(81-1^2)(81-2^2)(81-3^2)....(81-9^2)

= (81-1^2)(81-2^2)(81-3^2)....(81-81)

= (81-1^2)(81-2^2)(81-3^2)....0

= 0

a. 96 : 8 : 2 = 6

96 : ( 8 x 2 ) = 6

96 : 8 : 2 = 96 : ( 8 x 2 )

b. 81 : 9 : 3 = 3

81 : ( 9 x 3 ) = 3

81 : 9 : 3 = 81 : ( 9x 3)

tu hoc moi gioi

\(4^{x-5}=16\)

\(4^{x-5}=4^2\)

\(x-5=2\)

\(x=2+5\)

\(x=7\)

\(45-2^{x-1}=29\)

\(2^{x-1}=16\)

\(2^{x-1}=2^4\)

\(x-1=4\)

\(x=5\)

\(\left(2+x\right)^2=144\)

\(\left(2+x\right)^2=12^2\)

\(2+x=12\)

\(x=12-2\)

\(x=10\)

\(\left(x-5\right)^2=81\)

\(\left(x-5\right)^2=9^2\)

\(x-5=9\)

\(x=14\)

\(\left(13-x\right)^4=81\)

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

\(13-x=3\)

\(x=13-3\)

\(x=10\)

\(...4^{x-5}=4^2\Rightarrow x-5=2\Rightarrow x=7\)

\(...2^{x-1}=45-29=16\Rightarrow2^{x-1}=2^4\Rightarrow x-1=4\Rightarrow x=5\)

\(...\Rightarrow\left(2+x\right)^2=12^2\Rightarrow\left[{}\begin{matrix}2+x=12\\2+x=-12\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=-14\end{matrix}\right.\)

\(...\Rightarrow\left(x-5\right)^2=9^2\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

\(...\Rightarrow\left(13-x\right)^4=3^4\Rightarrow\left[{}\begin{matrix}13-x=3\\13-x=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=10\\x=16\end{matrix}\right.\)

Mình quên mũ 2 với mũ 4 là chia ra 2 trường hợp, sửa lại nhé:

\(\left(2+x\right)^2=144\)

 \(\left(2+x\right)^2=12^2\)

Th1: 

\(2+x=12\)

\(x=10\)

Th2: 

\(2+x=-12\)

\(x=-14\)

\(\left(x-5\right)^2=81\)

\(\left(x-5\right)^2=9^2\)

Th1: 

\(x-5=9\)

\(x=14\)

Th2: 

\(x-5=-9\)

\(x=-4\)

\(\left(13-x\right)^4=81\)

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

Th1: 

\(13-x=3\)

\(x=10\)

Th2: 

\(13-x=-3\)

\(x=16\)

Bài 2: 

x=13 nên x+1=14

\(f\left(x\right)=x^{14}-x^{13}\left(x+1\right)+x^{12}\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+14\)

\(=x^{14}-x^{14}-x^{13}+x^{13}-...+x^3+x^2-x^2-x+14\)

=14-x=1

x=13 nên x+1=14

f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14f(x)=x14−x13(x+1)+x12(x+1)−...+x2(x+1)−x(x+1)+14

=x14−x14−x13+x13−...+x3+x2−x2−x+14=x14−x14−x13+x13−...+x3+x2−x2−x+14

=14-x=1

=> \(\frac{x+2}{10^{10}}+\frac{x+2}{11^{11}}-\frac{x+2}{12^{12}}-\frac{x+2}{13^{13}}=0\)

=> (x + 2)(\(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\)) = 0

=> x + 2 = 0 (vì \(\frac{1}{10^{10}}+\frac{1}{11^{11}}-\frac{1}{12^{12}}-\frac{1}{13^{13}}\)= 0)

=> x = -2

Đề hiển thị lỗi. Bạn xem lại đề.