10. C nhé

C.24 x (8 - 2 ) = 24 x 8 - 24 x 2

Bài làm:

Ta có: 

Pt <=> \(\left(-8+x^2\right)^5=1\)

\(\Rightarrow-8+x^2=1\)

\(\Leftrightarrow x^2=9\)

\(\Rightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

Vậy \(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)

dời trả lời nhanh z

định giúp bạn mink kiếm điểm ai ngờ...:))

=x^2+16x+64+2(x^2+6x-16)+x^2-4x+4

=2x^2+12x+68+2x^2+12x-32

=4x^2+24x+36

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

\(=\left(x+8\right)^2-2\left(x+8\right)\left(x-2\right)+\left(x-2\right)^2\)

\(=\left[\left(x+8\right)-\left(x-2\right)\right]^2\)

\(=\left(x+8-x+2\right)^2\)

\(=10^2\)

\(=100\)

không đăng linh tinh

tào lao

kho..............wa...................troi................thi......................ret.....................ai..............tich...............ung.....................ho....................minh..................voi................ret............wa

ặc toán lớp 8 thì khó quá trời

Ta có: \(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

=>\(\frac{1}{x+2}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+8}+\frac{1}{x+8}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

=>\(\frac{1}{x+2}-\frac{1}{x+14}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

=>\(\frac{x+14-x-2}{\left(x+2\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)

=>\(\frac{x}{\left(x+2\right)\left(x+14\right)}=\frac{12}{\left(x+2\right)\left(x+14\right)}\)

=>x=12(nhận)