a)  \(2x^3+16=2\left(x^3+8\right)=2\left(x+2\right)\left(x^2-2x+4\right)\)

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

c) \(8a^3x-27b^3x=x\left(8a^3-27b^3\right)=x\left(2a-3b\right)\left(4a^2+6ab+9b^2\right)\)

p/s: chúc bạn học tốt

\(1+1=2\)

2222!!!

TC \(\left(ax+by\right)\left(bx+ay\right)-\left(a+b\right)^2\cdot xy\)

\(=\left(abx^2+a^2xy+b^2xy+aby^2-a^2xy-2abxy-b^2xy\right)\)

  \(=abx^2+aby^2-2abxy=ab\left(x-y\right)^2\)

Vi \(\left(x-y\right)^2\ge0\)(voi moi   x,y)

   va \(a,b\ge0\)(gt)

\(\Rightarrow ab\left(x-y\right)^2\ge0\)

\(\Rightarrow\left(ax+by\right)\left(bx+ay\right)\ge\left(a+b\right)^2\cdot xy\)

Ta có \(\left(a-\frac{1}{2}\right)^2\ge0\Rightarrow a^2+\frac{1}{4}\ge a\), tương tự, ta có 

\(b^2+\frac{1}{4}\ge b;c^2+\frac{1}{4}\ge c\)

Cộng 3 vế của 3  BĐT cùng chiều, ta có \(a^2+b^2+c^2+\frac{3}{4}\ge a+b+c\left(ĐPCM\right)\)

^.^

    x⁴ - 3x² +1

= x⁴ - 2x² + 1 - x²

= ( x² - 1 )² - x²

= ( x²- 1 +x ) ( x² - 1 - x ) 

umk, mình biết rồi. Tại đọc nhầm đề ấy mà

\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{\left(7\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}\)

\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)

\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}\)

\(=\frac{1}{4}\)

jup với mn

\(5x\left(x-1\right)=x-1\)

\(\Leftrightarrow5x=\frac{x-1}{x-1}\)

\(\Leftrightarrow5x=1\Leftrightarrow x=\frac{1}{5}\)

\(5x\left(x-1\right)=x-1\)

\(\Leftrightarrow5x\left(x-1\right)-x+1=0\)

\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}5x-1=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}5x=1\\x=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=1\end{cases}}\)