Ta có 5x² - (2x + 1).(x - 2) - x(3x + 3) + 7

= 5x² - (2x² - 3x - 2) - 3x² - 3x + 7

= 5x² - 2x² + 3x + 2 - 3x² - 3x + 7

= 9 \(\forall\)x

=> Biểu thức không phụ thuộc vào biến 

5x² - ( 2x + 1 )( x - 2 ) - x( 3x + 3 ) + 7

= 5x² - ( 2x² - 3x - 2 ) - 3x² - 3x + 7

= 5x² - 2x² + 3x + 2 - 3x² - 3x + 7

= 9

=> đpcm

Ta có :  \(\frac{a}{a+1}=\frac{a^2+a-a^2}{a+1}=\frac{a\left(a+1\right)}{a+1}-\frac{a^2}{a+1}=a-\frac{a^2}{a+1}\)

Tương tự và cộng theo vế ta được : \(P=a+b+c-\left(\frac{a^2}{a+1}+\frac{b^2}{b+1}+\frac{c^2}{c+1}\right)\)

\(=1-\left(\frac{a^2}{a+1}+\frac{b^2}{b+1}+\frac{c^2}{c+1}\right)\ge1-\frac{\left(a+b+c\right)^2}{a+b+c+3}=1-\frac{1}{4}=\frac{3}{4}\)

Dấu "=" xảy ra khi và chỉ khi \(a=b=c=\frac{1}{3}\)Vậy GTNN của P = 3/4 đạt được khi a=b=c=1/3

Địa chỉ : 189 . Tổ 8 , Khu 9 , Phường Hồng Hà , Phố Trần Quốc Toản .

Dây chuyền : Bạc , 81,92 %.

Điều kiện : Sự thật , yêu cầu . (đủ)

ta có \(2^n\)\(⋮\)2

=>\(2^n-1⋮1\)

=>\(2^n-1\)là hợp số

\(p^3+p^2+1\)

=\(p^2+2+p^3-1\)

=

\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)

\(=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)

\(=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)

\(=\frac{2^{10}.3^8.\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}\)

\(=\frac{-1}{3}\)

Ta có:

\(\frac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)

\(=\frac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)

\(=\frac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\)

\(=-\frac{2}{6}=-\frac{1}{3}\)

a) \(14\left(x-y\right)^2+21\left(y-x\right)\)

\(=14\left(x-y\right)^2-21\left(x-y\right)\)

\(=7\left(x-y\right)\left[2\left(x-y\right)-3\right]\)

\(=7\left(x-y\right)\left(2x-2y-3\right)\)

b) \(7x^5\left(y-3\right)-49x^4\left(3-y\right)^3\)

\(=7x^4\left(y-3\right)\left[x+7\left(y-3\right)^2\right]\)

\(=7x^4\left(y-3\right)\left(x+7y^2-42y+63\right)\)

c) \(\left(x^2-9\right)^2-x^2\left(x-3\right)^2\)

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

\(=\left(x-3\right)^2\left[\left(x+3\right)^2-x^2\right]\)

\(=\left(x-3\right)^2\left(x^2+6x+9-x^2\right)\)

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

d) \(\left(4x^2-1\right)^2-9\left(2x-1\right)^2\)

\(=\left(2x-1\right)^2\left(2x+1\right)^2-9\left(2x-1\right)^2\)

\(=\left(2x-1\right)^2\left[\left(2x+1\right)^2-9\right]\)

\(=\left(2x-1\right)^2\left(4x^2+4x+1-9\right)\)

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

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

Ta có: \(\left(x+3\right)^3-\left(x+1\right)^3=56\)

\(\Leftrightarrow x^3+9x^2+27x+27-x^3-3x^2-3x-1-56=0\)

\(\Leftrightarrow6x^2+24x-30=0\)

\(\Leftrightarrow x^2+4x-5=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}\)