gia tri cua a= (-3333)^4:(-1111)^4+( ; 3981)^5 .(-3981)^5 =
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Phân tích mẫu dạng 101 nhân với một số rồi quy đồng lên là xong !Gà!
Ta có :
\(A=\frac{7}{4}.\left(\frac{3333}{1212}+\frac{3333}{2020}+\frac{3333}{3030}+\frac{3333}{4242}\right)\)
\(A=\frac{7}{4}.\left[3333.\left(\frac{1}{1212}+\frac{1}{2020}+\frac{1}{3030}+\frac{1}{4242}\right)\right]\)
\(A=\frac{7}{4}.\left[3333.\left(\frac{1}{12.101}+\frac{1}{20.101}+\frac{1}{30.101}+\frac{1}{42.101}\right)\right]\)
\(A=\frac{7}{4}.\left[3333.\frac{1}{101}\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\right]\)
\(A=\frac{7}{4}.33.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(A=33.\left[\frac{7}{4}.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\right]\)
\(A=33.\left[\frac{7}{4}.\left(\frac{1}{3}-\frac{1}{7}\right)\right]\)
\(A=33.\left(\frac{7}{4}-\frac{4}{7}\right)\)
\(A=33.1\)
\(A=33\)
\(A=\frac{7}{4}\left(\frac{33}{12}+\frac{33}{20}+\frac{33}{30}+\frac{33}{42}\right)\)
\(A=\frac{7}{4}.33\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(A=\frac{7}{4}.33\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)
\(A=\frac{7}{4}.33\left(\frac{1}{3}-\frac{1}{7}\right)\)
\(A=\frac{7}{4}.33.\frac{4}{21}\)
\(A=11\)
1111+2222+3333+1111+2222+3333+1111+2222+3333+1111+2222+3333+1111+2222+3333=(1111+2222+3333)x5
=33330
a: ĐKXĐ: x<>2; x<>-2
b: \(A=\dfrac{3x\left(x-2\right)+2x+6}{2\left(x-2\right)\left(x+2\right)}=\dfrac{3x^2-6x+2x+6}{2\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{3x^2+4x+6}{2\left(x-2\right)\left(x+2\right)}\)
c: Khi x=-3 thì \(A=\dfrac{3\cdot\left(-3\right)^2-4\cdot3+6}{2\left(-3-2\right)\left(-3+2\right)}=\dfrac{21}{10}\)
\(=\left(\frac{-3333}{-1111}\right)^4+\left(\frac{3981}{-3981}\right)^5\)
\(=3^4+\left(-1\right)^5\)
\(=81-1\)
\(=80\)