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\(A=8x^3-\frac{1}{125}\)
\(A=\left(2x\right)^3-\left(\frac{1}{5}\right)^3\)
\(A=\left(2x-\frac{1}{5}\right)\left(4x^2+\frac{2}{5}x+\frac{1}{25}\right)\)
\(B=\left(x^2\right)^3-\left(\frac{1}{4}y\right)^3\)
\(B=\left(x^2-\frac{1}{4}y\right)\left(x^2+\frac{1}{4}x^2y+\frac{1}{16}y^2\right)\)
\(\frac{2}{x^2-2x}+\frac{1}{x}=\frac{x+2}{x-2}\)
\(\Leftrightarrow\frac{2}{x\left(x-2\right)}+\frac{1}{x}-\frac{x+2}{x-2}=0\)
\(\Leftrightarrow\frac{2}{x\left(x-2\right)}+\frac{x-2}{x\left(x-2\right)}+\frac{x\left(x+2\right)}{x\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{2+x-2+x^2+2x}{x\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{x^2+3x}{x\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{x\left(x+3\right)}{x\left(x-2\right)}=0\)
\(\Leftrightarrow\frac{x+3}{x-2}=0\)
\(\Rightarrow x+3=0\left(x-2\ne0\right)\)
\(\Leftrightarrow x=-3\)
Bài 1:
a: \(\dfrac{x-1}{x+1}-\dfrac{x+1}{x-1}+\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2-2x+1-x^2-2x-1+4}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-4x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{-4}{x+1}\)
b: \(=\dfrac{xy\left(x^2+y^2\right)}{x^4y}\cdot\dfrac{1}{x^2+y^2}=\dfrac{x}{x^4}=\dfrac{1}{x^3}\)
c: Đề thiếu rồi bạn
c: \(=\dfrac{1}{3x-2}-\dfrac{4}{3x+2}+\dfrac{3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{3x+2-12x+8+3x-6}{\left(3x-2\right)\left(3x+2\right)}\)
\(=\dfrac{-6x+4}{\left(3x-2\right)\left(3x+2\right)}=\dfrac{-2}{3x+2}\)
d: \(=\dfrac{x^2-4-x^2+10}{x+2}=\dfrac{6}{x+2}\)
e: \(=\dfrac{1}{2\left(x-y\right)}-\dfrac{1}{2\left(x+y\right)}-\dfrac{y}{\left(x-y\right)\left(x+y\right)}\)
\(=\dfrac{x+y-x+y-2y}{2\left(x-y\right)\left(x+y\right)}=\dfrac{0}{2\left(x-y\right)\left(x+y\right)}=0\)
a) \(\frac{x^2+5x+6}{x^2+7x+12}\)=\(\frac{x^2+2x+3x+6}{x^2+3x+4x+12}\)=\(\frac{x\left(x+2\right)+3\left(x+2\right)}{x\left(x+3\right)+4\left(x+3\right)}\)=\(\frac{\left(x+3\right)\left(x+2\right)}{\left(x+4\right)\left(x+3\right)}\)
b) \(\frac{7x^2+14x+7}{3x^2+3x}\)=\(\frac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}\)=\(\frac{7\left(x+1\right)^2}{3x\left(x+1\right)}\)=\(\frac{7\left(x+1\right)\left(x+1\right)}{3x\left(x+1\right)}\)=\(\frac{7\left(x+1\right)}{3x}\)
Lần lượt áp dụng bất đẳng thức Cô - si có 3 và 4 số, ta có:
\(\frac{a}{18}+\frac{b}{24}+\frac{2}{ab}\ge3.\sqrt[3]{\frac{a}{18}.\frac{b}{24}.\frac{2}{ab}}=\frac{1}{2}\)
\(\frac{a}{9}+\frac{c}{6}+\frac{2}{ac}\ge3.\sqrt[3]{\frac{a}{9}.\frac{c}{6}.\frac{2}{ac}}=1\)
\(\frac{b}{16}+\frac{c}{8}+\frac{2}{bc}\ge3.\sqrt[3]{\frac{b}{16}.\frac{c}{8}.\frac{2}{bc}}=\frac{3}{4}\)
\(\frac{a}{9}+\frac{b}{12}+\frac{c}{6}+\frac{8}{abc}\ge4.\sqrt[4]{\frac{a}{9}.\frac{b}{12}.\frac{c}{6}.\frac{8}{abc}}=\frac{4}{3}\)
\(\frac{13a}{18}+\frac{13b}{24}\ge2\sqrt{\frac{13a}{18}.\frac{13b}{24}}\ge2\sqrt{\frac{13.13.12}{18.24}}=\frac{13}{3}\)
\(\frac{13c}{24}+\frac{13b}{48}\ge2\sqrt{\frac{13c}{24}.\frac{13b}{48}}\ge2\sqrt{\frac{13.13.8}{24.48}}=\frac{13}{6}\)
Cộng vế với vế ta có:
\(a+b+c+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)+\frac{8}{abc}\ge\frac{121}{12}\)
Bạn tự nghĩ ra đấy à?