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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 $a=\dfrac{1}{3589};b=\dfrac{1}{297}$
$=>A=a(7+b)-(4-a)2b-7a-3ab$
$=>A=7a+ab-8b+2ab-7a-3ab$
$=>A=-8b=\dfrac{-8}{297}$
\(A=\dfrac{1}{3589}.7\dfrac{1}{297}-3\dfrac{3588}{3589}.\dfrac{2}{297}-\dfrac{7}{3589}-\dfrac{3}{3589.297}\)
\(A=\dfrac{1}{3589}.(7+\dfrac{1}{297})-(3+1-\dfrac{1}{3589}).\dfrac{2}{297}-\dfrac{7}{3589}-\dfrac{3}{3589.297}\)
\(A=\dfrac{1}{3589}.7+\dfrac{1}{3589}.\dfrac{1}{297}-\dfrac{6}{297}-\dfrac{2}{297}+\dfrac{2}{3589.297}-\dfrac{7}{3589}-\dfrac{3}{3589.297}\)
\(A=\dfrac{7}{3589}+\dfrac{1}{3589.297}-\dfrac{8}{297}+\dfrac{2}{3589.297}-\dfrac{7}{3589}-\dfrac{3}{3589.297}\)
\(A=0-\dfrac{8}{297}\)
\(A=-\dfrac{8}{297}\)
\(A=\frac{1}{3589}.7\frac{1}{297}-3\frac{3588}{3589}.\frac{2}{297}-\frac{7}{3589}-\frac{3}{3589.297}\)
\(A=\frac{1}{3589}.\left(7+\frac{1}{297}\right)-\left(4-\frac{1}{3589}\right).2.\frac{1}{297}-7.\frac{1}{3589}-3.\frac{1}{3689}.\frac{1}{297}\)
\(A=7.\frac{1}{3689}+\frac{1}{3589}.\frac{1}{297}-8.\frac{1}{297}+2.\frac{1}{3589}.\frac{1}{297}-7.\frac{1}{3589}\)
\(A=-8.\frac{1}{297}\)
\(A=\frac{-8}{297}\)
a) Đề ( \(x\ne\pm1\))
>\(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}=\frac{4}{\left(x+1\right)\left(x-1\right)}\\ \Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=4\\ \Leftrightarrow\left(x+1-x+1\right)\left(x+1+x-1\right)=4\\ \Leftrightarrow2.2x=4\Leftrightarrow x=1\left(kothỏa\right)\)
Vậy \(S=\varnothing\)
b) đề \(\left(x\ne-\frac{1}{2},\frac{1}{2}\right)\)
\(\frac{32x^2}{12\left(1-2x\right)\left(1+2x\right)}=\frac{-8x\left(1+2x\right)}{12\left(1-2x\right)\left(1+2x\right)}-\frac{3\left(1+8x\right)\left(1-2x\right)}{12\left(1-2x\right)\left(1+2x\right)}\\ \Leftrightarrow32x^2=-8x-16x^2-3-12x+48x^2\\ \Leftrightarrow20x+3=0\Leftrightarrow x=\frac{20}{3}\left(thỏadk\right)\)
Vậy \(S=\left\{\frac{20}{3}\right\}\)
\(\frac{2x-1}{5}-\frac{x-2}{3}=\frac{x+7}{15}\)
<=> \(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\)
<=> \(\frac{3\left(2x-1\right)}{5\cdot3}-\frac{5\left(x-2\right)}{3\cdot5}-\frac{x+7}{15}=0\)
<=> \(\frac{6x-3-5x+10-x-7}{15}=0\)
<=> \(\frac{-14}{15}=0\)
=> PT vô nghiệm