tìm a,b biết:
\(\frac{11}{17}<\frac{a}{b}<\frac{23}{29}và8b-9a=31\)
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\(\dfrac{1}{3\times7}+\dfrac{1}{7\times11}+\dfrac{1}{11\times15}+...+\dfrac{1}{a\times\left(a+4\right)}=\dfrac{50}{609}\)
\(\dfrac{1}{4}\times\left(\dfrac{4}{3\times7}+\dfrac{4}{7\times11}+...+\dfrac{4}{a\times\left(a+4\right)}\right)=\dfrac{50}{609}\)
\(\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{a}-\dfrac{1}{a\times4}=\dfrac{50}{609}\div\dfrac{1}{4}\)
\(\dfrac{1}{3}-\dfrac{1}{a\times4}=\dfrac{200}{609}\)
\(\dfrac{1}{a\times4}=\dfrac{1}{3}-\dfrac{200}{609}\)
\(\dfrac{1}{a\times4}=\dfrac{1}{203}\)
\(a\times4=203\)
\(a=\dfrac{203}{4}\)
\(\dfrac{1}{3\times7}\)+\(\dfrac{1}{7\times11}\)+\(\dfrac{1}{11\times15}\)+...+\(\dfrac{1}{a\times\left(a+4\right)}\) = \(\dfrac{50}{609}\)
4\(\times\)( \(\dfrac{1}{3\times7}\) +\(\dfrac{1}{7\times11}\)+\(\dfrac{1}{11\times15}\)+...+\(\dfrac{1}{a\times\left(a+4\right)}\)) = \(\dfrac{50}{609}\) \(\times\)4
\(\dfrac{4}{3\times7}\)+ \(\dfrac{4}{7\times11}\)+\(\dfrac{1}{11\times15}\)+...+\(\dfrac{4}{a\times\left(a+4\right)}\) = \(\dfrac{50}{609}\) \(\times\) 4
\(\dfrac{1}{3}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{11}\) + \(\dfrac{1}{11}\)-\(\dfrac{1}{15}\)+...+\(\dfrac{1}{a}\)-\(\dfrac{1}{a+4}\) = \(\dfrac{200}{609}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{a+4}\) = \(\dfrac{200}{609}\)
\(\dfrac{1}{a+4}\) = \(\dfrac{1}{3}\) - \(\dfrac{200}{609}\)
\(\dfrac{1}{a+4}\) = \(\dfrac{1}{203}\)
a + 4 = 203
\(a\) = 203 - 4
\(a\) = 199
Đáp số: \(a\) = 199
a) Có lẽ đề có vấn đề.
b) \(\frac{x-11}{y-10}=\frac{11}{10}\Rightarrow10\left(x-11\right)=11\left(y-10\right)\)
\(10x-110=11y-110\)
\(10x-11y-110+110=0\)
\(10x-11y=0\)
\(10x-\left(10y+y\right)=0\)
\(10x-10y-y=0\)
\(10\left(x-y\right)-y=0\)
TH1: x-y = -12
10 (-12) -y =0
-120 - y =0
y = -120
Thay y = -120 vào x-y = -12
x - (-120) = -12
x + 120 = -12
x= -12 - 120
x= -132
TH2: x-y = 12
10 * 12 -y = 0
120 - y =0
y = 120
Thay y= 120 vào x-y = 12
x - 120 = 12
x= 12 + 120
x= 132
Vậy nếu y= -120 thì x= -132
nếu y= 120 thì x= 132
Mình không biết cách để giới hạn a và b