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a) Tử số = 3^6.45^4−15^13.5^−9=3^6.(5.3^2)^4−(3.5)^13.5^−9=3^6.5^4.3^8−3^13.5^13.5^−9=3^14.5^4−3^13.5^4=3^13.5^4(3−1)=2.3^13.5^4
Mẫu số = 27^4.25^3+45^6=3^12.5^6+3^12.5^6=2.3^12.5^6
Phân số = 2.3^13.5^4/2.3^12.5^6=3/25
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\(A=\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{35.37}=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{35}-\frac{1}{37}\)
\(=\frac{1}{1}-\frac{1}{37}<1\text{ Vậy }A<1\)
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4 diem : 3 diem tao thanh 1 tam giac lon va 1 diem nam giau 3 dinh cua tam giac do se tao thanh 4 tam giac
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1: Xét ΔABC có AB=AC
nên ΔABC cân tại A
Suy ra: \(\widehat{ABC}=\widehat{ACB}\)
Xét ΔABH và ΔACH có
AB=AC
AH chung
BH=CH
Do đó: ΔABH=ΔACH
Suy ra: \(\widehat{AHB}=\widehat{AHC}\)
mà \(\widehat{AHB}+\widehat{AHC}=180^0\)
nên \(\widehat{AHB}=\widehat{AHC}=\dfrac{180^0}{2}=90^0\)
Do đó: AH\(\perp\)BC
\(A=\frac{27^2.64.125}{25.9^3.5^4}\)
\(A=\frac{3^3.2^6.5^3}{5^2.3^6.5^4}\)
\(A=\frac{2^6.5^3}{5^6.3^3}\)
\(A=\frac{2^6}{5^3.3^3}\)
\(A=\frac{64}{125.27}=\frac{64}{3375}\)
\(A=\frac{27^2.64.125}{25.9^3.5^4}\)
\(=\frac{\left(3^3\right)^2.2^6.5^3}{5^2.\left(3^2\right)^3.5^4}\)
\(=\frac{3^6.2^6.5^3}{5^2.3^6.5^4}\)
\(=\frac{2^6}{5^3}\)