x=1

y=2

Ta có: \(x^2-2xy+5y^2=y+1\)

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

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

Mà \(4y^2-4y-1=3y^2+\left(y^2-y\right)-1\)

\(=3y^2+y\left(y-1\right)-1\ge3\cdot1+0-1=2>0\)

\(\Rightarrow\left(x-y\right)^2+4y^2-y-1>0\)

=> pt vô nghiệm

x^4+4 
=x^4+4x^2-4x^2+4 
=x^4+4x^2+4-4x^2 
=(x^4+4x^2+4)-4x^2 
=(x^2+2)^2-(2x)^2 
=(x^2+2+2x)(x^2+2-2x)

tk cho m nha!

Rồi rồi :33

\(A=-x^2+6x+7\)

\(A=-\left(x^2-6x-7\right)\)

\(A=-\left(x^2-2\cdot x\cdot3+3^2-16\right)\)

\(A=-\left[\left(x-3\right)^2-16\right]\)

\(A=16-\left(x-3\right)^2\le16\forall x\)

Dấu "=" xảy ra \(\Leftrightarrow x-3=0\Leftrightarrow x=3\)

\(A=-x^2+6x+7\)

\(A=-\left(x^2-2.x.3+3^2-16\right)\)

\(A=-\left(x^2-2.x.3+3^2\right)+16\)

\(A=-\left(x-3\right)^2+16\)

Ta có :\(\left(x-3\right)^2\ge0\)

\(\Rightarrow-\left(x-3\right)^2\le0\)

\(\Rightarrow A\le16\)

\(\Rightarrow Max\)\(A=16\)

\(Khi:\left(x-3\right)=0\)

\(\Rightarrow x=3\)

a) \(4x-2=0\)

\(4x=2\)

\(x=\frac{1}{2}\)

b) \(x^2+5x-14=0\)

\(x^2+7x-2x-14=0\)

\(x\left(x+7\right)-2\left(x+7\right)=0\)

\(\left(x+7\right)\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+7=0\\x-2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-7\\x=2\end{cases}}\)

Vậy.........

a)4x-2=0                             

4x=0+2                                        

4x=2                                                              

x=2:4                                           

x=1/2

Bài 1 :

a) \(20xy-10x=10x\left(2y-1\right)\)

b) \(6x-6y+x^2-y^2\)

\(=6\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)

\(=\left(x-y\right)\left(x+y+6\right)\)

Bài 2 :

a) \(\frac{3x+5}{2x}-\frac{5}{2x}\)

\(=\frac{3x+5-5}{2x}\)

\(=\frac{3x}{2x}\)

\(=\frac{3}{2}\)

b) \(\frac{2x}{x+3}:\frac{x}{x^2-9}\)

\(=\frac{2x\left(x-3\right)\left(x+3\right)}{x\left(x+3\right)}\)

\(=2\left(x-3\right)\)