Question

1, Tìm giá trị lớn nhất của :

A=\(-x^2-5x+3\)

B=\(-2x^2-7x+9\)

C=\(-3x^2+5x-7\)

D=\(-4x^2+9x+3\)

E=\(-5x^2+10x+21\)

2, Tìm giá trị nhỏ nhất của :

A=\(x^2+5y^2+4xy+4x+10y+2010\)

B=\(x^2+15y^2+4x-10y-6xy+30\)

Answer 1

@Nguyễn Nhật Minh

@Aki Tsuki

@Phùng Khánh Linh

@Nào Ai Biết

@Nguyễn Thanh Hằng

@Mysterious Person

giúp mk với

Answer 2

Bài 1:

\(A=-x^2-5x+3=\frac{37}{4}-(x^2+5x+\frac{25}{4})\)

\(=\frac{37}{4}-(x+\frac{5}{2})^2\)

Vì \((x+\frac{5}{2})^2\geq 0\Rightarrow A=\frac{37}{4}-(x+\frac{5}{2})^2\leq \frac{37}{4}-0=\frac{37}{4}\)

Vậy A(max)\(=\frac{37}{4}\Leftrightarrow x=\frac{-5}{2}\)

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\(B=-2x^2-7x+9=\frac{121}{8}-2(x^2+\frac{7}{2}x+\frac{49}{16})\)

\(=\frac{121}{8}-2(x+\frac{7}{4})^2\)

Vì \((x+\frac{7}{4})^2\ge 0\Rightarrow B=\frac{121}{8}-2(x+\frac{7}{4})^2\leq \frac{121}{8}-2.0=\frac{121}{8}\)

Vậy B(max)\(=\frac{121}{8}\Leftrightarrow x=\frac{-7}{4}\)

Các câu còn lại bạn cũng làm tương tự.

Answer 3

Bài 1:

\(A=-x^2-5x+3=\frac{37}{4}-(x^2+5x+\frac{25}{4})\)

\(=\frac{37}{4}-(x+\frac{5}{2})^2\)

Vì \((x+\frac{5}{2})^2\geq 0\Rightarrow A=\frac{37}{4}-(x+\frac{5}{2})^2\leq \frac{37}{4}-0=\frac{37}{4}\)

Vậy A(max)\(=\frac{37}{4}\Leftrightarrow x=\frac{-5}{2}\)

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\(B=-2x^2-7x+9=\frac{121}{8}-2(x^2+\frac{7}{2}x+\frac{49}{16})\)

\(=\frac{121}{8}-2(x+\frac{7}{4})^2\)

Vì \((x+\frac{7}{4})^2\ge 0\Rightarrow B=\frac{121}{8}-2(x+\frac{7}{4})^2\leq \frac{121}{8}-2.0=\frac{121}{8}\)

Vậy B(max)\(=\frac{121}{8}\Leftrightarrow x=\frac{-7}{4}\)

Các câu còn lại bạn cũng làm tương tự.

Answer 4

Bài 2:

\(A=x^2+5y^2+4xy+4x+10y+2010\)

\(=(x^2+4y^2+4xy)+y^2+4x+10y+2010\)

\(=(x+2y)^2+4(x+2y)+2^2+(y^2+2y+1)+2005\)

\(=(x+2y+2)^2+(y+1)^2+2005\)

\(\geq 0+0+2005=2005\)

Vậy \(A_{\min}=2005\)

Dấu bằng xảy ra khi \(\left\{\begin{matrix} x+2y+2=0\\ y+1=0\end{matrix}\right.\Leftrightarrow x=0;y=-1\)

Answer 5

Bài 2:

\(A=x^2+5y^2+4xy+4x+10y+2010\)

\(=(x^2+4y^2+4xy)+y^2+4x+10y+2010\)

\(=(x+2y)^2+4(x+2y)+2^2+(y^2+2y+1)+2005\)

\(=(x+2y+2)^2+(y+1)^2+2005\)

\(\geq 0+0+2005=2005\)

Vậy \(A_{\min}=2005\)

Dấu bằng xảy ra khi \(\left\{\begin{matrix} x+2y+2=0\\ y+1=0\end{matrix}\right.\Leftrightarrow x=0;y=-1\)

Answer 6

2b)

\(B=x^2+15y^2+4x-10y-6xy+30\)

\(=(x^2+9y^2-6xy)+6y^2+4x-10y+30\)

\(=(x-3y)^2+4(x-3y)+4+6y^2+2y+26\)

\(=(x-3y+2)^2+6(y^2+\frac{1}{3}y+\frac{1}{36})+\frac{155}{6}\)

\(=(x-3y+2)^2+6(y+\frac{1}{6})^2+\frac{155}{6}\)

\(\geq 0+6.0+\frac{155}{6}=\frac{155}{6}\)

Vậy \(B_{\min}=\frac{155}{6}\Leftrightarrow \left\{\begin{matrix} x-3y+2=0\\ y+\frac{1}{6}=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{-5}{2}\\ y=\frac{-1}{6}\end{matrix}\right.\)

Answer 7

2b)

\(B=x^2+15y^2+4x-10y-6xy+30\)

\(=(x^2+9y^2-6xy)+6y^2+4x-10y+30\)

\(=(x-3y)^2+4(x-3y)+4+6y^2+2y+26\)

\(=(x-3y+2)^2+6(y^2+\frac{1}{3}y+\frac{1}{36})+\frac{155}{6}\)

\(=(x-3y+2)^2+6(y+\frac{1}{6})^2+\frac{155}{6}\)

\(\geq 0+6.0+\frac{155}{6}=\frac{155}{6}\)

Vậy \(B_{\min}=\frac{155}{6}\Leftrightarrow \left\{\begin{matrix} x-3y+2=0\\ y+\frac{1}{6}=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=\frac{-5}{2}\\ y=\frac{-1}{6}\end{matrix}\right.\)