Question

1) Rút gọn

a) A=\(\dfrac{x+\sqrt{xy}}{y+\sqrt{xy}}\)

b) B= \(\sqrt{\dfrac{\left(a-b\right)^3.b^3}{c}}\) .\(\sqrt{\dfrac{bc^3}{\left(a-b\right)}}\) ( với a-b>0, c<0)

c) C=(\(\sqrt{3+2\sqrt{2}}\) - \(\sqrt{3-2\sqrt{2}}\) ).(\(\sqrt{3-2\sqrt{2}}\) +\(\sqrt{3+2\sqrt{2}}\)

2) Giải phuong trình

a) \(\sqrt{x^2-4}\) -\(\sqrt{x-2}\) =0

b)\(\sqrt{3x^2+12x+16}\) +\(\sqrt{y^2-4y+13}\) =5

Answer 1

Biết đâu làm đó , sai thôi đừngg chửi nhé

1, Rút gọn

a) A = \(\dfrac{x+\sqrt{xy}}{y+\sqrt{xy}}\) = \(\dfrac{\left(\sqrt{x}\right)^2+\sqrt{xy}}{\left(\sqrt{y}\right)^2+\sqrt{xy}}\) = \(\dfrac{\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{y}+\sqrt{x}\right)}\) = \(\dfrac{\sqrt{x}}{\sqrt{y}}\)

b) B = \(\sqrt{\dfrac{\left(a-b\right)^3.b^3}{c}}\) . \(\sqrt{\dfrac{bc^3}{\left(a-b\right)}}\)

= \(\sqrt{\dfrac{\left(a-b\right)^3.b^3}{c}.\dfrac{bc^3}{\left(a-b\right)}}\) = \(\sqrt{\left(a-b\right)^2.b^4.c^2}\)

= \(\left|a-b\right|\) . \(\left|b^2\right|\) . \(\left|c\right|\)

= -(a -b) .b². c

Answer 2

bài 2:

a/ \(\sqrt{x^2-4}-\sqrt{x-2}=0\) đk: x≥2

<=> \(\sqrt{\left(x-2\right)\left(x+2\right)}-\sqrt{x-2}=0\)

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

<=>\(\left[{}\begin{matrix}\sqrt{x-2}=0\\\sqrt{x+2}-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)

vậy pt có 1 nghiệm x = 2

b/ \(\sqrt{3x^2+12x+16}+\sqrt{y^2-4y+13}=5\)

Ta có: \(\sqrt{3x^2+12x+16}+\sqrt{y^2-4y+13}=\sqrt{3\left(x^2+4x+4\right)+4}+\sqrt{\left(y^2-4y+4\right)+9}=\sqrt{3\left(x+2\right)^2+4}+\sqrt{\left(y-2\right)^2+9}\ge\sqrt{4}+\sqrt{9}=2+3=5\)

=> Dấu ''='' xảy ra khi x = -2; y = 2

Vậy pt có nghiệm x=-2; y = 2