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a)\(\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)=1-\sqrt{x^3}\)
b) \(\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)=\sqrt{x^3}+8\)
c)\(\left(\sqrt{x}-\sqrt{y}\right)\left(x+y+\sqrt{xy}\right)=\sqrt{x^3}-\sqrt{y^3}\)
d)\(\left(x+\sqrt{y}\right)\left(x^2+y-x\sqrt{y}\right)=x^3+\sqrt{y^3}\)
\(\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)=1-x\sqrt{x}\)
\(\left(\sqrt{x}+2\right)\left(x-2\sqrt{x}+4\right)=x\sqrt{x}+8\)
\(\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)=x\sqrt{x}-y\sqrt{y}\)
\(\left(x+\sqrt{y}\right)\left(x^2-x\sqrt{y}+y\right)=x^3+y\sqrt{y}\)
a: \(=4x-4x\sqrt{2}-2x\sqrt{2}+2x=6x-6x\sqrt{2}\)
b: \(=6x-4\sqrt{xy}+3\sqrt{xy}-2y=6x-\sqrt{xy}-2y\)
\(a,\left(4\sqrt{x}-\sqrt{2x}\right)\left(\sqrt{x}-\sqrt{2x}\right)=4x-4\sqrt{2}x-\sqrt{2}x+2x=6x-5\sqrt{2}x=\left(6-5\sqrt{2}\right)x\)
\(b,\left(2\sqrt{x}+\sqrt{y}\right)\left(3\sqrt{x}-2\sqrt{y}\right)=6x-4\sqrt{xy}+3\sqrt{xy}-2y=6x-4\sqrt{xy}-2y\)
\(\(b)\frac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\left(a,b\ge0;a,b\ne1\right)\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\left(a\sqrt{b}-b\sqrt{a}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab+1}\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\left(\sqrt{a}-\sqrt{b}\right)\left(\sqrt{ab}+1\right)}{\left(\sqrt{ab}-1\right)\left(\sqrt{ab}+1\right)}\)\)
\(\(=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{ab}-1\right)}\left(a,b\ge0.a,b\ne1\right)\)\)
_Minh ngụy_
\(\(c)\frac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)\)( tự ghi điều kiện )
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(\sqrt{x}-\sqrt{y}\right)^2.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{x}+y\sqrt{y}-\left(x\sqrt{x}+x\sqrt{y}-2x\sqrt{y}-2y\sqrt{x}+y\sqrt{x}+y\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}\)\)
\(\(=\frac{x\sqrt{y}+y\sqrt{x}}{\sqrt{x}+\sqrt{y}}\)\)( phá ngoặc và tính )
\(\(=\frac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}+\sqrt{y}}=\sqrt{xy}\)\)
_Minh ngụy_
a> c1: \(=1-\sqrt{x^3}=1-\sqrt{x^2.x}=1-x\sqrt{x}\)
c2 \(=1+\sqrt{x}+x-\sqrt{x}-x-x\sqrt{x}=1-x\sqrt{x}\)
b> c1: \(=\sqrt{x}\left(4-\sqrt{2}\right)\sqrt{x-\sqrt{2x}=\sqrt{x\left(x-\sqrt{2x}\right)}}\left(4-\sqrt{2}\right)\)
c2: \(=4\sqrt{x\left(x-\sqrt{2x}\right)}-\sqrt{2x\left(x-\sqrt{2x}\right)}=\sqrt{x\left(x-\sqrt{2x}\right)}\left(4-\sqrt{2}\right)\)
a) \(\frac{1}{\sqrt{x}-1}+\frac{1}{1+\sqrt{x}}=\frac{1+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(1+\sqrt{x}\right)}=\frac{2\sqrt{x}}{x-1}\)( x > 0 ; x ≠ 1 )
b) \(\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{4-x}=\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}+\frac{\sqrt{x}}{x-4}\)
\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2-2\sqrt{x}-4+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-6}{x-4}\)( x > 0 ; x ≠ 4 )
a) Với \(x>0\)và \(x\ne1\)ta có:
\(\frac{1}{\sqrt{x}-1}+\frac{1}{1+\sqrt{x}}+1\)
\(=\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\sqrt{x}+1+\sqrt{x}-1+x-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\sqrt{x}+1}{\sqrt{x}-1}\)
b) Với \(x>0\)và \(x\ne4\)ta có:
\(\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{4-x}=\frac{1}{\sqrt{x}+2}-\frac{2}{\sqrt{x}-2}-\frac{\sqrt{x}}{x-4}\)
\(=\frac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}-\frac{2\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\frac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\left(\sqrt{x}-2\right)-2\left(\sqrt{x}+2\right)+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\frac{\sqrt{x}-2-2\sqrt{x}-4+\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\frac{-6}{x-4}\)
Giải:
a) \(\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\)
\(=\left(\sqrt{x}\right)^3-1\)
Vậy ...
b) \(\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)\)
\(=\left(\sqrt{x}\right)^3+\left(\sqrt{y}\right)^3\)
Vậy ...
c) \(\left(2\sqrt{x}+\sqrt{y}\right)\left(3\sqrt{x}-2\sqrt{y}\right)\)
\(=6x+3\sqrt{xy}-4\sqrt{xy}-2y\)
\(=6x-\sqrt{xy}-2y\)
Vậy ...
\(a.\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)=x\sqrt{x}-1\)
\(b.\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)=x\sqrt{x}+y\sqrt{y}\)
\(c.\left(2\sqrt{x}+\sqrt{y}\right)\left(3\sqrt{x}-2\sqrt{y}\right)=6x-\sqrt{xy}-2y\)
a: \(=1-\left(\sqrt{x}\right)^3=1-x\sqrt{x}\)
b: \(=\left(\sqrt{x}\right)^3+2^3=x\sqrt{x}+8\)
c: \(=\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3=x\sqrt{x}-y\sqrt{y}\)
d: \(=x^3+\left(\sqrt{y}\right)^3=x^3+y\sqrt{y}\)