1/ 0, 71

2/ Tương tự 2 câu 1, 3 nhé!

3/ 11,25

Tick đúng nha! Thanks!

\(B=-\left|x^2-9\right|-\left(x-3\right)^2+2020\)

Ta thấy : \(\hept{\begin{cases}\left|x^2-9\right|\ge0\forall x\Rightarrow-\left|x^2-9\right|\le0\\\left(x-3\right)^2\ge0\forall x\Rightarrow-\left(x-3\right)^2\le0\end{cases}}\)

\(\Rightarrow-\left|x^2-9\right|-\left(x-3\right)^2\le0\)

\(\Rightarrow-\left|x^2-9\right|-\left(x-3\right)^2+2020\le2020\)

\(\Rightarrow B\le2020\)

Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}x^2-9=0\\x-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x^2=9\\x=3\end{cases}\Leftrightarrow}x=3}\)

....

/x+5/ là giá trị tyệt đối nha

Ai giả hộ với mik cần gấp

câu 1 sai đề

2. =9/3 vì căn x-5 lớn hơn hoặc bằng 0

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

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