7/12+6/9+3/8+5/12+5/15-5/8
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+) Số phần tử của tập hợp X là: $(30-1):1+1=30$ (phần tử)
+) Số phần tử của tập hợp T là: $(30-0):1+1=31$ (phần tử)
+ X = {1; 2; 3;...; 30}
Xét dãy số 1; 2; 3;...; 30
Đây là dãy số cách đều với khoảng cách là: 2 - 1 = 1
Dãy số trên có số số hạng là: (30 - 1) : 1 + 1 = 39 (số hạng)
Vậy tập hợp X có 30 hạng tử
+ T = {0; 1; 2; 3;...; 30}
Đây là dãy số cách đều với khoảng cách là: 1- 0 = 1
Số số hạng của dãy số trên là: (30 - 0) : 1 + 1 = 31 (số hạng)
Vậy tập hợp T có 31 hạng tử.
\(500-\left\{5\cdot\left[409-\left(2^3\cdot3-21\right)^2\right]-1724\right\}\\ =500-\left\{5\cdot\left[409-\left(8\cdot3-21\right)^2\right]-1724\right\}\\ =500-\left\{5\cdot\left[409-\left(24-21\right)^2\right]-1724\right\}\\ =500-\left[5\cdot\left(409-3^2\right)-1724\right]\\ =500-\left[5\cdot\left(409-9\right)-1724\right]\\ =500-\left(5\cdot400-1724\right)\\ =500-\left(2000-1724\right)\\ =500-276\\ =224\)
\(500-\left\{5\left[409-\left(2^3\times3-21\right)^2\right]-1724\right\}\)
\(=500-\left\{5\left[409-\left(24-21\right)^2\right]-1724\right\}\)
\(=500-\left\{5\left[409-9\right]-1724\right\}\)
\(=500-\left\{5.400-1724\right\}\)
\(=500-\left\{2000-1724\right\}\)
\(=500-2000+1724\)
\(=224\)
\(a.\dfrac{1}{2}-3x=-\dfrac{2}{5}\\ 3x=\dfrac{1}{2}+\dfrac{2}{5}\\ 3x=\dfrac{9}{10}\\ x=\dfrac{9}{10}:3\\ x=\dfrac{3}{10}\\ b.-x+\dfrac{1}{2}=-\dfrac{5}{6}\\ x=\dfrac{1}{2}+\dfrac{5}{6}\\ x=\dfrac{4}{3}\\ c.x+\dfrac{3}{5}=\left(-\dfrac{2}{5}\right)^2\\ x+\dfrac{3}{5}=\dfrac{4}{25}\\ x=\dfrac{4}{25}-\dfrac{3}{5}\\ x=-\dfrac{11}{25}\\ d.\dfrac{3}{7}+\dfrac{1}{7}:x=\dfrac{3}{14}\\ \dfrac{1}{7}:x=\dfrac{3}{14}-\dfrac{3}{7}=-\dfrac{3}{14}\\ x=\dfrac{1}{7}:-\dfrac{3}{14}=-\dfrac{2}{3}\\ e.-\dfrac{1}{3}\left(\dfrac{1}{7}-x\right)=\dfrac{1}{21}\\ \dfrac{1}{7}-x=\dfrac{1}{21}:-\dfrac{1}{3}=-\dfrac{1}{7}\\ x=\dfrac{1}{7}+\dfrac{1}{7}=\dfrac{2}{7}\\ h.\dfrac{1}{4}-3x+\dfrac{3}{2}=-0,75\\ \dfrac{1}{4}-3x+\dfrac{3}{2}=-\dfrac{3}{4}\\ 3x=\dfrac{1}{4}+\dfrac{3}{2}+\dfrac{3}{4}=\dfrac{5}{2}\\ x=\dfrac{5}{2}:3\\ x=\dfrac{5}{6}\\ i.\dfrac{2}{7}-\left(\dfrac{2}{3}+2x\right)=\dfrac{5}{7}\\ \dfrac{2}{3}+2x=\dfrac{2}{7}-\dfrac{5}{7}=-\dfrac{3}{7}\\ 2x=-\dfrac{3}{7}-\dfrac{2}{3}=-\dfrac{23}{21}\\ x=\dfrac{-23}{21}:2=-\dfrac{23}{42}\)
a: \(\dfrac{1}{2}-3x=-\dfrac{2}{5}\)
=>\(3x=\dfrac{1}{2}+\dfrac{2}{5}=\dfrac{5}{10}+\dfrac{4}{10}=\dfrac{9}{10}\)
=>\(x=\dfrac{9}{10}:3=\dfrac{9}{30}=\dfrac{3}{10}\)
b: \(-x+\dfrac{1}{2}=-\dfrac{5}{6}\)
=>\(-x=-\dfrac{5}{6}-\dfrac{1}{2}=-\dfrac{5}{6}-\dfrac{3}{6}=-\dfrac{8}{6}=-\dfrac{4}{3}\)
=>\(x=\dfrac{4}{3}\)
c: \(x+\dfrac{3}{5}=\left(-\dfrac{2}{5}\right)^2\)
=>\(x+\dfrac{3}{5}=\dfrac{4}{25}\)
=>\(x=\dfrac{4}{25}-\dfrac{3}{5}=\dfrac{4}{25}-\dfrac{15}{25}=-\dfrac{11}{25}\)
d: \(\dfrac{3}{7}+\dfrac{1}{7}:x=\dfrac{3}{14}\)
=>\(\dfrac{1}{7}:x=\dfrac{3}{14}-\dfrac{3}{7}=-\dfrac{3}{14}\)
=>\(x=-\dfrac{1}{7}:\dfrac{3}{14}=-\dfrac{1}{7}\cdot\dfrac{14}{3}=-\dfrac{2}{3}\)
e: \(-\dfrac{1}{3}\left(\dfrac{1}{7}-x\right)=\dfrac{1}{21}\)
=>\(\dfrac{1}{7}-x=\dfrac{1}{21}:\dfrac{-1}{3}=\dfrac{-1}{21}\cdot3=-\dfrac{1}{7}\)
=>\(x=\dfrac{1}{7}+\dfrac{1}{7}=\dfrac{2}{7}\)
h: \(\dfrac{1}{4}-3x+\dfrac{3}{2}=-0,75\)
=>\(-3x+\dfrac{5}{4}=-\dfrac{3}{4}\)
=>\(-3x=-\dfrac{3}{4}-\dfrac{5}{4}=-\dfrac{8}{4}=-2\)
=>\(x=\dfrac{-2}{-3}=\dfrac{2}{3}\)
i: \(\dfrac{2}{7}-\left(\dfrac{2}{3}+2x\right)=\dfrac{5}{7}\)
=>\(2x+\dfrac{2}{3}=\dfrac{2}{7}-\dfrac{5}{7}=-\dfrac{3}{7}\)
=>\(2x=-\dfrac{3}{7}-\dfrac{2}{3}=-\dfrac{9}{21}-\dfrac{14}{21}=-\dfrac{23}{21}\)
=>\(x=-\dfrac{23}{21}:2=-\dfrac{23}{42}\)
\(a.\dfrac{3}{7}\cdot\dfrac{5}{8}+\dfrac{3}{7}\cdot\dfrac{11}{8}+\dfrac{11}{7}=\dfrac{3}{7}\cdot\left(\dfrac{5}{8}+\dfrac{11}{8}\right)+\dfrac{11}{7}=\dfrac{3}{7}\cdot2+\dfrac{11}{7}=\dfrac{6}{7}+\dfrac{11}{7}=\dfrac{17}{7}\\ b.\dfrac{3}{8}\cdot19\dfrac{1}{3}-\dfrac{3}{8}\cdot\left(33\dfrac{1}{3}\right)=\dfrac{3}{8}\cdot\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)=\dfrac{3}{8}\cdot-14=\dfrac{-21}{8}\\ c.\dfrac{1}{3}\cdot\dfrac{5}{4}+\dfrac{1}{3}\cdot\dfrac{7}{4}-2022^0=\dfrac{1}{3}\cdot\left(\dfrac{5}{4}+\dfrac{7}{4}\right)-1=\dfrac{1}{3}\cdot\dfrac{12}{4}-1=\dfrac{1}{3}\cdot3-1=1-1=0\\ d.\dfrac{5}{13}+\left(-\dfrac{5}{17}\right)+\dfrac{-21}{41}+\dfrac{8}{13}+\dfrac{-20}{41}=\left(\dfrac{5}{13}+\dfrac{8}{13}\right)+\left(-\dfrac{5}{17}\right)+\left(\dfrac{-21}{41}+\dfrac{-20}{41}\right)=1+\left(-\dfrac{5}{17}\right)-1=-\dfrac{5}{17}\)
\(e.\dfrac{27}{13}:\dfrac{9}{7}+\dfrac{12}{13}:\dfrac{9}{7}=\dfrac{27}{13}\cdot\dfrac{7}{9}+\dfrac{12}{13}\cdot\dfrac{7}{9}=\dfrac{7}{9}\cdot\left(\dfrac{27}{13}+\dfrac{12}{13}\right)=\dfrac{7}{9}\cdot\dfrac{39}{13}=\dfrac{7}{9}\cdot3=\dfrac{7}{3}\\ g.\dfrac{8}{15}\cdot-\dfrac{4}{9}+\dfrac{8}{15}:\dfrac{-9}{5}-3\dfrac{2}{5}=\dfrac{8}{15}\cdot\dfrac{-4}{9}+\dfrac{8}{15}\cdot\dfrac{-5}{9}-\dfrac{17}{5}=\dfrac{8}{15}\cdot\left(\dfrac{-4}{9}+\dfrac{-5}{9}\right)-\dfrac{17}{5}=-\dfrac{8}{15}-\dfrac{17}{5}=-\dfrac{59}{15}\\ h.\left(-\dfrac{2}{3}+\dfrac{3}{13}\right):\dfrac{7}{8}+\left(-\dfrac{1}{3}+\dfrac{10}{13}\right):\dfrac{7}{8}=\left(-\dfrac{2}{3}+\dfrac{3}{13}\right)\cdot\dfrac{8}{7}+\left(-\dfrac{1}{3}+\dfrac{10}{13}\right)\cdot\dfrac{8}{7}=\dfrac{8}{7}\cdot\left(-\dfrac{2}{3}+\dfrac{3}{13}-\dfrac{1}{3}+\dfrac{10}{3}\right)=\dfrac{8}{7}\cdot\left(-1+1\right)=\dfrac{8}{7}\cdot0=0\)
\(a.\left(\dfrac{1}{2}\right)^2-\dfrac{3}{8}:-\dfrac{9}{2}=\dfrac{1}{4}-\dfrac{3}{8}\cdot\dfrac{-2}{9}=\dfrac{1}{4}+\dfrac{1}{12}=\dfrac{4}{12}=\dfrac{1}{3}\)
\(b.\dfrac{1}{2}:\left(\dfrac{4}{5}-\dfrac{1}{2}\right)=\dfrac{1}{2}:\dfrac{4\cdot2-5}{10}=\dfrac{1}{2}:\dfrac{3}{10}=\dfrac{1}{2}\cdot\dfrac{10}{3}=\dfrac{5}{3}\)
\(c.\left(-\dfrac{1}{3}\right)^2:\dfrac{5}{9}+\left(-1\right)^3=\dfrac{1}{9}:\dfrac{5}{9}-1=\dfrac{1}{9}\cdot\dfrac{9}{5}-1=\dfrac{1}{5}-1=-\dfrac{4}{5}\)
\(d.\left(\dfrac{3}{5}\right)^2-\left(\dfrac{4}{5}-6\dfrac{1}{2}\right)=\dfrac{9}{25}-\dfrac{4}{5}+6\dfrac{1}{2}=\dfrac{9}{25}-\dfrac{20}{25}+\dfrac{13}{2}=\dfrac{-11}{25}+\dfrac{13}{2}=\dfrac{303}{50}\)
a) \(\left(\dfrac{1}{2}\right)^2-\dfrac{3}{8}:\dfrac{-9}{2}=\dfrac{1}{4}-\dfrac{3}{8}\times\dfrac{2}{-9}\\ =\dfrac{1}{4}-\left(-\dfrac{1}{12}\right)=\dfrac{3}{12}+\dfrac{1}{12}=\dfrac{4}{12}=\dfrac{1}{3}\)
b) \(\dfrac{1}{2}:\left(\dfrac{4}{5}-\dfrac{1}{2}\right)=\dfrac{1}{2}:\dfrac{3}{10}\\ =\dfrac{1}{2}\times\dfrac{10}{3}=\dfrac{5}{3}\)
c) \(\left(-\dfrac{1}{3}\right)^2:\dfrac{5}{9}+\left(-1\right)^3=\dfrac{1}{9}\times\dfrac{9}{5}-1=\dfrac{1}{5}-1\\ =-\dfrac{4}{5}\)
TH1: \(-5\le x\le2\)
=> \(\left(2-x\right)-4\left(5+x\right)=-23\)
\(=>2-x-20-4x=-23\)
\(=>-5x-18=-23\\ =>-5x=-23+18\\ =>-5x=-5\\ =>x=\dfrac{-5}{-5}=1\left(tm\right)\)
TH2: \(x>2\)
\(\left(x-2\right)-4\left(5+x\right)=-23\)
\(=>x-2-20-4x=-23\\ =>-3x-22=-23\\ =>-3x=-1\\ =>x=-\dfrac{1}{-3}=\dfrac{1}{3}\left(ktm\right)\)
TH3: \(x< -5\)
\(\left(2-x\right)+4\left(5+x\right)=-23\\ =>2-x+20+4x=-23\\ =>3x+22=-23\\ =>3x=-45\\ =>x=-15\left(tm\right)\)
Vậy: ...
$|2-x|-4|5+x|=-23$ (1)
+, Với $x< -5\Rightarrow \begin{cases} |2-x|=2-x\\|5+x|=-5-x \end{cases}$, (1) trở thành:
$2-x-4(-5-x)=-23$
$\Rightarrow 2-x+20+4x=-23$
$\Rightarrow 3x+22=-23$
$\Rightarrow 3x=-23-22$
$\Rightarrow 3x=--45$
$\Rightarrow x=-15$ (tmdk)
+, Với $-5\le x\le 2\Rightarrow \begin{cases} |2-x|=2-x\\|5+x|=5+x \end{cases}$, (1) trở thành:
$2-x-4(5+x)=-23$
$\Rightarrow 2-x-20-4x=-23$
$\Rightarrow -5x-18=-23$
$\Rightarrow -5x=-23+18$
$\Rightarrow -5x==-5$
$\Rightarrow x=1$ (tmdk)
+, Với $x>2\Rightarrow \begin{cases} |2-x|=x-2\\|5+x|=5+x \end{cases}$, (1) trở thành:
$x-2-4(5+x)=-23$
$\Rightarrow x-2-20-4x=-23$
$\Rightarrow -3x-22=-23$
$\Rightarrow -3x=-23+22$
$\Rightarrow -3x=-1$
$\Rightarrow x=\frac13$ (loại)
Vậy: ...
1) 15.9^3 : 27 - 6^2.12 +18.41
= 15.729 : 27 - 36.12 +738
= 10935 : 27 - 432 + 738
= 405 - 432 + 738
= 711
2) 18.7^3 + 9.2^6 :8 - 12.7
= 18.343 + 9.64 :8 - 84
= 6174 + 72 - 84
= 6162
3) 23.2^5 + 288.4^2 :12
= 23.32 + 288.16 :12
= 736 + 384
= 1120
4) 12^2 : 36 + 8^2 : 4 - 5.3
= 144 : 36 + 64 : 4 - 15
= 4 + 16 - 15
= 5
\(x+10\%\cdot x+20\%\cdot x+30\%\cdot x=4,8\)
=>\(x\left(1+0,1+0,2+0,3\right)=4,8\)
=>\(x\cdot1,6=4,8\)
=>\(x=4.8:1.6=3\)
$\frac{7}{12}+\frac69+\frac38+\frac{5}{12}+\frac{5}{15}-\frac58$
$=\left(\frac{7}{12}+\frac{5}{12}\right)+\left(\frac69-\frac{5}{15}\right)+\left(\frac38-\frac58\right)$
$=1+\left(\frac13-\frac13\right)-\frac28$
$=1-\frac14=\frac34$