\(8^{3^{100}}-1=8^{300}-1=\left(8^{150}\right)^2-1^2=\left(8^{150}-1\right)\left(8^{150}+1\right)\left(ĐPCM\right)\)

A(x) = x(x-4) = 0 <=> x=0 or x=4

Áp dụng hằng đẳng thức nâng cao :
\(\left(x+y+z\right)^3=x^3+y^3+z^3+3\left(x+y+z\right)\left(xy+yz+xz\right)-3xyz\)

Vậy \(\left(x+y+z\right)^3-x^3-y^3-z^3=x^3+y^3+z^3+3\left(x+y+z\right)\left(xy+yz+xz\right)-3xyz-x^3-y^3-z^3 =3\left(x+y+z\right)\left(xy+yz+xz\right)-3xyz\)

Có dạng \(x^{3k+1}+x^{3n+2}+1\) thì có nhân tử chung là\(x^2+x+1\)

P(x)=\(\dfrac{5}{x-3}Max\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x-3}>0\\x-3>0,\\x-3Min\in Z\end{matrix}\right.\)

Vậy x-3=1 nên x=4 vậy max P(x)=5

Phần này khó chú ý nè bạn
Giải
Ta có f(x1+x2) = f(x1) + f(x2)
nên f(7) = f(3)+f(4)= f(2)+f(1) + f(2)+f(2) = f(1)+f(1)+f(1)+f(1)+f(1)+f(1)+f(1)=7

\(f\left(\dfrac{1}{7}\right)=\dfrac{1}{49}.f\left(7\right)=\dfrac{1}{49}.7=\dfrac{1}{7}\)

Ta có :\(f\left(\dfrac{5}{7}\right)=f\left(\dfrac{2}{7}\right)+f\left(\dfrac{3}{7}\right)=f\left(\dfrac{1}{7}\right)+f\left(\dfrac{1}{7}\right)+f\left(\dfrac{1}{7}\right)+f\left(\dfrac{2}{7}\right)=f\left(\dfrac{1}{7}\right)+f\left(\dfrac{1}{7}\right)+f\left(\dfrac{1}{7}\right)+f\left(\dfrac{1}{7}\right)+f\left(\dfrac{1}{7}\right)=\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{7}+\dfrac{1}{7}=\dfrac{5}{7}\)

Nhớ tick mình nha

a,\(c1\dfrac{4^2.4^3}{2^{10}}=\dfrac{4^5}{4^5}=1\)

\(c2\dfrac{4^2.4^3}{2^{10}}=\dfrac{2^4.2^6}{2^{10}}=1\)

cậu giải cho tớ đi ! lúc nãy tớ cho cậu thêm 3 sao rồi còn gì nữa ! nhá 

Tick mình nha

a, 4(x-3) = \(7^2-1^{10}=49-1=48\)

hay 4x-12= 48 \(\Rightarrow x=15\)

b,\(3^2\left(x+4\right)-5^2=5.2^2\Rightarrow9\left(x+4\right)-25=20\Rightarrow9\left(x+4\right)=45\Rightarrow x+4=5\Rightarrow x=1\)

c , 0:x = 0 \(\Leftrightarrow x\ne0\) thì luôn thỏa mãn

\(x\in\left(1;2;3;4;7\right)\)