dang it 1 thoi

108 trang nhé, tích giùm cho mình

Từ \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)

\(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)

\(=\dfrac{a+b}{b}\cdot\dfrac{b+c}{c}\cdot\dfrac{c+a}{a}\)

\(=\dfrac{-c}{b}\cdot\dfrac{-a}{c}\cdot\dfrac{-b}{a}=-\dfrac{abc}{abc}=-1\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:

\(VT=\dfrac{a^2}{x}+\dfrac{b^2}{y}+\dfrac{c^2}{z}\ge\dfrac{\left(a+b+c\right)^2}{x+y+z}=VP\)

Xảy ra khi \(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)

\(ax^3-3ax^2+3ax-a\)

\(=a\left(x^3-3x^2+3x-1\right)\)

\(=a\left(x-1\right)^3\)

đề nghiệm thực à :))

a)\(=3(x-y)(y-z)(z-x)\)

b)\(pt\Leftrightarrow 3(x-3)(x-2)=0\)

a)\(\Leftrightarrow\dfrac{x}{9}=\dfrac{y}{12}=\dfrac{z}{20}\Leftrightarrow\dfrac{2x}{18}=\dfrac{3y}{36}=\dfrac{z}{20}\)

Áp dụng tc dãy tỉ ....

b)\(\Leftrightarrow\dfrac{x}{\dfrac{3}{2}}=\dfrac{y}{\dfrac{4}{3}}=\dfrac{z}{\dfrac{5}{4}}\)

Áp dụng tc dãy tỉ ....

24 số,rồi tính tỏng 24 số đó là xong

\(2x-xy+y=-1\)

\(\Leftrightarrow2x-xy+y-2=-3\)

\(\Leftrightarrow-x\left(y-2\right)+\left(y-2\right)=-3\)

\(\Leftrightarrow-\left(x-1\right)\left(y-2\right)=-3\)

\(\Leftrightarrow\left(x-1\right)\left(y-2\right)=3\)