\(2\left[\left\{195+35\div7\right\}\div8+195\right]-400\)

= \(2\left[\left\{195+5\right\}\div8+195\right]-400\)

= \(2\left[200\div8+195\right]-400\)

= \(2\left[25+195\right]-400\)

= \(2.220-400\)

= \(440-400\)

= \(40\)

`(x.2^2)+4=10`

`(x.4)+4=10`

`x.4=10-4`

`x.4=6`

`x=6:4`

`x=3/2`

_________________

`2^2 . 5 : x = 10`

`4 . 5 : x = 10`

`20:x =10`

`x=20:10`

`x=2`

____________________________

`4.2^2 : x = 2`

`4.4 : x = 2`

`16: x = 2`

`x=16:2`

`x=8`

Số đó là 45.
Vì: 45 thêm số 0 vào giữa bằng 405
405 : 45 = 9 (thỏa mãn)

`5^2 +5x = 3`

`25+5x=3`

`5x=25-3`

`5x=22`

`x=22:5`

`x=22/5`

Vậy `x=22/5`

`5^2+5x=3`

`5x=3-5^2`

`5x=3-25`

`5x=-22`

\(x=\dfrac{-22}{5}\)

2++4+6+8+...+2n

= 210 2.(1+2+3+4+...+n)

= 210 1+2+3+4+...+n

= 210:2 1+2+3+4+...+n = 105

=> n.(n+1):2 = 105 n.(n+1)

= 105.2 n.(n+1)

= 210 Vì 14.15

= 210 => n = 14

a) \(\left(x-2\right)\left(x+1\right)=0\)

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

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

b) \(\left(x^2+7\right).\left(x^2-40\right)< 0\)

Ta có : \(x^2\ge0\)

\(\Rightarrow x^2+7>0\)

\(\Rightarrow x^2-40< 0\)

\(\Leftrightarrow-2\sqrt{10}< x< 2\sqrt{10}\)

c) Tương tự câu b 

\(\Rightarrow x^2-16=0\)

\(\Leftrightarrow x^2=16\Leftrightarrow x=\pm4\)

d) \(\left(x^2-7\right).\left(x^2-7\right)\left(x^2.16\right)< 0\)

\(\Leftrightarrow\left(x^2-7\right)^2.\left(x^2.16\right)< 0\)

\(\Leftrightarrow x^2.16< 0\)

\(\Leftrightarrow x^2< 0\)( Vô lí )

Vậy phương trình không có nghiệm