\(-1,62+\dfrac{2}{5}+x=7\\ =>-\dfrac{61}{50}+x=7\\ =>x=7--\dfrac{61}{50}\\ x=7+\dfrac{61}{50}\\ x=\dfrac{411}{50}\)

-1,62 + \(\dfrac{2}{5}\) + x = 7

                    x = 7 + 1,62 - \(\dfrac{2}{5}\)

                    x = 8,62 - 0,4

                    x = 8,22

           \(\dfrac{9}{5}\) = - 0,15 - x

- 0,15 - x = 1,8

          - x = 1,8 + 0,15

          - x = 1,95

            x = - 1,95

     Vậy ...

`@` `\text {Ans}`

`\downarrow`

\(-\dfrac{3}{5}-x=0,75\)

\(\Rightarrow x=-\dfrac{3}{5}-0,75\)

\(\Rightarrow x=-\dfrac{27}{20}\)

Vậy, `x = `\(-\dfrac{27}{20}\)

a, 40= 2³ x 5

60 = 2² x 3 x 5 

ƯCLN(40;60)= 2² x 5 = 20

b, 28 = 2² x 7

39 = 3 x 13

35 = 5 x 7

ƯCLN(28;39;35)=1

c, 48 = 2⁴ x 3

60 = 2² x 3 x 5

120 = 2³ x 3 x 5

ƯCLN(48;60;120)= 2² x 3 = 12

d, 30 = 2 x 3 x 5

75 = 3 x 5²

135= 3³ x 5

Vậy ƯCLN(30;75;135)= 3 x 5 = 15

\(2^{300}=\left(2^3\right)^{100}=8^{100}\)

\(3^{200}=\left(3^2\right)^{100}=9^{100}>8^{100}\)

\(\Rightarrow2^{300}< 3^{200}\)

`@` `\text {Ans}`

`\downarrow`

Ta có:

\(2^{300}=\left(2^3\right)^{100}=8^{100}\)

\(3^{200}=\left(3^2\right)^{100}=9^{100}\)

Vì `8 < 9 \Rightarrow `\(8^{100}< 9^{100}\)

`\Rightarrow `\(2^{300}< 3^{200}\)

___

`@` So sánh `2` lũy thừa cùng số mũ:

`a^m > b^m` khi `a > b.`

\(...\Rightarrow x=-13+\dfrac{5}{9}\)

\(\Rightarrow x=\dfrac{-117+5}{9}\)

\(\Rightarrow x=\dfrac{-112}{9}\)

\(...\Rightarrow x=-\dfrac{2}{7}-\dfrac{4}{9}\)

\(\Rightarrow x=-\dfrac{18}{36}-\dfrac{28}{36}\)

\(\Rightarrow x=-\dfrac{46}{36}=-\dfrac{23}{18}\)

2\(^{x+3}\)  + 2\(^x\) = 144

2\(^x\).( 2³ + 1) = 144

2^\(x\)(8 + 1) = 144

2\(^x\) . 9 = 144

2\(^x\)       = 144 : 9

2\(^x\)       = 16

2\(^x\)        = 2⁴

 \(x\)         = 4 

 Ta có \(\left(x+10\right)^4+\left(x-3\right)^4=\left[\left(x+10\right)^2\right]^2+\left[\left(3-x\right)^2\right]^2\)

\(\ge\dfrac{\left[\left(x+10\right)^2+\left(3-x\right)^2\right]^2}{2}\) \(\ge\dfrac{\left[\dfrac{\left(x+10+3-x\right)^2}{2}\right]^2}{2}\) \(=\dfrac{\left(\dfrac{13^2}{2}\right)^2}{2}\)\(=\dfrac{28561}{8}\) (áp dụng 2 lần bất đẳng thức \(a^2+b^2\ge\dfrac{\left(a+b\right)^2}{2}\))

 Suy ra \(P\le2000-\dfrac{28561}{8}=-\dfrac{12561}{8}\).

 Dấu "=" xảy ra \(\Leftrightarrow x+10=3-x\Leftrightarrow x=-\dfrac{7}{2}\)

 Vậy \(maxP=-\dfrac{12561}{8}\), max xảy ra khi \(x=-\dfrac{7}{2}\)

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