|x − 10|^10 + |x-11|^11= 1
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mk lười lắm tự chép đề bài nhé!
vì /x-10/ và /x-11/ sẽ có giá trị là một số nguyên hoặc số nguyên dương
=> có giá trị nhỏ nhất là 0
mà 0+1=1
và x-10>x-11
=>/x-10/=1=>/x-11/=0
=>x=11
vậy x=11
k nhé!
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(x-3)+(x-2)+(x-1)+...+10+11=11
⇒x-3+x-2+x-1+...+10+11=11
⇒3x+(-3-2-1+0+...+11)=11
⇒3x+60=11
⇒3x=-49
⇒x=-49/3
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\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
=> x + 1 = 0 ( vì 1/10 + 1/11 + 1/12 - 1/13 - 1/14 khac 0 )
=> x = -1
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Do \(\left|x-10\right|^{10}\ge0;\left|x-11\right|^{11}\ge0\)
Mà |x - 10| > |x - 11| => |x - 10|10 > |x - 11|11 và |x - 10|10 + |x - 11|11 = 1
=> |x - 10|10 = 1; |x - 11|11 = 0
=> |x - 10| = 1; |x - 11| = 0
=> x - 11 = 0
=> x = 11
Vậy x = 11
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x là số tự nhiên à bạn
To solve the equation ∣x−10∣10+∣x−11∣11=1|x − 10|^{10} + |x - 11|^{11} = 1, we can start by examining the properties of the absolute value function and the exponents.
- Case 1: x=10x = 10
- Substituting x=10x = 10, we have:
∣10−10∣10+∣10−11∣11=0+1=1|10 - 10|^{10} + |10 - 11|^{11} = 0 + 1 = 1This satisfies the equation, so x=10x = 10 is a solution.
- Case 2: x=11x = 11
- Substituting x=11x = 11, we have:
∣11−10∣10+∣11−11∣11=1+0=1|11 - 10|^{10} + |11 - 11|^{11} = 1 + 0 = 1This satisfies the equation, so x=11x = 11 is another solution.
For any other value of xx between 10 and 11, the term ∣x−10∣10|x - 10|^{10} and ∣x−11∣11|x - 11|^{11} will not sum to 1 because both terms will be positive and greater than zero, making the sum greater than 1. Outside the range of 10 to 11, one of the terms will be raised to a very high power, making the sum exceed 1.
Therefore, the solutions to the equation ∣x−10∣10+∣x−11∣11=1|x - 10|^{10} + |x - 11|^{11} = 1 are:
x=10x = 10\