10-18:x=2020^0
giúp em với
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: \(\left(x-1\right)^{2020}\ge0\forall x\)
\(\left|y-3\right|\ge0\forall y\)
Do đó: \(\left(x-1\right)^{2020}+\left|y-3\right|\ge0\forall x,y\)
Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-1=0\\y-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\end{matrix}\right.\)
Vậy: (x,y)=(1;3)
\(65-4^{x-3}=2020^0\\ \Rightarrow65-4^{x-3}=1\\ \Rightarrow4^{x-3}=64\\ \Rightarrow4^{x-3}=4^3\\ \Rightarrow x-3=3\\ \Rightarrow x=6\)
\(\Leftrightarrow x-\left[3-x+3+x-2\right]=0\)
=>x=4
\(2x^2-x-8=0\\ \Leftrightarrow\left(2x^2-x\right)-8=0\\ \Leftrightarrow x\left(2x-1\right)-8=0\\ \Leftrightarrow\left(x-8\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=\dfrac{1}{2}\end{matrix}\right.\)
a: 3(x+7)-2x+5>0
=>3x+21-2x+5>0
=>x+26>0
=>x>-26
Sửa đề: \(\dfrac{x+2}{18}-\dfrac{x+3}{8}< \dfrac{x-1}{9}-\dfrac{x-4}{24}\)
=>\(\dfrac{4\left(x+2\right)}{72}-\dfrac{9\left(x+3\right)}{72}< \dfrac{8\left(x-1\right)}{72}< \dfrac{3\left(x-4\right)}{72}\)
=>\(4\left(x+2\right)-9\left(x+3\right)< 8\left(x-1\right)-3\left(x-4\right)\)
=>\(4x+8-9x-27< 8x-8-3x+12\)
=>-5x-19<5x+4
=>-10x<23
=>\(x>-\dfrac{23}{10}\)
b: \(3x+2+\left|x+5\right|=0\left(1\right)\)
TH1: x>=-5
(1) trở thành: 3x+2+x+5=0
=>4x+7=0
=>\(x=-\dfrac{7}{4}\left(nhận\right)\)
TH2: x<-5
=>x+5<0
=>|x+5|=-x-5
Phương trình (1) sẽ trở thành:
\(3x+2-x-5=0\)
=>2x-3=0
=>2x=3
=>\(x=\dfrac{3}{2}\)
a) \(\left(x-5\right)\left(2x-3^2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\2x=9\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{9}{2}\end{matrix}\right.\)
b) \(2\left(3x-15\right)\left(5-x\right)=0\)
\(\Rightarrow6\left(x-5\right)\left(5-x\right)=0\Rightarrow x=5\)
(x - 5)(2x - 32) = 0
=> \(\left[\begin{array}{} x - 5 = 0\\ 2x - 3^{2} = 0 \end{array} \right.\)=> \(\left[\begin{array}{} x = 0 - 5 = -5\\ 2x = 0 - 3^{2} = 0 - 9 = -9 => x = \dfrac{9}{2} \end{array} \right.\)
\(a,\left(8-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}8-x=0\\x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-5\end{matrix}\right.\\ b,2x\left(x+81\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=0\\x+81=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-81\end{matrix}\right.\)
a)\(\left(8-x\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}8-x=0\\x+5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-5\end{matrix}\right.\)
b)\(2x\left(x+81\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x+81=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-81\end{matrix}\right.\)
\(\Rightarrow18:x=10-1=9\Rightarrow x=2\)
\(\Leftrightarrow10-18:x=1\)
\(\Leftrightarrow18:x=9\)
hay x=2