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21 tháng 9 2021
\(c,\Rightarrow\left[{}\begin{matrix}-2\left(x+2\right)+\left(4-x\right)=11\left(x< -2\right)\\2\left(x+2\right)+\left(4-x\right)=11\left(-2\le x\le4\right)\\2\left(x+2\right)+\left(x-4\right)=11\left(x>4\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{11}{3}\left(tm\right)\\x=3\left(tm\right)\\x=\dfrac{11}{3}\left(ktm\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{11}{3}\end{matrix}\right.\)
21 tháng 9 2021
\(a,\Rightarrow\left[{}\begin{matrix}x+\dfrac{5}{2}=3x+1\\x+\dfrac{5}{2}=-3x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{7}{8}\end{matrix}\right.\)
30 tháng 11 2016
a)\(-x^2\left(x^2-4\right)=-25\left(x^2-4\right)\)
\(\Leftrightarrow-x^2=-25\)
\(\Leftrightarrow x^2=25\)
\(\Leftrightarrow x=\pm5\)
NT
2
XO
25 tháng 10 2020
Ta có : (3x - 1)6 = (3x - 1)4
=> (3x - 1)6 - (3x - 1)4 = 0
=> (3x - 1)4[(3x - 1)2 - 1] = 0
=> \(\orbr{\begin{cases}\left(3x-1\right)^4=0\\\left(3x-1\right)^2=1\end{cases}}\Rightarrow\orbr{\begin{cases}3x-1=0\\3x-1\in\left\{1;-1\right\}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x\in\left\{\frac{2}{3};0\right\}\end{cases}}}\)
Vậy \(x\in\left\{0;\frac{1}{3};\frac{2}{3}\right\}\)là giá trị cần tìm
25 tháng 10 2020
(3x-1)6=(3x-1)4
=>(3x-1)6-(3x-1)4=0
=>(3x-1)4*(3x-1)2-(3x-1)4*1=0
=>(3x-1)4*[(3x-1)2-1]=0
=>(3x-1)4*(3x-1)2=0
=>(3x-1)4=0 hoặc (3x-1)2=0
+) (3x-1)4=0
+) (3x-1)2=0
Tự làm nốt nha
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
14 tháng 8 2023
a/
\(VT=\dfrac{\left(x+4\right)-\left(x+2\right)}{\left(x+2\right)\left(x+4\right)}+\dfrac{\left(x+8\right)-\left(x+4\right)}{\left(x+4\right)\left(x+8\right)}+\dfrac{\left(x+14\right)-\left(x+8\right)}{\left(x+8\right)\left(x+14\right)}=\)
\(=\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+8}+\dfrac{1}{x+8}-\dfrac{1}{x+14}=\)
\(=\dfrac{1}{x+2}-\dfrac{1}{x+14}=\dfrac{12}{\left(x+2\right)\left(x+14\right)}\)
\(\Rightarrow\dfrac{12}{\left(x+2\right)\left(x+14\right)}=\dfrac{x}{\left(x+2\right)\left(x+14\right)}\left(x\ne-2;x\ne-14\right)\)
\(\Rightarrow x=12\)
14 tháng 8 2023
\(\dfrac{x}{2023}+\dfrac{x+1}{2022}+...+\dfrac{x+2022}{1}+2023=0\)
\(\dfrac{1}{2023}x+\dfrac{1}{2022}x+\dfrac{1}{2022}\cdot1+...+\dfrac{1}{1}x+\dfrac{1}{1}\cdot2022+2023=0\)
\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)+\left(\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\right)=0\)
\(x\left(\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}\right)=\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023\)
\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2}{2021}+...+\dfrac{2022}{1}+2023}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)
\(x=\dfrac{\dfrac{1}{2022}+\dfrac{2022}{2022}+\dfrac{2}{2021}+\dfrac{2021}{2021}+...+\dfrac{2022}{1}+\dfrac{1}{1}}{\dfrac{1}{2023}+\dfrac{1}{2022}+...+\dfrac{1}{1}}\)
\(x=\dfrac{\dfrac{2023}{2022}+\dfrac{2023}{2021}+...+\dfrac{2023}{1}}{\dfrac{1}{2022}+\dfrac{1}{2021}+...+\dfrac{1}{1}}=2023\)
Vậy x = 2023
dễ mà
bn chia 2 trường hợp
TH1 : 3x - 4 = 6
TH2 : x + 2 = 6
Vậy ....
bn tự tính nha .. bn hỉu hăm ??