac{1}{4}\)+\(\dfrac{1}{4}\)-\(\dfrac{1}{6}\)+...+\(\dfrac{1}{x}\)-\(\dfrac{1}{x+2}\)
2A = \(\dfrac{1}{2}\)-\(\dfrac{1}{x+2}\)
mà A = \(\dfrac{1}{10}\)(đề bài)
nên 2A = \(\dfrac{2}{10}\) hay \(\dfrac{1}{2}\) - \(\dfrac{1}{x+2}\) = \(\dfrac{2}{10}\)
suy ra \(\dfrac{1}{x+2}\) = \(\dfrac{1}{2}\)-\(\dfrac{2}{10}\)=\(\dfrac{3}{10}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
a) Ta có: \(\dfrac{7^4\cdot3-7^3}{7^4\cdot6-7^3\cdot2}\)
\(=\dfrac{7^3\cdot\left(7\cdot3-1\right)}{7^3\cdot2\left(7\cdot3-1\right)}\)
\(=\dfrac{1}{2}\)
c) Ta có: \(E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}\)
\(\Leftrightarrow\dfrac{1}{3}\cdot E=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\)
\(\Leftrightarrow E-\dfrac{1}{3}\cdot E=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{100}}-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{101}}\right)\)
\(\Leftrightarrow E\cdot\dfrac{2}{3}=1-\dfrac{1}{3^{101}}\)
\(\Leftrightarrow E=\dfrac{3-\dfrac{3}{3^{101}}}{2}=\dfrac{1-\dfrac{1}{3^{100}}}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
3/4.8/9.15/16......9999/10000
= 3.8.15.....9999/4.9.16......10000
=101/50
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Áp dụng định lí tổng ba góc trong tam giác, ta có:
+)
\(\begin{array}{l}x + {120^o} + {35^o} = {180^o}\\ \Rightarrow x + {155^o} = {180^o}\\ \Rightarrow x = {180^o} - {155^o}\\ \Rightarrow x = {25^o}\end{array}\)
+)
\(\begin{array}{l}y + {70^o} + {60^o} = {180^o}\\ \Rightarrow y = {180^o} - {70^o} - {60^o}\\ \Rightarrow y = {50^o}\end{array}\)
+)
\(\begin{array}{l}z+ {90^o} + {55^o} = {180^o}\\ \Rightarrow z = {180^o} - {90^o} - {55^o}\\ \Rightarrow z = {35^o}\end{array}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
x=180-120-35=180-155=25 độ
y=180-70-60=50 độ
z=180-90-55=35 độ
![](https://rs.olm.vn/images/avt/0.png?1311)
*Bài làm:
~I) Tìm x:
➤Ta có: \(\frac{1}{2.4}\) + \(\frac{1}{4.6}\) + ... + \(\frac{1}{\left(2x-2\right)2x}\) = \(\frac{11}{48}\)
⇒ \(2\) . (\(\frac{1}{2.4}\) + \(\frac{1}{4.6}\) + ... + \(\frac{1}{\left(2x-2\right)2x}\)) = \(2\) . \(\frac{11}{48}\)
⇒ \(\frac{2}{2.4}\) + \(\frac{2}{4.6}\) + ... + \(\frac{2}{\left(2x-2\right)2x}\) = \(\frac{22}{48}\)
⇒ (\(\frac{1}{2}\) - \(\frac{1}{4}\)) + (\(\frac{1}{4}\) - \(\frac{1}{6}\)) + ... + (\(\frac{1}{2x-2}\) - \(\frac{1}{2x}\)) = \(\frac{22}{48}\)
⇒ \(\frac{1}{2}\) - \(\frac{1}{4}\) + \(\frac{1}{4}\) - \(\frac{1}{6}\) + \(\frac{1}{6}\) - ... - \(\frac{1}{2x-2}\) + \(\frac{1}{2x-2}\) - \(\frac{1}{2x}\) = \(\frac{22}{48}\)
⇒ \(\frac{1}{2}\) - \(\frac{1}{2x}\) = \(\frac{22}{48}\)
⇒ \(\frac{x}{x}\) . \(\frac{1}{2}\) - \(\frac{1}{2x}\) = \(\frac{22}{48}\)
⇒ \(\frac{x}{2x}\) - \(\frac{1}{2x}\) = \(\frac{22}{48}\)
⇒ \(\frac{x-1}{2x}\) = \(\frac{22}{48}\)
⇒ \(\frac{x-1}{2x}\) = \(\frac{22}{48}\)
⇒ \(x-1\) = \(\frac{22}{48}\) . \(2x\)
⇒ \(x-1\) = \(\frac{44x}{48}\)
⇒ \(x\) = \(\frac{44x}{48}\) + \(1\)
⇒ \(x\) = \(\frac{44x}{48}\) + \(\frac{48}{48}\)
⇒ \(x\) = \(\frac{44x+48}{48}\)
⇒ \(x\) = \(12\) (Chỗ này mình bấm máy tính nên hơi tắt;Bạn thông cảm)
*Vậy \(x\) = \(12\) .
\(6^x+4\cdot6^x=180\)
\(6^x\cdot\left(1+4\right)=36\cdot5\)
\(6^x\cdot5=36\cdot5\)
\(6^x=36\)
\(6^x=6^2\)
\(x=2\)