\(\left(\dfrac{x\sqrt{x}+1}{x-\sqrt{x}}-\dfrac{x\sqrt{x}+1}{x+\sqrt{x}}+\dfrac{x+1}{\sqrt{x}}\right)\) điều kiện xát định :(x > 0 ; x \(\ne\) 1 )

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

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

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

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

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

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

\(\left\{{}\begin{matrix}x+y=m+1\\x^2y+xy^2=2m^2-m-3\end{matrix}\right.\)\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x+y=m+1\\xy\left(x+y\right)=\left(2m-3\right)\left(m+1\right)\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left\{{}\begin{matrix}x+y=m+1\\xy=2m-3\end{matrix}\right.\)

theo hệ thức vi ét đảo ta có x ; y là nghiệm của phương trình :

\(x^2-\left(m+1\right)x+2m-3\)

\(\Delta\) = \(\left(m+1\right)^2-4\left(2m-3\right)\) = \(m^2+2m+1-8m+12\) = \(m^2-6m+13\)

= \(m^2-2.3.m+3^2+4\) = \(\left(m-3\right)^2+4\) \(\ge\) \(4>0\) \(\forall\)m

vậy hệ phương trình có 2 nghiệm phân biệt \(\forall\)m (đpcm)

bạn ơi sai đề rồi ; căn bật sao âm được

đổi 3 tấn 5 tạ =35 tạ

thửa thứ nhất thu hoạch được số thóc là:

(35-5):2=15 tạ

thửa thứ 2 thu được:

35-15=20 tạ

Đ/S:

\(5\sqrt{12a}+\sqrt{49a}-\sqrt{64a}\) = \(10\sqrt{3a}+7a-8a\) = \(10\sqrt{3a}-a\)

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

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

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

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

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

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

P = \(\dfrac{3x-3\sqrt{x}}{x^2-\sqrt{x}}\)

x = \(\sqrt{29+12\sqrt{5}}-\sqrt{29-12\sqrt{5}}\)

x = \(\sqrt{\left(2\sqrt{5}\right)^2+2.2\sqrt{5}.3+3^2}\) - \(\sqrt{\left(2\sqrt{5}\right)^2-2.2\sqrt{5}.3+3^2}\)

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

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

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

x = \(2\sqrt{5}+3-2\sqrt{5}+3\) = 6