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14 tháng 2 2022

x + 1/100 + x + 2/101 = x + 3/102 - 1

<=> x + 1/100 - 1 + x + 2/101 - 1 = x + 3/102 - 1 - 2

<=> x - 99/100 + x - 99/101 = x - 99/102 - 2

<=> x - 99/100 + x - 99/101 - x - 99/102 = -2

<=> (x - 99)(1/100 + 1/101 - 1/102) = -2

<=> x - 99 = -2/1/100 + 1/101 - 1/102

<=> x = -2/1/100 + 1/101 - 1/102 + 99

Bạn chịu khó bấm máy hộ mình, số to quá

27 tháng 3 2017

\(< =>\left(\dfrac{x-5}{100}-1\right)+\left(\dfrac{x-4}{101}-1\right)+\left(\dfrac{x-3}{102}-1\right)+3=\left(\dfrac{x-100}{5}-1\right)+\left(\dfrac{x-101}{4}-1\right)+\left(\dfrac{x-102}{3}-1\right)+3\)\(< =>\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}=\dfrac{x-105}{5}+\dfrac{x-105}{4}+\dfrac{x-105}{3}\)

\(< =>\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}\right)\) = 0

<=> x - 105 = 0

<=> x = 105

Vậy tập nghiệm của phương trình là S = \(\left\{105\right\}\)

27 tháng 3 2017

cảm ơn ban nhiều

9 tháng 2 2018

\(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)

\(\Leftrightarrow\dfrac{x-5}{100}-1+\dfrac{x-4}{101}-1+\dfrac{x-3}{102}-1=\dfrac{x-100}{5}-1+\dfrac{x-101}{4}-1+\dfrac{x-102}{3}-1\)

\(\Leftrightarrow\dfrac{x-5-100}{100}+\dfrac{x-4-101}{101}+\dfrac{x-3-102}{102}-\dfrac{x-100-5}{5}-\dfrac{x-101-4}{4}-\dfrac{x-102-3}{3}=0\)

\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}-\dfrac{x-105}{5}-\dfrac{x-105}{4}-\dfrac{x-105}{3}=0\)

\(\Leftrightarrow\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow x-105=0\) ( vì \(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}\ne0\) )

\(\Leftrightarrow x=105\)

Vậy tập nghiệm của pt là S ={ 105 }

b: \(\Leftrightarrow\left(\dfrac{29-x}{21}+1\right)+\left(\dfrac{27-x}{23}+1\right)+\left(\dfrac{25-x}{25}+1\right)+\left(\dfrac{23-x}{27}+1\right)+\left(\dfrac{21-x}{29}+1\right)=0\)

=>50-x=0

hay x=50

c: \(\Leftrightarrow\dfrac{x-2}{2001}+1=\dfrac{x-1}{2002}+\dfrac{x}{2003}\)

\(\Leftrightarrow\left(\dfrac{x-2}{2001}-1\right)=\left(\dfrac{x-1}{2002}-1\right)+\left(\dfrac{x}{2003}-1\right)\)

=>x-2003=0

hay x=2003

1 tháng 3 2022

9092 = 0

1 tháng 3 2022

cái này + mỗi phân số vs 1 á 

 

5 tháng 1 2018

a) \(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)

\(\Leftrightarrow\dfrac{x-5}{100}-1+\dfrac{x-4}{101}-1+\dfrac{x-3}{102}-1=\dfrac{x-100}{5}-1+\dfrac{x-101}{4}-1+\dfrac{x-102}{3}-1\)

\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}-\dfrac{x-105}{5}-\dfrac{x-105}{4}-\dfrac{x-105}{3}=0\)

\(\Leftrightarrow\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}\right)=0\)

\(\Leftrightarrow\left(x-105\right)=0;\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}\right)\ne0\)

\(\Leftrightarrow x=105\)

b) \(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=-5\)

\(\Leftrightarrow\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)

\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{50-x}{27}+\dfrac{50-x}{29}=0\)

\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{29}+\dfrac{1}{27}+\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}\right)=0\)

\(\Leftrightarrow50-x=0;\left(\dfrac{1}{29}+\dfrac{1}{27}+\dfrac{1}{25}+\dfrac{1}{23}+\dfrac{1}{21}\right)\ne0\)

\(\Leftrightarrow x=50\)

d: \(\Leftrightarrow x^3+6x^2+12x+8-x^3+6x^2-12x+8=12x^2-12x-8\)

\(\Leftrightarrow12x^2+16=12x^2-12x-8\)

=>-12x=24

hay x=-2

e: \(\left(x+5\right)\left(x+2\right)-3\left(4x-3\right)=\left(x-5\right)^2\)

\(\Leftrightarrow x^2+7x+10-12x+9=x^2-10x+25\)

=>-5x+19=-10x+25

=>5x=6

hay x=6/5

f: \(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)

=>x-105=0

hay x=105

7 tháng 7 2018

1)

\(\dfrac{x-5}{100}+\dfrac{x-4}{101}+\dfrac{x-3}{102}=\dfrac{x-100}{5}+\dfrac{x-101}{4}+\dfrac{x-102}{3}\)

\(\Leftrightarrow\dfrac{x-5}{100}+1+\dfrac{x-4}{101}+1+\dfrac{x-3}{102}+1=\dfrac{x-100}{5}+1+\dfrac{x-101}{4}+1+\dfrac{x-102}{3}+1\)

\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}=\dfrac{x-105}{5}+\dfrac{x-105}{4}+\dfrac{x-105}{3}+\dfrac{x-105}{2}\)

\(\Leftrightarrow\dfrac{x-105}{100}+\dfrac{x-105}{101}+\dfrac{x-105}{102}-\dfrac{x-105}{5}-\dfrac{x-105}{4}-\dfrac{x-105}{3}-\dfrac{x-105}{2}=0\)

\(\Leftrightarrow\left(x-105\right)\left(\dfrac{1}{100}+\dfrac{1}{101}+\dfrac{1}{102}-\dfrac{1}{5}-\dfrac{1}{4}-\dfrac{1}{3}-\dfrac{1}{2}\right)=0\)\(\Leftrightarrow105-x=0\)

\(\Leftrightarrow x=105\)

b)

\(\dfrac{29-x}{21}+\dfrac{27-x}{23}+\dfrac{25-x}{25}+\dfrac{23-x}{27}+\dfrac{21-x}{29}=0\)

\(\Leftrightarrow\dfrac{29-x}{21}+1+\dfrac{27-x}{23}+1+\dfrac{25-x}{25}+1+\dfrac{23-x}{27}+1+\dfrac{21-x}{29}+1=0\)

\(\Leftrightarrow\dfrac{50-x}{21}+\dfrac{50-x}{23}+\dfrac{50-x}{25}+\dfrac{20-x}{27}+\dfrac{50-x}{29}=0\)

\(\Leftrightarrow\left(50-x\right)\left(\dfrac{1}{21}+\dfrac{1}{23}+\dfrac{1}{25}+\dfrac{1}{27}+\dfrac{1}{29}\right)=0\)

\(\Leftrightarrow50-x=0\)

\(\Leftrightarrow x=50\)

7 tháng 7 2018

2)

\(\left(5x+1\right)^2=\left(3x-2\right)^2\)

\(\Leftrightarrow\left|5x+1\right|=\left|3x-2\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=3x-2\\5x+1=-3x+2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\\x=\dfrac{1}{8}\end{matrix}\right.\)

b) \(\left(x+2\right)^3=\left(2x+1\right)^3\)

\(\Leftrightarrow x^3+6x^2+12x+8=8x^3+12x^2+6x+1\)

\(\Leftrightarrow-7x^3-6x^2+6x+7=0\)

\(\Leftrightarrow-7x^3+7x^2-13x^2+13x-7x+7=0\)

\(\Leftrightarrow-7x^2\left(x-1\right)-13x\left(x-1\right)-7\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(-7x^2-13x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\-7x^2-13x-7=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x^2+\dfrac{13}{7}x+1\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\-7\left(x+\dfrac{13}{14}\right)^2-\dfrac{169}{196}=0\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow x=1\)

NV
7 tháng 4 2022

Xét hàm:

\(f\left(x\right)=\dfrac{1}{x}+\dfrac{1}{x^2}+...+\dfrac{1}{x^{100}}\)

\(\Rightarrow f'\left(x\right)=-\dfrac{1}{x^2}-\dfrac{2}{x^3}-\dfrac{3}{x^4}-...-\dfrac{100}{x^{101}}=-P\) (1)

Mặt khác \(f\left(x\right)\) là tổng cấp số nhân với \(\left\{{}\begin{matrix}n=100\\u_1=\dfrac{1}{x}\\q=\dfrac{1}{x}\end{matrix}\right.\)

\(\Rightarrow f\left(x\right)=u_1.\dfrac{1-q^{100}}{1-q}=\dfrac{1}{x}.\dfrac{1-\dfrac{1}{x^{100}}}{1-\dfrac{1}{x}}=\dfrac{1-\dfrac{1}{x^{100}}}{x-1}=\dfrac{x^{100}-1}{x^{101}-x^{100}}\)

\(\Rightarrow f'\left(x\right)=\dfrac{\left(x^{100}-1\right)'\left(x^{101}-x^{100}\right)-\left(x^{101}-x^{100}\right)'\left(x^{100}-1\right)}{\left(x^{101}-x^{100}\right)^2}=-\dfrac{x^{101}-101x^{100}+100}{x^{101}\left(x-1\right)^2}\) (2)

(1);(2) \(\Rightarrow P=\dfrac{x^{101}-101x^{100}+100}{x^{101}\left(x-1\right)^2}\)

26 tháng 9 2017

\(\left|x+\dfrac{1}{101}\right|+\left|x+\dfrac{2}{101}\right|+.....+\left|x+\dfrac{100}{101}\right|=101x\left(1\right)\)

VT(1) \(\ge0\) \(\Rightarrow VP\left(1\right)\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\left|x+\dfrac{1}{101}\right|+\left|x+\dfrac{2}{101}\right|+...+\left|x+\dfrac{100}{101}\right|=100x+\dfrac{5050}{101}=101x\\ \Rightarrow x=50\)

24 tháng 9 2017

tim gtln