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\(\left(-2\right)^4.3^{x-1}=324\)
\(16.3^{x-1}=324\)
\(3^{x-1}=\frac{324}{16}\)
\(3^x:3=\frac{81}{4}\)
\(3^x=\frac{243}{4}\left(vl\right)\)
(-2)4.3x-1=324 => 16.3x-1 = 324 => 3x-1 = 324:16 => 3x-1 = 81/4 => 3x:3 = 81/4 => 3x = 81/4.3 => 3x =243/4 (tự tính tiếp nhé bạn) ...
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\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}=\dfrac{x+1}{6}\)
\(\dfrac{x+1}{3}+\dfrac{x+1}{4}+\dfrac{x+1}{5}-\dfrac{x+1}{6}=0\)
\(\left(x+1\right)\left(\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}\right)=0\)
\(\)vì \(\dfrac{1}{3}>\dfrac{1}{6};\dfrac{1}{4}>\dfrac{1}{6};\dfrac{1}{5}>\dfrac{1}{6}=>\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}>0\)
\(=>x+1=0\)
\(=>x=-1\)
b,
\(\dfrac{x+1}{2020}+\dfrac{x+2}{2019}=\dfrac{x+3}{2018}+\dfrac{x+4}{2017}\)
\(\left(\dfrac{x+1}{2020}+1\right)+\left(\dfrac{x+2}{2019}+1\right)=\left(\dfrac{x+3}{2018}+1\right)+\left(\dfrac{x+4}{2017}+1\right)\)
\(\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}=\dfrac{x+2021}{2018}+\dfrac{x+2021}{2017}\)
\(=>\dfrac{x+2021}{2020}+\dfrac{x+2021}{2019}-\dfrac{x+2021}{2018}-\dfrac{x+2021}{2017}=0\)
\(=>\left(x+2021\right)\left(\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}\right)=0\)
Vì \(\dfrac{1}{2020}< \dfrac{1}{2018};\dfrac{1}{2019}< \dfrac{1}{2017}=>\dfrac{1}{2020}+\dfrac{1}{2019}-\dfrac{1}{2018}-\dfrac{1}{2017}< 0\)
\(=>x+2021=0\)
\(=>x=-2021\)
c,
\(\dfrac{x+2}{327}+\dfrac{x+3}{326}+\dfrac{x+4}{325}+\dfrac{x+5}{324}+\dfrac{x+349}{5}=0\)
\(\left(\dfrac{x+2}{327}+1\right)+\left(\dfrac{x+3}{326}+1\right)+\left(\dfrac{x+4}{325}+1\right)+\left(\dfrac{x+5}{324}+1\right)+\left(\dfrac{x+349}{5}-4\right)=0\)
\(\dfrac{x+329}{327}+\dfrac{x+329}{326}+\dfrac{x+329}{325}+\dfrac{x+329}{324}+\dfrac{x+329}{5}=0\)
\(=>\left(x+329\right)\left(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}\right)=0\)
Vì \(\dfrac{1}{327}+\dfrac{1}{326}+\dfrac{1}{325}+\dfrac{1}{324}+\dfrac{1}{5}>0\)
\(=>x+329=0\)
\(=>x=-329\)
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(x+1/5)^2 =26/25-17/25
<=> (x +1/5)^2 =(3/5)^2
<=> x+1/5=3/5
=> x= 2/5
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1.
(x + 1/5)² = 26/25 - 17/25
(x + 1/5)² = 9/25
Rút căn hai vế :
|(x + 1/5)| = 3/5
x = -4/5
hoặc
x = 2/5
2.
(x + 2) / 327 + (x + 3) / 326 + (x + 4) / 325 + (x + 5) / 324 + (x + 349) / 5 = 0
<=> (x + 2) / 327 +1+ (x + 3) / 326 +1+ (x + 4) / 325 +1+ (x + 5) / 324 +1+ (x + 349) / 5 -4 = 0
<=> (x+ 329)/327 + (x+ 329)/326 + (x+ 329)/325 + (x+ 329)//324 + (x+ 329)/5 =0
<=> (x+ 329).(1/327 + 1/ 326 + 1/325 + 1/324 +1/5) =0
Do (1/327 + 1/ 326 + 1/325 + 1/324 +1/5) >0 nên x+ 329 =0 => x= -329
Câu 1 chưa chắc đã đúng ( quên hết kiến thức lớp 6 rùi ) hihi
aaaaaaaa . chết rồi . cho mình sủa câu thứ nhất :
(x+1/5)2 + 17/25=26/25
( x + 1/5 ) 2 = 26/25 - 17/25
( x + 1/5 ) 2 = 3/52
x + 1/5 = 3/5
x = 2/5.
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\(3^{x+1}+3^{x+2}=324\\ \Rightarrow3^{x+1}\left(1+3\right)=324\\ \Rightarrow3^{x+1}=324:4=81=3^4\\ \Rightarrow x+1=4\Rightarrow x=3\)
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\(3^{x+1}+3^{x+2}=324\)
\(\Leftrightarrow\)\(3^x.3+3^x.3^2=324\)
\(\Leftrightarrow\)\(3^x\left(3+3^2\right)=324\)
\(\Leftrightarrow\)\(3^x\left(3+9\right)=324\)
\(\Leftrightarrow\)\(3^x.12=324\)
\(\Leftrightarrow\)\(3^x=\frac{324}{12}\)
\(\Leftrightarrow\)\(3^x=27\)
\(\Leftrightarrow\)\(3^x=3^3\)
\(\Leftrightarrow\)\(x=3\)
Vậy \(x=3\)
Chúc bạn học tốt ~
\(3^{x+1}+3^{x+2}=324\)
\(3^x.3+3^x.3^2=324\)
\(3^x.3+3^x.9=324\)
\(3^x.\left(3+9\right)=324\)
\(3^x.12=324\)
\(3^x=324:12\)
\(3^x=27\)
\(3^x=3^3\)
\(\Rightarrow x=3\)
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