\(\Leftrightarrow2x+1+2x+2+2x+3+....+2x+101=5757\)

\(\Leftrightarrow2x.101+\left[1+2+3+....+101\right]=5757\)

\(\Leftrightarrow202x+\frac{\left[101+1\right]101}{2}=5757\)

\(\Leftrightarrow202x+5151=5757\)

\(\Leftrightarrow202x=606\)

\(\Leftrightarrow x=3\)

Vậy x=3

a: =>\(\dfrac{2x-4}{2014}+\dfrac{2x-2}{2016}< \dfrac{2x-1}{2017}+\dfrac{2x-3}{2015}\)

=>\(\dfrac{2x-2018}{2014}+\dfrac{2x-2018}{2016}< \dfrac{2x-2018}{2017}+\dfrac{2x-2018}{2015}\)

=>2x-2018<0

=>x<2019

b: \(\Leftrightarrow\left(\dfrac{3-x}{100}+\dfrac{4-x}{101}\right)>\dfrac{5-x}{102}+\dfrac{6-x}{103}\)

=>\(\dfrac{x-3}{100}+\dfrac{x-4}{101}-\dfrac{x-5}{102}-\dfrac{x-6}{103}< 0\)

=>\(x+97< 0\)

=>x<-97

a.2^3x+1=128

=> 2^{3x + 1} = 2⁷

=> 3x + 1 = 7

=> 3x = 6

=> x = 2

vay_

a.2^3x +1=128

= 2^3x x 2=128

 =128:2=64=2 mũ 6

vậy x=2

b.(7x-11)³=2⁵+5²+200

(7x-11)³=257=6,3579 mũ 3

7x=17,3579

x=2,4797

c.(2x+1)+(2x+2)+(2x+3)+...+(2x+101)=5757

vế trái có 101 số hạng

VT =(2x +1+2x+101).101:2=(4x+102).101:2=5757

(4x+102).101 =5757.2=11514

(4x+102)=11514:101=114

4x=114-102=12

x=12:4=3

vậy x=3

Câu 2: 

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

Có \(VT\ge0\Rightarrow VP\ge0\Rightarrow x\ge0\)

do đó phương trình ban đầu tương đương với: 

\(x+\frac{1}{101}+x+\frac{2}{101}+...+x+\frac{100}{101}=101x\)

\(\Leftrightarrow100x+\left(\frac{1}{101}+\frac{2}{101}+...+\frac{100}{101}\right)=101x\)

\(\Leftrightarrow x=\frac{100.101}{2.101}=50\)

f(x) = -2x¹⁰¹ + 2x⁹⁹(x+1) + 2x⁹⁷(x+1) +.......+2x(x+1) -98

     =- 2x¹⁰¹ + (x+1)(2x⁹⁹+2x⁹⁷ +.......+2x) - 98

f(-1) = -2(-1) + 0                                       -98

     = 2-98 = - 96

Lời giải:

PT $\Leftrightarrow \frac{2x-1}{203}+1)+(\frac{2x-3}{205}+1)=(\frac{5-2x}{207}-1)-(\frac{2x}{101}+2)+5$

$\Leftrightarrow \frac{2x+202}{203}+\frac{2x+202}{205}=\frac{-(2x+202)}{207}-\frac{2x+202}{101}+5$
$\Leftrightarrow (2x+202)(\frac{1}{203}+\frac{1}{205}+\frac{1}{207}+\frac{1}{101})=5$

$\Leftrightarrow x=\frac{1}{2}[5: (\frac{1}{203}+\frac{1}{205}+\frac{1}{207}+\frac{1}{101})-202]$

Lời giải:

PT $\Leftrightarrow \frac{2x-1}{203}+1)+(\frac{2x-3}{205}+1)=(\frac{5-2x}{207}-1)-(\frac{2x}{101}+2)+5$

$\Leftrightarrow \frac{2x+202}{203}+\frac{2x+202}{205}=\frac{-(2x+202)}{207}-\frac{2x+202}{101}+5$
$\Leftrightarrow (2x+202)(\frac{1}{203}+\frac{1}{205}+\frac{1}{207}+\frac{1}{101})=5$

$\Leftrightarrow x=\frac{1}{2}[5: (\frac{1}{203}+\frac{1}{205}+\frac{1}{207}+\frac{1}{101})-202]$

1, a) 

Ta có:

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

Thay x=99 vào ta có:

\(\left(99+1\right)^2=100^2=10000\)

b) Ta có:

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

Thay x=101 vào ta có:

\(\left(101-1\right)^3=100^3=1000000\)

a,(2x+1)(y-3)=12

2x+1 và y-3 Ư(12)=

=>x=0,y=15

c) Ta có: \(36^{25}=\left(6^2\right)^{25}=6^{50}\)

\(25^{36}=\left(5^2\right)^{36}=5^{72}\)

Ta có: \(6^{50}=\left(6^5\right)^{10}=7776^{10}\)

mà \(5^{70}=\left(5^7\right)^{10}=78125^{10}\)

nên \(6^{50}< 5^{70}\)

mà \(5^{70}< 5^{72}\)

nên \(6^{50}< 5^{72}\)

hay \(36^{25}< 25^{36}\)