3x^2 - 10x +7
phân tích thành nhân tử
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\(=-3\left(x^2-\dfrac{10}{3}x+\dfrac{5}{3}\right)\\ =-3\left(x^2-2\cdot\dfrac{5}{3}x+\dfrac{25}{9}-\dfrac{10}{9}\right)\\ =\dfrac{10}{3}-3\left(x-\dfrac{5}{3}\right)^2\\ =\left[\sqrt{\dfrac{10}{3}}-\sqrt{3}\left(x-\dfrac{5}{3}\right)\right]\left[\sqrt{\dfrac{10}{3}}+\sqrt{3}\left(x-\dfrac{5}{3}\right)\right]\\ =\left(\dfrac{\sqrt{30}}{3}+\dfrac{5\sqrt{3}}{3}-x\sqrt{3}\right)\left(\dfrac{\sqrt{30}}{3}-\dfrac{5\sqrt{3}}{3}+x\sqrt{3}\right)\)
\(=\left(\dfrac{\sqrt{30}+5\sqrt{3}}{3}-x\sqrt{3}\right)\left(\dfrac{\sqrt{30}+5\sqrt{3}}{3}-x\sqrt{3}\right)\)
\(-3x^2+10x-5\)
\(=-3\left(x^2-\dfrac{10}{3}x+\dfrac{5}{3}\right)\)
\(=-3\left(x^2-2\cdot x\cdot\dfrac{5}{3}+\dfrac{25}{9}-\dfrac{10}{9}\right)\)
\(=-3\left(x-\dfrac{5+\sqrt{10}}{3}\right)\left(x-\dfrac{5-\sqrt{10}}{3}\right)\)
\(=x^3-3x^2+6x^2-18x+8x-24\\ =\left(x-3\right)\left(x^2+6x+8\right)\\ =\left(x-3\right)\left(x^2+2x+4x+8\right)\\ =\left(x-3\right)\left(x+2\right)\left(x+4\right)\)
\(x^3+3x^2-10x-24=\left(x^3-3x^2\right)+\left(6x^2-18x\right)+\left(8x-24\right)=x^2\left(x-3\right)+6x\left(x-3\right)+8\left(x-3\right)=\left(x-3\right)\left(x^2+6x+8\right)=\left(x-3\right)\left[\left(x^2+2x\right)+\left(4x+8\right)\right]=\left(x-3\right)\left[x\left(x+2\right)+4\left(x+2\right)\right]=\left(x-3\right)\left(x+2\right)\left(x+4\right)\)
\(3x^2+10x+3\)
\(=3x^2+x+9x+3\)
\(=x\left(3x+1\right)+3\left(3x+1\right)\)
\(=\left(3x+1\right)\left(x+3\right)\)
\(3x^2+10x+3=3x^2+9x+x+3=3x\left(x+3\right)+\left(x+3\right)\)
\(=\left(3x+1\right)\left(x+3\right)\)
chúc bn học tốt
a) \(x^2-3x=x\left(x-3\right)\)
b) \(10x\left(x-y\right)-8y\left(x-y\right)=2\left(x-y\right)\left(5x-4y\right)\)
c) \(x^2-9=\left(x-3\right)\left(x+3\right)\)
a,
\(y^2-x^2+10x-25\)
\(=y^2-\left(x^2-10x+25\right)\)
\(=y^2-\left(x-5\right)^2\)
\(=\left(y+x-5\right)\left(y-x+5\right)\)
a) \(y^2-x^2+10x-25=y^2-\left(x^2-10x+25\right)=y^2-\left(x^2-2.x.5+5^2\right)\)
\(=y^2-\left(x-5\right)^2=\left(y-x+5\right).\left(y+x-5\right)\)
b) \(\left(3x+1\right)^2=3x+1\Rightarrow\left(3x-1\right)^2-\left(3x+1\right)=0\)
\(\Rightarrow\left(3x+1\right)\left(3x+1-1\right)=0\Rightarrow\left(3x+1\right).3x=0\)
\(\Rightarrow\orbr{\begin{cases}3x+1=0\\3x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{3}\\x=0\end{cases}}}\)
giúp mình nhanh nha mình học có lớp 7 ,
\(3x^2-10x+7\)
\(=3x^2-3x-7x+7\)
\(=\left(3x^2-3x\right)-\left(7x-7\right)\)
\(=3x\left(x-1\right)-7\left(x-1\right)\)
\(=\left(3x-7\right)\left(x-1\right)\)