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`@` `\text {Ans}`
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\(\dfrac{253}{-254}=\dfrac{-253}{254}\)
\(-\dfrac{1234}{1232}=\dfrac{-617}{616}\)
Ta có: \(\dfrac{617}{616}>1\) ; \(\dfrac{253}{254}< 1\)
`=>`\(\dfrac{617}{616}>\dfrac{253}{254}\)
`=>`\(\dfrac{-617}{616}< \dfrac{-253}{254}\)
Vậy, \(\dfrac{253}{-254}>\dfrac{-1234}{1232}\).
\(MSC=254.1232\)
\(\dfrac{253}{-254}=\dfrac{-253}{254}=\dfrac{-253.1232}{254.1232}=\dfrac{\text{-311696}}{254.1232}\)
\(\dfrac{-1234}{1232}=\dfrac{-1234.254}{254.1232}=\dfrac{\text{-313436}}{254.1232}\)
mà \(\dfrac{\text{-313436}}{254.1232}< \dfrac{\text{-311696}}{254.1232}\)
\(\Rightarrow\dfrac{-1234}{1232}< \dfrac{253}{-254}\)
\(\)
\(11x^2-15x+4=0\)
\(\Leftrightarrow11x^2-11x-4x+4=0\)
\(\Leftrightarrow11x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(11x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\11x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{4}{11}\end{matrix}\right.\)
\(S=\left\{1,\dfrac{4}{11}\right\}\)
Đặt C(x)=0
\(\Leftrightarrow11x^2-15x+4=0\)
\(\Leftrightarrow11x^2-11x-4x+4=0\)
\(\Leftrightarrow11x\left(x-1\right)-4\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(11x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\11x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\11x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{4}{11}\end{matrix}\right.\)
Vậy: Nghiệm của đa thức \(C\left(x\right)=11x^2-15x+4\) là 1 và \(\dfrac{4}{11}\)
what?