\(f\left(x\right)=x^6-10x^5+10x^4-10x^3+10x^2-10x+10\)

\(f\left(x\right)=x^5\left(x-10\right)+x^3\left(x-10\right)+x\left(x-10\right)+10\)

\(f\left(x\right)=\left(x-10\right)\left(x^5+x^3+x\right)+10\)

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

\(\Rightarrow f\left(9\right)=9.\left(9-10\right)\left(9^4+9^2+1\right)+10\)

\(\Leftrightarrow f\left(9\right)=9.\left(-1\right).\left(6643\right)+10\)

\(\Leftrightarrow f\left(9\right)=-59777\)

P/s : làm cho zui thôi nha , sai đừng đáp đá 

\(x=9\)\(\Rightarrow x+1=10\)

\(\Rightarrow f\left(9\right)=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+\left(x+1\right)\)

               \(=x^6-x^6-x^5+x^5+.......-x+x+1=1\)

a) Ta có: \(P\left(x\right)=x^7-80x^6+80x^5-80x^4+...+80x+15\)

\(=x^7-x^6\left(x+1\right)+x^5\left(x+1\right)-...+x\left(x+1\right)+15\)

\(=x^7-x^7-x^6+x^6+x^5-...+x^2+x+15\)

\(=x+15\)

Thay x=79 vào biểu thức \(P\left(x\right)=x+15\), ta được:

\(P\left(79\right)=79+15=94\)

\(B=x^5-15x^4+16x^3-29x^2+13x\)

\(=x^5-14x^4-x^4+14x^3+2x^3-28x^2-x^2+14x-x+14-14\)

\(=x^4\left(x-14\right)-x^3\left(x-14\right)+2x^2\left(x-14\right)-x\left(x-14\right)-\left(x-14\right)-14\)

\(=\left(x^4-x^3+2x^2-x-1\right)\left(x-14\right)-14\)

Thay x = 14 => B = -14

Vậy...

phần còn lại tách ra làm tương tự nhé

cu tao to

Bài vừa dài vừa khó éo trả lời

\(\dfrac{3}{5x-5}-\dfrac{x+5}{10x-10}=\dfrac{6}{10\left(x-1\right)}-\dfrac{x+5}{10\left(x-1\right)}=\dfrac{6-\left(x+5\right)}{10\left(x-1\right)}\)

\(=\dfrac{-\left(x-1\right)}{10\left(x-1\right)}=-\dfrac{1}{10}\)

   \(\dfrac{3}{5x-5}-\dfrac{x+5}{10x-10}\)

= \(\dfrac{3}{5.(x-1)}\) - \(\dfrac{x+5}{2.5.(x-1)}\) 

= \(\dfrac{6-x-5}{10.(x-1)}\)

= \(\dfrac{1-x}{10.(x-1)}\)

= \(\dfrac{-1(x-1)}{10(x-1)}\)

= \(\dfrac{-1}{10}\)

\(đk:x\ne1\\ \dfrac{3}{5x-5}-\dfrac{x+5}{10x-10}\\ =\dfrac{3.2}{2.\left(5x-5\right)}-\dfrac{x+5}{10x-10}\\ =\dfrac{6}{10x-10}-\dfrac{x+5}{10x-10}\\ =\dfrac{6-x-5}{10x-10}\\ =\dfrac{1-x}{10x-10}\\ =\dfrac{-\left(x-1\right)}{10\left(x-1\right)}\\ =\dfrac{-1}{10}\)

D = \(x^{10}-25x^9+25x^8-25x^7+...+25x^2-25x+25\)với x = 24

thiếu 1 câu

A= với x = 4

= (x+1)(x+1)(x+1)

= x5−x4+x4+x3−x3+x2−x2+x−1

=x−1=4−1=3

Tương tự với các câu B,C,D

x\(\approx\)-0,84;-11,9

x=9

\(9^{14}-10.9^{13}+10.9^{12}-10.9^{11}+..+10.9^2-10.9+10\)

\(9^{14}-\left(9+1\right).9^{13}+\left(9+1\right).9^{12}+..+\left(9+1\right).9^2-\left(9+1\right)9+10\)

\(9^{14}-9^{14}-9^{13}+9^{13}+9^{12}-..+9^3+9^2-9^2-9+10=1\)

Vậy......

Cảm ơn