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a. Tại x=\(\frac{-1}{2}\), ta có:
\(\left(\frac{-1}{2}\right)^2+4.\left(\frac{-1}{2}\right)+3=\frac{1}{4}+\left(-2\right)+3=\frac{5}{4}\)
b. Ta có:
\(x^2+4x+3=0\)
\(\Rightarrow x^2+x+3x+3=0\)
\(\Rightarrow\left(x^2+x\right)+\left(3x+3\right)=0\)
\(\Rightarrow x\left(x+1\right)+3\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x+3\right)=0\)
\(\Rightarrow\hept{\begin{cases}x+1=0\\x+3=0\end{cases}\Rightarrow\hept{\begin{cases}x=-1\\x=-3\end{cases}}}\)
Vậy \(x=-1;x=-3\)
a/ \(M=x^4-xy^3+x^3y-y^4-1\)
\(\Leftrightarrow M=x^3\left(x+y\right)-y^3\left(x+y\right)-1\)
Mà \(x+y=0\)
\(\Leftrightarrow M=x^3.0-y^3.0-1\)
\(\Leftrightarrow M=-1\)
Vậy ...
b, Ta co: \(x^3+xy^2-x^2y-y^3+3\)
\(=\left(x^3-y^3\right)+\left(xy^2-x^2y\right)+3\)
\(=\left(x-y\right)^3+3xy\left(x-y\right)-xy\left(x-y\right)+3\)
= 3 ( vì x-y = 0)
C1: \(A=\left(\frac{36}{6}-\frac{4}{6}+\frac{3}{6}\right)-\left(\frac{150}{30}+\frac{50}{30}-\frac{45}{30}\right)-\left(\frac{18}{6}-\frac{14}{6}+\frac{15}{6}\right)\)
\(=\frac{35}{6}-\frac{155}{30}-\frac{19}{6}=\frac{35}{6}-\frac{31}{6}-\frac{19}{6}=-\frac{15}{6}=-2\frac{1}{2}\)
C2: \(6-\frac{2}{3}+\frac{1}{2}-5-\frac{5}{3}+\frac{3}{2}-3+\frac{7}{3}-\frac{5}{2}\)
\(=\left(6-5-3\right)-\left(\frac{2}{3}+\frac{5}{3}-\frac{7}{3}\right)+\left(\frac{1}{2}+\frac{3}{2}-\frac{5}{2}\right)\)
\(=-2-0-\frac{1}{2}=-2\frac{1}{2}\)