Chứng minh rằng :42/20. 24 +42/24 .28 +...+ 42/76. 80 < 1
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Đặt A=\(\dfrac{4^2}{20.24}+\dfrac{4^2}{24.28}+...+\dfrac{4^2}{76.80}\)
A=\(\dfrac{16}{20.24}+\dfrac{16}{24.28}+...+\dfrac{16}{76.80}\)
A=4.[\(\dfrac{4}{20.24}+\dfrac{4}{24.28}+...+\dfrac{4}{76.80}\)]
A=4.\(\left[\dfrac{1}{20}-\dfrac{1}{24}+\dfrac{1}{24}-\dfrac{1}{28}+...+\dfrac{1}{76}-\dfrac{1}{80}\right]\)
A=4.\(\left[\dfrac{1}{20}+\dfrac{1}{24}-\dfrac{1}{24}+\dfrac{1}{28}-\dfrac{1}{28}+...+\dfrac{1}{78}-\dfrac{1}{78}-\dfrac{1}{80}\right]\)
A=4.\(\left[\dfrac{1}{20}-\dfrac{1}{80}\right]\)
A=4.\(\dfrac{3}{80}\)
A=\(\dfrac{3}{20}\)<1
=>A<1
Tick mink nha
Mọi người tk mình đi mình đang bị âm nè!!!!!!Ai tk mình mình tk lại nha !!!
a) 57 + 28 = 85
24 + 67 = 91
46 + 39 = 85
b) 83 - 19 = 64
42 - 38 = 4
90 - 76 = 14
\(\begin{array}{*{20}{c}}{a)\,\,}\\{}\\{}\end{array}\begin{array}{*{20}{c}}{ + \begin{array}{*{20}{c}}{57}\\{28}\end{array}}\\\hline{\,\,\,\,85}\end{array}\) \(\begin{array}{*{20}{c}}{ + \begin{array}{*{20}{c}}{24}\\{67}\end{array}}\\\hline{\,\,\,\,91}\end{array}\) \(\begin{array}{*{20}{c}}{ + \begin{array}{*{20}{c}}{46}\\{39}\end{array}}\\\hline{\,\,\,\,85}\end{array}\)
\(\begin{array}{*{20}{c}}{b)\,\,}\\{}\\{}\end{array}\begin{array}{*{20}{c}}{ - \begin{array}{*{20}{c}}{83}\\{19}\end{array}}\\\hline{\,\,\,\,64}\end{array}\) \(\begin{array}{*{20}{c}}{ - \begin{array}{*{20}{c}}{42}\\{38}\end{array}}\\\hline{\,\,\,\,04}\end{array}\) \(\begin{array}{*{20}{c}}{ - \begin{array}{*{20}{c}}{90}\\{76}\end{array}}\\\hline{\,\,\,\,14}\end{array}\)
Cái này thì....mình mù tịt
Vì chưa học!!!!
Ai đồng ý thì cho mình xin 1 k!!!
\(76.\left(-42\right)+76.\left(-14\right)+24.\left(-56\right)\)
\(=76.\left(-42+-14\right)+24.\left(-56\right)\)
\(=76.\left(-56\right)+24.\left(-56\right)\)
\(=\left(76+24\right).-56\)
\(=100.-56\)
\(=-5600\)
Ta có: \(\frac{4^2}{20.24}+\frac{4^2}{24.28}+...+\frac{4^2}{76.80}\)
\(=4.\left(\frac{4}{20.24}+\frac{4}{24.28}+...+\frac{4}{76.80}\right)\)
\(=4.\left(\frac{1}{20}-\frac{1}{24}+\frac{1}{24}-\frac{1}{28}+...+\frac{1}{76}-\frac{1}{80}\right)\)
\(=4.\left(\frac{1}{20}-\frac{1}{80}\right)=4.\frac{3}{80}=\frac{3}{20}< 1\)
Vậy \(\frac{4^2}{20.24}+\frac{4^2}{24.28}+...+\frac{4^2}{76.80}< 1\)