Chủ đề này có 10 trả lời

[moster03]

Bài : Rút gọn biểu thức
a ) $\frac{1}{\sqrt{2}-\sqrt{3}}. \sqrt{\frac{3\sqrt{2}-2\sqrt{3}}{3\sqrt{2}+2\sqrt{3}}}$
b ) $\frac{\sqrt{3-2\sqrt{2}}}{\sqrt{17-12\sqrt{2}}}- \frac{\sqrt{3+2\sqrt{2}}}{\sqrt{17+12\sqrt{2}}}$
c ) $\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}$
d ) $\sqrt{6+2\sqrt{5}-\sqrt{29-12\sqrt{5}}}$
e ) $\frac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}$
f ) $\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$
g ) $\left ( 4+\sqrt{15} \right )\left ( \sqrt{10}-\sqrt{6} \right )\sqrt{4-\sqrt{15}}$
* h ) $\frac{\left ( \sqrt{x}+\sqrt{y} \right )^{2}-4\sqrt{xy}}{\sqrt{x}-\sqrt{y}}+\frac{y\sqrt{x}-x\sqrt{y}}{\sqrt{xy}}$
* i ) $\frac{\left ( \sqrt{x}+\sqrt{y} \right )^{2}}{x\sqrt{y}+y\sqrt{x}}\left ( \frac{x-y}{\sqrt{x}-\sqrt{y}}-\frac{x\sqrt{x}-y\sqrt{y}}{x-y} \right )$
* k ) $\frac{\left ( \sqrt{x^{2}+16}-4 \right )\left (\sqrt{x^{2}+16}+4 \right )\left ( x-2\sqrt{x}+4 \right )\sqrt{x+4\sqrt{x}+4}}{x\left ( x\sqrt{x}+8 \right )}$ với x > 0

[minhdat881439]

a)BT=$\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{\sqrt{6}(\sqrt{3}-\sqrt{2})}{\sqrt{6}(\sqrt{3}+\sqrt{2})}}=\frac{1}{\sqrt{2}-\sqrt{3}}\sqrt{\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}}=-1$

[L Lawliet]

h ) $\frac{\left ( \sqrt{x}+\sqrt{y} \right )^{2}-4\sqrt{xy}}{\sqrt{x}-\sqrt{y}}+\frac{y\sqrt{x}-x\sqrt{y}}{\sqrt{xy}}$

ĐKXĐ: $x>0$, $y>0$, $x\neq y$
$$\frac{\left ( \sqrt{x}+\sqrt{y} \right )^{2}-4\sqrt{xy}}{\sqrt{x}-\sqrt{y}}+\frac{y\sqrt{x}-x\sqrt{y}}{\sqrt{xy}}$$
$$=\frac{\left ( \sqrt{x}-\sqrt{y} \right )^2}{\sqrt{x}-\sqrt{y}}-\frac{\sqrt{xy}\left ( \sqrt{x}-\sqrt{y} \right )}{\sqrt{xy}}$$
$$=\sqrt{x}-\sqrt{y}-\sqrt{x}+\sqrt{y}=0$$

[triethuynhmath]

Chém thôi.
e)$=\sqrt{2}(\frac{3+\sqrt{5}}{5+\sqrt{5}}+\frac{3-\sqrt{5}}{5-\sqrt{5}})$
$=\sqrt{2}(\frac{(3+\sqrt{5})(5-\sqrt{5})+(3-\sqrt{5})(5+\sqrt{5})}{25-5})$
$=\sqrt{2}(\frac{10+2\sqrt{5}+10-2\sqrt{5}}{20})=\sqrt{2}$

Bài viết đã được chỉnh sửa nội dung bởi triethuynhmath: 16-07-2012 - 09:58

[minhdat881439]

b)BT=$\frac{\sqrt{2}-1}{(\sqrt{2}-1)^{2}}-\frac{\sqrt{2}+1}{(\sqrt{2}+1)^{2}}=\frac{1}{\sqrt{2}-1}-\frac{1}{\sqrt{2}+1}=2$

Bài viết đã được chỉnh sửa nội dung bởi minhdat881439: 16-07-2012 - 10:00

[triethuynhmath]

b)=$\frac{\sqrt{2}-1}{3-2\sqrt{2}}-\frac{\sqrt{2}+1}{3+2\sqrt{2}}=\frac{\sqrt{2}-1}{(\sqrt{2}-1)^2}-\frac{\sqrt{2}+1}{\sqrt{2}+1)^2}$
$=\frac{1}{\sqrt{2}-1}-\frac{1}{\sqrt{2}+1}=\frac{\sqrt{2}+1-\sqrt{2}+1}{2-1}=2$

[triethuynhmath]

c)$=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}=\sqrt{\sqrt{5}-\sqrt{5}+1}=1$
d)$=\sqrt{6+2\sqrt{5}-2\sqrt{5}+3}=\sqrt{9}=3$
g)$=\sqrt{(4+\sqrt{15})(4-\sqrt{15})}\sqrt{4+\sqrt{15}}(\sqrt{10}-\sqrt{6})=\sqrt{8+2\sqrt{15}}(\sqrt{5}-\sqrt{3})$
$=(\sqrt{5}+\sqrt{3})(\sqrt{5}-\sqrt{3})=2$

Bài viết đã được chỉnh sửa nội dung bởi triethuynhmath: 16-07-2012 - 10:09

[triethuynhmath]

f)$=\sqrt{2}(\frac{2+\sqrt{3}}{3+\sqrt{3}}+\frac{2-\sqrt{3}}{3-\sqrt{3}})$
$=\sqrt{2}(\frac{(2+\sqrt{3})(3-\sqrt{3})+(2-\sqrt{3})(3+\sqrt{3})}{9-3}$
$=\sqrt{2}(\frac{3+\sqrt{3}+3-\sqrt{3}}{6})=\sqrt{2}$
k)$=\frac{(x^2+16-16)(x-2\sqrt{x}+4)\sqrt{(\sqrt{x}+2)^2}}{x(\sqrt{x}+2)(x-2\sqrt{x}+4)}=x$

[triethuynhmath]

i)DKXD:$x,y>0.x\neq y$
$=\frac{(\sqrt{x}+\sqrt{y})^2}{(\sqrt{x}+\sqrt{y})\sqrt{xy}}(\frac{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})}{\sqrt{x}-\sqrt{y}}-\frac{(\sqrt{x}-\sqrt{y})(x+\sqrt{xy}+y)}{(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y}})$
$=(\frac{\sqrt{x}+\sqrt{y}}{\sqrt{xy}})(\sqrt{x}+\sqrt{y}-\frac{x+\sqrt{xy}+y}{\sqrt{x}+\sqrt{y}})$
$=\frac{(\sqrt{x}+\sqrt{y})^2-x-y-\sqrt{xy}}{\sqrt{xy}}=1$

[moster03]

Bài k làm rõ hơn được ko

[C a c t u s]

* k ) $\frac{\left ( \sqrt{x^{2}+16}-4 \right )\left (\sqrt{x^{2}+16}+4 \right )\left ( x-2\sqrt{x}+4 \right )\sqrt{x+4\sqrt{x}+4}}{x\left ( x\sqrt{x}+8 \right )}$ với x > 0

Bài k làm rõ hơn được ko

k. $\frac{\left ( \sqrt{x^{2}+16}-4 \right )\left (\sqrt{x^{2}+16}+4 \right )\left ( x-2\sqrt{x}+4 \right )\sqrt{x+4\sqrt{x}+4}}{x\left ( x\sqrt{x}+8 \right )}$
$=\frac{(x^2+16-16)(x-2\sqrt{x}+4)\sqrt{x+4\sqrt{x}+4}}{x(x\sqrt{x}+8)}$
$=\frac{x^2(x-2\sqrt{x}+4)\sqrt{(\sqrt{x}+2)^2}}{x(x\sqrt{x}+8)}$
$=\frac{x^2(x\sqrt{x}+2x-2.\sqrt{x}.\sqrt{x}-4\sqrt{x}+4\sqrt{x}+8)}{x(x\sqrt{x}+8)}$
$=\frac{x^2(x\sqrt{x}+8)}{x(x\sqrt{x}+8)}$
$=x$

Bài viết đã được chỉnh sửa nội dung bởi C a c t u s: 16-07-2012 - 11:16