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Tìm x biết:
5) {x\(^2\) - [16\(^2\) - ( 8\(^2\) - 9 . 7)\(^3\) - 17 . 15 ]\(^3\) - 5 . 3 ]\(^3\) = 1
\(\left\{x^2-\left[16^2-\left(8^2-9.7\right)^3-17.15\right]^3-5.3\right\}^3=1\)
\(\left\{x^2-\left[256-\left(64-9.7\right)^3-17.15\right]^3-5.3\right\}^3=1\)
\(\left\{x^2-\left[256-1^3-255\right]^3-15\right\}^3=1\)
\(\left\{x^2-\left[255-255\right]^3-15\right\}^3=1\)
\(\left\{x^2-0^3-15\right\}^3=1\)
\(\left\{x^2-15\right\}^3=1\)
\(\left\{4^2-15\right\}^3=1\)
\(\Rightarrow x=4\)
5: \(\Leftrightarrow\left\{x^2-\left[256-\left(64-63\right)^3-255\right]^3-15\right\}^3=1\)
\(\Leftrightarrow\left\{x^2-15\right\}^3=1\)
\(\Leftrightarrow x^2-15=1\)
=>x=4 hoặc x=-4
5\(\dfrac{8}{17}\):x + (-\(\dfrac{1}{17}\)) : x + 3\(\dfrac{1}{17}\) : 17\(\dfrac{1}{3}\)= \(\dfrac{4}{17}\)
\(\dfrac{93}{17}\).\(\dfrac{1}{x}\) + (-\(\dfrac{1}{17}\)) .\(\dfrac{1}{x}\) +\(\dfrac{3}{17}\)= \(\dfrac{4}{17}\)
\(\dfrac{1}{x}\).\(\dfrac{92}{17}\)=\(\dfrac{1}{17}\)
\(\dfrac{1}{1.4}\)+\(\dfrac{1}{4.7}\)+\(\dfrac{1}{7.10}\)+...+\(\dfrac{1}{x.\left(x+3\right)}\)=\(\dfrac{6}{19}\)
5\(\frac{8}{17}\) : x - (\(\frac{8}{17}\)) : x + 3\(\frac{1}{17}\): 17\(\frac13\) = \(\frac{4}{17}\)
(5\(\frac{8}{17}\) - \(\frac{8}{17}\)) : x + \(\frac{52}{17}\) : \(\frac{52}{3}\) = \(\frac{4}{17}\)
5:x + \(\frac{52}{17}\times\) \(\frac{3}{52}\) = \(\frac{4}{17}\)
5 : x + \(\frac{3}{17}\) = \(\frac{4}{17}\)
5 : x = \(\frac{4}{17}\) - \(\frac{3}{17}\)
5 : x = \(\frac{1}{17}\)
x = 5 : \(\frac{1}{17}\)
x = 5 x 17
x = 85
Vậy x = 85
1/1.4 + 1/4.7 + ...+1/x(x+3) = 6/19
3/1.4 + 3/4.7 +..+3/x(x+3) = 18/19
1/1 - 1/4+ 1/4 - 1/7 +...+1/x-1/(x+3) = 1 - 1/19
1- 1/(x+3) = 1 - 1/19
1/(x+3) = 1/19
x + 3 = 19
x = 19 - 3
x = 16
Vậy x = 16
a)\(x.6=-72\)
=> x = -12
b)\(\dfrac{7}{3}:x=\dfrac{-11}{45}\)
=> \(x=\dfrac{-105}{11}\)
c) \(x=\dfrac{-17}{9}\)
d)\(x=\dfrac{-3}{17}\)
e) \(x=\dfrac{7}{6}\)
f) \(\dfrac{39}{7}:x=11\)
=> \(x=\dfrac{39}{77}\)
a)
\(\dfrac{x}{8}=\dfrac{-9}{6}\)
=> x = \(\dfrac{8.\left(-9\right)}{6}\)
=> x = -12
15: \(\Leftrightarrow\left\{x^2-\left[8^2-\left(25-24\right)^3-63\right]^3-48\right\}=1\)
\(\Leftrightarrow x^2-48=1\)
=>x=7 hoặc x=-7
Ta có:
\(37 - 5 \left(\right. x + 9 \left.\right) = \mid - 17 \mid\)
👉 Vì \(\mid - 17 \mid = 17\)
⇒ \(37 - 5 \left(\right. x + 9 \left.\right) = 17\)
\(37 - 5 x - 45 = 17\) \(- 5 x - 8 = 17\) \(- 5 x = 25 \Rightarrow x = - 5\)
✅ Đáp án: x = -5
b)
\(- 2 < \mid x + 2 \mid < 2\)
👉 Vì \(\mid x + 2 \mid \geq 0\) nên điều kiện \(- 2 < \mid x + 2 \mid\) luôn đúng
⇒ chỉ cần:
\(\mid x + 2 \mid < 2\) \(- 2 < x + 2 < 2\) \(- 4 < x < 0\)
✅ Đáp án: \(- 4 < x < 0\)
c)
\(- 7 \mid x - 2013 \mid + 20120 = \mid + 1 \mid\)
👉 \(\mid + 1 \mid = 1\)
⇒
\(- 7 \mid x - 2013 \mid + 20120 = 1\) \(- 7 \mid x - 2013 \mid = 1 - 20120 = - 20119\) \(\mid x - 2013 \mid = \frac{20119}{7} = 2874.142857...\)
⇒
\(x - 2013 = \pm \frac{20119}{7}\) \(x = 2013 \pm \frac{20119}{7}\)
👉 Có 2 nghiệm:
- \(x = 2013 + \frac{20119}{7}\)
- \(x = 2013 - \frac{20119}{7}\)
(= khoảng 4887.14 và -861.14)
d)
\(21 + \left(\right. x - 10 \left.\right) = - 19\)
\(21 + x - 10 = - 19\) \(x + 11 = - 19\) \(x = - 30\)
✅ Đáp án: x = -30
✅ Tóm lại:
- a) \(x = - 5\)
- b) \(- 4 < x < 0\)
- c) \(x = 2013 \pm \frac{20119}{7}\)
- d) \(x = - 30\)
a: \(\Leftrightarrow-\dfrac{9}{46}+\dfrac{108}{46}-\dfrac{93}{23}:\left(\dfrac{13}{4}-\dfrac{5}{3}x\right)=1\)
\(\Leftrightarrow\dfrac{93}{23}:\left(\dfrac{13}{4}-\dfrac{5}{3}x\right)=\dfrac{53}{46}\)
\(\Leftrightarrow-\dfrac{5}{3}x+\dfrac{13}{4}=\dfrac{186}{53}\)
=>-5/3x=55/212
hay x=-33/212
c: \(\Leftrightarrow\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{18}{19}\)
\(\Leftrightarrow1-\dfrac{1}{x+3}=\dfrac{18}{19}\)
=>x+3=19
hay x=16