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Tổng của tử và mẫu ứng với số phần là: 23+77=100p
Tăng cả tử cả mẫu cùg 1 số thì tổng số phần ko đổi
Lúc sau thì tử là 23p, mẫu là 77p
=)Để tử là 23p=) tử là:(63452+36548):100.23=23000
=)số tự nhiên cần tìm:
63452-23000=40452
Cho lun cái k nha
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)
1) x/5 = 32/80
\(x : 5 = 32 : 80\) \(x = 5 \cdot \frac{32}{80}\) \(x = 5 \cdot \frac{2}{5}\) \(x = 2\)
2) 13/x = 26/30
\(13 : x = 26 : 30\) \(13 \cdot 30 = x \cdot 26\) \(390 = 26 x\) \(x = 390 : 26\) \(x = 15\)
3) -x/7 = 22/(-77)
\(- x : 7 = 22 : \left(\right. - 77 \left.\right)\) \(- x = 7 \cdot \frac{22}{- 77}\) \(- x = 7 \cdot \left(\right. - \frac{2}{7} \left.\right)\) \(- x = - 2\) \(x = 2\)
b)
1) x/9 = 28/36
\(x : 9 = 28 : 36\) \(x = 9 \cdot \frac{28}{36}\) \(x = 9 \cdot \frac{7}{9}\) \(x = 7\)
2) -10/x = 50/55
\(- 10 : x = 50 : 55\) \(- 10 \cdot 55 = 50 x\) \(- 550 = 50 x\) \(x = - 550 : 50\) \(x = - 11\)
3) 23/(-x) = 46/50
\(23 : \left(\right. - x \left.\right) = 46 : 50\) \(23 \cdot 50 = 46 \left(\right. - x \left.\right)\) \(1150 = - 46 x\) \(x = 1150 : \left(\right. - 46 \left.\right)\) \(x = - 25\)
✅ Kết quả cuối cùng
a)
- x = 2
- x = 15
- x = 2
b)
- x = 7
- x = -11
- x = -25
Nhân chéo nha :
\(\left(23+x\right).4=\left(40+x\right).3\)
Câu b cũng vậy
\(A=\frac{8}{9}\cdot\frac{15}{16}\cdot\frac{24}{25}\cdot...\cdot\frac{360}{361}\cdot\frac{399}{400}\)
\(A=\frac{2\cdot4\cdot3\cdot5\cdot4\cdot6\cdot...\cdot18\cdot20\cdot19\cdot21}{3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot...\cdot19\cdot19\cdot20\cdot20}\)
\(A=\frac{2\cdot21}{3\cdot20}\)
\(A=\frac{7}{10}\)
\(B=\frac{9}{8}\cdot\frac{16}{15}\cdot\frac{25}{24}\cdot...\cdot\frac{441}{440}\cdot\frac{484}{483}\)
\(B=\frac{3\cdot3\cdot4\cdot4\cdot5\cdot5\cdot...\cdot21\cdot21\cdot22\cdot22}{2\cdot4\cdot3\cdot5\cdot4\cdot6\cdot...\cdot20\cdot22\cdot21\cdot23}\)
\(B=\frac{3\cdot22}{2\cdot23}=\frac{33}{23}\)
\(C=\frac{17}{23}.\left(\frac{7}{61}+\frac{28}{61}+\frac{26}{61}\right)\)
\(C=\frac{17}{23}\cdot1=\frac{17}{23}\)
\(\frac{1}{20}+\frac{1}{44}+\frac{1}{77}+...+\frac{2}{x\left(x+3\right)}=\frac{101}{770}\)
\(\Rightarrow\)\(\frac{3}{2}.\left(\frac{1}{20}+\frac{1}{44}+\frac{1}{77}+...+\frac{2}{x\left(x+3\right)}\right)=\frac{101}{770}\).
\(\Rightarrow\)\(\frac{3}{40}+\frac{3}{88}+\frac{3}{154}+...+\frac{3}{x\left(x-3\right)}=\frac{303}{1540}\)
\(\Rightarrow\)\(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+...+\frac{3}{x\left(x+3\right)}=\frac{303}{1540}\)
\(\Rightarrow\)\(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+...+\frac{1}{x-1}\)
\(\Rightarrow\)\(\frac{1}{5}-\frac{1}{x+3}=\frac{303}{1540}\)
\(\Rightarrow\)\(\frac{1}{x+3}=\frac{1}{5}-\frac{303}{1540}\)
\(\Rightarrow\)\(\frac{1}{x+3}=\frac{308}{1540}-\frac{303}{1540}\)
\(\Rightarrow\)\(\frac{1}{x+3}=\frac{5}{1540}\)
\(\Rightarrow\)\(\frac{1}{x+3}=\)\(\frac{1}{308}\)
\(\Rightarrow\)\(x+3=308\)
\(\Rightarrow\)\(x=308-3\)
\(\Rightarrow\)\(x=305\)
a: \(\Leftrightarrow-\dfrac{9}{46}+\dfrac{108}{46}-\dfrac{93}{23}:\left(\dfrac{13}{4}-\dfrac{5}{3}x\right)=1\)
\(\Leftrightarrow\dfrac{93}{23}:\left(\dfrac{13}{4}-\dfrac{5}{3}x\right)=\dfrac{53}{46}\)
\(\Leftrightarrow-\dfrac{5}{3}x+\dfrac{13}{4}=\dfrac{186}{53}\)
=>-5/3x=55/212
hay x=-33/212
c: \(\Leftrightarrow\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{x\left(x+3\right)}=\dfrac{18}{19}\)
\(\Leftrightarrow1-\dfrac{1}{x+3}=\dfrac{18}{19}\)
=>x+3=19
hay x=16
\(\frac{63452+x}{36548+x}=\frac{23}{77}\)
\(\Rightarrow77\left(63452+x\right)=23\left(36548+x\right)\)
\(4885804+77x=840604+23x\)
\(77x-23x=840604-4885804\)
\(54x=-4045200\)
\(x=-4045200:54\)
\(x=-74911,11111\)
Vậy.....................................................
Ta có \(\frac{63452+x}{36528+x}=\frac{23}{77}\Leftrightarrow\frac{36528+x+26924}{36528+x}=\frac{23}{77}\)
\(\Leftrightarrow1+\frac{26924}{36528+x}=\frac{23}{77}\Leftrightarrow\frac{26924}{36528+x}=-\frac{54}{77}\)