\(P=3\left(x^2+\frac{7}{3}x-\frac{5}{3}\right)\)

\(P=3\left(x^2+2.x.\frac{7}{6}+\frac{49}{36}-\frac{109}{36}\right)\)

\(P=3\left(x+\frac{7}{6}\right)^2-\frac{109}{12}\)

\(P_{min}=-\frac{109}{12}\Leftrightarrow x==-\frac{7}{6}\)

tại sao lại có -109/12 vậy bạn?

\(a)\)\(\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)

\(\Leftrightarrow x^2-4x+4-x^2+9-6=0\)

\(\Leftrightarrow-4x+7=0\)

\(\Leftrightarrow x=\frac{7}{4}\)

Vậy\(x=\frac{7}{4}\)

\(b)\)\(4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)

\(\Leftrightarrow4\left(x^2-6x+9\right)-4x^2+1-10=0\)

\(\Leftrightarrow4x^2-24x+36-4x^2+1-10=0\)

\(\Leftrightarrow-24x+27=0\)

\(\Leftrightarrow x=\frac{9}{8}\)

Vậy\(x=\frac{9}{8}\)

\(c)\)\(\left(x-4\right)^2-\left(x-2\right)\left(x+2\right)=6\)

\(\Leftrightarrow x^2-8x+16-x^2+4-6=0\)

\(\Leftrightarrow-8x+14=0\)

\(\Leftrightarrow x=\frac{7}{4}\)

Vậy\(x=\frac{7}{4}\)

Linz

\(a.\left(x-2\right)^2-\left(x-3\right)\left(x+3\right)=6\)

\(x^2-4x+4-x^2+9-6=0\)

\(-4x=-7\)

\(x=\frac{7}{4}\)

\(b.4\left(x-3\right)^2-\left(2x-1\right)\left(2x+1\right)=10\)

\(4\left(x^2-6x+9\right)-4x^2+1=10\)

\(-24x=-27\Leftrightarrow x=\frac{9}{8}\)

Ta có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

=> \(\frac{1}{a}+\frac{1}{b}=\frac{1}{a+b+c}-\frac{1}{c}\)

=> \(\frac{a+b}{ab}=\frac{-\left(a+b\right)}{\left(a+b+c\right)c}\)

Nếu a + b = 0

=> a = -b

Khi đó \(\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}=\frac{1}{-b^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}=\frac{1}{c^{2017}}\)(1)

\(\frac{1}{a^{2017}+b^{2017}+c^{2017}}=\frac{1}{-b^{2017}+b^{2017}+c^{2017}}=\frac{1}{c^{2017}}\)(2)

Từ (1)(2) => \(\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}=\frac{1}{a^{2017}+b^{2017}+c^{2017}}\)(3)

Nếu a + b \(\ne\)0

=> ab = -(a + b + c).c

=> ab = -ac - bc - c²

=> ab + ac + bc+ c² = 0

=> a(b + c) + c(b + c) = 0

=> (a + c)(b + c) = 0

=> \(\orbr{\begin{cases}a+c=0\\b+c=0\end{cases}}\Rightarrow\orbr{\begin{cases}a=-c\\b=-c\end{cases}}\)

Tương tự (1);(2) thay a = -c vào đẳng thức ta được 

\(\hept{\begin{cases}\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}=\frac{1}{b^{2017}}\\\frac{1}{a^{2017}+b^{2017}+c^{2017}}=\frac{1}{b^{2017}}\end{cases}\Rightarrowđpcm}\)(4)

Với b = -c ta được 

\(\hept{\begin{cases}\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}=\frac{1}{a^{2017}}\\\frac{1}{a^{2017}+b^{2017}+c^{2017}}=\frac{1}{a^{2017}}\end{cases}}\Rightarrow\text{đpcm}\)(5)

Từ (3)(4)(5)

Vậy \(\frac{1}{a^{2017}}+\frac{1}{b^{2017}}+\frac{1}{c^{2017}}=\frac{1}{a^{2017}+b^{2017}+c^{2017}}\)