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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)
\(=-\sqrt{x-1}\left(\sqrt{x-1}-3\right)\left(\sqrt{x-1}+2\right)\)
1/ Thực hiện phép tính
a) √9−2√20+√12−2√35
\(=\sqrt{\left(\sqrt{5}-\sqrt{4}\right)^2}+\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}\)
\(=\sqrt{5}-\sqrt{4}+\sqrt{7}-\sqrt{5}=\sqrt{7}-\sqrt{4}=\sqrt{7}-2\)
Câu 3:
a: \(\sqrt{x}+\sqrt{y}-5\sqrt{xy}-5y\)
\(=\left(\sqrt{x}+\sqrt{y}\right)-5\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(1-5\sqrt{y}\right)\)
b: \(x\cdot\sqrt{x}-y\cdot\sqrt{y}\)
\(=\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3=\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)\)
Câu 2;
a: \(\sqrt{\frac{\left(x-2\right)^4}{\left(3-x\right)^2}}+\frac{x^2-4}{x-3}\)
\(=\frac{\left(x-2\right)^2}{\left|3-x\right|}+\frac{x^2-4}{x-3}\)
\(=\frac{-\left(x-2\right)^2}{x-3}+\frac{x^2-4}{x-3}=\frac{-x^2+4x-4+x^2-4}{x-3}=\frac{4x-8}{x-3}\)
b:Đặt B=\(4x-\sqrt8+\frac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)
\(=4x-2\sqrt2+\sqrt{\frac{x^2\left(x+2\right)}{x+2}}=4x-2\sqrt2+\sqrt{x^2}\)
\(=4x-2\sqrt2+\left|x\right|\)
TH1: x>=0
=>\(B=4x-2\sqrt2+x=5x-2\sqrt2\)
TH2: -2<x<0
=>|x|=-x
=>\(B=4x-2\sqrt2-x=3x-2\sqrt2\)
1/ \(x-6\sqrt{x}-8=\left(\sqrt{x}-3+\sqrt{17}\right)\left(\sqrt{x}-3-\sqrt{17}\right)\)
2/ Bài này làm gì còn phân tích được nữa.
\(A,ĐKXĐ:x;y\ge0\)
\(A=\sqrt{xy}-2\sqrt{y}-5\sqrt{x}+10\)
\(=\sqrt{y}\left(\sqrt{x}-2\right)-5\left(\sqrt{x}-2\right)\)
\(=\left(\sqrt{x}-2\right)\left(\sqrt{y}-5\right)\)
\(ĐKXĐ:x;y\ge0\)
\(B=a\sqrt{x}+b\sqrt{y}-\sqrt{xy}-ab\)
\(=\left(a\sqrt{x}-\sqrt{xy}\right)+\left(b\sqrt{y}-ab\right)\)
\(=\sqrt{x}\left(a-\sqrt{y}\right)+b\left(\sqrt{y}-a\right)\)
\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)
\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)
\(=\left(a-\sqrt{y}\right)\left(\sqrt{x}-b\right)\)
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 : \(x=\frac{1}{\sqrt{5}-2}=\frac{\sqrt{5}+2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}=\frac{\sqrt{5}+2}{5-4}=\sqrt{5}+2\)
\(y=\frac{1}{9+4\sqrt{5}}=\frac{9-4\sqrt{5}}{\left(9+4\sqrt{5}\right)\left(9-4\sqrt{5}\right)}=\frac{9-4\sqrt{5}}{81-80}=9-4\sqrt{5}=\left(\sqrt{5}-2\right)^2\)
\(\Rightarrow A=\left[\sqrt{5}+2-\sqrt{\left(\sqrt{5}-2\right)^2}\right]\left[\sqrt{5}+2-2\sqrt{\left(\sqrt{5}-2\right)^2}\right]\)
\(=\left(\sqrt{5}+2-\sqrt{5}-2\right)\left(\sqrt{5}+2-2\sqrt{5}+4\right)\)
\(=4\left(6-\sqrt{5}\right)\)
\(=24-4\sqrt{5}\)