1, \(a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)

\(=a^3+b^3+3a^3b+3ab^3+6a^2b^2\)

\(=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left(a^2+2ab+b^2\right)\)

\(=a^2-ab+b^2+3ab\left(a+b\right)^2\)

\(=a^2-ab+b^2+3ab\)

\(=a^2+2ab+b^2=\left(a+b\right)^2\)

\(=1\)

Vậy A = 1

Bài 2: ( đặt đề bài là A )

Đặt \(b+c-a=x,a+c-b=y,a+b-c=z\)

\(\Rightarrow a+b+c=x+y+z\)

\(\Leftrightarrow A=\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(x+z\right)-x^3-y^3-z^3\)

\(=3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)

\(=3.2c.2a.2b=24abc\)

Vậy...

Bài 3:

+) Xét p = 3 có: \(p^2+2=11\in P\) ( t/m )

+) Xét \(p\ne3\) thì:

+ \(p=3k+1\Rightarrow p^2+2=\left(3k+1\right)^2+2=9k^2+6k+3⋮3\notin P\)

+ \(p=3k+2\Rightarrow p^2+2=\left(3k+2\right)^2+2=9k^2+12k+6⋮3\notin P\)

Vậy p = 3

Bài 4:

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)

\(\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=4\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2c}{abc}+\dfrac{2a}{abc}+\dfrac{2b}{abc}=4\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2\left(a+b+c\right)}{abc}=4\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2=4\)

\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\)

\(\Rightarrowđpcm\)

\(x^8y^8+x^4y^4+1=x^8y^8+2x^4y^4+1-x^4y^4\)

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

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

x - 21 = 35

x=35+21

x=56

a, \(5\left(x+4\right)^2+4\left(x-5\right)^2-9\left(4+x\right)\left(x-4\right)\)

\(=5x^2+40x+80+4x^2-40x+100-9x^2+36\)

\(=216\)

\(\Rightarrowđpcm\)

b, \(\left(x+2y\right)^2+\left(2x-y\right)^2-5\left(x+y\right)\left(x-y\right)-10\left(y+3\right)\left(y-3\right)\)

\(=x^2+4xy+4y^2+4x^2-4xy+y^2-5x^2+5y^2-10y^2+30\)

\(=30\)

\(\Rightarrowđpcm\)

Ngay ở dưới có ng` làm rồi đó bạn

\(n^2+3⋮n-1\)

\(\Rightarrow n^2-n+n+3⋮n-1\)

\(\Rightarrow n\left(n-1\right)+n-1+4⋮n-1\)

\(\Rightarrow\left(n+1\right)\left(n-1\right)+4⋮n-1\)

\(\Rightarrow4⋮n-1\)

\(\Rightarrow n-1\in\left\{1;-1;2;-2;4;-4\right\}\)

\(\Rightarrow n\in\left\{2;0;3;-1;5;-3\right\}\)

Vậy...

\(\dfrac{x-241}{17}+\dfrac{x-220}{19}+\dfrac{x-195}{21}+\dfrac{x-166}{23}=10\)

\(\Rightarrow\dfrac{x-241}{17}-1+\dfrac{x-220}{19}-2+\dfrac{x-195}{21}-3+\dfrac{x-166}{23}-4=0\)

\(\Rightarrow\dfrac{x-258}{17}+\dfrac{x-258}{19}+\dfrac{x-258}{21}+\dfrac{x-258}{23}=0\)

\(\Rightarrow\left(x-258\right)\left(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}+\dfrac{1}{23}\right)=0\)

Mà \(\dfrac{1}{17}+\dfrac{1}{19}+\dfrac{1}{21}+\dfrac{1}{23}\ne0\)

\(\Rightarrow x-258=0\Rightarrow x=258\)

Vậy x = 258

Áp dụng bất đẳng thức AM - GM có:

\(\dfrac{2\sqrt{x}}{x+1}\le\dfrac{x+1}{x+1}=1\)

Dấu " = " khi x = 1

Vậy MAX \(\dfrac{2\sqrt{x}}{x+1}=1\) khi x = 1

a, \(3-2\left|4x-5\right|=\dfrac{2}{6}\)

\(\Rightarrow2\left|4x-5\right|=\dfrac{8}{3}\)

\(\Rightarrow\left|4x-5\right|=\dfrac{4}{3}\)

+) Xét \(x\ge\dfrac{5}{4}\) có:

\(4x-5=\dfrac{4}{3}\Rightarrow4x=\dfrac{19}{3}\Rightarrow x=\dfrac{19}{12}\) ( t/m )

+) Xét \(x< \dfrac{5}{4}\) có:

\(4x-5=\dfrac{-4}{3}\Rightarrow4x=\dfrac{11}{3}\Rightarrow x=\dfrac{11}{12}\) ( t/m )

Vậy...

b, tương tự

c, \(\left(7-3x\right)\left(2x+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}7-3x=0\\2x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=\dfrac{-1}{2}\end{matrix}\right.\)

Vậy...

d, \(2x\left(5-3x\right)>0\)

\(\Rightarrow\left\{{}\begin{matrix}2x>0\\5-3x>0\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}2x< 0\\5-3x< 0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x>0\\x< \dfrac{3}{5}\end{matrix}\right.\) hoặc \(\left\{{}\begin{matrix}x< 0\\x>\dfrac{3}{5}\end{matrix}\right.\) (loại )

Vậy \(0< x< \dfrac{3}{5}\)

e, tương tự

g, \(\left|3x+1\right|+\left|1-3x\right|=0\)

\(\Rightarrow\left|3x+1\right|+\left|3x-1\right|=0\)

+) Xét \(x\ge\dfrac{1}{3}\) có:

\(3x+1+3x-1=0\)

\(\Rightarrow6x=0\)

\(\Rightarrow x=0\) ( ko t/m )
+) Xét \(\dfrac{-1}{3}\le x< \dfrac{1}{3}\) có:

\(3x+1+1-3x=0\)

\(\Rightarrow2=0\) ( vô lí )

+) Xét \(x< \dfrac{-1}{3}\) có:

\(-3x-1+1-3x=0\)

\(\Rightarrow-6x=0\Rightarrow x=0\) ( ko t/m )

Vậy ko có giá trị x thỏa mãn đề bài

a, \(\dfrac{1}{36}a^2-\dfrac{1}{4}b^2=\left(\dfrac{1}{6}a-\dfrac{1}{2}b\right)\left(\dfrac{1}{6}a+\dfrac{1}{2}b\right)\)

b, \(x^2+9x+20=x^2+4x+5x+20\)

\(=x\left(x+4\right)+5\left(x+4\right)=\left(x+5\right)\left(x+4\right)\)

c, \(x^2+x-12=x^2+4x-3x-12\)

\(=x\left(x+4\right)-3\left(x+4\right)=\left(x-3\right)\left(x+4\right)\)