a)\(x^2-4x+y^2-2y+10=\left(x^2-4x+4\right)+\left(y^2-2y+1\right)+5\)

\(=\left(x-2\right)^2+\left(y-1\right)^2+5\ge5\)

Dấu "=" xảy ra khi x=2;y=1

b) tương tự câu a

c)\(x^2+2y^2-6x-8y+2xy+5=x^2+2y^2+2x\left(y-3\right)-8y+5\)

\(=x^2+2x\left(y-3\right)+\left(y^2-6x+9\right)+\left(y^2-2x+1\right)-5\)

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

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

Dấu "=" xảy ra khi x=2;y=1

k mk đi

ai k mk

mk k lại

thanks

a) \(2x^2+y^2+4x-2y-2xy+10\)

\(=x^2+x^2+y^2+4x-2y-2xy+4+6\)

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

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

.......................chắc không phải cách làm này đâu!

b) \(5x^2+y^2+2xy-4x\)

\(=x^2+4x^2+y^2+2xy-4x\)

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

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

a, \(2x^2\)+\(y^2\)+\(4x-2y-2xy+10\)\(=y^2\)\(-x^2\)\(-1+2x-2y-2xy+3x^2+2x+11\)\(=\left(y-x-1^{ }\right)^2\)\(+3\left(x^2+\frac{2}{3}x+\frac{1}{9}\right)+\frac{32}{3}\)\(=\left(y-x-1\right)^2+3\left(x+\frac{1}{3}\right)^2+\frac{32}{3}\)\(\ge\frac{32}{3}\)

VẬY GTNN CỦA BIỂU THỨC \(=\frac{32}{3}\)KHI \(y-x-1=0;x+\frac{1}{3}=0\Rightarrow x=\frac{-1}{3};y=\frac{2}{3}\)

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

Vậy GTNN của biểu thức =-1 khi x+y=0;2x-1=0\(\Rightarrow\)x=\(\frac{1}{2}\);y=\(\frac{-1}{2}\)

a) \(A=x^2+2y^2+2xy+4x+6y+19\)

\(=\left[\left(x^2+2xy+y^2\right)+2.\left(x+y\right).2+4\right]+\left(y^2+2y+1\right)+14\)

\(=\left[\left(x+y\right)^2+2\left(x+y\right).2+2^2\right]+\left(y+1\right)^2+14\)

\(=\left(x+y+2\right)^2+\left(y+1\right)^2+14\ge14\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x+y+2=0\\y=-1\end{cases}}\Leftrightarrow x=y=-1\)

b)Đề có gì đó sai sai...

c) Tương tự câu b,em cũng thấy sai sai...HÓng cao nhân giải ạ!

b) \(P=2x^2+y^2+2xy-2y-4\)

\(\Leftrightarrow2P=4x^2+2y^2+4xy-4y-8\)

\(\Leftrightarrow2P=\left(4x^2+4xy+y^2\right)+\left(y^2-4y+4\right)-12\)

\(\Leftrightarrow2P=\left(2x+y\right)^2+\left(y-2\right)^2-12\ge-12\forall x;y\)

Có \(2P\ge-12\Leftrightarrow P\ge-6\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}2x+y=0\\y-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-1\\y=2\end{cases}}}\)

bạn nhóm thành các bình phương nhé. còn dư 4xy với 1.

bạn trình bày cho mình đc ko?

\(\left\{{}\begin{matrix}4x^2+9y^2=9\\A=x-2y+3\end{matrix}\right.\)

Áp dụng bất đẳng thức Bunhiacopxki cho các cặp số \(\left(\dfrac{1}{2};2x\right);\left(-\dfrac{2}{3};3y\right)\)

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

\(\Rightarrow\left[\dfrac{1}{2}.2x+\left(-\dfrac{2}{3}\right).3y\right]^2\le\left(\dfrac{1}{4}+\dfrac{4}{9}\right)\left(4x^2+9y^2\right)=\dfrac{25}{36}.9\)

\(\Rightarrow x-2y\le\dfrac{5}{6}.3=\dfrac{5}{2}\)

\(\Rightarrow A=x-2y+3\le\dfrac{5}{2}+3\)

\(\Rightarrow A=x-2y+3\le\dfrac{11}{2}\)

Dấu "=" xảy ra khi và chỉ khi

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

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

\(\Rightarrow\dfrac{4x^2}{\dfrac{1}{4}}=\dfrac{9y^2}{\dfrac{4}{9}}=\dfrac{4x^2+9y^2}{\dfrac{1}{4}+\dfrac{4}{9}}=\dfrac{9}{\dfrac{25}{36}}=\dfrac{9.36}{25}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{9.36}{25}.\dfrac{1}{16}\\y^2=\dfrac{9.36}{25}.\dfrac{4}{36}=\dfrac{9.4}{25}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{3.6}{5}.\dfrac{1}{4}=\dfrac{9}{10}\\y=\dfrac{3.2}{5}=\dfrac{6}{5}\end{matrix}\right.\)

Vậy \(GTLN\left(A\right)=\dfrac{11}{2}\left(tạix=\dfrac{9}{10};y=\dfrac{6}{5}\right)\)

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