Vế trái có 101 số hạng

VT=(2x+1+2x+101).101:2

=(4x+102).101:2

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

(4x+102).101=5757.2

(4x+102).101=11514

4x+102=11514:101

4x+102=114

4x=114-102

4x=12

x=12:4

x=3

Vậy x thuộc {3}

ve trai co 101 so hang 

vt:(2x+1+2x101).101:2

=(4x+102).101:2

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: 

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

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

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

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

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:

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

1,(x-1)(y+5)=101

th1:x-1=101

<=>x=102

th2:y+5=101

<=>y=96

2,(x-2)(-y+5)=12

th1:x-2=12

<=>x=14

th2:-y+5=12

<=>-y=7

<=>y=-7

3,(x-5)(2x+6)<0

th1:x-5<0

<=>x<5

th2:2x+6<0

<=>2x<-6

<=>x<-3

Giúp mình vs.

\(\left|x-1\right|=2x-1\)

\(\Rightarrow\orbr{\begin{cases}x-1=2x-1\\x-1=-2x+1\end{cases}\left(đk:2x\ge\frac{1}{2}\right)}\)

\(\Rightarrow\orbr{\begin{cases}x=0\left(ktm\right)\\x=\frac{2}{3}\left(tm\right)\end{cases}}\)

vậy \(x=\frac{2}{3}\)

a) \(|2x+1|-19=-7\)

\(\Leftrightarrow|2x+1|=12\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=12\\2x+1=-12\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=11\\2x=-13\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{11}{2}\\x=\frac{-13}{2}\end{cases}}}\)

b) \(-28-7\cdot|-3x+15|=-70\)

\(\Leftrightarrow-7\cdot|-3x+15|=-42\)

\(\Leftrightarrow|-3x+15|=6\)

\(\Leftrightarrow\orbr{\begin{cases}-3x+15=6\\-3x+15=-6\end{cases}\Leftrightarrow\orbr{\begin{cases}-3x=-9\\-3x=-21\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=7\end{cases}}}\)