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\(A=\frac{1}{1\cdot2}+\frac{2}{2\cdot4}+\frac{3}{4\cdot7}+\frac{4}{7\cdot11}+...+\frac{10}{46\cdot56}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+...+\frac{1}{46}-\frac{1}{56}\)
\(A=1-\frac{1}{56}\)
\(A=\frac{55}{56}\)
\(B=\frac{4}{3\cdot7}+\frac{4}{7\cdot11}+\frac{4}{11\cdot15}+...+\frac{4}{23\cdot27}\)
\(B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+...+\frac{1}{23}-\frac{1}{27}\)
\(B=\frac{1}{3}-\frac{1}{27}\)
\(B=\frac{8}{27}\)
\(C=\frac{4}{3\cdot6}+\frac{4}{6\cdot9}+\frac{4}{9\cdot12}+...+\frac{4}{99\cdot102}\)
\(C=\frac{4}{3}\left(\frac{3}{3\cdot6}+\frac{3}{6\cdot9}+\frac{3}{9\cdot12}+...+\frac{3}{99\cdot102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{99}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\left(\frac{1}{3}-\frac{1}{102}\right)\)
\(C=\frac{4}{3}\cdot\frac{33}{102}\)
\(C=\frac{22}{51}\)
\(\left(2+2x\right)\cdot3\dfrac{5}{6}=46\)
\(\left(2+2x\right)\cdot\dfrac{23}{6}=46\)
\(2+2x=46:\dfrac{23}{6}\)
\(2+2x=46\cdot\dfrac{6}{23}\)
\(2+2x=12\)
\(2x=12-2\)
\(2x=10\)
\(x=10:2\)
\(x=5\)
Vậy \(x=5\).
(2+2x) . 3 5/6 = 46
=> (2+2x) . 23 = 46
=> 2 + 2x = 46 : 23
=> 2 + 2x = 2
=> 2x = 0
=>x = 0
Gọi số ghe chở 4 người là x (ghe) (ĐK: x < 10)
Khi đó số ghe chở 6 người là 10 - x (ghe)
Vì có 46 khách tham quan nên ta có pt: 4x + 6(10 - x) = 46
<=> 4x + 60 - 6x = 46
<=> -2x = -14
=> x = 7 (TM)
Vậy có 7 ghe chở 4 người và 10 - 7 = 3 (ghe) chở 6 người
`Answer:`
\(\frac{4}{20}+\frac{16}{42}+\frac{6}{15}-\left(-\frac{3}{21}\right)+\left(-\frac{10}{21}\right)+\frac{3}{20}\)
\(=\left(\frac{4}{20}+\frac{3}{20}+\frac{6}{15}\right)+\left(\frac{8}{21}+\frac{3}{21}-\frac{10}{21}\right)\)
\(=\frac{3}{4}+\frac{1}{21}\)
\(=\frac{67}{84}\)
\(\frac{42}{46}+\frac{250}{186}+\left(-\frac{2121}{2323}\right)+\left(-\frac{125125}{143143}\right)\)
\(=\frac{21}{23}+\frac{125}{93}+\left(-\frac{21}{23}\right)+\left(-\frac{125}{43}\right)\)
\(=\left(\frac{21}{23}+-\frac{21}{23}\right)+\left(\frac{125}{93}+-\frac{125}{43}\right)\)
\(=\frac{6250}{13299}\)
a)4^x=64
4^x=4^3
<=>x=3
b)3^x-1=27
3^x-1=3^3
<=>x-1=3
x=3+1
x=4
c)26+8x-6x=46
26+x(8-6)=46
26+2x=46
2x=46-26
2x=20
x=20:2
x=10
d)\(25\le5^x\le625\)
<=>\(5^2\le5^x\le5^4\)
<=>\(2\le x\le4\)
<=>\(x\in\left\{2;3;4\right\}\)
e)10+2x=4^5:4^3
10+2x=4^5-3
10+2x=4^2
10+2x=16
2x=16-10
2x=6
x=6:2
x=3
a) 4x = 64
\(\Rightarrow\)4x = 43
\(\Rightarrow\)x = 3
b) 3x - 1 = 27
\(\Rightarrow\)3x - 1 = 33
\(\Rightarrow\)x - 1 = 3
\(\Rightarrow\)x = 4
c) 26 + 8x - 6x = 46
\(\Rightarrow\)26 + 2x = 46
\(\Rightarrow\)2x = 46 - 26
\(\Rightarrow\)2x = 20
\(\Rightarrow\)x = 10
d) 25 \(\le\)5x \(\le\)625
\(\Rightarrow\)52\(\le\)5x\(\le\)54
\(\Rightarrow\)x \(\in\){ 2 ; 3 ; 4 }
e) 10 + 2x = 45 : 43
\(\Rightarrow\)10 + 2x = 42 = 16
\(\Rightarrow\)2x = 16 - 10
\(\Rightarrow\)2x = 6
\(\Rightarrow\)x = 3
a) \(\left(x+\frac{1}{4}-\frac{1}{3}\right):\left(2+\frac{1}{6}-\frac{1}{4}\right)=\frac{7}{46}\)
\(\left(x-\frac{1}{12}\right):\left(2-\frac{1}{12}\right)=\frac{7}{46}\)
\(\left(x-\frac{1}{12}\right):\frac{23}{12}=\frac{7}{46}\)
\(x-\frac{1}{12}=\frac{7}{46}.\frac{23}{12}\)
\(x-\frac{1}{12}=\frac{7}{24}\)
\(x=\frac{7}{24}+\frac{1}{12}\)
\(x=\frac{3}{8}\)
Vậy \(x=\frac{3}{8}\)
b) \(\frac{13}{15}-\left(\frac{13}{21}+x\right).\frac{7}{12}=\frac{7}{10}\)
\(\frac{13}{15}-\left(\frac{13}{21}+x\right)=\frac{7}{10}:\frac{7}{12}\)
\(\frac{13}{15}-\left(\frac{13}{21}+x\right)=\frac{7}{10}.\frac{12}{7}\)
\(\frac{13}{15}-\left(\frac{13}{21}+x\right)=\frac{6}{5}\)
\(\frac{13}{21}+x=\frac{13}{15}-\frac{6}{5}\)
\(\frac{13}{21}+x=\frac{-1}{3}\)
\(x=\frac{-1}{3}-\frac{13}{21}\)
\(x=\frac{-20}{21}\)
Vậy \(x=\frac{-20}{21}\)
10.(2x - 4) + 6 = 46
10.(2x - 4) = 46 - 6
10.(2x - 4) = 40
2x - 4 = 40 : 10
2x - 4 = 4
2x = 4 + 4
2x = 8
x = 8 : 2
x = 4
10(2x-4)+6=46
10(2x-4)=46-6
10(2x-4)=40
2x-4=40:10
2x-4=4
2x=4+4
2x=8
x=8:2
x=4
KING OF EASY