Trả lời đc câu b chưa bạn

nếu rồi cho mình lời giải nha

toán lớp mấy ???

cần giúp

A = \(x^2+9y^2+25+6xy-30y-10x-6xy+26\)

   = \(x^2-10x+25+9y^2-30y+25+1\)

   = \(\left(x-5\right)^2+\left(3y-5\right)^2+1\)

Có : \(\left(x-5\right)^2\ge0\forall x;\left(3y-5\right)^2\ge0\forall y\)

\(\Rightarrow A\ge1\)

Vậy GTNN của A là 1 \(\Leftrightarrow\hept{\begin{cases}x-5=0\\3y-5=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\\y=\frac{5}{3}\end{cases}}}\)

3, A=(x-3)^2+(x-11)^2

\(\Rightarrow\)(X^2-3^2)+(x^2-11^2)

\(\Rightarrow\)(X^2-9)+(X^2-121)

Ta có :X^2 \(\ge\)0 và X^2 \(\ge\)0

\(\Rightarrow\)X^2 - 9 \(\le\)-9 và X^2- 121 \(\le\)-121

\(\Rightarrow\)(X^2-9)+(X^2-121)\(\le\)-130

Dấu = xảy ra khi : X=0

Vậy : Min A = -130 khi x=0

Mình mới lớp 7 sai thì thôi nhé

Bài 1:

a: \(=x^2-3x+\dfrac{9}{4}+\dfrac{11}{4}=\left(x-\dfrac{3}{2}\right)^2+\dfrac{11}{4}>=\dfrac{11}{4}\)

Dấu '=' xảy ra khi x=3/2

b: \(=4x^2-4x+1+x^2+4x+4=5x^2+5>=5\)

Dấu '=' xảy ra khi x=0

Bài 2: 

a: \(=-\left(x^2-2x-4\right)=-\left(x^2-2x+1-5\right)=-\left(x-1\right)^2+5< =5\)

Dấu = xảy ra khi x=1

b: \(=-\left(x^2-4x+4\right)+4=-\left(x-2\right)^2+4< =4\)

Dấu '=' xảy ra khi x=2

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

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

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

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

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

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

\(A=4x^2-6x\left(x-y\right)+3y^2-12y+20\)

\(A=\left(2x\right)^2-2.2x.\frac{3}{2}y+\left(\frac{3}{2}y\right)^2-\frac{9}{4}y^2+3y^2-12y+20\)

\(A=\left(2x-\frac{3}{4}y\right)^2+\frac{3}{4}y^2-12y+432-432+20\)

\(A=\left(2x-\frac{3}{4}y\right)^2+3\left(\frac{1}{4}y^2-2.\frac{1}{2}.12+12^2\right)-432+20\)

\(\Rightarrow A=\left(2x-\frac{3}{4}y\right)^2+3\left(\frac{1}{2}y-12\right)^2-412\)

Ta có:\(\hept{\begin{cases}\left(2x-\frac{3}{4}y\right)^2\ge0\\\left(\frac{1}{2}y-12\right)^2\ge0\Rightarrow3\left(\frac{1}{2}y-12\right)^2\ge0\end{cases}}\)

\(\Rightarrow\left(2x-\frac{3}{4}y\right)^2+3\left(\frac{1}{2}y-12\right)^2-412\ge-412\)

\(\Rightarrow A_{min}=-412\)đạt được khi

i\(\hept{\begin{cases}\left(2x-\frac{3}{4}y\right)^2=0\\\left(\frac{1}{2}y-12\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}2x-\frac{3}{4}y=0\\\frac{1}{2}y-12=0\end{cases}\Leftrightarrow}\hept{\begin{cases}2x=\frac{3}{4}y\\\frac{1}{2}y=12\end{cases}\Leftrightarrow}\hept{\begin{cases}x=9\\y=24\end{cases}}}\)