Question

b1 : Phân tích đan thức thành nhân tử

a) x - 5 ( vs x > 0 )

b) 5 - 7\(x^2\) (vs x > 0)

b2 :Rút gọn bt

a) x - 4 + \(\sqrt{x^2-8x+16}\) vs x < 4

b) \(\sqrt{\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)^2}vs0\le x\le y\)

Answer 1

1.a) x-5(x>0)

= \(\left(\sqrt{x}-\sqrt{5}\right)\left(\sqrt{x}+\sqrt{5}\right)\)

b)5-7x^2(x>0)

=\(\left(\sqrt{5}-\sqrt{7x^2}\right)\left(\sqrt{5}+\sqrt{7x^2}\right)\\ =\left(\sqrt{5}-\sqrt{7}x\right)\left(\sqrt{5}+\sqrt{7}x\right)\)

2.a)\(x-4+\sqrt{x^2-8x+16}\left(x< 4\right)\\ =x-4+\sqrt{x^2-2.x.4+16}\\ =x-4+\sqrt{\left(x-4\right)^2}\\ =x-4+\left|x-4\right|\\ =x-4-x+4\left(vì\right)x< 4nên\left|x-4\right|< 0\)=0

b)\(\sqrt{\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)^2}\left(0\le x\le y\right)\\ =\sqrt{\left(\sqrt{x}-\sqrt{y}\right)^2}.\sqrt{\left(\sqrt{x}+\sqrt{y}\right)^2}\\ =\left|\sqrt{x}-\sqrt{y}\right|.\left|\sqrt{x}+\sqrt{y}\right|\\ =\left(\sqrt{y}-\sqrt{x}\right)\left(\sqrt{x}+\sqrt{y}\right)\left(vì\right)0\le x\le y\\ =y-x\)

Answer 2

1.

a. \(x-5=\left(\sqrt{x}-\sqrt{5}\right)\left(\sqrt{x}+\sqrt{5}\right)\)

b. \(5-7x^2=\left(\sqrt{5}-\sqrt{7}.x\right)\left(\sqrt{5}+\sqrt{7}.x\right)\)

2.

a. \(x-4+\sqrt{x^2-8x+16}=x-4+\sqrt{\left(x-4\right)^2}=x-4+\left|x-4\right|=x-4+4-x=0\)

b. \(\sqrt{\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{x}+\sqrt{y}\right)^2}=\sqrt{\left(x-y\right)^2}=\left|x-y\right|=y-x\)