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ax2+bx+c=a(x2+\(\dfrac{b}{a}\)x+\(\dfrac{c}{a}\))
=a(x2-(x1+x2)x+x1x2)
=a(x-x1)(x-x2)
Áp dụng:
Câu a: Ptr có 2 nghiệm là 5,6=>x2-11x+30=(x-5)(x-6)
Câu b: Ptr có 2 nghiệm là \(\dfrac{-2}{3}\),-4=>3x2+14x+8=3(x+\(\dfrac{2}{3}\))(x+4)
Câu c: Ptr có 2 nghiệm là \(\dfrac{2}{5}\),-2=>5x2+8x-4=5(x-\(\dfrac{2}{5}\))(x+2)
Câu d: Ptr có 2 nghiệm là 3+\(\sqrt{3}\),-2+\(\sqrt{3}\)=>
x2-(1+2\(\sqrt{3}\))x-3+\(\sqrt{3}\)=(x-3-\(\sqrt{3}\))(x+2-\(\sqrt{3}\))
a) \(x-2\sqrt{x-1}-4=\left(x-1\right)-2\sqrt{x-1}+1-4\)
\(=\left(\sqrt{x-1}-1\right)^2-4=\left(\sqrt{x-1}-3\right)\left(\sqrt{x-1}+1\right)\)
b) \(x-2\sqrt{x-6}-5-y^2=\left(x-6\right)-2\sqrt{x-6}+1-y^2\)
\(=\left(\sqrt{x-6}-1\right)^2-y^2=\left(\sqrt{x-6}-1+y\right)\left(\sqrt{x-6}-1-y\right)\)
c) \(x-2\sqrt{x-8}-7-a^2=\left(x-8\right)-2\sqrt{x-8}+1-a^2\)
\(=\left(\sqrt{x-8}-1\right)^2-a^2=\left(\sqrt{x-8}+a-1\right)\left(\sqrt{x-8}-a-1\right)\)
a) \(\left(\sqrt{x-1}-3\right)\left(\sqrt{x-1}+1\right)\)
b) \(\left(\sqrt{x-6}-1-y\right)\left(\sqrt{x-6}-1+y\right)\)
c) \(\left(\sqrt{x-8}-1-a\right)\left(\sqrt{x-8}-1+a\right)\)
Áp dụng định lí viet: \(x_1+x_2=-\frac{b}{a},x_1.x_2=\frac{c}{a}\)
\(ax^2+bx+c=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)=a\left(x^2-\left(x_1+x_2\right)x+x_1.x_2\right)=a\left[\left(x^2-x_1.x\right)-\left(x_2x-x_1x_2\right)\right]\)
=\(a\left[x\left(x-x_1\right)-x_2\left(x-x_1\right)\right]=a\left(x-x_1\right)\left(x-x_2\right)\)
\(A,ĐKXĐ:x;y\ge0\)
\(A=\sqrt{xy}-2\sqrt{y}-5\sqrt{x}+10\)
\(=\sqrt{y}\left(\sqrt{x}-2\right)-5\left(\sqrt{x}-2\right)\)
\(=\left(\sqrt{x}-2\right)\left(\sqrt{y}-5\right)\)
\(ĐKXĐ:x;y\ge0\)
\(B=a\sqrt{x}+b\sqrt{y}-\sqrt{xy}-ab\)
\(=\left(a\sqrt{x}-\sqrt{xy}\right)+\left(b\sqrt{y}-ab\right)\)
\(=\sqrt{x}\left(a-\sqrt{y}\right)+b\left(\sqrt{y}-a\right)\)
\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)
\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)
\(=\left(a-\sqrt{y}\right)\left(\sqrt{x}-b\right)\)