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\(\left|x+4\right|=2x-5\)
\(\Leftrightarrow\orbr{\begin{cases}x+4=2x-5\\x+4=-2x+5\end{cases}\Leftrightarrow\orbr{\begin{cases}x-2x=-5-4\\x+2x=5-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}-x=-9\\3x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=9\\x=\frac{1}{3}\end{cases}}}\)
Vậy x=9; x=\(\frac{1}{3}\)
giải
\(\Rightarrow\orbr{\begin{cases}x+4=2x-5\\x+4=-2x+5\end{cases}\Rightarrow\orbr{\begin{cases}x-2x=-5-4\\x+2x=5-4\end{cases}\Rightarrow}\orbr{\begin{cases}-x=-9\\3x=1\end{cases}\Rightarrow}\orbr{\begin{cases}x=9\\x=\frac{1}{3}\end{cases}}}\)
vậy pt có 2 nghiệm là \(9;\frac{1}{3}\)
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Bài giải
\(\frac{1}{2}\left(x+1\right)+\frac{1}{4}\left(x+3\right)=3\cdot\frac{1}{3}\cdot\left(x+20\right)\)
\(\frac{1}{2}\left[\left(x+1\right)+\frac{1}{2}\left(x+3\right)\right]=x+20\)
\(\frac{1}{2}\left[x+1+\frac{1}{2}x+\frac{3}{2}\right]=x+20\)
\(\frac{1}{2}\left[x\left(1+\frac{1}{2}\right)+1+\frac{3}{2}\right]=x+20\)
\(\frac{1}{2}\left[\frac{3}{2}x+\frac{5}{2}\right]=x+20\)
\(\frac{3}{4}x+\frac{5}{4}=x+20\)
\(\frac{3}{4}x-x=20-\frac{5}{4}\)
\(\frac{-1}{4}x=\frac{75}{4}\)
\(x=\frac{75}{4}\text{ : }\frac{-1}{4}\)
\(x=-75\)
\(\frac{1}{2}\left(x+1\right)+\frac{1}{4}\left(x+3\right)=3\cdot\frac{1}{3}\cdot\left(x+20\right)\)
\(\frac{1}{2}\left[\left(x+1\right)+\frac{1}{2}\left(x+3\right)\right]=x+20\)
\(\frac{1}{2}\left[x+1+\frac{1}{2}x+\frac{3}{2}\right]=x+20\)
\(\frac{1}{2}\left[x\left(1+\frac{1}{2}\right)+1+\frac{3}{2}\right]=x+20\)
\(\frac{1}{2}\left[\frac{3}{2}x+\frac{5}{2}\right]=x+20\)
\(\frac{3}{4}x+\frac{5}{4}=x+20\)
\(\frac{3}{4}x-x=20-\frac{5}{4}\)
\(\frac{-1}{4}x=\frac{75}{4}\)
\(x=\frac{75}{4}\text{ : }\frac{-1}{4}\)
\(x=-75\)
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\(\left|2x-3\right|=3-2x\)
\(ĐK:x\le\dfrac{3}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3-2x\\3-2x=3-2x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\0=0\left(đúng\right)\end{matrix}\right.\)
Vậy \(S=\left\{x\in R;x=\dfrac{3}{2}\right\}\)
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\(\text{∘}\) \(\text{Ans}\)
\(\downarrow\)
\(14x^2y^3-7xy^2\cdot\left(2x-3y\right)\)
`=`\(14x^2y^3-\left[7xy^2\cdot2x+7xy^2\cdot\left(-3y\right)\right]\)
`=`\(14x^2y^3-\left(14x^2y^2-21xy^3\right)\)
`=`\(14x^2y^3-14x^2y^2+21xy^3\)
\(\text{∘}\) \(\text{Kaizuu lv uuu.}\)
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\(\left(x-1\right)\left(x+1\right)\left(x+3\right)\)
\(=\left(x^2-1\right)\left(x+3\right)\)
\(=x^3+3x^2-x-3\)
Giúp vs mn ơi mik cần gấp ạ
b) -2x2 - 10x
c) -x2 + 3x
d) x^2 - 9x + 18