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a) \(B=3+3^2+3^3+...+3^{120}\)
\(B=3\cdot1+3\cdot3+3\cdot3^2+...+3\cdot3^{119}\)
\(B=3\cdot\left(1+3+3^2+...+3^{119}\right)\)
Suy ra B chia hết cho 3 (đpcm)
b) \(B=3+3^2+3^3+...+3^{120}\)
\(B=\left(3+3^2\right)+\left(3^3+3^4\right)+\left(3^5+3^6\right)+...+\left(3^{119}+3^{120}\right)\)
\(B=\left(1\cdot3+3\cdot3\right)+\left(1\cdot3^3+3\cdot3^3\right)+\left(1\cdot3^5+3\cdot3^5\right)+...+\left(1\cdot3^{119}+3\cdot3^{119}\right)\)
\(B=3\cdot\left(1+3\right)+3^3\cdot\left(1+3\right)+3^5\cdot\left(1+3\right)+...+3^{119}\cdot\left(1+3\right)\)
\(B=3\cdot4+3^3\cdot4+3^5\cdot4+...+3^{119}\cdot4\)
\(B=4\cdot\left(3+3^3+3^5+...+3^{119}\right)\)
Suy ra B chia hết cho 4 (đpcm)
c) \(B=3+3^2+3^3+...+3^{120}\)
\(B=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\left(3^7+3^8+3^9\right)+...+\left(3^{118}+3^{119}+3^{120}\right)\)
\(B=\left(1\cdot3+3\cdot3+3^2\cdot3\right)+\left(1\cdot3^4+3\cdot3^4+3^2\cdot3^4\right)+...+\left(1\cdot3^{118}+3\cdot3^{118}+3^2\cdot3^{118}\right)\)
\(B=3\cdot\left(1+3+9\right)+3^4\cdot\left(1+3+9\right)+3^7\cdot\left(1+3+9\right)+...+3^{118}\cdot\left(1+3+9\right)\)
\(B=3\cdot13+3^4\cdot13+3^7\cdot13+...+3^{118}\cdot13\)
\(B=13\cdot\left(3+3^4+3^7+...+3^{118}\right)\)
Suy ra B chia hết cho 13 (đpcm)
(-4;-3;-2;-1;0;1;2;3;4)
Ko có dấu ngoặc nhọn nên mik xài ngoặc tròn nha
Bài 1:
a)
\(\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\ \Leftrightarrow x-1=24\\ x=24+1\\ x=25\)
b)
\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\ \dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\ \dfrac{3x}{7}+1=\dfrac{1}{2}\\ \dfrac{3x}{7}=\dfrac{1}{2}-1\\ \dfrac{3x}{7}=\dfrac{-1}{2}\\ 3x=\dfrac{-1}{2}\cdot7\\ 3x=\dfrac{-7}{2}\\ x=\dfrac{-7}{2}:3\\ x=\dfrac{-7}{6}\)
c)
\(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\ x+\dfrac{7}{12}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{12}\\ x=\dfrac{1}{4}\)
d)
\(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\ \dfrac{-1}{6}x=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{-1}{6}\\ x=\dfrac{-7}{2}\)
e)
\(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\ \dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\ \dfrac{277}{690}-x=\dfrac{7}{46}\\ x=\dfrac{277}{690}-\dfrac{7}{46}\\ x=\dfrac{86}{345}\)
f)
\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\ \left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\ x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\ x-\dfrac{1}{12}=\dfrac{7}{24}\\ x=\dfrac{7}{24}+\dfrac{1}{12}\\ x=\dfrac{3}{8}\)
g)
\(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\ \dfrac{13}{21}+x=\dfrac{2}{7}\\ x=\dfrac{2}{7}-\dfrac{13}{21}\\ x=\dfrac{-1}{3}\)
h)
\(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{29}{24}\\ x=\dfrac{29}{24}:\dfrac{1}{2}\\ x=\dfrac{29}{12}\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\)
i)
\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)
a: x/5=32/80
nên x/5=2/5
hay x=2
13/x=26/30
nên 13/x=13/15
hay x=15
-x/7=22/-77
=>x/7=2/7
hay x=2
b: x/9=28/36
=>x/9=7/9
hay x=7
-10/x=50/55
=>-10/x=10/11
hay x=-11
a ) \(5\left(x^2\right)+7x+2\)
\(\Leftrightarrow5x^2+7x+2=0\)
\(\Leftrightarrow5x^2+5x+2x+2=0\)
\(\Leftrightarrow\left(5x+2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{5}\\x=-1\end{matrix}\right.\)
Vậy .............
b ) \(\dfrac{x+1}{17}+\dfrac{x+2}{16}=\dfrac{x+3}{15}+\dfrac{x+4}{14}\)
\(\Leftrightarrow\dfrac{x+1}{17}+1+\dfrac{x+2}{16}+1=\dfrac{x+3}{15}+1+\dfrac{x+4}{14}+1\)
\(\Leftrightarrow\dfrac{x+18}{17}+\dfrac{x+18}{16}=\dfrac{x+18}{15}+\dfrac{x+18}{14}\)
\(\Leftrightarrow\dfrac{x+18}{17}+\dfrac{x+18}{16}-\dfrac{x+18}{15}-\dfrac{x+18}{14}=0\)
\(\Leftrightarrow\left(x+18\right)\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)=0\)
Vì \(\left(\dfrac{1}{17}+\dfrac{1}{16}-\dfrac{1}{15}-\dfrac{1}{14}\right)\ne0\)
Ta có : \(x+18=0\Leftrightarrow x=-18\)
Vậy ......
c ) \(\dfrac{x-1}{x-3}=\dfrac{x-4}{x-7}\)
\(\Leftrightarrow\left(x-1\right)\left(x-7\right)=\left(x-3\right)\left(x-4\right)\)
\(\Leftrightarrow x^2-7x-x+7=x^2-4x-3x+12\)
\(\Leftrightarrow-x=5\)
\(\Leftrightarrow x=-5\)
Vậy ..
a,\(\frac{-2}{3}+\frac{1}{5}=\frac{-2.5}{3.5}+\frac{3.1}{3.5}=-\frac{10}{15}+\frac{3}{15}=-\frac{7}{15}\)
b,\(3\frac{11}{13}-5\frac{11}{3}=3+\frac{11}{13}-(5+\frac{11}{13})=\left(3-5\right)+\left(\frac{11}{13}-\frac{11}{13}\right)=-2+0=-2\)
c,\(1\frac{2}{3}:(-\frac{5}{3})=\frac{5}{3}:\left(-\frac{5}{3}\right)=\frac{5.3}{3.\left(-5\right)}=\frac{15}{-15}=-1\)
d,\(\frac{31}{17}+\frac{-5}{13}+\frac{-8}{13}-\frac{14}{17}=\left(\frac{31}{17}-\frac{14}{17}\right)+\left(\frac{-5}{13}+\frac{-8}{13}\right)=1+\left(-1\right)=0\)
Tìm x:
a,x+12=8
x=8-12
x=-4
b,\(\frac{2}{3}x+\frac{1}{2}=\frac{1}{10}\)
\(\frac{2}{3}x=\frac{1}{10}-\frac{1}{2}\)
\(\frac{2}{3}x=\frac{1}{10}-\frac{5}{10}\)
\(\frac{2}{3}x=\frac{2}{5}\)
\(x=\frac{2}{5}:\frac{2}{3}\)
\(x=\frac{6}{10}\)
\(x=\frac{3}{5}\)
a) \(\frac{-2}{3}+\frac{1}{5}=\frac{-10}{15}+\frac{3}{15}=\frac{-10+3}{15}=\frac{-7}{15}\)
b) \(3\frac{11}{13}-5\frac{11}{13}=\frac{50}{13}-\frac{76}{13}=\frac{50-76}{13}=\frac{-26}{13}=-2\)
c) \(1\frac{2}{3}:\frac{-5}{3}=\frac{5}{3}:\frac{-5}{3}=\frac{5}{3}\times\frac{3}{-5}=\frac{15}{-15}=-1\)
d) \(\frac{31}{17}+\frac{-5}{13}+\frac{-8}{13}-\frac{14}{17}\)
\(=\left(\frac{31}{17}-\frac{14}{17}\right)+\left(\frac{-5}{13}+\frac{-8}{13}\right)\)
\(=\left(\frac{31-14}{17}\right)+\left(\frac{-5-8}{13}\right)\)
\(=\frac{17}{17}+\frac{-13}{13}\)
\(=1+\left(-1\right)\)
\(=0\)
TÌM x
a) \(x+12=8\)
\(\Leftrightarrow x=8-12\)
\(\Leftrightarrow x=-4\)
b) \(\frac{2}{3}x+\frac{1}{2}=\frac{1}{10}\)
\(\Leftrightarrow\frac{2}{3}x=\frac{1}{10}-\frac{1}{2}=\frac{1}{10}-\frac{5}{10}=\frac{1-5}{10}\)
\(\Leftrightarrow\frac{2}{3}x=\frac{-4}{10}=\frac{-2}{5}\)
\(\Leftrightarrow x=\frac{-2}{5}:\frac{2}{3}=\frac{-2}{5}\times\frac{3}{2}\)
\(\Leftrightarrow x=\frac{-6}{10}=\frac{-3}{5}\)
a) x = 10.
b) x = 80.
c) x = 25.