a,  \(\left(2x+3\right)-\left(3x+1\right)=4\)

\(\Rightarrow2x+3-3x-1=4\)

\(\Rightarrow2x-3x=4+1-3\)

\(\Rightarrow-x=2\)

\(\Rightarrow x=-2\)

Vậy x = -2 

b, \(\left(3x-4\right)-\left(x-2\right)=x+5\)

\(\Rightarrow3x-4-x+2-x=5\)

\(\Rightarrow\left(3x-x-x\right)=5-2+4\)

\(\Rightarrow x=7\)

Vậy x = 7 

a, \(\left(2x+3\right)-\left(3x+1\right)=4\)

\(2x+3-3x-1=4\)

\(-x=2\)

\(\Rightarrow x=-2\)

a, \(\frac{23+x}{201-x}=\frac{3}{5}\)

\(\Rightarrow\left(23+x\right)5=3\left(201-x\right)\)

\(\Rightarrow115+5x=603-3x\)

\(\Rightarrow5x+3x=603-115\)

\(\Rightarrow8x=448\Rightarrow x=61\)

Vậy x = 81 

\(x+20=\frac{5}{7}\left(3x-20\right)\)

\(\Leftrightarrow7x+140=15x-100\)

\(\Leftrightarrow15x-7x=140+100\)

\(\Leftrightarrow8x=240\Rightarrow x=30\)

Vậy x = 30 

\(C\left(x\right)=x^2+x+3\)

Thay \(x=0\)vào đa thức \(C\left(x\right)=x^2+x+3\)

\(C\left(0\right)=0^2+0+3=3\)

vậy nghiệm của đa thức trên là \(3\) 

                                             hk tốt

\(C\left(x\right)=x^2+x+3=0\)

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

\(\Leftrightarrow x^2+x=-3\)(1)

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

Mà \(x^2\ge x\Rightarrow x^2+x\ge0\)[ trái với (1) ]

Vậy đa thức C(x) vô nghiệm

\(A=1+3+3^2+3^3+3^4+3^5+.....+3^{2017}\)

\(=1+3+\left(3^2+3^3+3^4+3^5\right)+.....+\left(3^{2014}+3^{2015}+3^{2016}+3^{2017}\right)\)

\(=4+3^2\left(1+3+3^2+3^3\right)+.....+3^{2014}\left(1+3+3^2+3^3\right)\)

\(=4+3^2\cdot40+....+3^{2014}\cdot40\)

\(=4+40\left(3^2+.....+3^{2014}\right)\) chia 40 dư 4.

\(\frac{3-x}{2016}-1=\frac{2-x}{2017}+\frac{1-x}{2018}\)

\(\Rightarrow\frac{3-x}{2016}-1+2=\frac{2-x}{2017}+\frac{1-x}{2018}+2\)(thêm 2 vô mỗi vế)

\(\Rightarrow\frac{3-x}{2016}+1=\left(\frac{2-x}{2017}+1\right)+\left(\frac{1-x}{2018}+1\right)\)

\(\Rightarrow\frac{2019-x}{2016}=\frac{2019-x}{2017}+\frac{2019-x}{2018}\)

\(\Rightarrow\left(2019-x\right)\cdot\frac{1}{2016}=\left(2019-x\right)\left(\frac{1}{2017}+\frac{1}{2018}\right)\)

\(\Rightarrow2019-x=0\)

\(\Rightarrow x=2019\)

Ta có:\(\frac{a}{a'}+\frac{b'}{b}=1\)

\(\Rightarrow ab+a'b'=a'b\)

\(\Rightarrow abc+a'b'c=a'bc\left(1\right)\)

Lại có:\(\frac{b}{b'}+\frac{c'}{c}=1\)

\(\Rightarrow bc+b'c'=b'c\)

\(\Rightarrow a'bc+a'b'c'=a'b'c\left(2\right)\)

Cộng vế theo vế của (1) và (2) ta được:

\(abc+a'b'c'=0\)

có \(x^2-2y=4^2-2\cdot8=16-16=\)0

do đó C=0

trả lời:

=7000

hok tốt 

mk ko muốn vào gr

Trả lời :

==7000

Ok

#Thiên_Hy