\(M=2x^2+9y^2-6xy-6x-12y+2028\\ =3\left(x^2-2xy+y^2\right)-\left(x^2+6x+9\right)+6\left(y^2-2y+1\right)+2025\\ =\left(x-y\right)^2-\left(x-3\right)^2+6\left(y-1\right)^2+2025\ge2025\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x=y\\x=3\\y=1\end{matrix}\right.\) (vô lí) nên dấu \("="\) ko thể xảy ra

\(N=x^2-4xy+5y^2+10x-22y+28\\ =\left(x^2+4y^2+25-4xy-20y+10x\right)+\left(y^2-2y+1\right)+2\\=\left(x-2y+5\right)^2+\left(y-1\right)^2+2\ge2\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x-2y=5\\y=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=1\end{matrix}\right.\)

\(M=2x^2+9y^2-6xy-6x-12y+2028=\left(x+2\right)^2-6y\left(x+2\right)+9y^2+\left(x-5\right)^2+1999=\left(x+2-3y\right)^2+\left(x-5\right)^2+2019\ge1999\)

\(ĐTXR\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=\dfrac{7}{3}\end{matrix}\right.\)

\(N=x^2-4xy+5y^2+10x-22y+28=\left(x+5\right)^2-4y\left(x+5\right)+4y^2+\left(y-1\right)^2+2=\left(x+5-2y\right)^2+\left(y-1\right)^2+2\ge2\)

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

9/12

\(\dfrac{9}{12}=\dfrac{21}{28}\)

9/12

D = 222...2 x 999...9

(50 c/s 2)      (50 c/s 9)

D = 222...2 x (1000...0 - 1)

(50 c/s 2)       (50 c/s 0)

D = 222...2000...0 - 222...2

(50 c/s 2) (50 c/s 0)(50 c/s 2)

D = 22...21777...78

 (49 c/s 2)  (49 c/s 7)

\(x^2=\left(2\sqrt{7}\right)^2\)

\(x=2\sqrt{7}\)

327 - 16 + 216 - 27 + 600          16 x 48 + 8 x 48 + 16 x 28

=(327 - 27) + (216 - 16) + 600      =16 x 48 + 16 x 24 + 16 x 28 

=300           + 200           + 600     =16 x (48 + 24 + 28)

=500                               + 600     =16 x 100

=1100                                            =1600

phương trình này nhìn từ đầu cũng bik vô nghiệm ko có x

\(\frac{x+1}{99}+\frac{x+2}{99}=\frac{x+10}{99}+\frac{x+20}{99}\)

Nhân 2 vế cho 99 ta được:

\(99.\left(\frac{x+1}{99}+\frac{x+2}{99}\right)=99.\left(\frac{x+10}{99}+\frac{x+20}{99}\right)\)

=>x+1+x+2=x+10+x+20

=>2x+3=2x+30

=>0x=27 (vô lí)

Vậy ko tìm dc x

Tích mình 3 mình tích lại 5 nhớ phải kết bạn nhé

\(a,1-\frac{9}{100}=\frac{91}{100}=\frac{910}{1000}\)

    \(1-\frac{99}{1000}=\frac{909}{1000}\)

Vì \(\frac{910}{1000}>\frac{909}{1000}\)nên\(\frac{9}{100}< \frac{99}{1000}\)

\(b,\frac{19}{21}< 1< \frac{27}{23}\)nên \(\frac{19}{21}< \frac{27}{23}\)

\(c,14< 28,99>100\)nên \(\frac{14}{99}< \frac{28}{100}\)

Câu c cách khác : \(\frac{28}{100}=\frac{14}{50}>\frac{14}{99}\)