=>\(x^2\)+ \(7x\)=3\(x^2\)+\(7x\)-5

=>\(-2x^2\)+5=0

=>2\(x^2\)-5=0

=>2\(x^2\)=5

=>\(x^2\)=\(\dfrac{5}{2}\)

=>\(x\)=-\(\sqrt{\dfrac{5}{2}}\)

=>\(x\)=+\(\sqrt{\dfrac{5}{2}}\)

`@` `\text {Ans}`

`\downarrow`

\(5^{x-2}-3^2=2^4-\left(6^8\div6^6-6^2\right)\)

`\Rightarrow`\(5^x\div5^2-9=16-\left(6^2-6^2\right)\)

`\Rightarrow`\(5^x\div5^2-9=16\)

`\Rightarrow`\(5^x\div5^2=25\)

`\Rightarrow`\(5^x=5^2\cdot5^2\)

`\Rightarrow`\(5^x=5^4\Rightarrow x=4\)

Vậy, `x = 4.`

5ˣ⁻² - 3² = 2⁴ - (6⁸ : 6⁶ - 6²)

5ˣ⁻² - 9 = 16 - (36 - 36)

5ˣ⁻² - 9 = 16

5ˣ⁻² = 16 + 9

5ˣ⁻² = 25

5ˣ⁻² = 5²

x - 2 = 2

x = 2 + 2

x = 4

3ˣ + 4² = 16

3ˣ + 16 = 16

3ˣ = 16 - 16

3ˣ = 0 (vô lý)

Vậy không tìm được x thỏa mabx yêu cầu

`3^x+4^2=16`

`=>3^x+16=16`

`=>3^x=16-16`

`=>3^x=0`

`=>x=1`

\(5x\left(x-3\right)=\left(x-2\right)\left(5x-1\right)-5\\ \Leftrightarrow5x^2-15x=5x^2-11x+2-5\\ \Leftrightarrow4x=3\\ \Leftrightarrow x=\dfrac{3}{4}\)

  \(\left(\dfrac{-1}{3}\right)^{-1}=-3\)

Vì \(\left(\dfrac{-1}{3}\right)^{-1}=\left(\dfrac{1}{-3}\right)^{-1}=\left(-3^{-1}\right)^{-1}=-3^{-1\times\left(-1\right)}=-3^1=-3\) 

=> \(\left(\dfrac{-1}{3}\right)^{-1}=-3\)

D = \(\dfrac{1}{1\times1981}\) + \(\dfrac{1}{2\times1982}\)+...+ \(\dfrac{1}{25\times2005}\)

D =\(\dfrac{1}{1980}\times\)( \(\dfrac{1980}{1\times1981}\)+ \(\dfrac{1980}{2\times1982}\)+....+ \(\dfrac{1980}{25\times2005}\))

D = \(\dfrac{1}{1980}\) \(\times\)(\(\dfrac{1}{1}\) - \(\dfrac{1}{1981}\) + \(\dfrac{1}{2}\) - \(\dfrac{1}{1982}\)+....+ \(\dfrac{1}{25}\) \(\times\) \(\dfrac{1}{2005}\))

D= \(\dfrac{1}{1980}\)[( \(\dfrac{1}{1}\) + \(\dfrac{1}{2}\) +....+ \(\dfrac{1}{25}\)) - ( \(\dfrac{1}{1981}\)+ \(\dfrac{1}{1982}\)+...+ \(\dfrac{1}{2005}\))]

E =\(\dfrac{1}{25}\times\)( \(\dfrac{1}{1\times26}\)+ \(\dfrac{1}{2\times27}\)+...+ \(\dfrac{1}{1980\times2005}\))

E =  \(\dfrac{1}{25}\). (\(\dfrac{25}{1\times26}\) + \(\dfrac{25}{2\times27}\)+....+ \(\dfrac{25}{1980\times2005}\))

E = \(\dfrac{1}{25}\).(\(\dfrac{1}{1}\)-\(\dfrac{1}{26}\)+\(\dfrac{1}{2}\)-\(\dfrac{1}{27}\)+...+\(\dfrac{1}{1980}\)-\(\dfrac{1}{2005}\))

E=\(\dfrac{1}{25}\)[\(\dfrac{1}{1}\)+...+ \(\dfrac{1}{25}\)+ (\(\dfrac{1}{26}\)+...+\(\dfrac{1}{1980}\)) - (\(\dfrac{1}{26}\)+...+\(\dfrac{1}{1980}\)) - (\(\dfrac{1}{1981}\)+..\(\dfrac{1}{2005}\))]

E = \(\dfrac{1}{25}\) .[\(\dfrac{1}{1}\)+\(\dfrac{1}{2}\)+...+\(\dfrac{1}{25}\) - (\(\dfrac{1}{1981}\)+\(\dfrac{1}{1982}\)+...+ \(\dfrac{1}{2005}\))]

\(\dfrac{D}{E}\) = \(\dfrac{\dfrac{1}{1980}}{\dfrac{1}{25}}\) = \(\dfrac{5}{396}\)

các anh chị giúp em với ạ

\(P=3x^2+x-2=3\left(x^2+\dfrac{1}{3}x+\dfrac{1}{9}\right)-\dfrac{5}{3}=3\left(x+\dfrac{1}{3}\right)^2-\dfrac{5}{3}\\ Vì:\left(x+\dfrac{1}{3}\right)^2\ge0\forall x\in R\\ Vậy:3\left(x+\dfrac{1}{3}\right)^2-\dfrac{5}{3}\ge\dfrac{5}{3}\forall x\in R\\ Vậy:min_P=\dfrac{5}{3}.khi.x=-\dfrac{1}{3}\)

Ai trả lời nhanh nhất mình tik cho