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a, \(A=x^2-x\sqrt{y}-2x\sqrt{y}+2y\)
\(=x\left(x-\sqrt{y}\right)-2\sqrt{y}\left(x-\sqrt{y}\right)\)
\(=\left(x-2\sqrt{y}\right)\left(x-\sqrt{y}\right)\)
\(a,\)\(A=x^2-3x\sqrt{y}+2y\)
\(=x^2-2x\sqrt{y}-x\sqrt{y}+2y\)
\(=x\left(x-2\sqrt{y}\right)-\sqrt{y}\left(x-2\sqrt{y}\right)\)
\(=\left(x-\sqrt{y}\right)\left(x-2\sqrt{y}\right)\)
\(b,\)Ta có : \(x=\frac{1}{\sqrt{5}-2}=\frac{\sqrt{5}+2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}=\frac{\sqrt{5}+2}{5-4}=\sqrt{5}+2\)
\(y=\frac{1}{9+4\sqrt{5}}=\frac{9-4\sqrt{5}}{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}=\frac{9-4\sqrt{5}}{81-80}=9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\)
\(\Rightarrow A=\left[\sqrt{5}+2-\sqrt{\left(\sqrt{5}-2\right)^2}\right]\left[\sqrt{5}+2-2\sqrt{\left(\sqrt{5}-2\right)^2}\right]\)
\(=\left(\sqrt{5}+2-\sqrt{5}-2\right)\left(\sqrt{5}+2-2\sqrt{5}+4\right)\)
\(=4\left(6-\sqrt{5}\right)\)
\(=24-4\sqrt{5}\)
a, \(x-\sqrt{x}\)= \(\sqrt{x}.\left(\sqrt{x}-1\right)\)
b, 3x+6\(\sqrt{x}\)= \(\sqrt{x}.\left(3\sqrt{x}+6\right)\)
c, x+2\(\sqrt{x}+1\)= \(\left(\sqrt{x}\right)^2+2\sqrt{x}+1=\left(\sqrt{x}+1\right)^2\)
d, \(3x-5\sqrt{x}+2=3x-3\sqrt{x}-2\sqrt{x}+2\)
=\(3\sqrt{x}.\left(\sqrt{x}-1\right)-2.\left(\sqrt{x}-1\right)\)
=\(\left(3\sqrt{x}-2\right).\left(\sqrt{x}-1\right)\)
úi sao bạn cũng là quản lý giống mình à, mình trả lời câu hỏi của bạn có được không nhỉ
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) \(\left(\sqrt{x-8}-1-a\right)\left(\sqrt{x-8}-1+a\right)\)
a) (\(\sqrt{x}\)-1)(\(\sqrt{x}\)+7)
b) (\(\sqrt{x}\)-4)(\(\sqrt{x}\)-2)
c) (\(\sqrt{x}\)+1)(3\(\sqrt{x}\)+2)
\(2x^2-3x\sqrt{x+3}+\left(x+3\right)\)
\(=2x^2-2x\sqrt{x+3}-x\sqrt{x+3}+\left(\sqrt{x+3}\right)^2\)
\(=2x\left(x-\sqrt{x+3}\right)-\sqrt{x+3}\left(x-\sqrt{x+3}\right)\)
\(=\left(2x-\sqrt{x+3}\right)\left(x-\sqrt{x+3}\right)\)
\(2x^2-3x\sqrt{x+3}+\left(x+3\right)\)
\(=2x^2-x\sqrt{x+3}-2x\sqrt{x+3}+\left(\sqrt{x+3}\right)^2\)
\(=x\left(2x-\sqrt{x+3}\right)-\sqrt{x+3}\left(2x-\sqrt{x+3}\right)\)
\(=\left(x-\sqrt{x+3}\right)\left(2x-\sqrt{x+3}\right)\)
Ta có : \(M=7\sqrt{x-1}-\sqrt{x^3-x^2}+x-1\)
\(=7\sqrt{x-1}-\sqrt{x^2\left(x-1\right)}+x-1\)
\(=7\sqrt{x-1}-x\sqrt{x-1}+\left(\sqrt{x-1}\right)^2\)
\(=\sqrt{x-1}\left(7-x+\sqrt{x-1}\right)\)
\(=\sqrt{x-1}\left(\sqrt{x-1}+2\right)\left(\sqrt{x-1}-3\right)\)
\(M=7\sqrt{x-1}-\sqrt{x^2\left(x-1\right)}+\left(\sqrt{x-1}\right)^2=\sqrt{x-1}\left(7-x+\sqrt{x-1}\right)\)
\(=\sqrt{x-1}\left(6-\left(x-1\right)+\sqrt{x-1}\right)\)( đến đây bạn có thể đặt \(\sqrt{x-1}=t\),t>=0 rồi giải)
\(=-\sqrt{x-1}\left(\sqrt{x-1}-3\right)\left(\sqrt{x-1}+2\right)\)
a ) \(x-y=\left(\sqrt{x}\right)^2-\left(\sqrt{y}\right)^2=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)\)
b ) \(x-2\sqrt{x}=\left(\sqrt{x}\right)^2-2\sqrt{x}=\sqrt{x}\left(\sqrt{x}-2\right)\)
Học tốt ~
a/ \(x-y=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)\)y )
b/ \(x-2\sqrt{x}\)
\(=\sqrt{x}\left(\sqrt{x}-2\right)\)
\(3x-7\sqrt{x}-20\)
\(=3x-12\sqrt{x}+5\sqrt{x}-20\)
\(=\left(\sqrt{x}-4\right)\left(3\sqrt{x}+5\right)\)