`@` `\text {Ans}`

`\downarrow`

\(\left(x-\dfrac{1}{5}\right)^2+1=3,5\div7\%\)

`=> (x-1/5)^2 + 1 = 3,5 \div 0,07`

`=> (x-1/5)^2 +1=50`

`=> (x-1/5)^2 = 49`

`=> (x-1/5)^2 = (+-7)^2`

`=>`\(\left[{}\begin{matrix}x-\dfrac{1}{5}=7\\x-\dfrac{1}{5}=-7\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=7+\dfrac{1}{5}\\x=-7+\dfrac{1}{5}\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=\dfrac{36}{5}\\x=-\dfrac{34}{5}\end{matrix}\right.\)

Vậy, `x={36/5; -34/5}.`

Mình cảm ơn:3!

\(\Leftrightarrow n^5+n^2-n^2+1⋮n^3+1\)

\(\Leftrightarrow-n^3+n⋮n^3+1\)

\(\Leftrightarrow n=1\)

Còn nửa phần dưới mình quên đăng ạ

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

b) \(=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)

c) \(=\sqrt{\left(2\sqrt{2}+3\right)^2}=2\sqrt{2}+3\)

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

e) \(=\sqrt{\left(4-\sqrt{6}\right)^2}=4-\sqrt{6}\)

f) \(=\sqrt{\left(3+\sqrt{7}\right)^2}=3+\sqrt{7}\)

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

m) \(=\sqrt{\left(2\sqrt{2}+\dfrac{1}{4}\right)^2}=2\sqrt{2}+\dfrac{1}{4}\)

x-1 = 18 hoac y+5= 18

x=19 hoac y=13

Sai rồi.

\(\Rightarrow\left[3\left(x+1\right)+8\right]⋮\left(x+1\right)\\ \Rightarrow x+1\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\\ \Rightarrow x\in\left\{-9;-5;-3;-2;0;1;3;7\right\}\)

Bạn nên chịu khó gõ đề ra khả năng được giúp sẽ cao hơn.

Câu h của em đây nhé

h, ( 1 + \(\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\)).(1 - \(\dfrac{3+\sqrt{3}}{\sqrt{3}+1}\))

= \(\dfrac{\sqrt{3}-1+3-\sqrt{3}}{\sqrt{3}-1}\).\(\dfrac{\sqrt{3}+1-3-\sqrt{3}}{\sqrt{3}+1}\)

= \(\dfrac{2}{\sqrt{3}-1}\).\(\dfrac{-2}{\sqrt{3}+1}\)

= \(\dfrac{-4}{2}\)

= -2

Hình như x=6 y=7 vì 6/6=1 7/7=1 nên x/6 = y/7

nhưng y lại lớn hơn 0 mà

Ta có : \(\frac{17}{11}=1+\frac{1}{1+\frac{1}{1+\frac{1}{5}}}\)

=> a = 5

loi giai: \(1+\frac{1}{1+\frac{1}{x+\frac{1}{a}}}=\frac{17}{11}\)

\(\frac{17}{11}=1+\frac{6}{11}=1+\frac{1}{1+\frac{5}{6}}=1+\frac{1}{1+\frac{1}{1+\frac{1}{5}}}\)

a=5