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21 tháng 8 2020

Các bạn giúp mình với ạ! TT

31 tháng 12 2016

\(x=\frac{4}{5};x=-\frac{5}{6};x=\frac{7}{15}\)

\(M+N=9\left(xz\right)^2+14\left(yz\right)^2=z^2\left(9x^2+14y^2\right)\)

31 tháng 12 2016

Câu 1 : \(x-\frac{1}{5}=\frac{3}{5}\)                                     

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

                      \(x=\frac{4}{5}\)

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

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

           \(x=-\frac{5}{6}\)

\(-x+\frac{1}{3}=-\frac{2}{15}\)

               \(x=\frac{1}{3}+\frac{2}{15}\)

               \(x=\frac{7}{15}\)

Câu 2 : Ta có : \(M+N=5x^2z^2+5y^2z+8x+4x^2z^2+9y^2z-8x\)

                                    \(=9x^2x^2+14y^2z\)

21 tháng 7 2019

#)Giải :

a) \(\left(5x+1\right)^2=\frac{36}{49}\Leftrightarrow\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\Leftrightarrow5x+1=\frac{6}{7}\Leftrightarrow5x=-\frac{1}{7}\Leftrightarrow x=-\frac{1}{35}\)

b) \(\left(x-\frac{2}{9}\right)^3=\left(\frac{2}{3}\right)^6\Leftrightarrow\left(x-\frac{2}{9}\right)^3=\left[\left(\frac{2}{3}\right)^2\right]^3\Leftrightarrow x-\frac{2}{9}=\left(\frac{2}{3}\right)^2=\frac{4}{9}\Leftrightarrow x=\frac{2}{3}\)

c) \(\left(8x-1\right)^{2x+1}=5^{2x+1}\Leftrightarrow8x-1=5\Leftrightarrow8x=6\Leftrightarrow x=\frac{6}{8}\)

21 tháng 7 2019

a) \(\left(5x+1\right)^2=\frac{36}{49}\)

 \(\left(5x+1\right)^2=\frac{6^2}{7^2}\)

\(\left(5x+1\right)^2=\left(\frac{6}{7}\right)^2\)

\(\Leftrightarrow5x+1=\frac{6}{7}\)

\(5x=\frac{6}{7}-1\)

\(5x=\frac{6}{7}-\frac{7}{7}\)

\(5x=-\frac{1}{7}\)

\(x=-\frac{1}{7}\div5\)

\(x=-\frac{1}{7}\times\frac{1}{5}\)

\(x=-\frac{1}{35}\)

Vậy \(x=-\frac{1}{35}\)

6 tháng 9 2020

Dài đấy :))

a) \(\left|x-1\right|-\left(-2\right)^3=9\cdot\left(-1\right)^{100}\)

\(\Leftrightarrow\left|x-1\right|-\left(-8\right)=9\cdot1\)

\(\Leftrightarrow\left|x-1\right|+8=9\)

\(\Leftrightarrow\left|x-1\right|=1\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=1\\x-1=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}\)

b) \(\frac{x-2}{-4}=\frac{-9}{x-2}\)( ĐKXĐ : \(x\ne2\))

\(\Leftrightarrow\left(x-2\right)\left(x-2\right)=-4\cdot\left(-9\right)\)

\(\Leftrightarrow\left(x-2\right)^2=36\)

\(\Leftrightarrow\left(x-2\right)^2=\left(\pm6\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=6\\x-2=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=8\\x=-4\end{cases}}\left(tmđk\right)\)

c) \(\frac{x-5}{3}=\frac{-12}{5-x}\)( ĐKXĐ : \(x\ne5\))

\(\Leftrightarrow\frac{x-5}{3}=\frac{-12}{-\left(x-5\right)}\)

\(\Leftrightarrow\frac{x-5}{3}=\frac{12}{x-5}\)

\(\Leftrightarrow\left(x-5\right)\left(x-5\right)=3\cdot12\)

\(\Leftrightarrow\left(x-5\right)^2=36\)

\(\Leftrightarrow\left(x-5\right)^2=\left(\pm6\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}x-5=6\\x-5=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=11\\x=-1\end{cases}}\left(tmđk\right)\)

d) \(8x-\left|4x+\frac{3}{4}\right|=x+2\)

\(\Leftrightarrow8x-x-2=\left|4x+\frac{3}{4}\right|\)

\(\Leftrightarrow7x-2=\left|4x+\frac{3}{4}\right|\)(*)

\(\left|4x+\frac{3}{4}\right|\ge0\Leftrightarrow4x+\frac{3}{4}\ge0\Leftrightarrow x\ge-\frac{3}{16}\)

Vậy ta xét hai trường hợp sau :

1. \(x\ge-\frac{3}{16}\)

(*) <=>\(7x-2=4x+\frac{3}{4}\)

\(\Leftrightarrow7x-4x=\frac{3}{4}+2\)

\(\Leftrightarrow3x=\frac{11}{4}\)

\(\Leftrightarrow x=\frac{11}{12}\)(tmđk)

2. \(x< -\frac{3}{16}\)

(*) <=> \(7x-2=-\left(4x+\frac{3}{4}\right)\)

\(\Leftrightarrow7x-2=-4x-\frac{3}{4}\)

\(\Leftrightarrow7x+4x=-\frac{3}{4}+2\)

\(\Leftrightarrow11x=\frac{5}{4}\)

\(\Leftrightarrow x=\frac{5}{44}\left(ktmđk\right)\)

Vậy x = 11/12

e) \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)

\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{x\left(x+1\right)}=\frac{2019}{2020}\)

\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2019}{2020}\)

\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4040}\)

\(\Leftrightarrow\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{x\left(x+1\right)}=\frac{2019}{4040}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2019}{4040}\)

\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2019}{4040}\)

\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{4040}\)

\(\Leftrightarrow x+1=4040\)

\(\Leftrightarrow x=4039\)

8 tháng 9 2020

ĐKXD là gì vậy