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

\(-\dfrac{5}{7}\left(\dfrac{2}{5}-x\right)=-\dfrac{7}{3}\\ \Rightarrow\dfrac{2}{5}-x=\left(-\dfrac{7}{3}\right):\left(-\dfrac{5}{7}\right)\\ \Rightarrow\dfrac{2}{5}-x=\left(-\dfrac{7}{3}\right).\left(-\dfrac{7}{5}\right)\\ \Rightarrow\dfrac{2}{5}-x=\dfrac{49}{15}\\ \Rightarrow x=\dfrac{2}{5}-\dfrac{49}{15}\\ \Rightarrow x=-\dfrac{43}{15}\)

`(7x-2^2):5=2`

`=>7x-4=2.5`

`=>7x=10+4`

`=>x=14:7`

`=>x=2`

\(\dfrac{1}{9}.3^5.3^x=3^{2x+1}\\ \Rightarrow3^{5+x}:9=3^{2x+1}\\ \Rightarrow3^{5+x}:3^2=3^{2x+1}\\ \Rightarrow3^{5+x-2}=3^{2x+1}\\ \Rightarrow3^{x+3}=3^{2x+1}\\ \Rightarrow x+3=2x+1\\ \Rightarrow2x+1-x-3=0\\ \Rightarrow x-2=0\\ \Rightarrow x=2\)

Câu 2:

Để `d` \\ `d'` thì:

`m^2 +1=m+1`

`<=>m^2 -m=0`

`<=>m(m-1)=0`

`<=>`\(\left[{}\begin{matrix}m=0\\m-1=0\end{matrix}\right.\)

`<=>`\(\left[{}\begin{matrix}m=0\\m=1\end{matrix}\right.\)

\(\dfrac{x-x^2}{5x^2-5}=\dfrac{x}{A}\\ \Leftrightarrow\dfrac{x\left(1-x\right)}{5\left(x^2-1\right)}=\dfrac{x}{A}\\ \Leftrightarrow\dfrac{x\left(1-x\right)}{-5\left(1-x^2\right)}=\dfrac{x}{A}\\ \Leftrightarrow\dfrac{x\left(1-x\right)}{-5\left(1-x\right)\left(1+x\right)}=\dfrac{x}{A}\\ \Leftrightarrow\dfrac{x}{-5\left(1+x\right)}=\dfrac{x}{A}\\ \Leftrightarrow A=-5\left(1+x\right)\\ \Leftrightarrow A=-5x-5\)

\(A=\dfrac{2}{1.2}+\dfrac{2}{2.3}+..+\dfrac{2}{99.100}\\ =2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\right)\\ =2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =2\left(1-\dfrac{1}{100}\right)\\ =2.\dfrac{99}{100}\\ =\dfrac{99}{50}\)

\(\sqrt{2x-1}-\dfrac{2}{3}=\dfrac{3}{4}\\ \Leftrightarrow\sqrt{2x-1}=\dfrac{17}{12}\\ \Leftrightarrow2x-1=\dfrac{289}{144}\\ \Leftrightarrow2x=\dfrac{433}{144}\\ \Leftrightarrow x=\dfrac{433}{288}\left(tm\right)\)

\(\sqrt{2-x}-\dfrac{3}{2}=0\\ \Leftrightarrow\sqrt{2-x}=\dfrac{3}{2}\\ \Leftrightarrow2-x=\dfrac{9}{4}\\ \Leftrightarrow x=-\dfrac{1}{4}\left(tm\right)\)

Vậy `S={-1/4}`

`3 2/5 = 3 + 2/5 = 3+0,4=3,4`