-_- Lúc này , mỗi 1 tbáo là 1 hivọng

Vậy mà 2 thôg báo r vẫn chẳg thấy câu trl đw

Nản ~~

haizz! Cài AdBlok đi -_-

\(\left(2y-3\right)^2-\left(y-1\right)^2=0\)

\(\Leftrightarrow\left(2y-3-y+1\right)\cdot\left(2y-3+y-1\right)=0\)

\(\Leftrightarrow\left(y-2\right)\left(3y-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y-2=0\\3y-4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}y=2\\y=\dfrac{4}{3}\end{matrix}\right.\)

\(a,\left(x+1\right)^3+\left(2-x\right)\left(4+2x+x^2\right)+3x\left(x+2\right)=17\)\(\Leftrightarrow x^3+3x^2+3x+1+8-x^3+3x^2+6x-17=0\)\(\Leftrightarrow6x^2+9x-8=0\)

\(\Leftrightarrow x^2+\dfrac{3}{2}x-\dfrac{4}{3}=0\)

\(\Leftrightarrow\left(x^2+\dfrac{3}{2}x+\dfrac{9}{16}\right)-\dfrac{9}{16}-\dfrac{4}{3}=0\)

\(\Leftrightarrow\left(x+\dfrac{3}{4}\right)^2=\dfrac{91}{48}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{3}{4}=\sqrt{\dfrac{91}{48}}\\x+\dfrac{3}{4}=-\sqrt{\dfrac{91}{48}}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{91}{48}}-\dfrac{3}{4}\\x=-\sqrt{\dfrac{91}{48}}-\dfrac{3}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-9+\sqrt{273}}{12}\\x=-\dfrac{9+\sqrt{273}}{12}\end{matrix}\right.\)

b, \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2-2\right)=15\)

\(\Leftrightarrow x^3+8-x^3+2x-15=0\)

\(\Leftrightarrow2x=7\Rightarrow x=\dfrac{7}{2}\)