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a: \(\left(3x-1\right)\left(-\dfrac{1}{2}x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\5-\dfrac{1}{2}x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=10\end{matrix}\right.\)
b: \(\dfrac{2}{3}x+\dfrac{1}{2}x=\dfrac{5}{2}:\dfrac{15}{4}=\dfrac{5}{2}\cdot\dfrac{4}{15}=\dfrac{20}{30}=\dfrac{2}{3}\)
=>7/6x=2/3
hay \(x=\dfrac{2}{3}:\dfrac{7}{6}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)
c: \(\left(\dfrac{44}{7}x+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)
\(\Leftrightarrow x\cdot\dfrac{44}{7}+\dfrac{3}{7}=\dfrac{-11}{7}:\dfrac{11}{5}=\dfrac{-5}{7}\)
\(\Leftrightarrow x\cdot\dfrac{44}{7}=-\dfrac{8}{7}\)
hay \(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)
1, a,\(\left(-7x^2\right)\left(3x^2-x-2\right)\)
\(=-21x^4+7x^3+14x^2\)
\(b,\left(2x^3-3x^2-10x+3\right):\left(x-3\right)\)
2,\(a,\left(x-3\right)\left(x^2+1\right)-\left(x-3\right)\left(x^2+3x+9\right)\)
\(=x^3+x-3x^2-3-x^3+27\)
\(=-3x^2+x+24\)
\(b,\left(2x+1\right)^2+\left(2x-1\right)^2+2\left(4x^2-1\right)\)
\(=4x^2+4x+1+4x^2-4x+1+8x^2-2\)
\(=24x^2\)
\(3,a,x^3-x^2-x+1\)
\(=x^2\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)^2\left(x+1\right)\)
\(b,3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(=3x\left(x+1\right)-10\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-10\right)\)
4, a. Bn kiểm tra lại đề bài nhé
b,\(4x^2-12xy+10y^2\)
\(=\left(4x^2-12xy+9y^2\right)+y^2\)
\(=\left(2x-3y\right)^2+y^2\ge0\forall x,y\)