Bài cuối:

Áp dụng BĐT AM - GM, ta có:

\(\dfrac{1}{a^2+bc}+\dfrac{1}{b^2+ac}+\dfrac{1}{c^2+ab}\)

\(\le\dfrac{1}{2a\sqrt{bc}}+\dfrac{1}{2b\sqrt{ac}}+\dfrac{1}{2c\sqrt{ab}}\)

\(=\dfrac{1}{2}\left(\dfrac{\sqrt{bc}+\sqrt{ac}+\sqrt{ab}}{abc}\right)\)

\(\le\dfrac{a+b+c}{2abc}\left(\text{đ}pcm\right)\)

Dấu "=" xảy ra khi a = b = c

\(\left(\dfrac{2\sqrt{x}-x}{x\sqrt{x}-1}-\dfrac{1}{\sqrt{x}-1}\right)\div\left(1-\dfrac{\sqrt{x}+2}{x+\sqrt{x}+1}\right)\)

\(=\left[\dfrac{\sqrt{x}\left(2-\sqrt{x}\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}-1}\right]\div\left[\dfrac{\left(x+\sqrt{x}+1\right)-\left(\sqrt{x}+2\right)}{x+\sqrt{x}+1}\right]\)

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

\(=\dfrac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x-1\right)}=\dfrac{1}{x-1}\)

\(\sqrt{9x^2-6x+1}+\sqrt{25-30x+9x^2}\)

\(=\sqrt{\left(3x-1\right)^2}+\sqrt{\left(5-3x\right)^2}\)

\(=\left|3x-1\right|+\left|5-3x\right|\)

\(\ge\left|3x-1+5-3x\right|=4\)

ta có sơ đồ :

số bé : I-----I

số lớn : I------I------I------I--I

[ chỗ ngắn ứng với 7 đơn vị ] 

số bé là :

[ 257 - 7 ] : [ 3 -1 ] x 1 = 125

số lớn là : 

125 + 257 = 382

P(x) = x⁴ - 8x + 63

P(x) phân tích đươc thành nhân tử có dạng (x² + ax + b)(x² + cx + d)

= x⁴ + (a + c)x³ + (b + ac + d)x² + (ad + bc)x + bd

Đồng nhất đa thức trên với đa thức đã cho, ta có:

\(\left\{{}\begin{matrix}a+c=0\\b+ac+d=0\\ad+bc=-8\\bd=63\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=-4\\b=7\\c=4\\d=9\end{matrix}\right.\)

Vậy P(x) = (x² - 4x + 7)(x² + 4x + 9)

Q(x) = 2x⁴ - 7x³ + 17x² - 20x + 14

Q(x) phân tích đươc thành nhân tử có dạng (x² + ax + b)(2x² + cx + d)

= 2x⁴ + (2a + c)x³ + (2b + ac + d)x² + (ad + bc)x + bd

Đồng nhất đa thức trên với đa thức đã cho, ta có:

\(\left\{{}\begin{matrix}2a+c=-7\\2b+ac+d=17\\ad+bc=-20\\bd=14\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=-2\\b=2\\c=-3\\d=7\end{matrix}\right.\)

Vậy Q(x) = (x² - 2x + 2)(2x² - 3x + 7)

2x² + 2y² + z² + 2xy + 2yz + 2xz + 10x + 6y + 34 = 0

<=> [x² + y² + z² + 2(xy + yz + xz)] + (x² + 10x + 25) + (y² + 6y + 9) = 0

<=> (x + y + z)² + (x + 5)² + (y + 3)² = 0

\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=0\\x+5=0\\y+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-5\\y=-3\\z=8\end{matrix}\right.\)

2x² + 2y² + z² + 2xy + 2xz + 2yz + 2x + 4y + 5 = 0

<=> (x² + y² + z² + 2xy + 2yz + 2xz) + (x² + 2x + 1) + (y² + 4y + 4) = 0

<=> (x + y + z)² + (x + 1)² + (y + 2)² = 0

\(\Leftrightarrow\left\{{}\begin{matrix}x+y+z=0\\x+1=0\\y+2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=-2\\z=3\end{matrix}\right.\)

\(M=\left(\dfrac{x\sqrt{x}-1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}\right)\left(1-\dfrac{3-\sqrt{x}}{\sqrt{x}+1}\right)\)

\(=\left[\dfrac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right]\)\(\left[\dfrac{\left(\sqrt{x}+1\right)-\left(3-\sqrt{x}\right)}{\sqrt{x}+1}\right]\)

\(=\left[\dfrac{\left(x+\sqrt{x}+1\right)-\left(x-\sqrt{x}+1\right)}{\sqrt{x}}\right]\times\dfrac{2\sqrt{x}-2}{\sqrt{x}+1}\)

\(=\dfrac{2\sqrt{x}\times2\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\)

\(=\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}+1}\)

\(\sqrt{28-6\sqrt{3}}\)

\(=\sqrt{\left(3\sqrt{3}-1\right)^2}\)

\(=3\sqrt{3}-1\)

\(\sqrt{6-\sqrt{20}}\)

\(=\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(=\sqrt{5}-1\)

\(\sqrt{2x+3+2\sqrt{\left(x+1\right)\left(x+2\right)}}\)

\(=\sqrt{\left(\sqrt{x+2}+\sqrt{x+1}\right)^2}\)

\(=\sqrt{x+2}+\sqrt{x+1}\)

\(\sqrt{2x+2-2\sqrt{x^2+2x-3}}\)

\(=\sqrt{\left(x-1\right)-2\sqrt{\left(x-1\right)\left(x+3\right)}+\left(x+3\right)}\)

\(=\sqrt{\left(\sqrt{x+3}-\sqrt{x-1}\right)^2}\)

\(=\left|\sqrt{x+3}-\sqrt{x-1}\right|\)

\(\sqrt{21-6\sqrt{6}}+\sqrt{21+6\sqrt{6}}\)

\(=\sqrt{\left(3\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(3\sqrt{2}-\sqrt{3}\right)^2}\)

\(=3\sqrt{2}+\sqrt{3}+3\sqrt{2}-\sqrt{3}\)

\(=6\sqrt{2}\)