2.So sánh \(\dfrac{37}{51}\) và \(\dfrac{43}{49}\)
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Ta có:
\(\dfrac{37}{-49}< 0;\dfrac{-12}{-35}=\dfrac{12}{35}>0\)
\(\Rightarrow\dfrac{37}{-49}< \dfrac{-12}{-35}\)
Vậy...
Lời giải:
\(A=1.3.5.7...99=\frac{1.2.3.4...99.100}{2.4.6.8.100}=\frac{1.2.3...99.100}{(1.2)(2.2)(3.2)...(50.2)}\)
\(=\frac{1.2.3...99.100}{(1.2.3...50).2^{50}}=\frac{51.52...100}{2^{50}}=\frac{51}{2}.\frac{52}{2}....\frac{100}{2}=B\)
\(\frac{43-x}{57}+\frac{46-x}{54}=\frac{49-x}{51}+\frac{52-x}{48}\)
\(\Leftrightarrow\left(\frac{43-x}{57}+1\right)+\left(\frac{46-x}{54}+1\right)=\left(\frac{49-x}{51}+1\right)+\left(\frac{52-x}{48}+1\right)\)
\(\Leftrightarrow\frac{43-x+57}{57}+\frac{46-x+54}{54}=\frac{49-x+51}{51}+\frac{52-x+48}{48}\)
\(\Leftrightarrow\frac{100-x}{57}+\frac{100-x}{54}=\frac{100-x}{51}+\frac{100-x}{48}\)
\(\Leftrightarrow\frac{100-x}{57}+\frac{100-x}{54}-\left(\frac{100-x}{51}+\frac{100-x}{48}\right)=0\)
\(\Leftrightarrow\left(100-x\right)\left[\left(\frac{1}{57}+\frac{1}{54}\right)-\left(\frac{1}{51}+\frac{1}{48}\right)\right]=0\) (*)
Vì\(\frac{1}{57}< \frac{1}{51},\frac{1}{54}< \frac{1}{48}\Rightarrow\left(\frac{1}{57}+\frac{1}{54}\right)< \left(\frac{1}{51}+\frac{1}{48}\right)\)
\(\Rightarrow\left(\frac{1}{57}+\frac{1}{54}\right)-\left(\frac{1}{51}+\frac{1}{48}\right)< 0\)
Phương trình (*) xảy ra khi: \(100-x=0\Leftrightarrow x=100\)
Vậy phương trình có nghiệm duy nhất là x = 100
a) -23/49 = -1081/2303
-25/47 = -1225/2303
=> -1081/2303 > -1225/2303 hay -23/49 > -25/47
b) -317/633 = -235531/470319
-371/743 = -234843/470319
=> -235531/470319 < -234843/470319 hay -317/633 < -371/743
Ta có:
\(\dfrac{37}{51}< \dfrac{43}{51}\)
\(\dfrac{43}{51}< \dfrac{43}{49}\)
Do đó \(\dfrac{37}{51}< \dfrac{43}{49}\)
.