\(\left(2x-3\right)\left(4x^2+6x+9\right)-4x\left(2x^2-1\right)\)

\(=8x^3-27-8x^3+4x\\ =8x^3-8x^3+4x-27\\ =4x-27\)

\(=2x^2\left(x-1\right)-4x\left(x-1\right)=\left(x-1\right)\left(2x^2-4x\right)=2x\left(x-2\right)\left(x-1\right)\)

\(x^2+4y^2-4x-4y+5=0\)

\(\Leftrightarrow x^2-4x+4+4y^2-4y+1=0\)

\(\Leftrightarrow\left(x^2-2\cdot x\cdot2+2^2\right)+\left[\left(2y\right)^2-2\cdot2y\cdot1+1^2\right]=0\)

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

\(\Leftrightarrow\hept{\begin{cases}x-2=0\\2y-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\y=\frac{1}{2}\end{cases}}}\)

Vậy....

nhom (x2-4x+5)+(4y2-4y)

\(x^2+4y^2-4x-4y+5=0\)

\(\Leftrightarrow\)\(\left(x^2-4x+4\right)+\left(4y^2-4y+1\right)=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}\left(x-2\right)^2=0\\\left(2y-1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-2=0\\2y-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\y=\frac{1}{2}\end{cases}}\)

Vậy x = 2 ; y = \(\frac{1}{2}\)

giải hpt: \(\hept{\begin{cases}x^2+2y-4x=0\\4x^2-4xy^2+y^4-2y+4=0\end{cases}}\)

Cộng hai vế lại với nhau ta có: 

\(4x^2-4xy^2+y^4+x^2-4x+4=0\)

\(\Leftrightarrow\left(2x-y\right)^2+\left(x-2\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}2x-y^2=0\\x-2=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y^2=4\end{cases}}}\)

\(\Rightarrow\hept{\begin{cases}x=2;y=2\left(tm\right)\\x=2;y=-2\end{cases}}\)

Thay x,y vào pt và tính

=> x=2 và y=2 thỏa mãn 

=>(x;y)=(2;2) (t/m)

@Linh: Làm nhầm rồi 

HPT\(\hept{\begin{cases}x^2+2y-4x=0\\4x^2-4xy^2+y^4-2y+4=0\end{cases}}\)

Cộng vế với vế của hai phương trình, ta được:

\(HPT\Leftrightarrow5x^2-4xy^2+y^2-4x+4=0\)

\(\Leftrightarrow\left(4x^2-4xy^2+y^2\right)+\left(x^2-4x+4\right)=0\)

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

\(\Leftrightarrow\hept{\begin{cases}2x-y=0\\x-2=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=4\end{cases}}\)

Vậy \(\left(x;y\right)=\left(2;4\right)\)
 

2x+4x+6x+...+2014x=2015*2016

2x+4x+6x+...+2014x=4062240

x.(2+4+6+...+2014)=4062240

x.1015056=4062240

x=4062240:1015056

x=4,001986097

Theo đề, ta có:  \(\dfrac{1+2x}{18}=\dfrac{1+4x}{34}\)

\(\Leftrightarrow34\left(1+2x\right)=18\left(1+4x\right)\)

\(\Leftrightarrow34+68x=18+72x\)

\(\Leftrightarrow34-18=72x-68x\)

\(\Leftrightarrow16=4x\)

\(\Leftrightarrow x=4\)

Khi \(x=4\) vào ta có:   \(\dfrac{1+4.4}{34}=\dfrac{1+6.4}{2y^2}\Leftrightarrow\dfrac{1}{2}=\dfrac{25}{2y^2}\) 

\(\Leftrightarrow2y^2=50\)

\(\Leftrightarrow y^2=50\)

\(\Leftrightarrow y=\pm5\)

ĐK:\(x\ge2\)

\(\sqrt{x-2}\times\left(x^2-4x+3\right)=0\)

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

\(\Leftrightarrow\hept{\begin{cases}x=1\left(loai\right)\\x=2\left(tm\right)\\x=3\left(tm\right)\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}\left(tm\right)}\)