a) Áp dụng Cauchy Schwars ta có:

\(M=\frac{a^2}{a+1}+\frac{b^2}{b+1}+\frac{c^2}{c+1}\ge\frac{\left(a+b+c\right)^2}{a+b+c+3}=\frac{9}{6}=\frac{3}{2}\)

Dấu "=" xảy ra khi: a = b = c = 1

b) \(N=\frac{1}{a}+\frac{4}{b+1}+\frac{9}{c+2}\ge\frac{\left(1+2+3\right)^2}{a+b+c+3}=\frac{36}{6}=6\)

Dấu "=" xảy ra khi: x=y=1

c) \(P=\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}\ge\frac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}=\frac{9}{2.3}=\frac{3}{2}\)

Dấu "=" xảy ra khi: x=y=1

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

\(=7\cdot\sqrt{\left(a-b\right)^2+4ab}\)

\(=7\cdot\sqrt{7^2+4\cdot60}=119\)

\(a>b>0\Rightarrow a+b>0\)

\(\left(a+b\right)^2=\left(a-b\right)^2+4ab=7^2+4.60=289\Rightarrow a+b=17\)

\(\Rightarrow a^2-b^2=\left(a-b\right)\left(a+b\right)=7.17=119\)

\(a^2+b^2=\left(a-b\right)^2+2ab=7^2+2.60=169\)

\(\Rightarrow a^4+b^4=\left(a^2+b^2\right)^2-2\left(ab\right)^2=169^2-2.60^2=21361\)

Đây nhé! Tích giúp c nha

\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)

\(\Rightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)

\(\Rightarrow ab+bc+ca=-5\)

\(\Rightarrow\left(ab+bc+ca\right)^2=25\)

\(\Rightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2+2abc\left(a+b+c\right)=25\)

\(\Rightarrow\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2=25\)

\(\Rightarrow a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left[\left(ab\right)^2+\left(bc\right)^2+\left(ca\right)^2\right]\)

\(=10^2-2.25=50\)

Ta có: a+b+c=0 ⇒(a+b+c)²=0

Hay a²+b²+c²+2ab+2bc+2ca=0

1+2(ac+bc+ca)=0

ab+bc+ca=\(\dfrac{-1}{2}\)

\(\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=100\left(1\right)\)

\(\left(ab+bc+ca\right)^2=a^2b^2+b^2c^2+c^2a^2+b^2ac+c^2ab+a^bc=a^2b^2+b^2c^2+c^2+a^2+2abc\left(a+b+c\right)=a^2b^2+b^2c^2+c^2a^2=25\)

hay \(2\left(a^2b^2+b^2c^2+c^2a^2\right)=50\left(2\right)\)

Từ (1) và (2) ⇒a⁴+b⁴+c⁴=50

\(a,a^2+b^2=\left(a+b\right)^2-2ab=9^2-2\cdot20=41\\ b,a^4+b^4=\left(a^2+b^2\right)^2-2a^2b^2=41^2-2\left(ab\right)^2\\ =1681-2\cdot400=881\\ c,\left(a-b\right)^2=a^2+b^2-2ab=41-2\cdot20=1\\ \Rightarrow a-b=1\\ \Rightarrow C=a^2-b^2=\left(a-b\right)\left(a+b\right)=9\cdot1=9\)

Ta có 

+) Nếu  thì 

Ta có : 

+ Nếu  làm tương tự

a)Có \(a^2+1\ge2a\) với mọi a; \(b^2+1\ge2b\) với mọi b

Cộng vế với vế \(\Rightarrow a^2+b^2+2\ge2\left(a+b\right)\)

Dấu = xảy ra <=> a=b=1

b) Áp dụng BĐT bunhiacopxki có:

\(\left(x+y\right)^2\le\left(1+1\right)\left(x^2+y^2\right)\Leftrightarrow\left(x+y\right)^2\le2\)

\(\Leftrightarrow-\sqrt{2}\le x+y\le\sqrt{2}\)

\(\Rightarrow\left(x+y\right)_{max}=\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x+y=\sqrt{2}\\x=y\end{matrix}\right.\)\(\Leftrightarrow x=y=\dfrac{\sqrt{2}}{2}\)

\(\left(x+y\right)_{min}=-\sqrt{2}\Leftrightarrow\left\{{}\begin{matrix}x+y=-\sqrt{2}\\x=y\end{matrix}\right.\)\(\Leftrightarrow x=y=-\dfrac{\sqrt{2}}{2}\)

c) \(S=\dfrac{1}{ab}+\dfrac{1}{a^2+b^2}=\dfrac{1}{a^2+b^2}+\dfrac{1}{2ab}+\dfrac{1}{2ab}\)

Với x,y>0, ta có: \(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\) (1)

Thật vậy (1) \(\Leftrightarrow\dfrac{y+x}{xy}\ge\dfrac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)\(\Leftrightarrow\left(x-y\right)^2\ge0\) (lđ)

Áp dụng (1) vào S ta được:

\(S\ge\dfrac{4}{a^2+b^2+2ab}+\dfrac{1}{2ab}\)

Lại có: \(ab\le\dfrac{\left(a+b\right)^2}{4}\) \(\Leftrightarrow2ab\le\dfrac{\left(a+b\right)^2}{2}\Leftrightarrow2ab\le\dfrac{1}{2}\)\(\Rightarrow\dfrac{1}{2ab}\ge2\)

\(\Rightarrow S\ge\dfrac{4}{\left(a+b\right)^2}+2=6\)

\(\Rightarrow S_{min}=6\Leftrightarrow a=b=\dfrac{1}{2}\)

Ta có: a + b + c = 0

\(\Rightarrow\) (a + b + c)² = 0

\(\Leftrightarrow\) a² + b² + c² + 2ab + 2bc + 2ac = 0

\(\Leftrightarrow\) 2009 + 2(ab + bc + ac) = 0

\(\Leftrightarrow\) ab + bc + ac = \(\dfrac{-2009}{2}\)

\(\Leftrightarrow\) (ab + bc + ac)² = \(\left(\dfrac{-2009}{2}\right)^2\)

\(\Leftrightarrow\) a²b² + b²c² + a²c² + 2abc(a + b + c) = \(\left(\dfrac{-2009}{2}\right)^2\)

\(\Leftrightarrow\) a²b² + b²c² + c²a² = \(\left(\dfrac{-2009}{2}\right)^2\)    (Vì a + b + c = 0)

Lại có: a² + b² + c² = 2009

\(\Rightarrow\) (a² + b² + c²)² = 2009²

\(\Leftrightarrow\) a⁴ + b⁴ + c⁴ + 2(a²b² + b²c² + c²a²) = 2009²

\(\Leftrightarrow\) a⁴ + b⁴ + c⁴ + 2.\(\dfrac{2009^2}{4}\) = 2009²

\(\Leftrightarrow\) a⁴ + b⁴ + c⁴ = 2009² - \(\dfrac{2009^2}{2}\) = 2018040,5

Chúc bn học tốt!