Question

[Bích Nguyễn Ngọc]

b) \(\left|x-\frac{1}{3}\right|\)+\(\frac{4}{5}\)= \(\left|\left(-3,2\right)+\frac{2}{5}\right|\)

c) \(\frac{x-1}{2013}+\frac{x-2}{2012}\)\(-\frac{x-3}{2011}\)\(=\frac{x-4}{2010}\)

d) \(\left(x-7\right)^{x+1}-\left(x-7\right)^{x+11}=0\)

k) \(3x+x^2=0\)

m) \(x^2-2x-3\left(x-2\right)=0\)

Answer 1

[nguyễn minh phương]

co ghi dau ma biet

Answer 2

[Nguyễn Tiến Đạt]

mk ko chép lại đề nhé bn

b, 

=>\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\left|-\frac{14}{5}\right|\)

=>\(\left|x-\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\) \(\Rightarrow\left|x-\frac{1}{3}\right|=2\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{1}{3}=-2\\x-\frac{1}{3}=2\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{5}{3}\\x=\frac{7}{3}\end{cases}}}\)

c,\(\Rightarrow\frac{x-1}{2013}+\frac{x-2}{2012}-\frac{x-3}{2011}-\frac{x-4}{2010}=0\)

=> \(\frac{x-1}{2013}-1+\frac{x-2}{2012}-1-\left(\frac{x-3}{2011}-1+\frac{x-4}{2010}-1\right)=0\)

=>\(\frac{x-2014}{2013}+\frac{x-2014}{2012}-\frac{x-2014}{2011}-\frac{x-2014}{2010}=0\)

=.\(\left(x-2014\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\right)=0\)

Do \(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\ne0\)=> x-2014=0

=> x=2014

d,\(\left(x-7\right)^{x-1}-\left(x-7\right)^{x+11}=0\)

=>\(\left(x-7\right)^{x-1}.\left[1-\left(x-7\right)^{x+12}\right]=0\)

=> \(\orbr{\begin{cases}\left(x-7\right)^{x-1}=0\\1-\left(x-7\right)^{x+12}=0\end{cases}}\)

=> \(\orbr{\begin{cases}x-7=0\\\left(x-7\right)^{x+12}=0\end{cases}}\)

=>x=7 hoặc x-7=1 hoặc x+12=0

=> x=7 hoặc x=8 hoặc x=-12

Vậy x=7, x=8, x=-12

k,3x+x²=0

=> x(3+x)=0

=>\(\orbr{\begin{cases}x=0\\3+x=0\end{cases}}\)

=>\(\orbr{\begin{cases}x=0\\x=-3\end{cases}}\)

m, x²-2x-3(x-2)=0

=> x(x-2)-3(x-2)=0

=> (x-3)(x-2)=0

=>\(\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=3\\x=2\end{cases}}\)

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