\(\frac{x+18}{2018}+\frac{x+17}{2017}+\frac{x+16}{2016}=3\)

\(\Rightarrow\frac{x+18}{2018}-1+\frac{x+17}{2017}-1+\frac{x+16}{2016}-1=3-3\)

\(\Rightarrow\frac{x+18-2018}{2018}+\frac{x+17-2017}{2017}+\frac{x+16-2016}{2016}=0\)

\(\Rightarrow\frac{x-2000}{2018}+\frac{x-2000}{2017}+\frac{x-2000}{2016}=0\)

\(\Rightarrow\left(x-2000\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\right)=0\)

Vì \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\ne0\)

=> x - 2000 = 0

=> x            = 2000

Ta có : 

\(\frac{x+18}{2018}+\frac{x+17}{2017}+\frac{x+16}{2016}=3\)

\(\Leftrightarrow\)\(\left(\frac{x+18}{2018}-1\right)+\left(\frac{x+17}{2017}-1\right)+\left(\frac{x+16}{2016}-1\right)=3-3\) ( trừ hai vế cho 3 ) 

\(\Leftrightarrow\)\(\frac{x-2000}{2018}+\frac{x-2000}{2017}+\frac{x-2000}{2016}=0\)

\(\Leftrightarrow\)\(\left(x-2000\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\right)=0\)

Vì \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}\ne0\)

Nên \(x-2000=0\)

\(\Rightarrow\)\(x=2000\)

Vậy \(x=2000\)

Chúc bạn học tốt ~ 

a) Ta có : \(16^{17}=\left(2^4\right)^{17}=2^{68}\)

                 \(8^{18}=\left(2^3\right)^{18}=2^{54}\)

Vì \(2^{68}>2^{54}\Rightarrow16^{17}>8^{18}\)

b) Ta có:  \(3^{555}=\left(3^5\right)^{111}=243^{111}\)

                \(5^{333}=\left(5^3\right)^{111}=125^{111}\)

Vì \(243^{111}>125^{111}\Rightarrow3^{555}>5^{333}\)
c) Ta có : \(2017^2=2017\cdot2017=2017\cdot2016+2017\)                                       
                \(2016\cdot2018=2016\cdot\left(2017+1\right)=2016\cdot2017+2016\)
Vì 2016 < 2017 nên 2016*2017 + 2017 > 2016*2017 + 2016
Vậy \(2017^2>2016\cdot2018\)
Lưu ý : dấu \(\left(\cdot\right)\)là dấu nhân nha bạn 

2, 

a) \(315-\left(135-x\right)=215\)

\(\Rightarrow135-x=315-215\)

\(\Rightarrow135-x=100\)

\(\Rightarrow x=135-100\)

\(\Rightarrow x=35\)

b) \(x-320:32=25\cdot16\)

\(\Rightarrow x-10=5^2\cdot4^2\)

\(\Rightarrow x-10=20^2\)

\(\Rightarrow x-10=400\)

\(\Rightarrow x=410\)

c) \(3\cdot x-2018:2=23\)

\(=3\cdot x-1009=23\)

\(\Rightarrow3\cdot x=1032\)

\(\Rightarrow x=1032:3\)

\(\Rightarrow x=344\)

d) \(280-9\cdot x-x=80\)

\(\Rightarrow280-x\cdot\left(9+1\right)=80\)

\(\Rightarrow280-10\cdot x=80\)

\(\Rightarrow10\cdot x=280-80\)

\(\Rightarrow10\cdot x=200\)

\(\Rightarrow x=20\)

e) \(38\cdot x-12\cdot x-x\cdot16=40\)

\(\Rightarrow x\cdot\left(38-12-16\right)=40\)

\(\Rightarrow x\cdot10=40\)

\(\Rightarrow x=40:10\)

\(\Rightarrow x=4\)

bài 1 cậu ko giải giúp mình dc à

Lời giải:

a.

PT $\Leftrightarrow (x+3)^2=2016^{2020}-17^{91}+9$

Ta thấy: $2016^{2020}-17^{91}+9\equiv 0-(-1)^{91}+0\equiv -1\equiv 2\pmod 3$

Mà 1 scp thì chia $3$ chỉ dư $0$ hoặc $1$ nên pt vô nghiệm.

b.

$x^2=2016(y-1)^2-2017^{2019}\equiv 0-1^{2019}\equiv 3\pmod 4$
Mà 1 scp chia $4$ chỉ dư $0$ hoặc $1$ nên vô lý.

Vậy pt vô nghiệm.

c.

$(x-1)^2=2017^{2017}+1\equiv 1^{2017}+1\equiv 2\pmod 4$
Mà 1 scp khi chia cho $4$ chỉ dư $0$ hoặc $1$ nên vô lý

Vậy pt vô nghiệm

d.

$(x+2)^2=2018^{10}+4\equiv (-1)^{10}+1\equiv 2\pmod 3$

Mà 1 scp khi chia $3$ dư $0$ hoặc $1$ nên vô lý

Vậy pt vô nghiệm.

Ta có:  \(\frac{x-18}{2018}=\frac{x-17}{2017}\)

\(\Rightarrow\left(x-18\right).2017=\left(x-17\right).2018\)( tính chất của 2 tỉ số bằng nhau )

\(2017x-2017.18=2018x-2018.17\)

\(2018.17-2017.18=2018x-2017x\)

\(\left(2017+1\right).17-2017.\left(17+1\right)=x\)

\(2017.17+17-2017.17-2017=x\)

\(x=-2000\)

Vậy \(x=-2000\)

\(\frac{x+1}{99}+\frac{x+2}{98}=\frac{x-1}{101}+\frac{x-2}{102}\)

\(\Rightarrow\left(\frac{x+1}{99}+1\right)+\left(\frac{x+2}{98}+1\right)=\left(\frac{x-1}{101}+1\right)+\left(\frac{x-2}{102}+1\right)\) ( cộng cả 2 vế thêm 2 )

\(\frac{x+100}{99}+\frac{x+100}{98}=\frac{x+100}{101}+\frac{x+100}{102}\)

\(\Rightarrow\frac{x+100}{99}+\frac{x+100}{98}-\frac{x+100}{101}-\frac{x+100}{102}=0\)

\(\left(x+100\right).\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{101}-\frac{1}{100}\right)=0\)

Ta có: \(\frac{1}{99}+\frac{1}{98}-\frac{1}{101}-\frac{1}{100}\ne0\)

\(\Rightarrow x+100=0\)

​​​\(x=-100\)​

Vậy \(x=-100\)

a, \(\frac{x-18}{2018}=\frac{x-17}{2017}\)

=>\(\frac{x-18}{2018}+1=\frac{x-17}{2017}+1\)

=>\(\frac{x-18+2018}{2018}=\frac{x-17+2017}{2017}\)

=>\(\frac{x+2000}{2018}=\frac{x+2000}{2017}\)

=>\(\frac{x+2000}{2018}-\frac{x+2000}{2017}=0\)

=>\(\left(x+2000\right)\left(\frac{1}{2018}-\frac{1}{2017}\right)=0\)

Mà \(\frac{1}{2018}-\frac{1}{2017}\ne0\)

=>x+2000=0 => x=-2000

b, 

=>\(\frac{x+1}{99}+1+\frac{x+2}{98}+1=\frac{x-1}{101}+1+\frac{x-2}{102}+1\)

=>\(\frac{x+1+99}{99}+\frac{x+2+98}{98}=\frac{x-1+101}{101}+\frac{x-2+102}{102}\)

=>\(\frac{x+100}{99}+\frac{x+100}{98}=\frac{x+100}{101}+\frac{x+100}{102}\)

=>\(\frac{x+100}{99}+\frac{x+100}{98}-\frac{x+100}{101}-\frac{x+100}{102}=0\)

=>\(\left(x+100\right)\left(\frac{1}{99}+\frac{1}{98}-\frac{1}{101}-\frac{1}{102}\right)=0\)

Mà \(\frac{1}{99}+\frac{1}{98}-\frac{1}{101}-\frac{1}{102}\ne0\)

=>x+100=0 => x=-100

\(\frac{x-1}{2018}=\frac{x-17}{2017}\)

\(\Rightarrow\frac{x-18}{2018}=\frac{x-17}{2017}=\frac{\left(x-18\right)-\left(x-17\right)}{2018-2017}=-\frac{1}{1}=-1\)

\(\Rightarrow x-18=-2018\)

\(\Rightarrow x=-2000\)

Uyên ơi m đừng mất dậy đừng chửi con gái h đú vc

Các đề bài trên khi chuyển vế đều bị mất đi x nên không có x thỏa mãn

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