\(4^{n+3}+4^{n+2}-4^{n+1}-4^n\)
\(4^{n+3}+4^{n+2}-4^{n+1}-4^n\)
\(4^{n+3}+4^{n+2}-4^{n+1}-4^n\\ =4^n.4^3+4^n.4^2-4^n.4-4^n\\ =4^n\left(4^3-4^2-4-1\right)\\ =4^n.43\)
|x| + |x+2| = 0
\(\left|x\right|>=0\forall x\)
\(\left|x+2\right|>=0\forall x\)
Do đó: \(\left|x\right|+\left|x+2\right|>=0\forall x\)
Dấu '=' xảy ra khi x=0 và x+2=0
=>x=0 và x=-2
=>\(x\in\varnothing\)
-4/5.x-(0,25-x)=-13/3
`# \text {nKaiz}`
`-4/5*x - (0,25 - x) = -13/3`
`=> -4/5x - 0,25 + x = -13/3`
`=> -4/5x + x = -13/3 + 0,25`
`=> 1/5x = -49/12`
`=> x = -49/12 \div 1/5`
`=> x = -245/12`
Vậy, `x = -245/12.`
tính giúp mik vs ạ
tính hợp lí:
1/12345 - 3 3/4 + 0,125 + 3 3/4 - 1/8 - 1/12345
\(\dfrac{1}{12345}-3\dfrac{3}{4}+0,125+3\dfrac{3}{4}-\dfrac{1}{8}-\dfrac{1}{12345}\)
\(=\left(\dfrac{1}{12345}-\dfrac{1}{12345}\right)+\left(-3\dfrac{3}{4}+3\dfrac{3}{4}\right)+\left(0,125-\dfrac{1}{8}\right)\)
\(=0+0+\left(\dfrac{1}{8}-\dfrac{1}{8}\right)\)
\(=0\)
#Toru
B\(=-1-\dfrac{1}{3}-\dfrac{1}{6}-\dfrac{1}{10}-\dfrac{1}{15}-...-\dfrac{1}{1225}\)
\(B=-\left(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{1225}\right)\)
\(\dfrac{1}{2}B=-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+...+\dfrac{1}{2450}\right)\)
\(\dfrac{1}{2}B=-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{2.4}+\dfrac{1}{4.5}+...+\dfrac{1}{49.50}\right)\)
\(\dfrac{1}{2}B=-\left(1-\dfrac{1}{50}\right)\)
\(\dfrac{1}{2}B=-1+\dfrac{1}{50}\)
\(\dfrac{1}{2}B=\dfrac{-49}{50}\)
\(B=\dfrac{-49}{25}\)
\(B=-\dfrac{2}{2}-\dfrac{2}{6}-\dfrac{2}{12}-...-\dfrac{2}{2450}\)
\(=-2\left(\dfrac{1}{2}+\dfrac{1}{6}+...+\dfrac{1}{2450}\right)\)
\(=-2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
=-2*49/50
=-49/25
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2022}\right)\)
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)....\left(1-\dfrac{1}{2022}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}\cdot...\cdot\dfrac{2021}{2022}\)
\(=\dfrac{1\cdot2\cdot3\cdot...\cdot2021}{2\cdot3\cdot4\cdot...\cdot2022}\)
\(=\dfrac{1}{2022}\)
Tính giá trị biểu thức
\(\dfrac{5}{27}+\dfrac{7}{17}+0,25-\dfrac{5}{27}+\dfrac{7}{17}\)
=5/27-5/27+7/17+7/17+1/4
=14/17+1/4
=56/68+17/68
=73/68
B = 3/2 - 2/21 - { 7/12 - [ 15/21 - ( 1/3 - 5/4 ) - ( 2/7 + 1/3 )]}
Giải nhanh giúp mình với ạ mình đang cần gấp
\(=\dfrac{3}{2}-\dfrac{2}{21}-\dfrac{7}{12}+\left[\dfrac{15}{21}-\dfrac{1}{3}+\dfrac{5}{4}-\dfrac{2}{7}-\dfrac{1}{3}\right]\)
=11/12-2/21+5/7-2/3+5/4-2/7
=11/12-2/3+5/4-2/21+3/7
=11/12-8/12+15/12-2/21+9/21
=18/12+7/21
=3/2+1/3
=9/6+2/6=11/6
12\(\dfrac{3}{5}\) : \(\dfrac{-5}{7}\) + 2\(\dfrac{2}{5}\) : \(\dfrac{-5}{7}\)
\(=\left(12+\dfrac{3}{5}\right)\cdot\dfrac{-7}{5}+\left(2+\dfrac{2}{5}\right)\cdot\dfrac{-7}{5}\)
=-7/5(12+3/5+2+2/5)
=-15*7/5=-21
`= (12 3/5 + 2 2/5) : -5/7`
`= 15 : -5/7`
`= -21`
10:
\(=\dfrac{1}{5}\left(\dfrac{5}{4\cdot9}+\dfrac{5}{9\cdot14}+...+\dfrac{5}{49\cdot54}\right)\cdot\dfrac{108}{1-\left(3+5+...+51\right)}\)
\(=\dfrac{1}{5}\cdot\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+...+\dfrac{1}{49}-\dfrac{1}{54}\right)\cdot\dfrac{108}{1-25\cdot\dfrac{\left(51+3\right)}{2}}\)
\(=\dfrac{1}{5}\cdot\dfrac{50}{216}\cdot\dfrac{108}{1-25\cdot27}\)
\(=\dfrac{10}{2}\cdot\dfrac{1}{1-675}=-\dfrac{5}{674}\)