\(C=x^2+3x\)

\(=x^2+2.x.\frac{3}{2}+\frac{9}{4}-\frac{9}{4}\)

\(=\left(x+\frac{3}{2}\right)^2-\frac{9}{4}\)

Vì \(\left(x+\frac{3}{2}\right)^2\ge0;\forall x\)

\(\Rightarrow\left(x+\frac{3}{2}\right)^2-\frac{9}{4}\ge0-\frac{9}{4};\forall x\)

Hay \(C\ge-\frac{9}{4};\forall x\)

Dấu"="xảy ra \(\Leftrightarrow\left(x+\frac{3}{2}\right)^2=0\)

                     \(\Leftrightarrow x=\frac{-3}{2}\)

Vậy \(C_{min}=\frac{-9}{4}\)\(\Leftrightarrow x=\frac{-3}{2}\)

A= (x² -1)( x² +2 )(x² +3)(x² +6)

=> A= [ (x²-1)(x² +6)] [ (x² +2)(x² +3)]

=> A= (x⁴ + 5x² -6)(x⁴ +5x²+6)

=> A= (x⁴ +5x²)² - 6² = ( x⁴ + 5x²)² -36

Min A= -36 khi x⁴ +5x² =0 => x²( x² +5) =0 => x=0  hoặc x = căn 5

luộn nha , ko có căn 5 đâu bạn, vì x² + 5 luôn lớn hơn 0

GIải hộ mình bài 4 câu a nhé <3

không biết bó tay

Ta có: b² = ac => \(\frac{a}{b}=\frac{b}{c}\); c² = bd => \(\frac{b}{c}=\frac{c}{d}\);  d² = ce => \(\frac{c}{d}=\frac{d}{e}\); e² = df => \(\frac{d}{e}=\frac{e}{f}\)

\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{e}=\frac{e}{f}\)\(\Rightarrow\frac{a^5}{b^5}=\frac{b^5}{c^5}=\frac{c^5}{d^5}=\frac{d^5}{e^5}=\frac{e^5}{f^5}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{a^5}{b^5}=\frac{b^5}{c^5}=\frac{c^5}{d^5}=\frac{d^5}{e^5}=\frac{e^5}{f^5}=\frac{a^5+b^5+c^5+d^5+e^5}{b^5+c^5+d^5+e^5+f^5}\)(1)

Lại có: \(\frac{a^5}{b^5}=\frac{a}{b}.\frac{a}{b}.\frac{a}{b}.\frac{a}{b}.\frac{a}{b}=\frac{a}{b}.\frac{b}{c}.\frac{c}{d}.\frac{d}{e}.\frac{e}{f}=\frac{a}{f}\)(2)

Từ (1), (2) \(\Rightarrow\frac{a^5+b^5+c^5+d^5+e^5}{b^5+c^5+d^5+e^5+f^5}=\frac{a}{f}\)(đpcm)

\(a^{102}+b^{102}=\left(a^{101}+b^{101}\right)\left(a+b\right)-ab\left(a^{100}+b^{100}\right)\)

Mà \(a^{100}+b^{100}=a^{101}+b^{101}=a^{102}+b^{102}\)

Khi đó:

\(a^{102}+b^{102}=\left(a^{102}+b^{102}\right)\left(a+b-ab\right)\)

\(\Rightarrow a+b-ab=1\)

\(\Rightarrow\left(a-1\right)\left(b-1\right)=0\)

\(\Rightarrow a=1;b=1\)

\(\Rightarrow a^{2010}+b^{2010}=2\)