\(a,=\sqrt{x}\left(\sqrt{y}-\sqrt{x}\right)\\ b,=\left(\sqrt{x}-\sqrt{y}\right)^2\\ c,=\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\\ d,=\sqrt{x}\left(\sqrt{y}+2\right)-3\left(\sqrt{y}+2\right)\\ =\left(\sqrt{x}-3\right)\left(\sqrt{y}+2\right)\)

gợi ý nè

1) \(ab+c=ab+c\left(a+b+c\right)\)....

2) nhiều cách lắm nhưng tớ chỉ đưa ra 2 cách ...có vẻ hay

đặt \(\sqrt{x}=a,\sqrt{y}=b\)

=>a³+b³=a⁴+b⁴=a⁵+b⁵

c₁: ta có: \(\left(a^3+b^3\right)\left(a^5+b^5\right)=\left(a^4+b^4\right)^2\)......

c₂: a⁵+b⁵=(a+b)(a⁴+b⁴)-ab(a³+b³)

=> 1=(a+b)-ab .......

3) try use UCT

4) tính sau =))

\(\sqrt{x+2-3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\)

\(\Leftrightarrow\sqrt{\left(2x-5\right)-6\sqrt{2x-5}+9}+\sqrt{\left(2x-5\right)-2\sqrt{2x-5}+1}=4\)

\(\Leftrightarrow\sqrt{\left(\sqrt{2x-5}-3\right)^2}+\sqrt{\left(\sqrt{2x-5}-1\right)^2}=4\)

\(\Leftrightarrow\left|\sqrt{2x-5}-3\right|+\left|\sqrt{2x-5}-1\right|=4\)

Đến đây lập bảng xét dấu là xong.

. . .

\(\sqrt{x}+\sqrt{y-z}+\sqrt{z-x}=\dfrac{1}{2}\left(y+3\right)\)

\(\Leftrightarrow2\sqrt{x}+2\sqrt{y-z}+2\sqrt{z-x}=y+3\)

\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-z-2\sqrt{y-z}+1\right)+\left(z-x-2\sqrt{z-x}+1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-z}-1\right)^2+\left(\sqrt{z-x}-1\right)^2=0\)

Tự làm tiếp nhé.

a, \(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\dfrac{\sqrt{3}.\sqrt{5}-\sqrt{3}.\sqrt{2}}{\sqrt{5}.\sqrt{7}-\sqrt{7}.\sqrt{2}}\)

\(=\dfrac{\sqrt{3}.\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}.\left(\sqrt{5}-\sqrt{2}\right)}=\dfrac{\sqrt{3}}{\sqrt{7}}\)

b, \(\dfrac{2\sqrt{15}-2\sqrt{10}+\sqrt{6}-3}{2\sqrt{5}-2\sqrt{10}-\sqrt{3}+\sqrt{6}}\)

\(=\dfrac{2.\sqrt{5}.\sqrt{3}-2.\sqrt{2}.\sqrt{5}-\sqrt{3}.\sqrt{3}+\sqrt{2}.\sqrt{3}}{2.\sqrt{5}-2.\sqrt{2}.\sqrt{5}-\sqrt{3}+\sqrt{2}.\sqrt{3}}\)

\(=\dfrac{2\sqrt{5}\left(\sqrt{3}-\sqrt{2}\right)-\sqrt{3}.\left(\sqrt{3}-\sqrt{2}\right)}{2\sqrt{5}.\left(1-\sqrt{2}\right)-\sqrt{3}.\left(1-\sqrt{2}\right)}\)

\(=\dfrac{\left(2\sqrt{5}+\sqrt{3}\right).\left(\sqrt{3}-\sqrt{2}\right)}{\left(2\sqrt{5}-\sqrt{3}\right).\left(1-\sqrt{2}\right)}=\dfrac{\sqrt{3}-\sqrt{2}}{1-\sqrt{2}}\)

c, \(\dfrac{x+\sqrt{xy}}{y+\sqrt{xy}}=\dfrac{\sqrt{x}.\sqrt{x}+\sqrt{x}.\sqrt{y}}{\sqrt{y}.\sqrt{y}+\sqrt{x}.\sqrt{y}}\)

\(=\dfrac{\sqrt{x}.\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y}.\left(\sqrt{x}+\sqrt{y}\right)}=\dfrac{\sqrt{x}}{\sqrt{y}}\)

Chúc bạn học tốt!!!

d) \(\dfrac{\sqrt{a}+a\sqrt{b}-\sqrt{b}-b\sqrt{a}}{ab-1}\) = \(-\dfrac{\sqrt{a}\left(1+\sqrt{ab}\right)-\sqrt{b}\left(1+\sqrt{ab}\right)}{1-ab}\)

= \(-\dfrac{\left(\sqrt{a}-\sqrt{b}\right)\left(1+\sqrt{ab}\right)}{\left(1+\sqrt{ab}\right)\left(1-\sqrt{ab}\right)}\) = \(-\dfrac{\sqrt{a}-\sqrt{b}}{1-\sqrt{ab}}\) = \(\dfrac{\sqrt{b}-\sqrt{a}}{1-\sqrt{ab}}\)

a) \(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}\)

\(=\dfrac{\left(\sqrt{15}-\sqrt{6}\right)\cdot\left(\sqrt{35}+\sqrt{14}\right)}{21}\)

\(=\dfrac{\sqrt{525}+\sqrt{210}-\sqrt{210}-\sqrt{84}}{21}\)

\(=\dfrac{5\sqrt{21}-2\sqrt{21}}{21}\)

\(=\dfrac{3\sqrt{21}}{21}\)

\(=\dfrac{\sqrt{21}}{7}\)

\(A,ĐKXĐ:x;y\ge0\)

\(A=\sqrt{xy}-2\sqrt{y}-5\sqrt{x}+10\)

\(=\sqrt{y}\left(\sqrt{x}-2\right)-5\left(\sqrt{x}-2\right)\)

\(=\left(\sqrt{x}-2\right)\left(\sqrt{y}-5\right)\)

\(ĐKXĐ:x;y\ge0\)

\(B=a\sqrt{x}+b\sqrt{y}-\sqrt{xy}-ab\)

\(=\left(a\sqrt{x}-\sqrt{xy}\right)+\left(b\sqrt{y}-ab\right)\)

\(=\sqrt{x}\left(a-\sqrt{y}\right)+b\left(\sqrt{y}-a\right)\)

\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)

\(=\sqrt{x}\left(a-\sqrt{y}\right)-b\left(a-\sqrt{y}\right)\)

\(=\left(a-\sqrt{y}\right)\left(\sqrt{x}-b\right)\)

\(ĐKXĐ:x;y\ge0\)

\(C=\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}\)

\(=\left(\sqrt{x^3}+\sqrt{x^2y}\right)-\left(\sqrt{y^3}+\sqrt{xy^2}\right)\)

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

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

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

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

a)  \(1+\sqrt{3}+\sqrt{5}+\sqrt{15}\)

\(=\left(1+\sqrt{3}\right)+\sqrt{5}\left(1+\sqrt{3}\right)\)

\(=\left(1+\sqrt{3}\right)\left(1+\sqrt{5}\right)\)

b)  \(\sqrt{10}+\sqrt{14}+\sqrt{15}+\sqrt{21}\)

\(=\sqrt{5}\left(\sqrt{2}+\sqrt{3}\right)+\sqrt{7}\left(\sqrt{2}+\sqrt{3}\right)\)

\(=\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{5}+\sqrt{7}\right)\)

c)  \(\sqrt{35}-\sqrt{15}+\sqrt{14}-\sqrt{6}\)

\(=\sqrt{5}\left(\sqrt{7}-\sqrt{3}\right)+\sqrt{2}\left(\sqrt{7}-\sqrt{3}\right)\)

\(=\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{5}+\sqrt{2}\right)\)

e)  \(xy+y\sqrt{x}+\sqrt{x}+1\)

\(=y\sqrt{x}\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)\)

\(=\left(\sqrt{x}+1\right)\left(y\sqrt{x}+1\right)\)

g)  \(3+\sqrt{x}+9-x\)

\(=\left(3+\sqrt{x}\right)+\left(3+\sqrt{x}\right)\left(3-\sqrt{x}\right)\)

\(=\left(3+\sqrt{x}\right)\left(4-\sqrt{x}\right)\)

Bài 1:

a. ta có \(\dfrac{x\sqrt{x}+y\sqrt{y}}{\sqrt{x}+\sqrt{y}}-\left(\sqrt{x}-\sqrt{y}\right)^2\)

= \(\dfrac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-\sqrt{xy}+y\right)}{\sqrt{x}+\sqrt{y}}-x+2\sqrt{xy}-y\)

= \(x-\sqrt{xy}+y-x+2\sqrt{xy}-y\)

=\(\sqrt{xy}\)

b.ĐK: x ≠ 1

Ta có: A= \(\sqrt{\dfrac{x+2\sqrt{x}+1}{x-2\sqrt{x}+1}}\)=\(\sqrt{\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)^2}}\)=\(\dfrac{\sqrt{x}+1}{\left|\sqrt{x}-1\right|}\)

*Nếu \(\sqrt{x}-1\ge0\Rightarrow\sqrt{x}\ge1\)

⇒ A = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

*Nếu \(\sqrt{x}-1< 0\Rightarrow\sqrt{x}< 1\)

⇒ A=\(\dfrac{\sqrt{x}+1}{-\sqrt{x}+1}\)

c.Ta có:

\(a,\dfrac{x+2\sqrt{x}-3}{\sqrt{x}-1}\)

\(\Leftrightarrow\dfrac{x+3\sqrt{x}-\sqrt{x}-3}{\sqrt{x}-1}\)

\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+3\right)-\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)

\(\Rightarrow\sqrt{x}+3\)

\(b,\dfrac{4y+3\sqrt{y}-7}{4\sqrt{y}+7}\)

\(\Leftrightarrow\dfrac{4y+7\sqrt{y}-4\sqrt{y}-7}{4\sqrt{y}+7}\)

\(\Leftrightarrow\dfrac{\sqrt{y}.\left(4\sqrt{y}\right)-\left(4\sqrt{y}+7\right)}{4\sqrt{y}+7}\)

\(\Leftrightarrow\dfrac{\left(4\sqrt{y}+7\right).\left(\sqrt{y}-1\right)}{4\sqrt{y}+7}\)

\(\Rightarrow\sqrt{y}-1\)

\(c,\dfrac{x\sqrt{y}-y\sqrt{x}}{\sqrt{x}-\sqrt{y}}\)

\(\Leftrightarrow\dfrac{\sqrt{xy}.\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}\)

\(\Rightarrow\sqrt{xy}\)

\(d,\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\)

\(\Leftrightarrow\dfrac{x+\sqrt{x}-4\sqrt{x}-4}{x+3\sqrt{x}-4\sqrt{x}-12}\)

\(\Leftrightarrow\dfrac{\sqrt{x}.\left(\sqrt{x}+1\right)-4\left(\sqrt{x}+1\right)}{\sqrt{x}.\left(x+3\right)-4\left(\sqrt{x}+3\right)}\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right).\left(\sqrt{x}-4\right)}\)

\(\Leftrightarrow\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

\(\Rightarrow\dfrac{x-2\sqrt{x}-3}{x-9}\)

\(e,\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+\sqrt{4}}\)

\(\Leftrightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{1+2}\)

\(\Rightarrow\dfrac{1+\sqrt{x}+\sqrt{y}+\sqrt{xy}}{3}\)

\(f,\sqrt{8-2\sqrt{15}}+\sqrt{5}+\sqrt{3}\)

\(\Leftrightarrow\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}+\sqrt{5}+\sqrt{3}\)

\(\Leftrightarrow\sqrt{5}-\sqrt{3}+\sqrt{5}+\sqrt{3}\)

\(\Rightarrow2\sqrt{5}-\sqrt{3}+\sqrt{3}\)

\(g,\sqrt{9-2\sqrt{4}}-\sqrt{9+2\sqrt{14}}\)

\(\Leftrightarrow\sqrt{9-2\times2}-\sqrt{\left(\sqrt{2}+\sqrt{7}\right)^2}\)

\(\Leftrightarrow\sqrt{9-4}-\left(\sqrt{2}+\sqrt{7}\right)\)

\(\Rightarrow\sqrt{5}-\sqrt{2}-\sqrt{7}\)