\(\left|x^2-2x\right|=x\)

\(\Leftrightarrow x^2-2x=\pm x\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x=x\\x^2-2x=-x\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-x=0\\x^2-2x+x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-3x=0\\x^2-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x\left(x-3\right)=0\\x\left(x-1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-3=0\\x=0\\x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=3\\x=1\end{matrix}\right.\)

Vậy...........

b)\(x^3+3x^2y-4xy^2-12y^3\)

\(=\left(x^3+3x^2y\right)-\left(4xy^2+12y^3\right)\)

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

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

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

a)\(2x^2-7xy+5y^2\)

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

\(=2x\left(x-y\right)-5y\left(x-y\right)\)

\(=\left(x-y\right)\left(2x-5y\right)\)

nếu đề là \(\dfrac{31-2x}{x+32}\) thì bạn nên ghi (31-2x)/(x+23)

chứ bạn ghi 31-2x/x+23 thì mn tưởng \(31-\dfrac{2x}{2}+23\) Nguyễn Văn Đức nếu đúng như vậy thì bạn nên xem lại, còn ko thì cho mình xin lỗi.

a)

a=9

b=5

b)

a=8

b=2

x=1

\(A=3+3^2+3^3+3^4+.......+3^{100}\)

\(\Leftrightarrow3A=3^2+3^3+3^4+.......+3^{101}\)

\(\Leftrightarrow3A-A=\left(3^2+3^3+.......+3^{101}\right)-\left(3+3^2+3^3+.......+3^{100}\right)\)

\(\Leftrightarrow2A=3^{101}-3\)

\(\Leftrightarrow2A+3=3^{101}=3^n\)

Vậy \(n=101\) để \(2A+3=3^n\)

\(\dfrac{3}{2}x+3=\dfrac{5}{3}x-2\)

\(\Leftrightarrow\dfrac{3}{2}x+3-\left(\dfrac{5}{3}x-2\right)=0\)

\(\Leftrightarrow\dfrac{3}{2}x+3-\dfrac{5}{3}x+2=0\)

\(\Leftrightarrow5-\dfrac{1}{6}x=0\)

\(\Leftrightarrow\dfrac{1}{6}x=5\)

\(\Leftrightarrow x=30\)

Vậy x=30

bài này thi violympic à Nguyễn Thị Uyển Nhi

Ta có:

\(\dfrac{x^3}{8}=\dfrac{y^3}{27}=\dfrac{z^3}{64}=\dfrac{x}{\sqrt[3]{8}}=\dfrac{y}{\sqrt[3]{27}}=\dfrac{z}{\sqrt[3]{64}}=\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)và \(x^2+2y^2-3z^2=-650\)

Áp dụng tính chất dãy tỉ số bằng nhau:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x^2}{2^2}=\dfrac{2y^2}{2.3^2}=\dfrac{3z^2}{3.4^2}=\dfrac{x^2+2y^2-3y^2}{4+18-48}=\dfrac{-650}{-26}=25\)

\(\dfrac{x}{2}=25\Rightarrow x=25.2=50\)

\(\dfrac{y}{3}=25\Rightarrow y=25.3=75\)

\(\dfrac{z}{4}=25\Rightarrow z=25.4=100\)

Vậy \(x=50;y=75;z=100\)