a) \(A=2x^2-\dfrac{1}{3}y\)

A= \(\left(2-\dfrac{1}{3}\right)\)\(x^2y\)

A=\(\dfrac{5}{3}\)\(x^2y\)

Tại \(x=2;y=9\) ta có

A=\(\dfrac{5}{3}\).(2)\(^2\).9 = \(\dfrac{5}{3}\).4 .9 = 60

Vậy tại \(x=2;y=9\) biểu thức A= 60

b) P=\(2x^2+3xy+y^2\)            (\(y^2\) là 1\(y^2\) nha bạn)

P=\(\left(2+3+1\right)\left(x^2.x\right)\left(y.y^2\right)\)

P= 6\(x^3y^3\)

Tại \(x=-\dfrac{1}{2};y=\dfrac{2}{3}\) ta có

P= 6.\(\left(-\dfrac{1}{2}\right)^3.\left(\dfrac{2}{3}\right)^3\) = 6.\(\left(-\dfrac{1}{8}\right).\dfrac{8}{27}\) = \(-\dfrac{2}{9}\)

Vậy tại \(x=-\dfrac{1}{2};y=\dfrac{2}{3}\) biểu thức P= \(-\dfrac{2}{9}\)

c)\(\left(-\dfrac{1}{2}xy^2\right).\left(\dfrac{2}{3}x^3\right)\)

=\(\left((-\dfrac{1}{2}).\dfrac{2}{3}\right)\left(x.x^3\right).y^2\)

=\(-\dfrac{1}{3}\)\(x^4y^2\)

Tại \(x=2;y=\dfrac{1}{4}\)ta có

\(-\dfrac{1}{3}\).\(\left(2\right)^4.\left(\dfrac{1}{4}\right)^2=-\dfrac{1}{3}.16.\dfrac{1}{16}=-\dfrac{1}{3}\)

\(\)Vậy \(x=2;y=\dfrac{1}{4}\) biểu thức \(\left(-\dfrac{1}{2}xy^2\right).\left(\dfrac{2}{3}x^3\right)\)= \(-\dfrac{1}{3}\)

CHÚC BẠN HỌC TỐT NHA

x^2(x + 2) + 4(x + 2) = 0

(x^2 + 4)(x + 2) =0

=>  x^2 + 4 = 0 hoặc x + 2 = 0

Ta có : x^2 >= 0 => x^2 + 4 >= 4 mà x^2 + 4 = 0 => Vô lí

Vậy x + 2 = 0 => x = -2

Vậy x = -2

Bạn kia giải hơi khó nhìn nên t giải lại.

\(x^2\left(x+2\right)+4\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x^2+4\right)=0\)

\(\Rightarrow\hept{\begin{cases}x^2+4=0\\x+2=0\end{cases}}\Rightarrow\hept{\begin{cases}x^2\ge0\Rightarrow x^2+4\ge4\\x=-2\end{cases}}\)

Xét trường hợp \(x^2\ge0\Rightarrow x^2+4\ge4\)

Mà \(x^2+4=0\)(vô lý)

Suy ra phương trình có nghiệm là (-2)

\(A=\frac{3x^2+3xy+3y^2-2x^2-4xy-2y^2}{x^2+xy+y^2}=3-\frac{2\left(x+y\right)^2}{x^2+xy+y^2}\le3\)

\(A=\frac{\frac{1}{3}x^2+\frac{1}{3}xy+\frac{1}{3}y^2+\frac{2}{3}x^2-\frac{4}{3}xy+\frac{2}{3}y^2}{x^2+xy+y^2}=\frac{1}{3}+\frac{\frac{2}{3}\left(x-y\right)^2}{x^2+xy+y^2}\ge\frac{1}{3}\)

cam on ban nha

Ta có:

\(\dfrac{P}{Q}=\dfrac{R}{S}\Leftrightarrow1+\dfrac{P}{Q}=1+\dfrac{R}{S}\Leftrightarrow\dfrac{Q+P}{Q}=\dfrac{R+S}{S}\)

=> ĐPCM

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I passed all my exams so my parents are so delighted

Tom is playing computer games right now although he has an important text tomorrow morning

I pass all the exam, so my parents are so delighted.

Tom is playing computer games right now although he has an important text tomorrow morning.