Question

[Nguyễn Hải An]

Help me

Answer 1

[Phương An]

Bài 4:

a)

\(M=x+\sqrt{2-x}=-\left(2-x\right)+\sqrt{2-x}+2\)

Đặt \(\sqrt{2-x}=m\left(m\ge0\right)\)

\(\Rightarrow M=-m^2+m+2\)

\(=-\left(m^2-m+\dfrac{1}{4}\right)+\dfrac{1}{4}+2\)

\(=\dfrac{9}{4}-\left(m-\dfrac{1}{2}\right)^2\le\dfrac{9}{4}\)

Dấu "=" xảy ra khi \(m=\dfrac{1}{2}\Leftrightarrow\sqrt{2-x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{7}{4}\)

b)

\(5x^2+9y^2-12xy+8=24\left(2y-x-3\right)\)

\(\Leftrightarrow5x^2+24x+9y^2-48y-12xy+80=0\)

\(\Leftrightarrow\left(4x^2+9y^2+64-12xy-48y+32x\right)+\left(x^2-8x+16\right)=0\)

\(\Leftrightarrow\left(2x-3y+8\right)^2+\left(x-4\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=\dfrac{16}{3}\end{matrix}\right.\) (loại)

Vậy . . .

Answer 2

[Phương An]

Bài 2:

a)

\(M=\dfrac{x^5}{30}-\dfrac{x^3}{6}+\dfrac{2x}{15}\)

\(=\dfrac{x^5-5x^3+4x}{30}\)

\(=\dfrac{x\left(x^4-5x^2+4\right)}{30}\)

\(=\dfrac{x\left(x^2-4\right)\left(x^2-1\right)}{30}\)

\(=\dfrac{x\left(x-2\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)}{30}\)

Suy ra nếu x nguyên thì M cũng nguyên ^.^

Bài 3:

a) Chứng minh \(VP\ge VT\) dùng Cauchy Shwarz dạng Engel.

b) Xét \(M=2a^2+2b^2+2\)

\(=\left(a^2+1\right)+\left(b^2+1\right)+\left(a^2+b^2\right)\)

\(\ge2a+2b+2ab\) (áp dụng bđt AM - GM)

\(\Rightarrow a^2+b^2+1\ge a+b+ab\left(\text{đ}pcm\right)\)