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)

\(65-4^{x-3}=2020^0\\ \Rightarrow65-4^{x-3}=1\\ \Rightarrow4^{x-3}=64\\ \Rightarrow4^{x-3}=4^3\\ \Rightarrow x-3=3\\ \Rightarrow x=6\)

x=6 nha bạn

tick cho mk

\(\Leftrightarrow x-\left[3-x+3+x-2\right]=0\)

=>x=4

\(2x^2-x-8=0\\ \Leftrightarrow\left(2x^2-x\right)-8=0\\ \Leftrightarrow x\left(2x-1\right)-8=0\\ \Leftrightarrow\left(x-8\right)\left(2x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=\dfrac{1}{2}\end{matrix}\right.\)

2x^2-x-8=0

Δ=(-1)^2-4*2*(-8)

=1+8+8=65>0

Vì Δ>0 nên phương trình có 2 nghiệm phân biệt là;

\(\left\{{}\begin{matrix}x_1=\dfrac{1-\sqrt{65}}{4}\\x_2=\dfrac{1+\sqrt{65}}{4}\end{matrix}\right.\)

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

=> \(\left[\begin{array}{} x - 5 = 0\\ 2x - 3^{2} = 0 \end{array} \right.\)=> \(\left[\begin{array}{} x = 0 - 5 = -5\\ 2x = 0 - 3^{2} = 0 - 9 = -9 => x = \dfrac{9}{2} \end{array} \right.\)

b) 2(3x - 15)(5 - x) = 0

      (6x - 30)(5 - x) = 0

=> \(\begin{cases} 6x - 30 = 0\\ 5 - x =0 \end{cases} \) => \(\begin{cases} 6x = 0 + 30 = 30 => x = 5\\ x = 0 + 5 = 5 \end{cases} \)

25

\(a,\left(8-x\right)\left(x+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}8-x=0\\x+5=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-5\end{matrix}\right.\\ b,2x\left(x+81\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x=0\\x+81=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-81\end{matrix}\right.\)

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)\(2x\left(x+81\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x=0\\x+81=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-81\end{matrix}\right.\)

Viết dưới dạng latex để mik dễ hỗ trợ bn nhé

\(\sqrt{3x+15}=\sqrt{10} \)

\( \sqrt{4(1-x)^{2}}-6=0\)

a: ta có: \(\sqrt{3x+15}=\sqrt{10}\)

\(\Leftrightarrow3x+15=10\)

hay \(x=-\dfrac{5}{3}\)