a) \(\left(x-5\right)^{12}=\left(x-5\right)^{10}\)
\(\Rightarrow\left(x-5\right)^{12}-\left(x-5\right)^{10}=0\)
\(\Rightarrow\left(x-5\right)^{10}\left[\left(x-5\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)^{10}=0\\\left(x-5\right)^2-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}\left(x-5\right)^{10}=0^{10}\\\left(x-5\right)^2=0+1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\\left(x-5\right)^2=1\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0+5\\\left(x-5\right)^2=1^2\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x-5=\pm1\end{cases}}\)
\(\Rightarrow x=5;\orbr{\begin{cases}x-5=1\\x-5=-1\end{cases}}\)
\(\Rightarrow x=5;\orbr{\begin{cases}x=1+5\\x=-1+5\end{cases}}\)
\(\Rightarrow x=5;\orbr{\begin{cases}x=4\\x=6\end{cases}}\)
Vậy x = 4 hoặc x = 5 hoặc x = 6
\(a)\left(x-5\right)^{12}=\left(x-5\right)^{10}\)
\(\Leftrightarrow\left(x-5\right)^{12}-\left(x-5\right)^{10}=0\)
\(\Leftrightarrow\left(x-5\right)^{10}\left[\left(x-5\right)^2-1\right]=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x-5\right)^{10}=0\\\left(x-5\right)^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\\left(x-4\right)\left(x-6\right)=0\end{cases}}\)
[ ra \(\left(x-4\right)\left(x-6\right)\)do \(\left(x-5\right)^2-1=\left(x-5-1\right)\left(x-5+1\right)=\left(x-6\right)\left(x-4\right)\)]
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=4;x=6\end{cases}}\)
_Minh ngụy_
Câu b thiếu '= ......"
\(\(c)\frac{1}{2.4}+\frac{1}{4.6}+.....+\frac{1}{2x\left(2x+2\right)}=\frac{3}{16}\)\)
\(\(\Rightarrow\frac{1}{4.\left(1.2\right)}+\frac{1}{4.\left(2.3\right)}+......+\frac{1}{4\left(x-1\right).x}=\frac{3}{16}\)\)
\(\(\Leftrightarrow\frac{1}{4}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{2.3}+....+\frac{1}{x\left(x-1\right)}\right)=\frac{3}{16}\)\)
\(\(\Rightarrow\frac{2-1}{1.2}+\frac{3-2}{2.3}+.......+\frac{x-\left(x-1\right)}{x\left(x-1\right)}=\frac{3}{4}\)\)
\(\(\Leftrightarrow1-\frac{1}{2}+\frac{1}{2}-........+\frac{1}{x-1}-\frac{1}{x}=\frac{3}{4}\)\)
\(\(\Leftrightarrow1-\frac{1}{x}=\frac{3}{4}\)\)
\(\(\Leftrightarrow\frac{1}{x}=\frac{1}{4}\)\)
\(\(\Rightarrow x=4\left(t/m:x\ne0\right)\)\)
_Minh ngụy_
\(d)\left(x-3\right)\left(x+5\right)< 0\)
Trường hợp 1 :
\(\hept{\begin{cases}x-3>0\\x+5< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>3\\x< -5\end{cases}}\Rightarrow x\in\left\{\varnothing\right\}}\)
Trường hợp 2:
\(\hept{\begin{cases}x-3< 0\\x+5>0\end{cases}\Rightarrow\hept{\begin{cases}x< 3\\x>-5\end{cases}\Rightarrow}-5< x< 3}\)
Vậy \(-5< x< 3\)là các giá trị cần tìm
_Minh ngụy_
\(e)\frac{x+4}{x-2}< 0\left(x\ne0\right)\)
Trường hợp 1 :
\(\hept{\begin{cases}x+4< 0\\x-2>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -4\\x>2\end{cases}}}\Rightarrow x\in\left\{\varnothing\right\}\)
Trường hợp 2:
\(\hept{\begin{cases}x+4>0\\x-2< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-4\\x< 2\end{cases}\Rightarrow-4< x< 2}}\)
Kết hợp với điều kiện :\(x\ne0\Rightarrow-4< x< 2\)và \(x\ne0\)là các giá trị cần tìm
_Minh ngụy_
Giờ rảnh mới giải được >>
\(b)\left(x-3\right)^{2020}+\left(2y+1\right)^{2018}=0\)
Ta có :
\(\hept{\begin{cases}\left(x-3\right)^{2020}\ge0\forall x\\\left(2y+1\right)^{2018}\ge0\forall x\end{cases}}\)
Để \(\left(x-3\right)^{2020}+\left(2y+1\right)^{2018}=0\)cần :
\(\hept{\begin{cases}x-3=0\\2y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=-\frac{1}{2}\end{cases}}\)
Vậy \(\hept{\begin{cases}x=3\\y=-\frac{1}{2}\end{cases}}\)là các giá trị cần tìm
_Minh ngụy_
