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Lời giải:
$5^x+5^{x+1}+5^{x+2}+5^{x+3}=1+2+3+...+87+88-4^2$
$5^x(1+5+5^2+5^3)=88.89:2-16$
$5^x.156=3900$
$5^x=3900:156=25=5^2$
$\Rightarrow x=2$
Bài 1:
a: =42x53+47x42
=42x100
=4200
b: =1152-374-1152-65+374=-65
c: =(-1)+(-1)+...+(-1)+211
=211-105
=106
3\(x^2\).(5\(x\) + 1) + 6\(x^3\).(5\(x\) + 2) = 9\(x^3\) .(5\(x\) + 3)
15\(x^3\) + 3\(x^2\) + 30\(x^4\) + 12\(x^3\) = 45\(x^4\) + 27\(x^3\)
(15\(x^3\) + 12\(x^3\)) + 3\(x^2\) + 30\(x^4\) - 45\(x^4\) - 27\(x^3\) = 0
27\(x^3\) + 3\(x^2\) - 15\(x^4\) - 27\(x^3\) = 0
3\(x^2\) - 15\(x^4\) = 0
3\(x^2\).(1 - 5\(x^2\)) = 0
\(\left[{}\begin{matrix}x^2=0\\1-5x^2=0\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\5x^2=1\end{matrix}\right.\)
\(\left[{}\begin{matrix}x=0\\x=\mp\dfrac{\sqrt{5}}{5}\end{matrix}\right.\)
a: Ta có: \(142-\left[50-\left(2^3\cdot10-2^3\cdot5\right)\right]\)
\(=142-\left[50-80+40\right]\)
=142-10
=132
a. 142 - [50-(23.10-23.5)]
= 142 - [ 50 - ( 8 . 10 - 8 . 5 )]
= 142 - [ 50 - ( 80 - 40 )]
= 142 - [ 50 - 40 ]
= 142 - 10 = 132
b. 375 : { 32[ 4+(5.32 - 42 )]} - 14
= 375 : { 32[4+(5.9 - 42 )]} - 14
= 375 : { 32[4 + ( 45 - 42 )]} - 14
= 375 : {32[4+3]} - 14
= 375 : 224 - 14
c.{210 : [ 16+3.(6+3.2^2)]} - 3
= {210 : [ 16 + 3.(6+3.4)]} - 3
= { 210 :[16+3.(6+12)]}-3
= {210 : [ 16+3.18)]} - 3
= { 210 : [ 16 + 54]} - 3
= { 210 : 70 } - 3
= 30 - 3 = 27
d.500-{5.[409-(2^3 . 3-21^2)] - 1724}
= 500-{5.[409-(8 . 63 - 21^2] - 1724}
= 500 - { 5.[409- (504 - 441) - 1724}
= 500 -{ 5.[ 409 - 63 ] - 1724}
= 500 - { 5.346 - 1724}
= 500 - { 1720 - 1724 }
= 500 - 6
= 494
`@` `\text {Ans}`
`\downarrow`
`1)`
\(2x+\dfrac{1}{2}=\dfrac{5}{3}\)
`\Rightarrow`\(2x=\dfrac{5}{3}-\dfrac{1}{2}\)
`\Rightarrow`\(2x=\dfrac{7}{6}\)
`\Rightarrow`\(x=\dfrac{7}{6}\div2\)
`\Rightarrow`\(x=\dfrac{7}{12}\)
Vậy, `x = 7/12`
`2)`
\(\dfrac{1}{7}+\dfrac{4}{5}x=\dfrac{5}{3}\)
`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{5}{3}-\dfrac{1}{7}\)
`\Rightarrow`\(\dfrac{4}{5}x=\dfrac{32}{21}\)
`\Rightarrow`\(x=\dfrac{32}{21}\div\dfrac{4}{5}\)
`\Rightarrow`\(x=\dfrac{40}{21}\)
Vậy, `x = 40/21`
`3)`
\(\dfrac{3}{5}-\dfrac{3}{5}x=\dfrac{1}{7}\)
`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{3}{5}-\dfrac{1}{7}\)
`\Rightarrow`\(\dfrac{3}{5}x=\dfrac{16}{35}\)
`\Rightarrow`\(x=\dfrac{16}{35}\div\dfrac{3}{5}\)
`\Rightarrow`\(x=\dfrac{16}{21}\)
Vậy, `x = 16/21`
`4)`
\(\dfrac{5}{6}-3x=\dfrac{3}{4}\)
`\Rightarrow`\(3x=\dfrac{5}{6}-\dfrac{3}{4}\)
`\Rightarrow`\(3x=\dfrac{1}{12}\)
`\Rightarrow`\(x=\dfrac{1}{12}\div3\)
`\Rightarrow`\(x=\dfrac{1}{36}\)
Vậy, `x = 1/36`
`5)`
\(\dfrac{5}{3}-\dfrac{1}{2}x=\dfrac{3}{7}\)
`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{5}{3}-\dfrac{3}{7}\)
`\Rightarrow`\(\dfrac{1}{2}x=\dfrac{26}{21}\)
`\Rightarrow`\(x=\dfrac{26}{21}\div\dfrac{1}{2}\)
`\Rightarrow`\(x=\dfrac{52}{21}\)
Vậy, `x = 52/21`
`6)`
\(5x+\dfrac{1}{2}=\dfrac{2}{3}\)
`\Rightarrow`\(5x=\dfrac{2}{3}-\dfrac{1}{2}\)
`\Rightarrow`\(5x=\dfrac{1}{6}\)
`\Rightarrow`\(x=\dfrac{1}{6}\div5\)
`\Rightarrow`\(x=\dfrac{1}{30}\)
Vậy, `x = 1/30.`
1: =>25x-15-6x+12=11-5x
=>19x-3=11-5x
=>24x=14
=>x=7/12
2: =>8-12x-5+10x=4-6x
=>4-6x=-2x+3
=>-4x=-1
=>x=1/4
\(5^x+5^{x+1}+5^{x+2}+5^{x+3}=1+2+3+...+87+88-4^2\)
=>\(5^x+5^x\cdot5+5^x\cdot25+5^x\cdot125=88\cdot\dfrac{\left(88+1\right)}{2}-16\)
=>\(156\cdot5^x=44\cdot89-16=3900\)
=>\(5^x=\dfrac{3900}{156}=25\)
=>x=2