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k hộ mik nha alan walker ngầu lol
Ta có :\(\frac{x-5}{4}=\frac{-9}{5-x}\)
\(\Rightarrow\frac{x-5}{4}=\frac{\left(-9\right).\left(-1\right)}{\left(5-x\right).\left(-1\right)}\)
\(\Rightarrow\frac{x-5}{4}=\frac{9}{x-5}\)
=>(x-5).(x-5)=9.4
\(\Rightarrow\left(x-5\right)^2=36\)
\(\Rightarrow\orbr{\begin{cases}x-5=6\\x-5=-6\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=11\\x=-1\end{cases}}\)
Vậy \(x=\left(11;-1\right)\)
Ta có: \(5.\left|x\right|=\left|-2017\right|+\left|13\right|\)
\(5.\left|x\right|=2017+13\)
\(5.\left|x\right|=2030\)
\(\left|x\right|=406\)
\(\Rightarrow x\in\left\{-406;406\right\}\)
Sắp thi rùi, chúc bn hok "tuốt" nha^^
\(5^{x+2}-5^x=3.10^3\)
\(\Rightarrow5^x\left(5^2-1\right)=3000\)
\(\Rightarrow5^x.24=3000\Rightarrow5^x=125=5^3\)
\(\Rightarrow x=3\)
\(5^{x+2}-5^x=3\cdot10^3\)
\(\Leftrightarrow5^x\cdot24=3000\)
\(\Leftrightarrow x=3\)
a)2-x=17-(-5)
2-x=17+5
2-x=22
x=2-20
x=-20
b)x-12=-9-15
x-12=-24
x=(-24)+12
x=-12
c)11-(15+11)=x-(25-9)
11-26=x-16
-15=x-16
x=-15+16
x=1
d)x-25=(7-x)-(25+7)
x-25=(7-x)-32
2x-25=7-32
2x-25=-25
2x=(-25)+25
2x=0
x=0:2
x=0
tick cho mk nha bạn
a)2-x=17-(-5)
2-x=17+5
2-x=22
x=2-20
x=-20
b)x-12=-9-15
x-12=-24
x=(-24)+12
x=-12
c)11-(15+11)=x-(25-9)
11-26=x-16
-15=x-16
x=-15+16
x=1
d)x-25=(7-x)-(25+7)
x-25=(7-x)-32
2x-25=7-32
2x-25=-25
2x=(-25)+25
2x=0
x=0:2
x=0
\(2^5.x-2^3.x+x=\left(7^5:7^5\right).2\)
\(\Rightarrow32.x-8.x+x=1.2\)
\(\Rightarrow\left(32-8+1\right).x=2\)
\(\Rightarrow25.x=2\)
\(\Rightarrow x=2:25\)
\(\Rightarrow x=\frac{2}{25}\)
Vậy \(x=\frac{2}{25}\)
Vì 15= 16 nên x - 5 = 1
x= 1+5
x=6
\(\left(x-5\right)^5=\left(x-5\right)^6\)
\(\Rightarrow\left(x-5\right)^6-\left(x-5\right)^5=0\)
\(\Rightarrow\left(x-5\right)^5.\left(x-5-1\right)=0\)
\(\Rightarrow\left(x-5\right)^5.\left(x-6\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x-5\right)^5=0\\x-6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)