Thay x = 0 ; y = 1/4 ta được 

\(0^6+2.0^2.\left(\dfrac{1}{4}\right)^3-0^5-\left(\dfrac{1}{4}\right)^5+\dfrac{0.1}{4}-\dfrac{1}{4}.0^5-6.0=-\left(\dfrac{1}{4}\right)^5\)

3, \(C\left(x\right)=\left(x-6\right)^2+\left(x-2\right)^2+100\)

\(=x^2-12x+36+x^2-4x+4+100=2x^2-16x+140\)

\(=2\left(x^2-8x+16-16\right)+140=2\left(x-4\right)^2+108\ge108\)

Dấu ''='' xảy ra khi x = 4 

4, \(D\left(x\right)=9x^2-12x+4+16x^2+40x+25+999=25x^2+28x+29+999=25\left(x^2+\dfrac{2.14}{25}x+\dfrac{196}{625}-\dfrac{196}{625}\right)+29+999\)

\(=25\left(x+\dfrac{14}{25}\right)^2+\dfrac{25504}{25}\ge\dfrac{25504}{25}\)

Dấu ''='' xảy ra khi x = -14/25 

7, \(\left\{{}\begin{matrix}\dfrac{1}{x-1}-\dfrac{2}{y}=2\\\dfrac{2}{x-1}-\dfrac{1}{y}=-1\end{matrix}\right.\)

Đặt \(\dfrac{1}{x-1}=a;\dfrac{1}{y}=b\)

\(\left\{{}\begin{matrix}a-2b=2\\2a-b=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=-3\\b=-5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x-1}=-3\\\dfrac{1}{y}=-5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=-\dfrac{1}{3}\\y=-\dfrac{1}{5}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{1}{5}\end{matrix}\right.\)

-mấy câu sau tương tự nhé 

\(u\left(x\right)=2x^3-\left(m-3\right)x+16-m^2\)

\(u\left(x\right)'=6x^2-\left(m-3\right)\)

TH1 

\(\left\{{}\begin{matrix}u'\left(x\right)\ge0\\u\left(0\right)\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x^2-m+3\ge0\\16-m^2\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\le6x^2+3\\16-m^2\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\le3\\m\le-4;m\ge4\end{matrix}\right.\Rightarrow m\le-4\)

Do m thuộc ( -2022 ; 2023 ) => -2022 < m < 2023 

=> -2022 < m =< -4 => 2018 số 

TH2 : \(\left\{{}\begin{matrix}u'\left(x\right)\le0\\u\left(0\right)\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\ge27\\-4\le m\le4\end{matrix}\right.\)(loại) 

=> Có 2018 giá trị nguyên m 

Cho \(u\left(x\right)=mx^3-mx^2+16x-32\)

\(u'\left(x\right)=3mx^2-2xm+16\)

TH1 \(\left\{{}\begin{matrix}u\left(x\right)'\le0\\u\left(2\right)\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3mx^2-2xm+16\le0\\4m\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\le-\dfrac{16}{3x^2-2x}\\m\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\le-16\\m\ge0\end{matrix}\right.\)

loại 

TH2 : \(\left\{{}\begin{matrix}u'\left(x\right)\ge0\\u\left(2\right)\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\ge-\dfrac{16}{3x^2-2x}\\4m\le0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}m\ge-2\\m\le0\end{matrix}\right.\Rightarrow-2\le m\le0\)

Mà m nguyên => m = -2 ; -1 ; 0 

\(3x^2-6xy=3x\left(x-2y\right)\)

4.3

a, \(\left(3a-2b\right)^2=9a^2-12ab+4b^2\)

b, \(\left(4-3a\right)^2=16-24a+9a^2\)

c, \(\left(5a-4b\right)^2=25a^2-40ab+16b^2\)

d, \(\left(x+\dfrac{1}{2}y\right)^2=x^2+xy+\dfrac{1}{4}y^2\)

e, \(\left(\dfrac{9x}{8}+\dfrac{7y}{9}\right)^2=\dfrac{81x^2}{64}+\dfrac{7}{4}x+\dfrac{49y^2}{81}\)

f, \(\left(2x+\dfrac{3y}{4}\right)^2=4x^2+3xy+\dfrac{9y^2}{16}\)

g, \(\left(\dfrac{8x}{11}+\dfrac{5y}{3}\right)^2=\dfrac{64x^2}{121}+\dfrac{80}{33}xy+\dfrac{25y^2}{9}\)

h, \(\left(-\dfrac{x}{2}+\dfrac{3y}{7}\right)^2=\dfrac{x^2}{4}-2x.\dfrac{3y}{7}+\dfrac{9y^2}{49}=\dfrac{x^2}{4}-\dfrac{6xy}{7}+\dfrac{9y^2}{49}\)

- mấy ý sau tương tự nhé 

\(\left\{{}\begin{matrix}\left(x+1\right)\left(y-1\right)=xy-1\\\left(x-3\right)\left(y-3\right)=xy-3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}xy-x+y=xy-1\\xy-3x-3y+9=xy-3\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-x+y=-1\\-3x-3y=-12\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-3x+3y=-3\\-3x-3y=-12\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}-6x=-15\\y=-1+x\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=\dfrac{2}{3}\end{matrix}\right.\)

1, \(3x^2-6xy=3x\left(x-2xy\right)\)

2, \(x^3-4x=x\left(x^2-4\right)=x\left(x-2\right)\left(x+2\right)\)

3, bn xem lại nhé

4, \(x^2-4x+4-9y^2=\left(x-2\right)^2-\left(3y\right)^2=\left(x-2-3y\right)\left(x-2+3y\right)\)

5, \(x^3-x+2y-8y^3=\left(x-2y\right)\left(x^2+2xy+4y^2\right)-\left(x-2y\right)=\left(x-2y\right)\left(x^2+2xy+4y^2-1\right)\)

6, \(xy+z^2+xz+yz=x\left(y+z\right)+z\left(y+z\right)=\left(x+z\right)\left(y+z\right)\)

7, \(64x^3-27y^3=\left(4x\right)^3-\left(3y\right)^3=\left(4x-3y\right)\left(16x^2+12xy+9y^2\right)\)

8, \(x^3+y^3+x+y=\left(x+y\right)\left(x^2-xy+y^2\right)+\left(x+y\right)=\left(x+y\right)\left(x^2-xy+y^2+1\right)\)

9, \(4x^2-y^2+4y-4=4x^2-\left(y-2\right)^2=\left(2x-y+2\right)\left(2x+y-2\right)\)

Diện tích toàn phần

\(S_{tp}=S_{xq}+S_{đáy}=52.12+\dfrac{8+18}{2}.12=780cm^2\)