mk có cách chứng minh khác:

\(a\sqrt{ac}+b\sqrt{ba}+c\sqrt{cb}\le\dfrac{a^2+b^2+c^2+ab+bc+ca}{2}\)

áp dụng bất đẳng thức cauchy cho 2 số dương \(\dfrac{a^3}{b}\) và \(ab\) ta có : \(\dfrac{a^3}{b}+ab\ge2\sqrt{\dfrac{a^3}{b}.ab}=2\sqrt{a^4}=2a^2\) (1)

tương tự ta cũng có:\(\dfrac{b^3}{c}+bc\ge2b^2\) (2)

\(\dfrac{c^3}{a}+ca\ge2c^2\) (3)

từ(1);(2);(3) \(\Rightarrow\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}\ge2\left(a^2+b^2+c^2\right)-ab-bc-ca\)(*)

còn chứng minh :

\(2\left(a^2+b^2+c^2\right)-ab-bc-ca\ge\dfrac{a^2+b^2+c^2+ab+bc+ca}{2}\)

\(\Leftrightarrow3\left(a^2+b^2+c^2\right)-3\left(ab+bc+ca\right)\ge0\)

\(\Leftrightarrow3\left(a^2+b^2+c^2-ab-bc-ca\right)\ge0\) (**)

(luôn đúng vì \(a^2+b^2+c^2\ge ab+bc+ca\) )

từ (*);(**) suy ra đpcm

dấu "=" xảy ra khi a=b=c

a/ \(5x\left(x-3\right)^2-5\left(x-1\right)^3+15\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow5x\left(x^2-6x+9\right)-5\left(x^3-3x^2+3x-1\right)+15\left(x^2-4\right)=5\)

\(\Leftrightarrow5x^3-30x^2+45x-5x^3+15x^2-15x+5+15x^2-60-5=0\)

\(\Leftrightarrow30x-60=0\)

\(\Leftrightarrow30x=60\)

\(\Leftrightarrow x=2\)

vậy x=2

b/ \(\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)

\(\Leftrightarrow3x-4x^2+6-8x=x^2+4x+4\)

\(\Leftrightarrow x^2+4x^2+4x+18x-3x+4-6=0\)

\(\Leftrightarrow5x^2+9x-2=0\)

\(\Leftrightarrow\left(5x-1\right)\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-2\end{matrix}\right.\)

vậy \(x=\dfrac{1}{5}\) hoặc \(x=-2\)

c/ \(x^2\left(x^2+1\right)-x^2-1=0\)

\(\Leftrightarrow x^2\left(x^2+1\right)-\left(x^2+1\right)=0\)

\(\Leftrightarrow\left(x^2+1\right)\left(x^2-1\right)=0\)

vì x²+1 >0 nên x² - 1 = 0 \(\Rightarrow x^2=1\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

vậy \(x=1\) hoặc \(x=-1\)

\(a\left(b^2-c^2\right)+b\left(c^2-a^2\right)+c\left(a^2-b^2\right)\) \(=ab^2-ac^2+bc^2-ba^2+ca^2-cb^2\)

\(=\left(ab^2-cb^2\right)+\left(ca^2-ac^2\right)+\left(bc^2-ba^2\right)\)

\(=b^2\left(a-c\right)+ac\left(a-c\right)-b\left(a-c\right)\left(a+c\right)\)

\(=\left(a-c\right)\left(b^2+ac-ab-bc\right)\)

\(=\left(a-c\right)\left(\left(b^2-bc\right)-\left(ab-ac\right)\right)\)

\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)

Xét riêng (x + y)⁴ = [(x + y)²]²

= [x²+2xy+y²]²

= x⁴ +4x²y² + y⁴ + 4x³y + 2x²y²+4xy³
Vậy (x + y)⁴ +x⁴ + y⁴ = x⁴ +4x²y² + y⁴ + 4x³y + 2x²y²+4xy³+ x⁴ + y⁴
= 2x⁴ + 2y⁴ + 6x²y² + 4x³y + 4xy³
= 2(x⁴ + y⁴ + 3x²y² +2x³y + 2xy³)
= 2(x⁴ + y⁴ + x²y² + 2x³y + 2xy³ + 2x²y²)
= 2(x² + xy + y²)²

\(\left(x+1\right)^3-\left(x-1\right)^3-9x^2+3x\)

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

= \(x^3+3x^2+3x+1-x^3+3x^2-3x+1-9x^2+3x\)

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

kẻ AK vuông góc BC

AH vuông góc AD

= 105\(^o\), = 60\(^o\)

= 15\(^o\)

ta có \(^0\)

AK = AK= \(\dfrac{\sqrt{3}}{2}\)

suy ra KC =

AC²= AK² + KC² = \(\dfrac{3}{4}+\dfrac{21+12\sqrt{3}}{4}\)

AC² =

\(\dfrac{1}{CA^2}=\dfrac{1}{6+3\sqrt{3}}\) (1)

Xét tam giác ABH có : AB=1 (gt)

suy ra BH = 2 và AH =

suy ra DC=

\(\dfrac{AD^2}{AC^2}=\dfrac{EB^2}{BC^2}\) ( TA LÉT)

suy ra AD²=\(\dfrac{6+3\sqrt{3}}{7+4\sqrt{3}}\)= \(6-3\sqrt{3}\)

suy ra \(\dfrac{1}{AD^2}=\dfrac{1}{6-3\sqrt{3}}\) (2)

cộng (1) và (2) suy ra ta có đpcm

ta có : \(\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a}=\dfrac{a^3}{b}+bc+\dfrac{b^3}{c}+ca+\dfrac{c^3}{a}+ab-\left(ac+bc+ab\right)\)

\(=\dfrac{a^3}{b}+bc+\dfrac{b^3}{c}+ca+\dfrac{c^3}{a}+ab-\left(\dfrac{ab}{2}+\dfrac{bc}{2}+\dfrac{ab}{2}+\dfrac{ac}{2}+\dfrac{bc}{2}+\dfrac{ac}{2}\right)\)

\(\ge2.\sqrt{\dfrac{a^3}{b}.bc}+2\sqrt{\dfrac{b^3}{c}.ca}+2\sqrt{\dfrac{c^3}{a}.ab}-2\sqrt{\dfrac{ab.bc}{4}}-2\sqrt{\dfrac{ab.ac}{4}}-2\sqrt{\dfrac{bc.ac}{4}}\)

\(\ge2a\sqrt{ac}+2b\sqrt{ba}+2c\sqrt{cb}-b\sqrt{ac}-a\sqrt{bc}-c\sqrt{ab}=a\sqrt{ac}+b\sqrt{ba}+c\sqrt{cb}\left(ĐPCM\right)\)

Bài 2:

a/ đặt A= \(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\)

\(\Leftrightarrow A^2=3+\sqrt{5}+3-\sqrt{5}-2\sqrt{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}\)

\(\Leftrightarrow A^2=6-2\sqrt{9-5}\)

\(\Leftrightarrow A^2=6-2\sqrt{4}=6-4=2\)

\(\Rightarrow A=\sqrt{2}\)

\(\Rightarrow\)\(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}\) = \(\sqrt{2}\)

\(\Rightarrow\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}}-\sqrt{2}=\sqrt{2}-\sqrt{2}=0\)

b/ \(\left(\sqrt{12}+\sqrt{75}+\sqrt{27}\right):\sqrt{15}\)

\(=\dfrac{\sqrt{12}}{\sqrt{15}}+\dfrac{\sqrt{75}}{\sqrt{15}}+\dfrac{\sqrt{27}}{\sqrt{15}}=\sqrt{\dfrac{12}{15}}+\sqrt{\dfrac{75}{15}}+\sqrt{\dfrac{27}{15}}\)

\(=\dfrac{2\sqrt{5}}{5}+\sqrt{5}+\dfrac{3\sqrt{5}}{5}=\left(\dfrac{2\sqrt{5}}{5}+\dfrac{3\sqrt{5}}{5}\right)+\sqrt{5}\)

\(=\sqrt{5}+\sqrt{5}=2\sqrt{5}\)

c/ \(\left(12\sqrt{20}-8\sqrt{200}+7\sqrt{450}\right):\sqrt{10}\)

\(=\left(24\sqrt{5}-80\sqrt{2}+105\sqrt{2}\right):\sqrt{10}\)

\(=\left(24\sqrt{5}+25\sqrt{2}\right):\sqrt{10}=\dfrac{24\sqrt{5}}{\sqrt{10}}+\dfrac{25\sqrt{2}}{\sqrt{10}}\)

\(=12\sqrt{2}+5\sqrt{5}\)