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M = 4⁰ + 4¹ + 4² + 4³ + ........ + 4²³

=> 4M = 4¹ + 4² + 4³ + 4⁴ + ............ + 4²⁴

=> 4M - M = (4¹ + 4² + 4³ + 4⁴ + ............ + 4²⁴) - (4⁰ + 4¹ + 4² + 4³ + ........ + 4²³)

=> 3M = 4²⁴ - 1

=> 3M + 1 = 4²⁴ = 4⁷.4¹⁷ > 6.3⁷

ĐKXĐ:\(x\ne-1;1\)

Xét \(x< -1\)

PT\(\Leftrightarrow\dfrac{-1}{x+1}+\dfrac{2}{x-1}=\dfrac{1}{x^2-1}\)

\(\Leftrightarrow\dfrac{-x+1+2x+2}{x^2-1}=\dfrac{1}{x^2-1}\)

\(\Leftrightarrow x+3=1\)

\(\Leftrightarrow x=-2\left(tm\right)\)

Xét x>-1

PT\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{2}{x-1}=\dfrac{1}{x^2-1}\)

\(\Leftrightarrow\dfrac{x-1+2x+2}{x^2-1}=\dfrac{1}{x^2-1}\)

\(\Leftrightarrow3x+1=1\)

\(\Leftrightarrow x=0\left(tm\right)\)

Vậy ...

BĐT\(\Leftrightarrow\left(x^2-7x+6\right)\left(x^2-7x+12\right)+9\ge0\)

Đặt \(x^2-7x+9=t\)

BĐT\(\Leftrightarrow\left(t-3\right)\left(t+3\right)+9\ge0\)

\(\Leftrightarrow t^2-9+9\ge0\)

\(\Leftrightarrow t^2\ge0\)(luôn đúng)

\(\Rightarrowđpcm\)

a)\(A=\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}\)

\(A=\dfrac{a^3}{abc}+\dfrac{b^3}{abc}+\dfrac{c^3}{abc}\)

\(A=\dfrac{a^3+b^3+c^3}{abc}\)

\(A=\dfrac{3abc}{abc}=3\)(vì a+b+c=0)

b)Ta có: a+b+c=0

\(\Rightarrow\left\{{}\begin{matrix}a=-b-c\\b=-c-a\\c=-a-b\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a^2=\left(b+c\right)^2\\b^2=\left(c+a\right)^2\\c^2=\left(a+b\right)^2\end{matrix}\right.\)

\(\Rightarrow B=\dfrac{a^2}{\left(b+c\right)^2-b^2-c^2}+\dfrac{b^2}{\left(a+c\right)^2-c^2-a^2}+\dfrac{c^2}{\left(a+b\right)^2-a^2-b^2}\)

\(\Rightarrow B=\dfrac{a^2}{2bc}+\dfrac{b^2}{2ca}+\dfrac{c^2}{2ab}\)

\(\Rightarrow B=\dfrac{a^3+b^3+c^3}{2abc}\)

\(\Rightarrow B=\dfrac{3abc}{2abc}=\dfrac{3}{2}\)(vì a+b+c=0)