Tìm x biết:
\(\left(x-7\right)^{x+1}=\left(x-7\right)^{x+11}\)
tìm xϵZ biết:
\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
\(\Leftrightarrow\left(x-7\right)^{x+1}\left[1-\left(x-7\right)^{10}\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-7\right)^{x+1}=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-7=0\\x-7=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=8\end{matrix}\right.\)
Tìm x biết \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
=> (x-7)^x+1 + 1.[1-(x-7)^10] = 0
=> x-7 = 0 hoặc 1-(x-7)^10 = 0
=> x=7 hoặc x = 8 hoặc x = 6
k mk nha
tìm x biết
\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
=> (x-7)x+1(1 - (x-7)10) = 0
=> (x-7)x+1 = 0
=> x-7 = 0 => x = 7
hoặc 1 - (x-7)10 = 0
=> (x-7)10 = 1
=> x-7 = 1
=> x = 8
\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
\(\Leftrightarrow\left(x-7\right)^{x+1}.\left[1-\left(x-7\right)^{10}\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-7\right)^{x+1}=0\\1-\left(x-7\right)^{10}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\\left(x-7\right)^{10}=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=8\end{cases}}\)
\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
\(\Leftrightarrow\left(x-7\right)^{x+1}.\left[1-\left(x-y\right)^{10}\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-7^{x+1}=0\\1-\left(x-y\right)^{10}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\\left(x-y\right)^{10}=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=8\end{cases}}\)
Cái này ms đúng
Tìm x, biết:
a)\(3\left|x+4\right|-\left|2x+1\right|-5\left|x+3\right|+\left|x-9\right|=5\)
b)\(\left|\frac{11}{5}-x\right|+\left|x+\frac{1}{5}\right|+\frac{41}{5}=1,2\)
c)\(2\left|x+\frac{7}{2}\right|+\left|x\right|-\frac{7}{2}=\left|\frac{11}{5}-x\right|\)
Mình cần gấp bài này !!!
1/Tìm x, biết :
a/ \(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}+\frac{x}{\left(x+3\right)\left(x+34\right)}\)
b/ \(\frac{3}{\left(x-4\right)\left(x-7\right)}+\frac{6}{\left(x-7\right)\left(x-13\right)}+\frac{15}{\left(x-13\right)\left(x-28\right)}-\frac{1}{x-28}=\frac{-1}{20}\)
Tìm x
\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
\(\Leftrightarrow\left(x-7\right)\left(x-6\right)\left(x-8\right)=0\)
hay \(x\in\left\{6;7;8\right\}\)
\(\Leftrightarrow\left(x-7\right)^{x+1}-\left(x-7\right)^{x+1}\left(x-7\right)^{10}=0\)
\(\Leftrightarrow\left(x-7\right)^{x+1}.\left(1-\left(x-7\right)^{10}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\\left(x-7\right)^{10}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\\left[{}\begin{matrix}x-7=1\Rightarrow x=8\\x-7=-1\Rightarrow x=6\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x=\left\{6;7;8\right\}\)
tìm x , biết: \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)
(x-7)x+1(1-(x-7)10)=0
=>(x-7)x+1=0 hoặc 1-(x-7)10=0
x-7=0 hoặc (x-7)10=1
x=7 hoặc /x-7/=1(/x-7/ là giá trị tuyệt đối của x-7)
x=7 hoặc x-7=1 hoặ x-7=-1
x=7 hoặc x-8 hoặc x=6
tick cho minh nha!!!
Tìm x biết
\(\left(x-7\right)^{x+11}-\left(x-7\right)^{x+1}=0\)
\(\left(x-7\right)^{x+11}-\left(x-7\right)^{x+1}=0\)
\(\left(x-7\right)^{x+1}.\left(x-7\right)^{10}-\left(x-7\right)^{x+1}=0\)
\(\left(x-7\right)^{x+1}\left[\left(x-7\right)^{10}-1\right]=0\)
Th1 \(x-7=0\Rightarrow x=7\)
Th2 \(\left(x-7\right)^{10}-1=0\)
\(\left(x-7\right)^{10}=1\Rightarrow x-7=0\Leftrightarrow x=7\)
Vậy x=7
Tìm x biết
\(\left(x-7\right)^{x+1}-\left(x-7^{ }\right)^{x+11}=0\)
Tìm x biết :\(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)\(0\)