Question

[Dương Nguyễn Phương Anh]

Giúp tui nhen 

Answer 1

[Nguyễn Ngọc Huy Toàn]

b.\(g\left(x\right)=\left(0,2-\dfrac{3}{5}x\right)\left(x^2-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}0,2-\dfrac{3}{5}x=0\\x^2-5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=\pm\sqrt{5}\end{matrix}\right.\)

d.\(p\left(x\right)=x^3+4x^2=0\)

\(\Leftrightarrow x^2\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

e.\(q\left(x\right)=3\left(x+5x^2\right)-15\left(x^2-x+6\right)=0\)

\(\Leftrightarrow3x+15x^2-15x^2+15x-90=0\)

\(\Leftrightarrow18x-90=0\)

\(\Leftrightarrow x=5\)

g.\(g\left(x\right)=\dfrac{2x-1}{3}-\dfrac{x-3}{4}=0\)

\(\Leftrightarrow\dfrac{4\left(2x-1\right)-3\left(x-3\right)}{12}=0\)

\(\Leftrightarrow4\left(2x-1\right)-3\left(x-3\right)=0\)

\(\Leftrightarrow8x-4-3x+9=0\)

\(\Leftrightarrow5x+5=0\)

\(\Leftrightarrow x=-1\)

h,\(i\left(x\right)=x^2+2x+9=0\)

Ta có:\(x^2+2x+9=\left(x^2+2x+1\right)+8=\left(x+1\right)^2+8>8>0;\forall x\)

\(\Rightarrow\) đa thức vô nghiệm

Answer 2

[Dương Nguyễn Phương Anh]

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