Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(xy-y\sqrt{x}+\sqrt{x}-1\\ =\left(xy-y\sqrt{x}\right)+\left(\sqrt{x}-1\right)\\ =\sqrt{x}y\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)\\ =\sqrt{x}y\left(\sqrt{x}-1\right)\)
14) \(\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{2^2-2\cdot2\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\left(\sqrt{2-\sqrt{3}}\right)\cdot\left(\sqrt{2-\sqrt{3}}\right)\)
i) \(=x\left(x-5\right)+3\left(x-5\right)=\left(x-5\right)\left(x+3\right)\)
j) \(=x\left(x-1\right)+8\left(x-1\right)=\left(x-1\right)\left(x+8\right)\)
k) \(=x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x+3\right)\)
l) \(=x\left(x-3\right)-2\left(x-3\right)=\left(x-3\right)\left(x-2\right)\)
m) \(=x\left(x^2-4x+4\right)=x\left(x-2\right)^2\)
\(=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-1-y\right)\left(x-1+y\right)\)
= \(2^2-2.2.\left(\sqrt{5}\right)^2+\left(\sqrt{5}\right)^2\)
\(=\left(2-\sqrt{5}\right)^2\)
\(9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\)