`@` `\text {Ans}`
`\downarrow`
`(2^2+1) \times (x+14) = 5^2 \times 4 + (2^5 + 3^2 + 7^2) \div 2`
` \Rightarrow (4+1) \times (x+14) = 5^2\times 2^2 + ( 32 + 9 + 49) \div 2`
`\Rightarrow 5 \times (x+14) = (5*2)^2 + (32+58) \div 2`
`\Rightarrow 5 \times (x+14) = 10^2+90 \div 2`
`\Rightarrow 5 \times (x+14) = 100 + 45`
`\Rightarrow 5 \times (x+14) = 145`
`\Rightarrow x+14 = 145 \div 5`
`\Rightarrow x+14=29`
`\Rightarrow x=29-14`
`\Rightarrow x=15`
Vậy, `x=15.`
\(\left(2^2+1\right)\cdot\left(x+14\right)=5^2\cdot4+\left(2^5+3^2+7^2\right):2\)
\(\Rightarrow\left(4+1\right)\cdot\left(x+14\right)=25\cdot4+\left(32+9+49\right):2\)
\(\Rightarrow5\cdot\left(x+14\right)=100+45\)
\(\Rightarrow5x+70=145\)
\(\Rightarrow5x=75\)
\(\Rightarrow x=\dfrac{75}{5}=15\)