55×9624 ai giai dum mik ik
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\(x^2+4y^2+3x-6y=\left(x^2+3x\right)-\left(4y^2+6y\right)=x\left(x+3\right)-2y\left(2y+3\right)\)
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\(\frac{9}{12}\)\(+\frac{4}{7}\)\(=\frac{9:3}{12:3}\)\(+\frac{4}{7}\)\(=\frac{3}{4}\)\(+\frac{4}{7}\)\(=\frac{3\times7}{4\times7}\)+ \(\frac{4\times4}{7\times4}\)= \(\frac{21}{28}\)+ \(\frac{16}{28}\)=\(\frac{21+16}{28}\)=\(\frac{37}{28}\)
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= [(x^3+3x^2)-(2x^2+6x)+(3x+9)] : (x+3)
= (x+3).(x^2-2x+3) : (x+3) = x^2-2x+3
k mk nha
x^2-2x+3 nhé
nhớ đặt tính ra chia nha
tốn 1%calo não của mjk đó
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\(x\in\left(\infty;-\infty\right)\)
\(\left(1-x\right)^3=-\left(x-1\right)^3\)
\(-\left(x-1\right)^3=2^5.3\)
\(1-2\sqrt[3]{12}\)
Sau đó bạn tự\(\Rightarrow\)X nha
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We have \(3x^2+8x^3+x^4+9-8x^3-3x^2\)
\(=\left(3x^2-3x^2\right)+\left(8x^3-8x^3\right)+\left(x^4-9\right)\)
\(=x^4-9\)
If my answer is right, I hope you k for me =)) =.='
key: \(3x^2+8x^3+x^4+9+\left(-2x\right)^3-3x^2\)
\(=3x^2+8x^3+x^4+9-2x^3-3x^2\)
\(=\left(3x^2-3x^2\right)+\left(8x^3-2x^3\right)+x^4+9\)
\(=4x^3+x^4+9\)
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