1) Ta có: \(3\left(x-1\right)-5x=9\left(x+4\right)-20\)

\(\Leftrightarrow-2x-3=9x+16\)

\(\Leftrightarrow-11x=19\)

hay \(x=-\dfrac{19}{11}\)

2: Ta có: \(4\left(3x+2\right)-3\left(x-4\right)=9x+20\)

\(\Leftrightarrow12x+8-3x+12-9x-20=0\)

\(\Leftrightarrow0x=0\)(luôn đúng

Xét B = \(\frac{201+202+203}{202+203+204}\)

         = \(\frac{201}{202+203+204}\)\(+\)\(\frac{202}{202+203+204}\)\(+\)\(\frac{203}{202+203+204}\)

Vì 202 < 202 + 203 + 204

=> \(\frac{201}{202}\)> \(\frac{201}{202+203+204}\)( 1 )

Vì 203 < 202 + 203 + 204

=> \(\frac{202}{203}\)>\(\frac{202}{202+203+204}\)( 2 )

Vì 204 < 202 + 203 + 204

=> \(\frac{203}{204}\)> \(\frac{203}{202+203+204}\)( 3 )
Cộng vế với vế của ( 1 ), ( 2 ) và ( 3 )

=> \(\frac{201}{202}+\frac{202}{203}+\frac{203}{204}\)> \(\frac{201+202+203}{202+203+204}\)

=> A > B

Vậy A > B

Ta có : \(\frac{x+4}{200}+\frac{x+3}{201}=\frac{x+2}{202}+\frac{x+1}{203}\)

=> \(\frac{x+4}{200}+\frac{x+3}{201}-\frac{x+2}{202}-\frac{x+1}{203}=0\)

=> \(\frac{x+4}{200}+1+\frac{x+3}{201}+1-\frac{x+2}{202}-1-\frac{x+1}{203}-1=0\)

=> \(\frac{x+204}{200}+\frac{x+204}{201}-\frac{x+204}{202}-\frac{x+204}{203}=0\)

=> \(\left(x+204\right)\left(\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\right)=0\)

=> \(x+204=0\)

=> \(x=-204\)

Vậy phương trình có tập nghiệm là S = { -204 }

a, 202²⁰³=(101.2)²⁰³

=101²⁰³.2²⁰³

=101²⁰².2²⁰².202

b, 203²⁰²=(101,5.2)²⁰²

=101,5²⁰².2²⁰²

còn lại dễ

b, 1990¹⁰+1990⁹=1990⁹.1990+1990⁹=1990⁹.(1990+1)=1990⁹.1991

1991²⁰=1991¹⁹.1991

=>1990¹⁰+1990⁹<1991²⁰

c, 11¹⁹⁷⁹<11¹⁹⁸⁰=(11³)⁶⁶⁰=1331⁶⁶⁰

37¹³²⁰=(37²)⁶⁶⁰=1369⁶⁶⁰

=>11¹⁹⁷⁹<37¹³²⁰

<=> (2-x/201 + 1) + (x/203 - 1) = (1-x/202 + 1) + (1-1)

<=> 203-x/201 + x-203/203 = 203-x/202

<=> 203-x/201 - 203-x/203 - 203-x/202 = 0

<=> (203-x).(1/201-1/203-1/202) = 0

<=> 203-x = 0 ( vì 1/201-1/203-1/202 khác 0 )

<=> x=203

Vậy x=203

k mk nha

mik cần gấp

Xét B =  \(\frac{201+202+203}{202+203+204}\)

          = \(\frac{201}{202+203+204}\)+ \(\frac{202}{202+203+204}\)+ \(\frac{203}{202+203+204}\)

 Vì 202 < 202 + 203 + 204 nên \(\frac{201}{202}\)>\(\frac{201}{202+203+204}\)(1)

Vì 203 < 202 + 203 + 204 nên \(\frac{202}{203}\)> \(\frac{202}{202+203+204}\)(2)

Vì 204 < 202 + 203 + 204 nên \(\frac{202}{203}\)>\(\frac{202}{202+203+204}\)(3)

Cộng vế vơi vế của (1) , (2) và (3)

=>\(\frac{201}{202}+\frac{202}{203}+\frac{203}{204}\)> \(\frac{201+202+203}{202+203+204}\)

=> A > B

Vậy A > B

\(\frac{x+5}{200}+\frac{x+4}{201}=\frac{x+3}{202}+\frac{x+2}{203}\)

=> \(\left(1+\frac{x+5}{200}\right)+\left(1+\frac{x+4}{201}\right)=\left(1+\frac{x+3}{202}\right)+\left(1+\frac{x+2}{203}\right)\)

=> \(\frac{x+205}{200}+\frac{x+205}{201}=\frac{x+205}{202}+\frac{x+205}{203}\)

=> \(\frac{x+205}{200}+\frac{x+205}{201}-\frac{x+205}{202}-\frac{x+205}{203}=0\)

=> \(\left(x+205\right).\left(\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\right)=0\)

Do \(\frac{1}{200}>\frac{1}{202};\frac{1}{201}>1-\frac{1}{203}\)

=> \(\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\ne0\)

=> \(x+205=0\)

=> \(x=-205\)

\(\frac{x+5}{200}+\frac{x+4}{201}=\frac{x+3}{202}+\frac{x+2}{203}\)

\(=>\frac{x+5+200}{200}+\frac{x+4+201}{201}-\frac{x+3+202}{202}-\frac{x+2+203}{203}=0\)

\(=>\frac{x+205}{200}+\frac{x+205}{201}-\frac{x+205}{202}-\frac{x+205}{203}=0\)

\(=>\left(x+205\right).\left(\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\right)=0\)

\(Do:\frac{1}{200}+\frac{1}{201}-\frac{1}{202}-\frac{1}{203}\ne0\)

\(=>x+205=0\)

\(=>x=-205\)

vế phải đâu?

ko có vế pk

Ủa ko có vế phải thì mình làm bằng niềm tin à? :D

202^203<203^202

Ta có :

202²⁰³ = 8 242 408¹⁰¹ ( 1 )

203²⁰² = 42 209¹⁰¹       ( 2 )

Từ ( 1 ) và ( 2 ) suy ra 202²⁰³ < 203²⁰²

vì cơ số 202<cơ số 203 nên 202^203<203^202