a, \(x^2\) + 4\(x\) + 10

= ( \(x^2\) + 4\(x\) + 4) + 6

= (\(x\) + 2)² + 6

vì (\(x\) + 2)² ≥ 0 

⇒ (\(x\) + 2)² + 6 ≥ 6 > 0 vậy đa thức đã cho vô nghiệm (đpcm)

b, \(x^2\) - 2\(x\) + 5

= (\(x^2\) - 2\(x\) + 1) + 4 

= (\(x\) - 1)² + 4

Vì (\(x\) - 1)² ≥ 0 ⇒ (\(x\) -1)² + 4≥ 4 > 0

Vậy đa thức đã cho vô nghiệm (đpcm)

a, (\(\dfrac{9}{4}\))⁵ : (\(\dfrac{1}{4}\))⁵

= (\(\dfrac{9}{4}\) : \(\dfrac{1}{4}\))⁵

= 9⁵ 

= 59049

b, 18² : 9²

= (18:9)²

= 2²

= 4

c, [(-2)\(^4\)]³ 

= 2¹²

= 4096

d, 5⁷.(\(\dfrac{1}{5}\))⁷

= (5.\(\dfrac{1}{5}\))⁷

= 1⁷

= 1

e, (6,5)³: (6,5)²

= 6,5

3.(\(x\) - 2)⁴ = 45

   ( \(x\) - 2)⁴ = 45: 3

   (\(x\) - 2)⁴ = 15

    \(\left[{}\begin{matrix}x-2=\sqrt[4]{15}\\x-2=-\sqrt[4]{15}\end{matrix}\right.\)

    \(\left[{}\begin{matrix}x=2+\sqrt[4]{15}\\x=2-\sqrt[4]{15}\end{matrix}\right.\)

\(\dfrac{6}{7}< 1< \dfrac{7}{4}\Rightarrow0>-\dfrac{6}{7}>-\dfrac{7}{4}\left(1\right)\)

\(\dfrac{8}{13}=\dfrac{8.3}{13.3}=\dfrac{24}{39}< 0< \dfrac{2}{3}=\dfrac{2.13}{3.13}=\dfrac{26}{39}\left(2\right)\)

\(\left(1\right),\left(2\right)\Rightarrow-\dfrac{7}{4}< -\dfrac{6}{7}< 0< \dfrac{8}{13}< \dfrac{2}{3}\)

a, 5ⁿ⁺¹ - 5^n-1 = 125⁴.2³.3

5^n-1.(5² - 1) = 125⁴.24

5^n-1.24         = 125⁴.24

5^n-1             = 125⁴

5^n-1             = (5³)⁴

5^n-1            = 5¹²

n - 1           = 12

n                = 12 + 1

n                = 13

b,2^2n-1 + 2²ⁿ⁺² = 3.2¹¹

   2^2n-1.(1 + 2³) = 3.2¹¹

  2^2n-1.9 = 3.2¹¹

 2^2n-1      = 2¹¹: 3

2²ⁿ        = 2¹² : 3 (xem lại đề bài em nhá)

\(\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{1}{x.\left(2x+1\right)}=\dfrac{1}{10}\)

\(\Leftrightarrow\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{2x.\left(2x+1\right)}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{2x.\left(2x+1\right)}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{2x}-\dfrac{1}{2x+1}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x+1}=\dfrac{1}{20}\)

\(\Leftrightarrow\dfrac{1}{2x+1}=\dfrac{9}{20}\)

\(\Leftrightarrow2x+1=\dfrac{20}{9}\Leftrightarrow x=\dfrac{11}{18}\)

Em giải như XYZ olm em nhé

Sau đó em thêm vào lập luận sau:

\(x\) = \(\dfrac{11}{18}\)

Vì \(\in\) N* 

Vậy \(x\in\) \(\varnothing\)

\(\dfrac{1}{15}\) + \(\dfrac{1}{21}\) + \(\dfrac{1}{28}\) + \(\dfrac{1}{36}\) +...+ \(\dfrac{2}{x\left(x+1\right)}\) = \(\dfrac{11}{40}\) (\(x\in\) N*)

\(\dfrac{1}{2}\).(\(\dfrac{1}{15}\)+\(\dfrac{1}{21}\)+\(\dfrac{1}{28}\)+\(\dfrac{1}{36}\)+.....+ \(\dfrac{2}{x\left(x+1\right)}\)) = \(\dfrac{11}{40}\) \(\times\) \(\dfrac{1}{2}\)

\(\dfrac{1}{30}\) + \(\dfrac{1}{42}\) + \(\dfrac{1}{56}\) + \(\dfrac{1}{72}\)+...+ \(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{11}{80}\)

\(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\) + \(\dfrac{1}{7.8}\)+...+ \(\dfrac{1}{x\left(x+1\right)}\) = \(\dfrac{11}{80}\)

\(\dfrac{1}{5}\) - \(\dfrac{1}{6}\) + \(\dfrac{1}{6}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{8}\) + \(\dfrac{1}{8}\)-\(\dfrac{1}{9}\)+...+ \(\dfrac{1}{x}\)-\(\dfrac{1}{x+1}\) = \(\dfrac{11}{80}\)

\(\dfrac{1}{5}\) - \(\dfrac{1}{x+1}\) = \(\dfrac{11}{80}\)

         \(\dfrac{1}{x+1}\) = \(\dfrac{1}{5}\) - \(\dfrac{11}{80}\)

           \(\dfrac{1}{x+1}\) = \(\dfrac{1}{16}\)

            \(x\) + 1 = 16

            \(x\)       = 16 - 1

             \(x\)     = 15 

Bạn xem lại đề

? tam giác ABCD

- \(\dfrac{1}{4}\) = \(-\dfrac{1.3}{4.3}\) = \(\dfrac{-3}{12}\)

- \(\dfrac{1}{5}\) = \(\dfrac{-1.3}{3.5}\) = \(\dfrac{-3}{15}\)

Ba số hữu tỉ nằm giữa hai số hữu tỉ - \(\dfrac{1}{4}\); - \(\dfrac{1}{5}\) là ba số hữu tỉ nằm giữa hai số hữu tỉ: - \(\dfrac{3}{12}\) và - \(\dfrac{3}{15}\) 

Đó lần lượt là các số hữu tỉ sau: 

-\(\dfrac{3}{13};\) - \(\dfrac{3}{14}\);