tim x biet
6^x+8^x=10^x
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a) \(\left(x-6\right)^{10}=\left(x-6\right)^8\)
\(\Rightarrow\left(x-6\right)^{10}-\left(x-6\right)^8=0\)
\(\Rightarrow\left(x-6\right)^8.\left[\left(x-6\right)^2-1\right]=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-6\right)^8=0\\\left(x-6\right)^2-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-6=0\\\left(x-6\right)^2=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0+6\\x-6=\pm1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x-6=1\\x-6=-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\\x=7\\x=5\end{matrix}\right.\)
Vậy \(x\in\left\{6;7;5\right\}.\)
Chúc em học tốt!
Troi a gio hoi ten moi bao bai ha
Thoi giup to di choc to se bao ten cho
\(\frac{1}{6}=\frac{x}{18}\Rightarrow x=3\)
\(\frac{x}{8}=-\frac{1}{4}\Rightarrow x=-2\)
\(\frac{4}{-5}=\frac{x}{10}\Rightarrow x=-8\)
\(\frac{11}{5}=-\frac{22}{x}\Rightarrow x=-10\)
\(\frac{x}{8}=\frac{8}{x}\Rightarrow x^2=64\Rightarrow x=8\)
\(\frac{x}{-11}=-\frac{11}{x}\Rightarrow x^2=121\Rightarrow x=11\)
#H
a) (10-x)2:3+3=-9
(10-x)2:3=-9-3
(10-x)2:3=-11
(10-x)2=-11.3
(10-x)2=-33
Vô lí vì -33 ko chuyển đc mũ 2
Vậy x e 0
b) |x-6|.7-4=31
|x-6|=(31+4):7
|x-6|=5
\(\Rightarrow\orbr{\begin{cases}x-6=5\Rightarrow5+6=11\\x-6=-5\Rightarrow x=-5+6=1\end{cases}}\)
Vậy.....................................................
a) \(\frac{\left(10-x\right)^2}{3}+3=-9\)
\(\Leftrightarrow\frac{\left(10-x\right)^2}{3}=-12\)
\(\Leftrightarrow\left(10-x\right)^2=-36\)( vô nghiệm vì (10-x)2 \(\ge\)0 với mọi x thuộc Z)
b) |x-6|.7-(-2)2=-8+39
<=> |x-6|.7-4=31
<=> |x-6|.7=35
<=> |x-6|=5
\(\Leftrightarrow\orbr{\begin{cases}x-6=5\\x-6=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=11\\x=1\end{cases}}}\)
a) \(\dfrac{x}{15}\)=\(\dfrac{x-2}{30}\)
=>30x=15 (x-2)
=>30x=15x-30
=>30x-15x=-30
=>15x =-30
=>x =-2
b)\(\dfrac{7}{8}\)<\(\dfrac{x}{80}\)<\(\dfrac{9}{10}\)
=>\(\dfrac{70}{80}\)<\(\dfrac{x}{80}\)<\(\dfrac{72}{80}\)
=> 70<x<72
=>x=71
c)\(\dfrac{1}{6}\)<\(\dfrac{5}{x}\)<\(\dfrac{1}{4}\)
=>\(\dfrac{5}{30}\)<\(\dfrac{5}{x}\)<\(\dfrac{5}{20}\)
=>20<x<30
=>x={21;22;23;24;25;26;27;28;29}
\(2(x-5)-3(x+6)=10+x\)
\(\Rightarrow2x-10-3x+18=10x\)
\(\Rightarrow2x-10-3x-18=10+x\)
\(\Rightarrow-x-28=10+x\)
\(\Rightarrow-x-x=10+28\)
\(\Rightarrow-2x=38\)
\(\Rightarrow x=-19\)
Vậy ..
\(2\left(x-5\right)-\left(3x+6\right)=10+x\)
\(\Rightarrow2x-10-3x-18=10+x\)
\(\Rightarrow-x-28=10+x\)
\(\Rightarrow-x=38+x\)
\(\Rightarrow-2x=38\Leftrightarrow x=-19\)
x=2
cho minh xin cach lam nha