1. \(x^2+2y^2+2xy-2y+1=0\)

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

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

Có: \(\left(x+y\right)^2\ge0;\left(y-1\right)^2\ge0\)

Mà theo bài ra: \(\left(x+y\right)^2+\left(y-1\right)^2=0\)

\(\Rightarrow\hept{\begin{cases}\left(x+y\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x+y=0\\y-1=0\end{cases}}\Rightarrow\hept{\begin{cases}x+y=0\\y=1\end{cases}}\Rightarrow\hept{\begin{cases}x=-1\\y=1\end{cases}}\)

a) (x-1)*(x+2)-(x-3)*(-x+4)=19

\(\Leftrightarrow x^2+2x-x-2-\left(-x^2+4x+3-12\right)=19\)

\(\Leftrightarrow x^2+2x-x-2+x^2-4x-3+12=19\)

\(\Leftrightarrow2x^2-3x+7-19=0\)

\(\Leftrightarrow2x^2-3x-12=0\)

Đề sai??

b) (2x -1)*(3x+5)-(6x-1)*(6x+1)=(-17)

\(\Leftrightarrow6x^2+10x-3x-5-\left(36x^2+6x-6x-1\right)=-17\)

\(\Leftrightarrow6x^2+10x-3x-5-36x^2-6x+6x+1=-17\)

\(\Leftrightarrow-30x^2+7x-4+17=0\)

\(\Leftrightarrow-30x^2+7x+13=0\)

???

c) (x+1)*(x+1)-(x-1)*(x-1)=9

\(\Leftrightarrow\left(x+1\right)^2-\left(x-1\right)^2=9\)

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

\(\Leftrightarrow2x.2=9\)

\(\Leftrightarrow x=\frac{9}{4}\)

a) \(x:\frac{1}{2}+\frac{3}{4}=1\frac{19}{20}\)

   \(x:\frac{1}{2}+\frac{3}{4}=\frac{39}{20}\)

\(x:\frac{1}{2}=\frac{39}{20}-\frac{3}{4}\)

\(x:\frac{1}{2}=\frac{39}{20}-\frac{15}{20}\)

\(x:\frac{1}{2}=\frac{24}{20}\)

\(x=\frac{24}{20}.\frac{1}{2}\)

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

\(x:\frac{1}{2}+\frac{3}{4}=1\frac{19}{20}\)

\(x:\frac{1}{2}=1\frac{19}{20}-\frac{3}{4}\)

\(x:\frac{1}{2}=\frac{39}{20}-\frac{3}{4}\)

\(x:\frac{1}{2}=\frac{39}{20}-\frac{15}{20}\)

\(x:\frac{1}{2}=\frac{24}{20}\)

\(x=\frac{24}{20}\times\frac{1}{2}\)

\(x=\frac{24}{40}\)

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

a: =2/5-3/5+3/7=3/7-1/5

=15/35-7/35

=8/35

b: =>5/7:x=4/3

=>x=5/7:4/3=5/7*3/4=15/28

c: =>x-1/3=15/8:4/5=15/8*5/4=75/32

=>x=75/32+1/3=257/96

d: =>2x+1/8=2/7

=>2x=9/56

=>x=9/112

e: =>2x=10/3-5/4-3/4=10/3-2=4/3

=>x=2/3

\(a,\dfrac{2}{5}+\dfrac{3}{7}+\left(-\dfrac{3}{5}\right)\\ =\dfrac{2}{5}+\dfrac{3}{7}-\dfrac{3}{5}\\=\left(\dfrac{2}{5}-\dfrac{3}{5}\right)+\dfrac{3}{7}\\ =-\dfrac{1}{5}+\dfrac{3}{7}\\ =-\dfrac{7}{35}+\dfrac{15}{35}\\ =\dfrac{8}{35}\\ b,1-\dfrac{5}{7}:x=-\dfrac{1}{3}\\ =>\dfrac{5}{7}:x=1-\left(-\dfrac{1}{3}\right)\\ =>\dfrac{5}{7}:x=1+\dfrac{1}{3}\\ =>\dfrac{5}{7}:x=\dfrac{3}{3}+\dfrac{1}{3}\\ =>\dfrac{5}{7}:x=\dfrac{4}{3}\\ =>x=\dfrac{5}{7}:\dfrac{4}{3}\\ =>x=\dfrac{5}{7}.\dfrac{3}{4}\\ =>x=\dfrac{15}{28}\\ c,\dfrac{4}{5}\left(x-\dfrac{1}{3}\right)=\dfrac{15}{8}\\ =>x-\dfrac{1}{3}=\dfrac{15}{8}:\dfrac{4}{5}\\ =>x-\dfrac{1}{3}=\dfrac{15}{8}.\dfrac{5}{4}\\ =>x-\dfrac{1}{3}=\dfrac{75}{32}\\ =>x=\dfrac{75}{32}+\dfrac{1}{3}\\ =>x=\dfrac{257}{96}\)

\(d,\dfrac{2}{3}:\left(2x+\dfrac{1}{8}\right)=\dfrac{7}{3}\\ =>2x+\dfrac{1}{8}=\dfrac{2}{3}:\dfrac{7}{3}\\ =>2x+\dfrac{1}{8}=\dfrac{2}{3}.\dfrac{3}{7}\\ =>2x+\dfrac{1}{8}=\dfrac{2}{7}\\ =>2x=\dfrac{2}{7}-\dfrac{1}{8}\\ =>2x=\dfrac{16}{56}-\dfrac{7}{56}\\ =>2x=\dfrac{9}{56}\\ =>x=\dfrac{9}{56}:2\\ =>x=\dfrac{9}{112}\\ e,2x+\dfrac{3}{4}=\dfrac{10}{3}-\dfrac{5}{4}\\ =>e,2x+\dfrac{3}{4}=\dfrac{40}{12}-\dfrac{15}{12}\\ =>2x+\dfrac{3}{4}=\dfrac{25}{12}\\ =>2x=\dfrac{25}{12}-\dfrac{3}{4}\\ =>2x=\dfrac{25}{12}-\dfrac{9}{12}\\ =>2x=\dfrac{16}{12}\\ =>2x=\dfrac{4}{3}\\ =>x=\dfrac{4}{3}:2\\ =>x=\dfrac{4}{6}\\ =>x=\dfrac{2}{3}\)

\(x\left(x+1\right)^4+x\left(x+1\right)^3+x\left(x+1\right)^2+\left(x+1\right)^2\)

\(=\left(x+1\right)^2\left[x\left(x+1\right)^2+x\left(x+1\right)+x+1\right]\)

\(=\left(x+1\right)^2\left[x\left(x+1\right)\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\right]\)

\(=\left(x+1\right)^2\left\{\left(x+1\right)\left[x\left(x+1\right)+x+1\right]\right\}\)

\(=\left(x+1\right)^2\left\{\left(x+1\right)\left[x^2+x+x+1\right]\right\}\)

\(=\left(x+1\right)^2\left[\left(x+1\right)\left(x^2+2x+1\right)\right]\)

\(=\left(x+1\right)^2\cdot\left(x+1\right)^3\)

\(=\left(x+1\right)^5\left(đpcm\right)\)

thanks bonking