Ta có: \(L=\left(-\dfrac{3}{4}x^5y^4\right)\left(xy^2\right)\left(-\dfrac{8}{9}x^2y^5\right)\)

\(=\dfrac{2}{3}x^8y^{11}\)

\(\Rightarrow\) Bậc của L là: 19

Hệ số: \(\dfrac{2}{3}\)

\(L=\left(\dfrac{-3}{4}x^5y^4\right)\left(xy^2\right)\left(\dfrac{-8}{9}x^2y^5\right)\)

\(=\left(\dfrac{-3}{4}.\dfrac{-8}{9}\right)\left(x^5.x.x^2\right)\left(y^4.y^2.y^5\right)\)

\(=\dfrac{2}{3}.x^8.y^{11}\)

\(\Rightarrow\)Bậc của đơn thức L là 19 và hệ số là \(\dfrac{2}{3}\)

Vậy...

E + F = (5xy - \(\dfrac{2}{3}\)x\(^2\)y + xyz\(^2\) - 1) + (2x\(^2\)y - xyz\(^2\) - \(\dfrac{2}{5}\)xy + x + \(\dfrac{1}{2}\))

= 5xy - \(\dfrac{2}{3}\)x\(^2\)y + xyz\(^2\) - 1 + 2x\(^2\)y -xyz\(^2\) - \(\dfrac{2}{5}\)xy + x + \(\dfrac{1}{2}\)

= (5xy - \(\dfrac{2}{5}\)xy) + (\(\dfrac{-2}{3}\)x\(^2\)y + 2x\(^2\)y) + (xyz\(^2\) - xyz\(^2\)) + (-1 + \(\dfrac{1}{2}\)) + x

= \(\dfrac{23}{5}\)xy + \(\dfrac{4}{3}\) x\(^2\)y - \(\dfrac{1}{2}\) + x

E - F = (5xy - \(\dfrac{2}{3}\)x\(^2\)y + xyz\(^2\) - 1) - (2x\(^2\)y - xyz\(^2\) - \(\dfrac{2}{5}\)xy + x + \(\dfrac{1}{2}\))

= 5xy - \(\dfrac{2}{3}\)x\(^2\)y + xyz\(^2\) - 1 - 2x\(^2\)y + xyz\(^2\) + \(\dfrac{2}{5}\)xy - x - \(\dfrac{1}{2}\)

= (5xy + \(\dfrac{2}{5}\)xy) + (\(\dfrac{-2}{3}\)x\(^2\)y - 2x\(^2\)y) + (xyz\(^2\) + xyz\(^2\))+ (-1 - \(\dfrac{1}{2}\)) - x

= \(\dfrac{27}{5}\)xy - \(\dfrac{8}{3}\)x\(^2\)y + 2xyz\(^2\) - \(\dfrac{3}{2}\) - x

Vậy E - F = \(\dfrac{27}{5}\)xy - \(\dfrac{8}{3}\)x\(^2\)y + 2xyz\(^2\) - \(\dfrac{3}{2}\) - x

Bài1:

\(M=\dfrac{9-x}{4-x}=1+\dfrac{5}{4-x}\)

Để M đạt giá trị lớn nhất thì 4-x phải đặt giá trị nhỏ nhất

=>4-x đạt giá trị là số nguyên dương nhỏ nhất có thể

=>4-x=1

=>x=3

Thay x=3 vào M,ta có:

\(M=\dfrac{9-3}{4-3}=\dfrac{6}{1}=6\)

Vậy....

Bài2:

\(\left(x-2\right)^2+\left(2y-1\right)^2\)

Với mọi x;y thì \(\left(x-2\right)^2>=0;\left(2y-1\right)^2>=0\)

=>\(\left(x-2\right)^2+\left(2y-1\right)^2>=0\)

Để \(\left(x-2\right)^2+\left(2y-1\right)^2=0\) thì

\(\left(x-2\right)^2=0\) và \(\left(2y-1\right)^2=0\)

=>\(x-2=0\) và \(2y-1=0\)

=>\(x=2vay=\dfrac{1}{2}\)

Vậy....

\(M=\dfrac{9-x}{4-x}=\dfrac{5+4-x}{4-x}=\dfrac{5}{4-x}+\dfrac{4-x}{4-x}=\dfrac{5}{4-x}+1\)Để \(max_M\) thì \(\dfrac{5}{x-4}\) phải là số nguyên lớn nhất có thể

Vậy \(\dfrac{5}{x-4}=5\Rightarrow x=3\)

Thay vào biểu thức:

\(max_M=\dfrac{9-3}{4-3}=6\)

\(\left(x-2\right)^2+\left(2y-1\right)^2=0\)

\(\left\{{}\begin{matrix}\left(x-2\right)^2\ge0\\\left(2y-1\right)^2\ge0\end{matrix}\right.\)

\(\Rightarrow\left(x-2\right)^2+\left(2y-1\right)^2\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left(x-2\right)^2=0\\\left(2y-1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=\dfrac{1}{2}\end{matrix}\right.\)

1.

a)\(\left(\dfrac{1}{2}\cdot\left(-2\right)\cdot\dfrac{-1}{3}\right)\cdot\left(x^2\cdot x^2\cdot x^2\right)\cdot\left(y^2\cdot y^3\right)\cdot z\)

\(\dfrac{1}{3}x^6y^5z\)

Deg=12

Mấy câu kia tương tự nha cố gắng lên!

a) Ta có: \(-2xy^2\cdot\left(x^3y-2x^2y^2+5xy^3\right)\)

\(=-2x^4y^3+4x^3y^4-10x^2y^5\)

b) Ta có: \(\left(-2x\right)\cdot\left(x^3-3x^2-x+1\right)\)

\(=-2x^4+6x^3+2x^2-2x\)

c) Ta có: \(3x^2\left(2x^3-x+5\right)\)

\(=6x^5-3x^3+15x^2\)

d) Ta có: \(\left(-10x^3+\frac{2}{5}y-\frac{1}{3}z\right)\cdot\left(-\frac{1}{2}xy\right)\)

\(=5x^4y-\frac{1}{5}xy^2+\frac{1}{6}xyz\)

e) Ta có: \(\left(3x^2y-6xy+9x\right)\cdot\left(-\frac{4}{3}xy\right)\)

\(=-4x^3y^2+8x^2y^2-12x^2y\)

f) Ta có: \(\left(4xy+3y-5x\right)\cdot x^2y\)

\(=4x^3y^2+3x^2y^2-5x^3y\)

a/ \(\left|-x\right|=1,5\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1,5\\x=-1,5\end{matrix}\right.\)

Vậy .....

b/ \(\left|x+\dfrac{1}{2}\right|=2\dfrac{1}{2}\)

\(\Leftrightarrow\left|x+\dfrac{1}{2}\right|=\dfrac{5}{2}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{5}{2}\\x+\dfrac{1}{2}=-\dfrac{5}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

Vậy ....

c/ \(\left|0,5-x\right|=\left|-0,5\right|\)

\(\left|0,5-x\right|=0,5\)

\(\Leftrightarrow\left[{}\begin{matrix}0,5-x=0,5\\0,5-x=-0,5\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy ...

Cảm ơn bn nha.