7/12xy^2+11/18x^3y
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a: \(=\dfrac{6}{3}\cdot x\cdot\dfrac{y^2}{y}=2xy\)
b: \(=\dfrac{62}{2}\cdot\dfrac{x^4}{x^3}\cdot\dfrac{y^3}{y^2}=31xy\)
c: \(=\dfrac{-18}{6}\cdot\dfrac{x^4}{x^2}\cdot\dfrac{y^3}{y}=-3x^2y^2\)
d: \(=\dfrac{27}{9}\cdot\dfrac{x^5}{x^3}\cdot\dfrac{y^6}{y^3}=3x^2y^3\)
e: \(=\dfrac{18}{12}\cdot\dfrac{x^3}{x}\cdot\dfrac{y^4}{y^3}=\dfrac{3}{2}x^2y\)
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\(\dfrac{5x-1+x+1}{3x^2y}=\dfrac{6x}{3x^2y}=\dfrac{2}{xy}\)
\(\dfrac{21x^2+22y}{36x^3y^2}\)
\(\dfrac{x\left(4x-7\right)+7x-16}{\left(x+2\right)\left(4x-7\right)}=\dfrac{4x^2-16}{\left(x+2\right)\left(4x-7\right)}=\dfrac{4\left(x-2\right)\left(x+2\right)}{\left(x+2\right)\left(4x-7\right)}=\dfrac{4x-8}{4x-7}=1-\dfrac{1}{4x-7}\)
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an có 10000000 quả cam an cho mẹ gấp đôi rồi an co ba số quả lớn hơn mẹ 200 vậy an còn bao nhiêu quả cam
a) \(\frac{5x-1}{3x^2y}+\frac{x-1}{3x^2y}=\frac{5x-1+x-1}{3x^2y}=\frac{6x}{3x^2y}=\frac{2}{xy}\)
b) \(\frac{7}{12xy^2}+\frac{11}{18x^3y}=\frac{7\left(\frac{3}{2}x^2\right)}{18x^3y^2}+\frac{11y}{18x^3y^2}=\frac{10,5x^2+11y}{18x^3y^2}\)
c) \(\frac{x}{x+2}+\frac{7x-16}{\left(x+2\right)\left(4x-7\right)}=\frac{x\left(4x-7\right)}{\left(x+2\right)\left(4x-7\right)}+\frac{7x-16}{\left(x+2\right)\left(4x-7\right)}\)
\(=\frac{4x^2-7x+7x-16}{\left(x+2\right)\left(4x-7\right)}=\frac{4x^2-16}{\left(x+2\right)\left(4x-7\right)}\)
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M=9x2+6y2+18x−12xy−12y−27
=(9x2−12xy+4y2)+( 18x−12y)+9+2y2−36
=[(3x)2 −2.3x.2y+(2y)2]+(18x−12y)+ 9+2y2− 36
=(3x−2y)2+2.(3x−2y) .3+32+2y2−36
=(3x−2y+3)2+2y2−36
∀x;y ta có :
(3x−2y+3)2≥0
2y2≥0
⇒(3x−2y+3)2+2y2≥0
⇒(3x-2y+3)2+2y2-36≥-36
⇒M≥-36
Dấu = xảy ra ⇔{3x−2y+3=02y2=0
⇔{x=-1 y=0
Vậy MinM=-36⇔{x=-1 y=0
Do đó : M≥−36
⇒ Chọn đáp án D
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Từ phương trình \(\left(2\right)\): \(3x+4y=0\Leftrightarrow y=-\dfrac{3}{4}x\)
Thế vào phương trình \(\left(1\right)\) ta được:
\(\left(18x^2+\dfrac{9}{2}x-17\right)\left(21x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3\pm\sqrt{553}}{24}\\x=\pm\dfrac{\sqrt{21}}{21}\end{matrix}\right.\)
\(x=\dfrac{-3+\sqrt{553}}{24}\Rightarrow y=\dfrac{3-\sqrt{553}}{32}\)
\(x=\dfrac{-3-\sqrt{553}}{24}\Rightarrow y=\dfrac{3+\sqrt{553}}{32}\)
\(x=\dfrac{\sqrt{21}}{21}\Rightarrow y=-\dfrac{\sqrt{21}}{28}\)
\(x=-\dfrac{\sqrt{21}}{21}\Rightarrow y=\dfrac{\sqrt{21}}{28}\)
Vậy ...
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bạn ơi đề là như nào vậy
\(\frac{7}{12}\)xy2 + \(\frac{11}{18}\)x3y là giải phương trình hay tìm nghiệm của phương trình vậy bn
\(\frac{7}{12xy^2}+\frac{11}{18x^3y}=\frac{21x^2}{36x^3y^2}+\frac{22y}{36x^3y^2}=\frac{21x^2+22y}{36x^3y^2}\)