Ta có: 

\(3< \dfrac{5}{2x-1}< 10;đk:2x-1>0\Rightarrow x>\dfrac{1}{2}\)

\(\Rightarrow\dfrac{30}{10}< \dfrac{30}{6\left(2x-1\right)}< \dfrac{30}{3}\\ \Rightarrow3< 6\left(2x-1\right)< 10\\ \Rightarrow3< 12x-6< 10\)

\(\Rightarrow\left\{{}\begin{matrix}12x-6>3\\12x-6< 10\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>\dfrac{9}{12}=\dfrac{3}{4}\\x< \dfrac{16}{12}=\dfrac{4}{3}\end{matrix}\right.\)

\(\Rightarrow\dfrac{3}{4}< x< \dfrac{4}{3}\Rightarrow x\inℤ.thì.x=1\)

Vậy có một giá trị nguyên của x thõa mãn bài toán là x =1

\(3< \dfrac{5}{2}x-1< 10\Rightarrow4< \dfrac{5}{2}x< 11\)

\(\Rightarrow\dfrac{8}{5}< x< \dfrac{22}{5}\Rightarrow x=\left\{2;3;4\right\}\) có 3 số nguyên x thỏa mãn

\(2^5.5^5-10^6:2^5.5^5\)

\(=2^5.5^5\left(1-10^6\right)\)

GT| m⊥r; n⊥r                                                                                                          | H₃ = 50°                                                                                                       KL| 1) m//n                                                                                                              | 2) I₁ = ?, I₂ = ?

1) Chứng minh:

    Có m⊥r

           n⊥r

⇒ m//n (cùng ⊥ với r)

2) Có I₂ = H₃ = 50° (2 góc SLT)

Có I₁ + I₂ = 180° (2 góc kề bù)

⇒ I₁ = 180° - I₂

⇒ I₁ = 180° - 50° = 130°

đổi lại tên nhé mình đánh nhầm 

Đặt \(A=4+4^2+4^3+...+4^{40}\)

\(4A=4^2+4^3+...+4^{41}\)

\(\Rightarrow4A-A=3A=4^{41}-4\)

\(\Rightarrow A=\dfrac{4^{41}-4}{3}\)

Tick mik nha

\(M=\left(100-1\right)\left(100-2^2\right)\left(100-3^2\right)...\left(100-50^2\right)\\ =\left(100-1\right)\left(100-2^2\right)\left(100-3^2\right)...\left(100-10^2\right)...\left(100-50^2\right)\\ =\left(100-1\right)\left(100-2^2\right)\left(100-3^2\right)...\left(100-100\right)...\left(100-50^2\right)\)

\(=\left(100-1\right)\left(100-2^2\right)\left(100-3^2\right)...0...\left(100-50^2\right)\\ =0\)

M = (100 – 1).(100 – 22). (100 – 32)…(100 – 502)

M = (100 – 1).(100 – 22). (100 – 32)… (100 – 92) .(100 – 102) .(100 – 112) …(100 – 502)

M = (100 – 1).(100 – 22). (100 – 32)… (100 – 92). (100 – 100) .(100 – 112) …(100 – 502)

M = (100 – 1).(100 – 22). (100 – 32)… (100 – 92) .0.(100 – 112) …(100 – 502)

M = 0

Vậy M = 0

\(\left(x+1\right)^2=\dfrac{81}{10}\\ =>x+1=\pm\sqrt{\dfrac{81}{10}}=\pm\dfrac{9}{\sqrt{10}}=\pm\dfrac{9\sqrt{10}}{10}\\ =>x=\dfrac{-10\pm9\sqrt{10}}{10}\)

\(\left(x+1\right)^2=\dfrac{81}{10}\)

<=>\(\left(x+1\right)=\pm\sqrt{\dfrac{81}{10}}\)

<=> \(\left[{}\begin{matrix}x=\dfrac{-10+9\sqrt{10}}{10}\\x=-\dfrac{10+9\sqrt{10}}{10}\end{matrix}\right.\)

a) x - 3/4 = -7/6

            x = -7/6 + 3/4

            x = -5/12

b) x + 5/6 = 1/12

             x = 1/12 - 5/6

             x = -9/12

c) 5/4  + x = 2/3

              x = 2/3 - 5/4

              x = -7/12

d) 5/3 - x = 3/7

            x = 5/3 - 3/7

            x = 26/21