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a) \(\dfrac{7}{13}\)\(\times\)\(\dfrac{7}{15}\)-\(\dfrac{5}{12}\)\(\times\)\(\dfrac{21}{39}+\dfrac{49}{91}\)\(\times\)\(\dfrac{8}{15}\)
= \(\dfrac{7}{13}\)\(\times\)\(\dfrac{7}{15}\)-\(\dfrac{5}{12}\times\dfrac{7}{13}+\dfrac{7}{13}\times\dfrac{8}{15}\)
= \(\dfrac{7}{13}\left(\dfrac{7}{15}-\dfrac{5}{12}+\dfrac{8}{15}\right)\)
= \(\dfrac{7}{13}\times\dfrac{7}{12}\)
= \(\dfrac{49}{156}\)
b) \(\left(\dfrac{12}{199}+\dfrac{23}{200}-\dfrac{34}{201}\right)\times\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
= \(\left(\dfrac{12}{199}+\dfrac{23}{200}-\dfrac{34}{201}\right)\times0\)
= 0

\(5\left|x\right|=7-\left(-8\right)\)
<=> 5|x|=15
<=> |x|=3
<=> \(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
*) -2|x-3|=-10
<=> |x-3|=5
<=> \(\orbr{\begin{cases}x-3=5\\x-3=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-2\end{cases}}}\)
*) |x-9|=|-91|
<=> |x-9|=91
<=> \(\orbr{\begin{cases}x-9=91\\x-9=-91\end{cases}\Leftrightarrow\orbr{\begin{cases}x=100\\x=-82\end{cases}}}\)
5|x|=7-(-8)
5|x|=7+8
5|x|=15
|x|=15:5
|x|=3
=> x=+3
vậy x=+3
-2|x-3|=-10
|x-3|=-10:(-2)
|x-3|=5
\(\Rightarrow\orbr{\begin{cases}x-3=5\\x-3=-5\end{cases}\Rightarrow\orbr{\begin{cases}x=8\\x=-2.\end{cases}}}\)
Vậy x=8 hoặc x=-2
|x-9|=|-91|
|x-9|=91
\(\Rightarrow\orbr{\begin{cases}x-9=91\\x-9=-91\end{cases}}\Rightarrow\orbr{\begin{cases}x=100\\x=-82\end{cases}}\)
Vậy x=100 hoặc x=-82

\(\frac{x}{15}=\frac{3}{5}+\frac{-2}{3}\)
\(\frac{x}{15}=\frac{-1}{15}\)
=> \(x=-1\)
\(\frac{x}{182}=\frac{-6}{14}\cdot\frac{35}{91}\)
\(\frac{x}{182}=\frac{-15}{91}\)
=> \(91x=182\cdot\left(-15\right)\)
=> \(91x=-2730\)
=> \(x=-30\)
g, \(\frac{x}{15}=\frac{3}{5}+\frac{-2}{3}\Leftrightarrow\frac{x}{15}=\frac{3}{5}-\frac{2}{3}\Leftrightarrow\frac{x}{15}=-\frac{1}{15}\)
\(\Leftrightarrow x=-1\)
h, \(\frac{x}{182}=\frac{-6}{14}.\frac{35}{91}\Leftrightarrow\frac{x}{182}=-\frac{15}{91}\Leftrightarrow\frac{x}{182}=\frac{-30}{182}\)
\(\Leftrightarrow x=-30\)

P/s: Bài này chỉ tính được giá trị "gần đúng" của biểu thức thôi nhé!
\(\frac{21}{54}+\frac{3}{75}:\frac{\left(\frac{39}{65}+0,415-\frac{33}{600}\right)\frac{21}{91}}{7^2-18,25+13\frac{15}{36}-16\frac{17}{102}}\)
\(\Leftrightarrow\frac{21}{54}+\frac{3}{75}:\frac{\left(0,6+0,415-0,055\right)0,23}{49-18,25+\frac{483}{36}-\frac{1649}{102}}\)
\(\Leftrightarrow\frac{21}{54}+\frac{3}{75}:\frac{\left(0,6+0,415-0,055\right)0,23}{49-18,25+13,41-16,1}\)
\(\Leftrightarrow\frac{21}{54}+\frac{3}{75}:\frac{0,96.0,23}{28,06}\Leftrightarrow\frac{21}{54}+\frac{3}{75}:\frac{0,2208}{28,06}\)
\(\Leftrightarrow\frac{21}{54}+\left(\frac{3}{75}:\frac{0,2208}{28,06}\right)\Leftrightarrow\frac{21}{54}+\left(\frac{3}{75}.\frac{28,06}{0,2208}\right)=\frac{21}{54}+\frac{61}{12}=\frac{197}{36}\)
P/s: Giải bài này mà mệt cả đầu =((

Bài 1 :
a, \(2^x+2^{x+1}=24\)
\(\Rightarrow2^x.1+2^x.2=24\)
\(\Rightarrow2^x\left(1+2\right)=24\)
\(\Rightarrow2^x=24\div3\)
\(\Rightarrow2^x=8=2^3\)
Vậy : x = 3
b, \(x^2-x=0\)
\(\Rightarrow x.x-x.1=0\)
\(\Rightarrow x\left(x-1\right)=0\)
Để : \(x\left(x-1\right)=0\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
Vậy x = 1
Bài 2 :
a, \(Q=3+3^3+3^5+...+3^{101}\)
\(\Rightarrow9Q=3^3+3^5+3^7+...+3^{103}\)
\(\Rightarrow9Q-Q=\left(3^3+3^5+3^7+...+3^{103}\right)-\left(3+3^3+3^5+...+3^{101}\right)\)
\(\Rightarrow8Q=3^{103}-3\)
\(\Rightarrow Q=\frac{3^{103}-3}{8}\)
b, \(Q=3+3^3+3^5+...+3^{101}\)
\(\Rightarrow Q=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+...+\left(3^{97}+3^{99}+3^{101}\right)\)
\(\Rightarrow Q=\left(3+3^3+3^5\right)+3^6\left(3+3^3+3^5\right)+...+3^{96}\left(3+3^3+3^5\right)\)
\(\Rightarrow Q=1.273+3^6.273+...+3^{96}.273\)
\(\Rightarrow Q=\left(1+3^6+...+3^{96}\right)273\)
Vì : \(1+3^6+...+3^{96}\in N\) ; \(273=3.91\Rightarrow Q⋮91\)
Vậy ...
Phương An
soyeon_Tiểubàng giải
Võ Đông Anh Tuấn
Nguyễn Huy Tú
Trương Hồng Hạnh
Nguyễn Đình Dũng
Nguyễn Huy Thắng
Trần Quỳnh Mai
Nguyễn Thanh Vân
Nguyễn Thị Thu An
Hoàng Lê Bảo Ngọc
Silver bullet
Nguyễn Anh Duy
Lê Nguyên Hạo
Nguyễn Phương HÀ
3/x=3/7
x=3*7/3
x=7
\(\frac{3}{x}\) = \(\frac{39}{91}\)
\(\frac{3}{x}=\frac37\)
\(x=7\)
Vậy \(x=7\)