5x-2x+33
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\(2x=3y=5z\Rightarrow\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{15}=\dfrac{y}{10}=\dfrac{z}{6}=\dfrac{x-y+z}{15-10+6}=\dfrac{-33}{11}=-3\)
\(\Rightarrow\left\{{}\begin{matrix}x=\left(-3\right).15=-45\\y=\left(-3\right).10=-30\\z=\left(-3\right).6=-18\end{matrix}\right.\)
Theo tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{5}}=\dfrac{x-y+z}{\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{5}}=\dfrac{-33}{\dfrac{11}{30}}=-90\)
Do đó: x=-45; y=-30; z=-18
d) \(2x^3+3x^2+3x+1=2x^3+x^2+2x^2+x+2x+1\)
\(=x^2\left(2x+1\right)+x\left(2x+1\right)+\left(2x+1\right)=\left(2x+1\right)\left(x^2+x+1\right)\)
e) \(2x^3-5x^2+5x-3=2x^3-3x^2-2x^2+3x+2x-3\)
\(=x^2\left(2x-3\right)-x\left(2x-3\right)+\left(2x-3\right)=\left(2x-3\right)\left(x^2-x+1\right)\)
a: \(\left|3x-2\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=4\\3x-2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{2}{3}\end{matrix}\right.\)
b: Ta có: \(\left|5x-3\right|=\left|x-7\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-3=x-7\\5x-3=7-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=-4\\6x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{5}{3}\end{matrix}\right.\)
b)\(\frac{2x-1}{x-1}+\frac{x}{\left(x-1\right)\left(x-2\right)}=\frac{6x-2}{x-2}\)
\(< =>\left(2x-1\right)\left(x-2\right)+x-\left(6x-2\right)\left(x-1\right)=0\)
\(< =>2x^2-x-4x+2+x-\left(6x^2-2x-6x+2\right)=0\)
\(< =>2x^2-4x+2-6x^2+8x-2=0\)
\(< =>-4x^2+4x=0\)
\(< =>-4x\left(x+1\right)=0\)
\(TH1:-4x=0< =>x=0\) \(TH2:x+1=0< =>x=-1\)
\(S=\left\{-1;0\right\}\)
c) \(x^3-9x^2+6x+16=x^3-8x^2-x^2+8x-2x+16\)
\(=x^2\left(x-8\right)-x\left(x-8\right)-2\left(x-8\right)=\left(x-8\right)\left(x^2-x-2\right)=\left(x-8\right)\left(x-2\right)\left(x+1\right)\)
d) \(2x^3+3x^2+3x+1=\left(2x+1\right)\left(x^2+x+1\right)\)
e) \(2x^3-5x^2+5x-3=\left(2x-3\right)\left(x^2-x+1\right)\)
đè kì vậy bạn
5x-2x+33 hay 5x-2x=33 vậy bn