\(A=\frac{1}{3589}.7\frac{1}{297}-3\frac{3588}{3589}.\frac{2}{297}-\frac{7}{3589}-\frac{3}{3589.297}\)=?
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\(A=\frac{1}{3589}.7\frac{1}{297}-3\frac{3588}{3589}.\frac{2}{297}-\frac{7}{3589}-\frac{3}{3589.297}\)
\(A=\frac{1}{3589}.\left(7+\frac{1}{297}\right)-\left(4-\frac{1}{3589}\right).2.\frac{1}{297}-7.\frac{1}{3589}-3.\frac{1}{3689}.\frac{1}{297}\)
\(A=7.\frac{1}{3689}+\frac{1}{3589}.\frac{1}{297}-8.\frac{1}{297}+2.\frac{1}{3589}.\frac{1}{297}-7.\frac{1}{3589}\)
\(A=-8.\frac{1}{297}\)
\(A=\frac{-8}{297}\)
Thay 999990=a vào biểu thức A ta được
\(A=\left(a+4\right)\left(a+9\right)\left(a+2\right)-\left(a+6\right)\left(a+1\right)\left(a+8\right)\)
\(=\left(a^2+13a+36\right)\left(a+2\right)-\left(a^2+7a+6\right)\left(a+8\right)\)
\(=a^3+2a^2+13a^2+26a+36a+72-a^3-8a^2-7a^2-56a-6a-48\)
\(=24\)
Thay b=44440 vào B ta được
\(B=\left(b+3\right)\left(b+8\right)\left(b+1\right)-\left(b+5\right)b\left(b+7\right)\)
\(=\left(b^2+11b+24\right)\left(b+1\right)-\left(b^2+5b\right)\left(b+7\right)\)
\(=b^3+b^2+11b^2+11b+24b+24-b^3-7b^2-5b^2-35b\)
\(=24\)
Vậy A=B (=24)
Sửa lại :) bài dưới vô tình gõ sai ~
Đặt $a=\dfrac{1}{3589};b=\dfrac{1}{297}$
$=>A=a(7+b)-(4-a)2b-7a-3ab$
$=>A=7a+ab-8b+2ab-7a-3ab$
$=>A=-8b=\dfrac{-8}{297}$
\(A=\dfrac{1}{3589}.7\dfrac{1}{297}-3\dfrac{3588}{3589}.\dfrac{2}{297}-\dfrac{7}{3589}-\dfrac{3}{3589.297}\)
\(A=\dfrac{1}{3589}.(7+\dfrac{1}{297})-(3+1-\dfrac{1}{3589}).\dfrac{2}{297}-\dfrac{7}{3589}-\dfrac{3}{3589.297}\)
\(A=\dfrac{1}{3589}.7+\dfrac{1}{3589}.\dfrac{1}{297}-\dfrac{6}{297}-\dfrac{2}{297}+\dfrac{2}{3589.297}-\dfrac{7}{3589}-\dfrac{3}{3589.297}\)
\(A=\dfrac{7}{3589}+\dfrac{1}{3589.297}-\dfrac{8}{297}+\dfrac{2}{3589.297}-\dfrac{7}{3589}-\dfrac{3}{3589.297}\)
\(A=0-\dfrac{8}{297}\)
\(A=-\dfrac{8}{297}\)
Ta có:
\(A=\frac{1}{358}.\left(7+\frac{1}{297}\right)-\left(4-\frac{1}{358}\right).2.\frac{1}{297}-7.\frac{1}{358}-\frac{3}{297}.\frac{1}{358}\)
\(=\frac{1}{358}.\left(7+\frac{1}{297}-7-\frac{3}{297}\right)-\left(4-\frac{1}{358}\right).\frac{2}{297}\)
\(=\frac{1}{358}.\left(-\frac{2}{297}\right)-\frac{2}{297}.\left(4-\frac{1}{358}\right)\)
\(=\left(-\frac{2}{297}\right)\left(\frac{1}{358}+4-\frac{1}{358}\right)\)
\(=\left(-\frac{2}{297}\right)\left(-4\right)\)
\(=\frac{8}{297}\)
Vậy giá trị biểu thức A là \(\frac{8}{297}\)
\(A=\dfrac{1}{358}.\left(7+\dfrac{1}{297}\right)-\left(4-\dfrac{1}{358}\right).2.\dfrac{1}{297}-7.\dfrac{1}{358}-3.\dfrac{1}{297}.\dfrac{1}{359}\)
\(A=7.\dfrac{1}{358}+\dfrac{1}{297}.\dfrac{1}{358}-4.2.\dfrac{1}{297}+2.\dfrac{1}{297}.\dfrac{1}{358}-7.\dfrac{1}{358}-3.\dfrac{1}{297}.\dfrac{1}{359}\)
\(A=\left(7.\dfrac{1}{358}-7.\dfrac{1}{358}\right)+\left(\dfrac{1}{297}.\dfrac{1}{358}+2.\dfrac{1}{297}.\dfrac{1}{358}-3.\dfrac{1}{297}.\dfrac{1}{358}\right)-4.2.\dfrac{1}{297}\)
\(A=0+0+\dfrac{-8}{297}\)
\(A=\dfrac{-8}{297}\)
Chúc bạn học tốt!!!
A= \(\dfrac{1}{358}\left(7+\dfrac{1}{297}\right)-\left(4-\dfrac{1}{358}\right).2.\dfrac{1}{297}-7.\dfrac{1}{358}-3.\dfrac{1}{297}.\dfrac{1}{358}\)
A= \(\dfrac{7}{358}+\dfrac{1}{358.297}-\dfrac{8}{297}+\dfrac{2}{358.297}-\dfrac{7}{358}-\dfrac{3}{358.297}\)
A= \(-\dfrac{8}{297}\)