=\(\left(3x^2-4x-13-4x^2+9\right)\left(3x^2-4x-13+4x^2-9\right)-\left(x+2\right)^4\)

=\(\left(-x^2-4x-4\right)\left(7x^2-4x-22\right)-\)\(\left(x+2\right)^{^{ }2.2}\)

=\(-\left(x+2\right)^2\left(7x^2-4x-22\right)-\left(x+2\right)^2\left(x+2\right)^2\)

=\(-\left(x+2\right)^2\)\(\left(7x^2-4x-22-x^2-4x-4\right)\)

\(-\left(x+2\right)^2\)(\(6x^2-8x-26\))

đề yêu cầu làm j z bạn?

Bạn xem trả lời ở đây nhé

Câu hỏi của Bùi Thị Hoài - Toán lớp 9 - Học toán với OnlineMath

Ta có: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)

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

\(=2\left(a^2+b^2+c^2+2ab+2ac+2bc\right)-6ab-6bc-6ac\)

\(=2\left(a+b+c\right)^2-6\left(ab+bc+ac\right)\)

\(=2.6^2-6.12=0\)

Mà : \(\left(a-b\right)^2\ge0;\left(b-c\right)^2\ge0;\left(a-c\right)^2\ge0\)

nên \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\)

Do đó: \(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)

<=> \(\hept{\begin{cases}\left(a-b\right)^2=0\\\left(a-c\right)^2=0\\\left(c-a\right)^2=0\end{cases}}\Leftrightarrow a=b=c\)

Vậy \(\left(a-b\right)^{2012}+\left(b-c\right)^{2013}+\left(c-a\right)^{2014}=0\)

Em mới học lớp 6 có gì sai  sót mong anh chỉ bảo !

\(\frac{x-7}{x^2+1}=\frac{x+6}{x^2+x+1}\)

\(\Leftrightarrow\frac{x-7}{x^2+1}=\frac{x+6}{x^2+x+1},x\inℝ\)

\(\Leftrightarrow\left(x-7\right).\left(x^2+x+1\right)=\left(x+6\right).\left(x^2+1\right)\)

\(\Leftrightarrow\left(x-7\right).\left(x^2+x+1\right)-\left(x+6\right).\left(x^2+1\right)=0\)

\(\Leftrightarrow x^3+x^2+x-7.x^2-7.x-7-\left(x^3+x+6.x^2+6\right)=0\)

\(\Leftrightarrow x^3+x^2+x-7.x^2-7.x-7-x^3-x-6.x^2-6=0\)

\(\Leftrightarrow-12.x^2-7.x-13=0\)

\(\Leftrightarrow12.x^2+7.x+13=0\)

\(\Leftrightarrow x=\frac{-7\pm\sqrt{7^2-4.12.13}}{2.12}\)

\(\Leftrightarrow x=\frac{-7\pm\sqrt{49-624}}{24}\)

\(\Leftrightarrow x=\frac{-7\pm\sqrt{-575}}{24}\)

Vậy x \(\notinℝ\)

 tìm ĐKXĐ mak e