125: (2\(x\) - 3\(x\)) = 5⁵: 5³

  125: (-\(x\))  = 25

           \(x\)    = - 125 : 25

           \(x\)    = - 5

\(16^3\cdot8^4=\left(2^4\right)^3\cdot\left(2^3\right)^4=2^{12}\cdot2^{12}=2^{24}\)

\(\left(3\cdot x-16\right)\cdot7^3=2\cdot7\)

\(\Rightarrow3\cdot x-16=\dfrac{2\cdot7}{7^3}\)

\(\Rightarrow3\cdot x-16=\dfrac{2}{7^2}\)

\(\Rightarrow3\cdot x-16=\dfrac{2}{49}\)

\(\Rightarrow3\cdot x=\dfrac{2}{49}+16\)

\(\Rightarrow3\cdot x=\dfrac{786}{49}\)

\(\Rightarrow x=\dfrac{786}{49}:3\)

\(\Rightarrow x=\dfrac{262}{49}\)

\(\left(3x-16\right)\cdot7^3=2\cdot7\)

\(\left(3x-16\right)\cdot343=14\)

\(1029x-5488=14\)

\(1029x=14+5488=5502\)

\(x=\dfrac{5502}{1029}=\dfrac{262}{49}\)

\(201^{60}=\left(201^4\right)^{15}=1944810000^{15}\)

\(398^{45}=\left(398^3\right)^{15}=63044792^{15}\)

Do \(1944810000>63044792\)

\(\Rightarrow1944810000^{15}>63044792^{15}\)

\(\Rightarrow201^{60}>398^{45}\)

Ta có:

\(201^{60}>200^{60};398^{45}< 400^{45}\)

\(200^{60}=\left(2.100\right)^{60}=2^{60}.100^{60}=2^{60}.\left(10^2\right)^{60}\)

\(=2^{60}.10^{120}=2^{60}.10^{30}.10^{90}\)

\(400^{45}=\left(2.100\right)^{45}=2^{45}.100^{45}=2^{45}.\left(10^2\right)^{45}\)

\(=2^{45}.10^{90}\)

Mà \(2^{60}.10^{30}.10^{90}>2^{45}.10^{90}\)

\(\Rightarrow200^{60}>400^{45}\)

\(\Rightarrow201^{60}>200^{60}>400^{45}>398^{45}\)

\(\Rightarrow201^{60}>398^{45}\)

\(2\left[\left(7-3^{2021}:3^{2020}\right):2^2+99\right]-100\)

\(=2\left[\left(7-3^{2021-2020}\right):4+99\right]-100\)

\(=2\left[\left(7-3\right):4+99\right]-100\)

\(=2\left(4:4+99\right)-100\)

\(=2\left(1+99\right)-100\)

\(=2\cdot100-100\)

\(=100\cdot\left(2-1\right)\)

\(=100\)

\(\dfrac{x}{2}+\dfrac{x}{3}=5\)

\(\dfrac{3x}{6}+\dfrac{2x}{6}=\dfrac{30}{6}\)

\(3x+2x=30\)

\(5x=30\)

\(x=\dfrac{30}{5}=6\)

\(\left(5-x\right)+\left(7-x\right)+\left(9-x\right)+...+\left(95-x\right)+\left(97-x\right)=47\times45\)
\(5-x+7-x+9-x+...+95-x+97-x=2115\)
\(\left(5+7+9+...+95+97\right)-\left(x+x+x+...+x+x\right)=2115\)
Đặt \(A=5+7+9+...+95+97\)
Số số hạng của A (hay số lần x) là:
\(\left(97-5\right):2+1=47\)(số)
Tổng của A là:
\(\dfrac{\left(5+97\right)\times47}{2}=2397\)
\(\Rightarrow2397-47x=2115\)
\(\Rightarrow47x=282\)
\(\Rightarrow x=6\)
Vậy x = 6
#kễnh

\(\left(5-x\right)+\left(7-x\right)+...+\left(97-x\right)=47\cdot45\)

\(5-x+7-x+...+97-x=2115\)

\(\left(5+7+...+97\right)-\left(x+x+...+x\right)=2115\)

Số phần tử của phần bị trừ và phần hiệu: \(\dfrac{\left(97-5\right)}{2}+1=47\)

\(\left(97+5\right)\cdot\dfrac{47}{2}-47x=2115\)

\(2397-47x=2115\)

\(47x=2397-2115=282\)

\(x=282:47=6\)

\(E=\left\{k\in\mathbb{N}|x=8k,0\le8k\le80\right\}\)