Question

[Cybrog Bình]

Mik khá là lú về các bài hằng đẳng thức đáng nhớ mong người giúp bài này cái.

Answer 1

[Yen Nhi]

a) \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)

\(\Rightarrow\left(4x^2-4x+1\right)+\left(x^2+6x+9\right)-5\left(x^2-49\right)=0\)

\(\Rightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)

\(\Rightarrow2x+255=0\)

\(\Rightarrow2x=-255\)

\(\Rightarrow x=\frac{-255}{2}\)

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

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

\(\Rightarrow x\left(3x+2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\3x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-2}{3}\end{cases}}\)

Answer 2

[Yen Nhi]

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

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

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

\(\Rightarrow\left(x+3\right)^2=9\)

\(\Rightarrow x^2+6x+9=9\)

\(\Rightarrow x^2+6x=0\)

\(\Rightarrow x\left(x+6\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=-6\end{cases}}\)

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

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

\(\Rightarrow\left(2x-2\right)^2=1\)

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