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(x+1)+(x+3)+...+(x+99)=0
Tổng các số hạng là: (99+1):2=50 (số hạng)
=> (x+1)+(x+3)+...+(x+99)=0 <=> 50.x+(1+3+5+...+99) = 0
<=> 50.x+=0 <=> 50.x+2500=0 => x=-2500/50=-50
1.
\(\left(\frac{3}{1\times3}+\frac{3}{3\times5}+\frac{3}{5\times7}+...+\frac{3}{97\times99}\right)-x:\frac{3}{2}=\frac{7}{3}\\
\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{97\times99}\right):\frac{3}{2}-x:\frac{3}{2}=\frac{7}{3}\\\left[\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\right)-x\right]:\frac{3}{2}=\frac{7}{3}\\
\left(1-\frac{1}{99}\right)-x=\frac{7}{3}\times\frac{3}{2}\\
\frac{98}{99}-x=\frac{7}{2}\\
x=\frac{98}{99}-\frac{7}{2}=\frac{-497}{198}\)
2.\(\frac{x}{y}=\frac{4}{3}\Rightarrow\hept{\begin{cases}x=4a\\y=3a\\x-y=4a-3a=a\end{cases}}\\ \left(x-y\right)^{2015}=5^{2015}\Rightarrow x-y=5\\ \Rightarrow a=5\Rightarrow\hept{\begin{cases}x=4\times5=20\\y=3\times5=15\end{cases}}\)
1.
c. \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)
2.
a. \(45-5\left(y+1\right)=10\)
\(\Rightarrow5\left(y+1\right)=35\)
\(\Rightarrow y+1=7\)
\(\Rightarrow y=6\)
b. \(y:2+y:2=15\)
\(\Rightarrow\frac{1}{2}y+\frac{1}{2}y=15\)
\(\Rightarrow y=15\)
Bài 1 :
\(a,12,5\times32\times8\)
\(=\left(12,5\times8\right)\times32\)
\(=100\times32\)
\(=3200\)
\(b,20,9+20,9\times99\)
\(=20,9\times\left(1+99\right)\)
\(=20,9\times100\)
\(=2090\)
\(c,\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{50}{50}-\frac{1}{50}\)
\(=\frac{49}{50}\)
Bài 2 :
\(a,45-5\times\left(y+1\right)=10\)
\(5\times\left(y+1\right)=45-10\)
\(5\times\left(y+1\right)=35\)
\(y+1=35\div5\)
\(y+1=7\)
\(y=7-1\)
\(y=6\)
\(b,y\div2+y\div2=15\)
\(y\times\frac{1}{2}+y\times\frac{1}{2}=15\)
\(2\times\left(y\times\frac{1}{2}\right)=15\)
\(y=15\)
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