\(Q=\dfrac{x-3\sqrt{x}-4}{x-\sqrt{x}-12}\left(ĐK:x\ge0;x\ne16\right)\\ =\dfrac{x-4\sqrt{x}+\sqrt{x}-4}{x-4\sqrt{x}+3\sqrt{x}-12}\\ =\dfrac{\sqrt{x}\left(\sqrt{x}-4\right)+\left(\sqrt{x}-4\right)}{\sqrt{x}\left(\sqrt{x}-4\right)+3\left(\sqrt{x}-4\right)}\\ =\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-4\right)}\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)

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

ta có 5^x+5^x*25=650 suy ra 5^x *26=650 ,5^x=25 suy ra x=2

Bài này mình biết làm nhưng hơi dài nên nhác viết

1/20+1/4+2/5=7/10 nha bn!

\(x=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\\ =\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{4-2\sqrt{3}}}\\ =\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{3+1+2\sqrt{3}}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{3+1-2\sqrt{3}}}\\ =\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\\ =\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{3}+1}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{3}+1}\\ =\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{3+\sqrt{3}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{3-\sqrt{3}}\\ =\dfrac{2\sqrt{2}+\sqrt{6}}{3+\sqrt{3}}+\dfrac{2\sqrt{2}-\sqrt{6}}{3-\sqrt{3}}\\ =\dfrac{\left(2\sqrt{2}+\sqrt{6}\right)\left(\sqrt{3}-1\right)}{\sqrt{3}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}+\dfrac{\left(2\sqrt{2}-\sqrt{6}\right)\left(\sqrt{3}+1\right)}{\sqrt{3}\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\\ =\dfrac{2\sqrt{6}+3\sqrt{2}-2\sqrt{2}-\sqrt{6}+2\sqrt{6}-3\sqrt{2}+2\sqrt{2}-\sqrt{6}}{2\sqrt{3}}\\ =\dfrac{4\sqrt{6}-2\sqrt{6}}{2\sqrt{3}}=\dfrac{\sqrt{6}\left(4-2\right)}{2\sqrt{3}}=\dfrac{2\sqrt{6}}{2\sqrt{3}}=\sqrt{2}\)

\(y=\dfrac{3+\sqrt{5}}{\sqrt{10}+\sqrt{3+\sqrt{5}}}-\dfrac{3-\sqrt{5}}{\sqrt{10}+\sqrt{3-\sqrt{5}}}\\ =\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{5}+\sqrt{6+2\sqrt{5}}}-\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{5}+\sqrt{6-2\sqrt{5}}}\\ =\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{5}+\sqrt{5+1+2\sqrt{5}}}-\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{5}+\sqrt{5+1-2\sqrt{5}}}\\ =\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{5}+\sqrt{\left(\sqrt{5}+1\right)^2}}-\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{5}+\sqrt{\left(\sqrt{5}-1\right)^2}}\\ =\dfrac{\sqrt{2}\left(3+\sqrt{5}\right)}{2\sqrt{5}+\sqrt{5}+1}-\dfrac{\sqrt{2}\left(3-\sqrt{5}\right)}{2\sqrt{5}+\sqrt{5}-1}\\ =\dfrac{3\sqrt{2}+\sqrt{10}}{\sqrt{5}+1}-\dfrac{3\sqrt{2}-\sqrt{10}}{\sqrt{5}-1}\\ =\dfrac{\left(3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{5}-1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}-\dfrac{\left(3\sqrt{2}-\sqrt{10}\right)\left(\sqrt{5}+1\right)}{\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}\\ =\dfrac{3\sqrt{10}+5\sqrt{2}-3\sqrt{2}-\sqrt{10}-3\sqrt{10}+5\sqrt{2}-3\sqrt{2}+\sqrt{10}}{4}\\ =\dfrac{4\sqrt{2}}{4}=\sqrt{2}\)

Vậy \(x=y\)

=6/18+3/18+1/18=10/18=5/9

question 1: x=13x6=78

question 3:60 mm

question 4:x=32x5=160

question 7:16

Bạn phantuananh làm đúng rồi.Vì với mọi m y luôn cố định nên cho m = 2 thì y0 = 1

Viết lại (d): 2 = (x+1)m - 2x. Cho m = 0 thì x0 = -1

Ta có:2,25=1,5x1,5

Vì A=B nên A=1,5;B=1,5

Còn là:

150-15=135

k nha