M=\(\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{6}\right)x\left(1-\frac{1}{10}\right)x\left(1-\frac{1}{15}\right)x\left(1-\frac{1}{21}\right)x\left(1-\frac{1}{28}\right)\)
Những câu hỏi liên quan
1.Tìm x :
a,frac{1}{5.6}+frac{1}{6.7}+...+frac{1}{xleft(x+1right)}frac{13}{90}
b,frac{1}{1.4}+frac{1}{4.7}+frac{1}{7.10}+...+frac{1}{xleft(x+3right)}frac{49}{148}
c,frac{7}{left(x+3right)left(x+10right)}+frac{11}{left(x+10right)left(x+21right)}+frac{1}{left(x+21right)left(x+34right)}frac{x}{left(x+3right)left(x+34right)}
d,frac{3}{left(x-4right)left(x-7right)}+frac{6}{left(x-7right)left(x-13right)}+frac{15}{left(x-13right)left(x-28right)}-frac{1}{x-38}frac{-1}{20}
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1.Tìm x :
a,\(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{13}{90}\)
b,\(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{49}{148}\)
c,\(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}\)\(+\frac{1}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
d,\(\frac{3}{\left(x-4\right)\left(x-7\right)}+\frac{6}{\left(x-7\right)\left(x-13\right)}\)\(+\frac{15}{\left(x-13\right)\left(x-28\right)}\)\(-\frac{1}{x-38}=\frac{-1}{20}\)
a, \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{x\left(x+1\right)}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{5}-\frac{1}{x+1}=\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{5}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{18}{90}-\frac{13}{90}\)
⇒ \(\frac{1}{x+1}=\frac{1}{18}\)
⇒ x + 1 = 18
⇒ x = 17
Vậy x = 17
b, \(\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{x\left(x+3\right)}=\frac{49}{148}\)
⇒ \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}=\frac{49.3}{148}\)
⇒ \(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(1-\frac{1}{x+3}=\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=1-\frac{147}{148}\)
⇒ \(\frac{1}{x+3}=\frac{1}{148}\)
⇒ x + 3 = 148
⇒ x = 145
Vậy x = 145
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Mình cần gấp bài này !!!
1/Tìm x, biết :
a/ \(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}+\frac{x}{\left(x+3\right)\left(x+34\right)}\)
b/ \(\frac{3}{\left(x-4\right)\left(x-7\right)}+\frac{6}{\left(x-7\right)\left(x-13\right)}+\frac{15}{\left(x-13\right)\left(x-28\right)}-\frac{1}{x-28}=\frac{-1}{20}\)
tìm x,biết:a)frac{2}{left(x+2right)left(x+4right)}+frac{4}{left(x+4right)left(x+8right)}+frac{6}{left(x+8right)left(x+14right)}frac{x}{left(x+2right)left(x+14right)}b)frac{x+1}{10}+frac{x+1}{11}+frac{x+1}{12}frac{x+1}{13}+frac{x+1}{14}c)left(x+2right)^2frac{38}{25}+frac{9}{10}-frac{11}{15}+frac{13}{21}-frac{15}{28}+frac{17}{36}-...+frac{197}{4851}-frac{199}{4950}giúp tớ với,huhu
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tìm x,biết:
a)\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
b)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
c)\(\left(x+2\right)^2=\frac{38}{25}+\frac{9}{10}-\frac{11}{15}+\frac{13}{21}-\frac{15}{28}+\frac{17}{36}-...+\frac{197}{4851}-\frac{199}{4950}\)
giúp tớ với,huhu
Tìm x: \(\frac{3}{\left(x-4\right)\left(x-7\right)}+\frac{6}{\left(x-7\right)\left(x-13\right)}+\frac{15}{\left(x-13\right)\left(x-28\right)}+\frac{1}{x-28}=\frac{-1}{20}\)
Tìm x
Xem chi tiết
a,\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|=4x\)
b,\(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
a) Dễ thấy VT > 0;mà VT=VP
=>VP > 0 => 4x > 0=> x > 0
=>\(\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{3}\right|=x+\frac{1}{3};\left|x+\frac{1}{6}\right|=x+\frac{1}{6}\)
=>BT đầu tương đương \(\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{3}\right)+\left(x+\frac{1}{6}\right)=4x\)
\(=>3x+1=4x=>x=1\)
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a) Để đẳng thức xảy ra thì: x>0 (vì: \(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|>0\) )
Khi đó: \(\left|x+\frac{1}{2}\right|=x+\frac{1}{2};\left|x+\frac{1}{3}\right|=x+\frac{1}{3};\left|x+\frac{1}{6}\right|=x+\frac{1}{6}\)
=>\(x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{6}=4x\)
<=>x=1
Vậy x=1
b)Điều kiện: \(x\ne-3;-10;-21;-34\)
\(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
<=>\(\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
<=>\(\frac{1}{x+3}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
=>x+34-x-3=x
<=>x=31 (nhận)
Vậy x=31
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a,\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|=4x\)
Ta có: \(\begin{cases}\left|x+\frac{1}{2}\right|\ge0\\\left|x+\frac{1}{3}\right|\ge0\\\left|x+\frac{1}{6}\right|\ge0\end{cases}\)
\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|\ge0\)
\(\Rightarrow4x\ge0\)
\(\Rightarrow\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{3}\right|+\left|x+\frac{1}{6}\right|=x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{6}\)
Khi đó, ta có: \(x+\frac{1}{2}+x+\frac{1}{3}+x+\frac{1}{6}=4x\)
\(\Rightarrow3x+1=4x\)
\(\Rightarrow x=1\)
b) Từ đề suy ra:
\(\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
\(\Rightarrow\frac{1}{x+3}-\frac{1}{x+34}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
\(\Rightarrow\frac{x+34}{\left(x+3\right)\left(x+34\right)}-\frac{x+3}{\left(x+3\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
\(\Rightarrow\frac{31}{\left(x+3\right)\left(x+34\right)}=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
\(\Rightarrow x=31\)
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Xem thêm câu trả lời
Câu 1 : Thực hiện các phép tính sau :
\(\left(\frac{1}{6}-1\right)\left(\frac{1}{10}-1\right)\left(\frac{1}{15}-1\right)\left(\frac{1}{21}-1\right)\left(\frac{1}{28}-1\right)\)
Câu 2 : Tìm x biết
\(\frac{\left|x\right|}{x-7}\le0\)
Tính:
a) \(A=\left(1-\frac{1}{15}\right)x\left(1-\frac{1}{21}\right)x\left(1-\frac{1}{28}\right)x.........x\left(1-\frac{1}{1275}\right)\)
b) \(B=\left(1+\frac{1}{1x3}\right)x\left(1+\frac{1}{2x4}\right)x\left(1+\frac{1}{3x5}\right)x............x\left(1+\frac{1}{99x101}\right)\)
\(A=\left(1-\frac{1}{15}\right).\left(1-\frac{1}{21}\right).\left(1-\frac{1}{28}\right)......\left(1-\frac{1}{1275}\right)\)
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\(\frac{3}{\left(x-4\right)\left(x-7\right)}+\frac{6}{\left(2-7\right)\left(x-13\right)}+\frac{15}{\left(x-13\right)\left(x-28\right)}-\frac{1}{x-28}=-\frac{1}{20}\)
\(\frac{3}{\left(x-4\right)\left(x-7\right)}+\frac{6}{\left(2-7\right)\left(x-13\right)}+\frac{15}{\left(x-13\right)\left(x-28\right)}-\frac{1}{x-28}=-\frac{1}{20}\)
ghi sai để nên mik ko giải dc