1, \(2x^3-50x=0\Leftrightarrow2x\left(x^2-25\right)=0\Leftrightarrow x=0;x=\pm5\)

2, \(5x^2-4\left(x^2-2x+1\right)-5=0\)

\(\Leftrightarrow5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left[5\left(x+1\right)-4\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+9\right)=0\Leftrightarrow x=-9;x=1\)

3, \(6x\left(x-2\right)=x-2\Leftrightarrow\left(6x-1\right)\left(x-2\right)=0\Leftrightarrow x=\frac{1}{6};x=2\)

4, \(7\left(x-2020\right)^2-x+2020=0\Leftrightarrow7\left(x-2020\right)^2-\left(x-2020\right)=0\)

\(\Leftrightarrow\left(x-2020\right)\left[7\left(x-2020\right)-1\right]=0\Leftrightarrow x=2020;x=\frac{14141}{7}\)

5, \(x^2-10x=-25\Leftrightarrow x^2-10x+25=0\Leftrightarrow\left(x-5\right)^2=0\Leftrightarrow x=5\)

6, \(x^2-2x-3=0\Leftrightarrow\left(x+1\right)\left(x-3\right)=0\Leftrightarrow x=-1;x=3\)

\(1,\)

\(2x^3-50x=0\)

\(\Leftrightarrow2x\left(x^2-25\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x^2-25=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)

\(2,\)

\(5x^2-4\left(x^2-2x+1\right)-5=0\)

\(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)

\(\Leftrightarrow x^2+8x-9=0\)

\(\Leftrightarrow x^2-x+9x-9=0\)

\(\Leftrightarrow x\left(x-1\right)+9\left(x-1\right)=0\)

\(\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+9=0\\x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-9\\x=1\end{cases}}\)

\(3,\)

\(6x\left(x-2\right)=x-2\)

\(\Leftrightarrow6x\left(x-2\right)-\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(6x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{6}\end{cases}}\)

\(4,\)

\(7\left(x-2020\right)^2-x+2020=0\)

\(\Leftrightarrow7\left(x-2020\right)^2-\left(x-2020\right)=0\)

\(\Leftrightarrow\left(x-2020\right)[7\left(x-2020\right)-1]=0\)

\(\Leftrightarrow\left(x-2020\right)[7x-14141]=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2020\\7x=14141\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=\frac{14141}{7}\end{cases}}\)

\(5,\)

\(x^2-10x=-25\)

\(\Leftrightarrow x^2-10x+25=0\)

\(\Leftrightarrow\left(x-5\right)^2=0\)

\(\Leftrightarrow x-5=0\)

\(\Leftrightarrow x=5\)

\(6,\)

\(x^2-2x-3=0\)

\(\Leftrightarrow x^2-3x+x-3=0\)

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

\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

Hello bạn, mk cx tên Mai nek.

\(\frac{2}{5}.\left(x-1\right)+1=\frac{3}{5}\)

\(\Rightarrow\frac{2}{5}\left(x+1\right)=\frac{3}{5}-1\)

\(\Rightarrow\frac{2}{5}\left(x+1\right)=-\frac{2}{5}\)

\(\Rightarrow x+1=-\frac{2}{5}:\frac{2}{5}\)

\(\Rightarrow x+1=-1\)

\(\Rightarrow x=-1-1\)

\(\Rightarrow x=-2\)

\(\left(\frac{2}{7}\times x+1\right)\times\left(3-\frac{1}{2}\times x\right)=0\)

\(TH1:\frac{2}{7}\times x+1=0\)

\(\frac{2}{7}\times x=-1\)

\(x=-\frac{2}{7}\)

\(TH2:3-\frac{1}{2}\times x=0\)

\(\frac{1}{2}\times x=3\)

\(x=\frac{3}{2}\)

Vậy \(x\in\left\{\frac{3}{2};-\frac{2}{7}\right\}\)

\(\frac{5}{4}\times x+1=\frac{1}{2}x+\frac{3}{4}\)

\(\frac{5}{4}x-\frac{1}{2}x=\frac{3}{4}-1\)

\(\left(\frac{5}{4}-\frac{1}{2}\right)x=-\frac{1}{4}\)

\(\frac{3}{4}x=-\frac{1}{4}\)

\(x=-\frac{1}{4}\times\frac{4}{3}\)

\(x=-\frac{1}{3}\)

Vậy \(x\in\left\{-\frac{1}{3}\right\}\)

1. \(\dfrac{2019}{2020}-\left(\dfrac{2019}{2020}-\dfrac{2020}{2021}\right)\)

\(=\dfrac{2019}{2020}-\dfrac{2019}{2020}+\dfrac{2020}{2021}\)

\(=0+\dfrac{2020}{2021}=\dfrac{2020}{2021}\)

Giải:

1) \(\dfrac{2019}{2020}-\left(\dfrac{2019}{2020}-\dfrac{2020}{2021}\right)\)  

\(=\dfrac{2019}{2020}-\dfrac{2019}{2020}+\dfrac{2020}{2021}\) 

\(=\left(\dfrac{2019}{2020}-\dfrac{2019}{2020}\right)+\dfrac{2020}{2021}\) 

\(=0+\dfrac{2020}{2021}\) 

\(=\dfrac{2020}{2021}\) 

2) \(\dfrac{2}{9}+\dfrac{7}{9}:\left(\dfrac{42}{5}-\dfrac{7}{5}\right)\) 

\(=\dfrac{2}{9}+\dfrac{7}{9}:7\) 

\(=\dfrac{2}{9}+\dfrac{1}{9}\) 

\(=\dfrac{1}{3}\) 

3) \(\dfrac{3}{4}+\dfrac{x}{4}=\dfrac{5}{8}\) 

            \(\dfrac{x}{4}=\dfrac{5}{8}-\dfrac{3}{4}\) 

            \(\dfrac{x}{4}=\dfrac{-1}{8}\)  

\(\Rightarrow x=\dfrac{4.-1}{8}=\dfrac{-1}{2}\) 

4) \(\left|3x+1\right|-\dfrac{1}{4}=\dfrac{-1}{4}\) 

            \(\left|3x-1\right|=\dfrac{-1}{4}+\dfrac{1}{4}\) 

            \(\left|3x-1\right|=0\) 

             \(3x-1=0\) 

                    \(3x=0+1\) 

                    \(3x=1\) 

                      \(x=1:3\) 

                      \(x=\dfrac{1}{3}\) 

Chúc bạn học tốt!

4) \(\left|3x+1\right|-\dfrac{1}{4}=\dfrac{-1}{4}\) 

            \(\left|3x+1\right|=\dfrac{-1}{4}+\dfrac{1}{4}\) 

            \(\left|3x+1\right|=0\) 

              \(3x+1=0\) 

                    \(3x=0-1\) 

                    \(3x=-1\) 

                      \(x=-1:3\) 

                      \(x=\dfrac{-1}{3}\)

CÂU NÀO CÁC BẠN LÀM ĐC THÌ GIÚP MK NHA !!!! 

a, x=-505

b, x=35/8 hoac -37/8

nhung cau con lai thi tong tu

a. \(\left|4x+2020\right|=0\)

\(\Rightarrow4x+2020=0\)

\(\Rightarrow4x=-2020\)

\(\Rightarrow x=-505\)

b. \(\left|2x+\frac{1}{4}\right|+\left|-5\right|=\left|-14\right|\)

\(\Rightarrow\left|2x+\frac{1}{4}\right|+5=14\)

\(\Rightarrow\left|2x+\frac{1}{4}\right|=9\)

\(\Rightarrow\orbr{\begin{cases}2x+\frac{1}{4}=9\\2x+\frac{1}{4}=-9\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=\frac{35}{4}\\2x=-\frac{37}{4}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{35}{8}\\x=-\frac{37}{8}\end{cases}}\)

c. \(\left|2020-5x\right|-\left|3\right|=-\left|-8\right|\)

\(\Rightarrow\left|2020-5x\right|-3=-8\)

\(\Rightarrow\left|2020-5x\right|=-5\left(vl\right)\)

=> x vô nghiệm

d. \(\left|x^2+4x\right|=0\)

\(\Rightarrow x^2+4x=0\)

\(\Rightarrow x\left(x+4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x+4=0\end{cases}\Rightarrow}\orbr{\begin{cases}x=0\\x=-4\end{cases}}\)

chẳng có đề bài biết làm ntn

\(\dfrac{2}{5}\times\dfrac{1}{7}+\dfrac{2}{7}\times\dfrac{2}{5}\)

\(=\dfrac{2}{5}\times\left(\dfrac{1}{7}+\dfrac{2}{7}\right)\)

\(=\dfrac{2}{5}\times\dfrac{3}{7}\)

\(=\dfrac{6}{35}\)

\(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)

\(x+\dfrac{1}{6}=\dfrac{3}{4}\)

\(x=\dfrac{9}{12}-\dfrac{2}{12}\)

\(x=\dfrac{7}{12}\)

\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2020}\right)+x=\dfrac{1}{2}\)

\(\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2019}{2020}+x=\dfrac{1}{2}\)

\(\dfrac{1}{2020}+x=\dfrac{1}{2}\)

\(x=\dfrac{1}{2}-\dfrac{1}{2020}\)

\(x=\dfrac{1010}{2020}-\dfrac{1}{2020}\)

\(x=\dfrac{1009}{2020}\)

\(\dfrac{2}{5}\times\dfrac{1}{7}+\dfrac{2}{7}\times\dfrac{2}{5}\)

\(=\dfrac{2}{5}\times\left(\dfrac{1}{7}+\dfrac{2}{7}\right)\)

\(=\dfrac{2}{5}\times\dfrac{3}{7}\)

\(=\dfrac{6}{35}\)

\(x+\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}\)

\(\Rightarrow\dfrac{1}{2}\times\dfrac{1}{3}=\dfrac{3}{4}-x\)

\(\Rightarrow\dfrac{3}{4}-x=\dfrac{1}{6}\)

\(\Rightarrow x=\dfrac{3}{4}-\dfrac{1}{6}=\dfrac{7}{12}\)

\(\left(1-\dfrac{1}{2}\right)\times\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{1}{4}\right)\times...\times\left(1-\dfrac{1}{2020}\right)+x=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{2}\times\dfrac{2}{3}\times\dfrac{3}{4}\times...\times\dfrac{2019}{2020}+x=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1\times2\times3\times4\times...\times2019}{2\times3\times4\times5\times...\times2020}+x=\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{2020}+x=\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{1}{2}-\dfrac{1}{2020}=\dfrac{1009}{2020}\)

a: \(\dfrac{7}{4}+\dfrac{-3}{5}=\dfrac{35-12}{20}=\dfrac{23}{20}\)

d: \(\left(-\dfrac{1}{4}\right)^2\cdot\dfrac{4}{11}+\dfrac{7}{11}\cdot\left(-\dfrac{1}{4}\right)^2=\dfrac{1}{16}\)

\(\dfrac{7}{4}+\dfrac{-3}{5}=\dfrac{35}{20}+\dfrac{-12}{20}=\dfrac{23}{20}\)

a: \(A=1-\dfrac{2\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}{4\left(25-\dfrac{2}{2018}+\dfrac{1}{2019}-\dfrac{1}{2020}\right)}\)

=1-2/4=1/2

b: \(B=\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=5\cdot\dfrac{-6}{9}=-\dfrac{10}{3}\)

c: x-y=0 nên x=y

\(C=x^{2020}-x^{2020}+y\cdot y^{2019}-y^{2019}\cdot y+2019\)

=2019