tìm x,biết\(\frac{x}{52}=\frac{-14}{72}\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)\(x=\dfrac{-14\cdot52}{72}\\ x=\dfrac{-91}{9}\)
b)\(x=\dfrac{120\cdot7.2}{70}\\ x=\dfrac{432}{35}\)
c)\(x=\dfrac{2\dfrac{2}{3}\cdot8.5}{5}\\ x=\dfrac{68}{15}\)
d)\(x=\dfrac{4\dfrac{2}{5}\cdot9.5}{8}\\ x=\dfrac{209}{40}\)
\(\frac{x-2}{12}+\frac{x-2}{20}+\frac{x-2}{30}+\frac{x-2}{42}+\frac{x-2}{56}+\frac{x-2}{72}=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\left(\frac{3}{9}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\cdot\frac{2}{9}=\frac{16}{9}\)
\(x-2=\frac{16}{9}:\frac{2}{9}\)
\(x-2=\frac{16}{9}\cdot\frac{9}{2}\)
\(x-2=8\)
\(x=8+2\)
\(x=10\)
Vậy \(x=10\)
\(\left(x-2\right)\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=\)\(=\frac{16}{9}\)
\(\left(x-2\right)\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}+\frac{1}{8}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(\left(x-2\right)\left(\frac{2}{9}\right)=\frac{16}{9}\)
2(x-2)=16
x-2=8
x=10
\(\frac{72-x}{3}=\frac{x-18}{5}\)
\(\Rightarrow\frac{\left(72-x\right).5}{15}=\frac{3\left(x-18\right)}{15}\)
\(\Rightarrow\left(72-x\right).5=3\left(x-18\right)\)
\(\Rightarrow360-5x=3x-54\)
\(\Rightarrow-3x-5x=-360-54\)
\(\Rightarrow-8x=-414\)
\(\Rightarrow x=51,75\)
+) Vì \(\frac{72-x}{3}=\frac{x-18}{5}\)
\(\Rightarrow5\left(72-x\right)=3\left(x-18\right)\)
\(5.72-5x=3x-3.18\)
\(360-5x=3x-54\)
\(-5x-3x=-54-360\)
\(-8x=-414\)
\(x=-414:\left(-8\right)\)
\(x=51,75\)
Vậy x = 51,75
=> x-2.(1/12+1/20+1/30+1/42+1/56+1/72)=16/9
Đặt : Sáng = 1/12+1/20+1/30+1/42+1/56+1/72
=> Sáng = 1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9
=> Sáng = 1.(1/3-1/4+1/4-1/5+...+1/8-1/9
=> Sáng = 91.(1/3-1/9)
=> Sáng = 2/9
Thay Sáng vô biểu thức 1/12+1/20+1/30+1/42+1/56+1/72
Ta được :
x-2.2/9=16/9
giờ thì tự làm nha
=> x-2.(1/12+1/20+1/30+1/42+1/56+1/72)=16/9
Đặt : Sáng = 1/12+1/20+1/30+1/42+1/56+1/72
=> Sáng = 1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9
=> Sáng = 1.(1/3-1/4+1/4-1/5+...+1/8-1/9
=> Sáng = 91.(1/3-1/9)
=> Sáng = 2/9
Thay Sáng vô biểu thức 1/12+1/20+1/30+1/42+1/56+1/72
Ta được :
x-2.2/9=16/9
giờ thì tự làm nha
Ai k mk mk k lại
ta có:$\frac{x-1}{12}+\frac{x-1}{20}+\frac{x-1}{30}+\frac{x-1}{42}+\frac{x-1}{56}+\frac{x-1}{72}=\frac{16}{9}$
=> x+1(1/12+1/20+1/30+1/42+1/56+1/72)=16/9
=> x+1.2/9=16/9
=> x+1 = (16/9):(2/9)
=> x+1 = 8
=> x = 9
thông cảm mình ko đánh được dấu ngoặc tròn
[x-1].[1/12+1/20+1/30+1/42+1/56+1/72] =16/9
[x-1].[1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9]=16/9
[x-1].[1/3-1/9]=16/9
[x-1].2/9=16/9
x-1=16/9:2/9
x-1=8
x=7
Vậy x=7
<=> (x-18/74 - 1)+(x-20/72 - 1)+(x-22/70 - 1) = 0
<=> x-92/74 + x-92/72 + x-92/70 = 0
<=> (x-92).(1/74+1/72+1/70) = 0
<=> x-92 = 0 ( vì 1/74 + 1/72 + 1/70 > 0 )
<=> x=92
Vậy S = {92}
Tk mk nha
Ta có :
\(\frac{x-18}{74}+\frac{x-20}{72}+\frac{x-22}{70}=3\)
\(\Leftrightarrow\)\(\left(\frac{x-18}{74}-1\right)+\left(\frac{x-20}{72}-1\right)+\left(\frac{x-22}{70}-1\right)=3-3\)
\(\Leftrightarrow\)\(\frac{x-92}{74}+\frac{x-92}{72}+\frac{x-92}{70}=0\)
\(\Leftrightarrow\)\(\left(x-92\right)\left(\frac{1}{74}+\frac{1}{72}+\frac{1}{70}\right)=0\)
Vì \(\left(\frac{1}{74}+\frac{1}{72}+\frac{1}{70}\right)\ne0\)
\(\Rightarrow\)\(x-92=0\)
\(\Rightarrow\)\(x=92\)
Vậy \(x=92\)
Chúc bạn học tốt
-91/9 nha bạn.
thank you