\(\Leftrightarrow x^4\left(x-1\right)-4x^3\left(x-1\right)+4x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(x^4-4x^3+4x-1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[x^3\left(x-1\right)-3x^2\left(x-1\right)-3x\left(x-1\right)+\left(x-1\right)\right]\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x^3-3x^2-3x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2\left[\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\right]=0\)

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

- Khi x - 1 = 0 thì x = 1

- Khi x + 1 = 0 thì x = -1

- Khi \(x^2-4x+1=0\Leftrightarrow\left(x-2\right)^2=3\Leftrightarrow\orbr{\begin{cases}x=\sqrt{3}+2\\x=-\sqrt{3}+2\end{cases}}\)

Pt có tậo nghiệm là: \(S=\left\{1;-1;\sqrt{3}+2;-\sqrt{3}+2\right\}\)

\(5^6-25^3=\left(5^2\right)^3-25^3=25^3-25^3=0\)

\(\Rightarrow\frac{\left(1^6-29^3\right)\left(2^6-28^3\right)\left(3^6-27^3\right)\left(4^6-26^3\right)\left(5^6-25^3\right).....\left(10^6-20^3\right)}{\left(1^6+29^3\right)\left(2^6+28^3\right)\left(3^6+27^3\right)\left(4^6+26^3\right)\left(5^6+25^3\right).....\left(10^6+20^3\right)}=0\)

Nhớ vẽ hình nha.

a) \(0,25x^3+x^2+x=0\)

\(\Leftrightarrow x\left(0,25x^2+x+1\right)=0\)

\(\Leftrightarrow x\left[\left(\frac{1}{2}x\right)^2+2\cdot\frac{1}{2}x\cdot1+1^2\right]=0\)

\(\Leftrightarrow x\left(\frac{1}{2}x+1\right)^2=0\)

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

Vậy....

b) \(\frac{2-x}{2007}-1=\frac{1-x}{2008}-\frac{x}{2009}\)

\(\Leftrightarrow\frac{2-x}{2007}-1+2=\frac{1-x}{2008}+1+\frac{-x}{2009}+1\)

\(\Leftrightarrow\frac{2-x+2007}{2007}=\frac{1-x+2008}{2008}+\frac{-x+2009}{2009}\)

\(\Leftrightarrow\frac{2009-x}{2007}=\frac{2009-x}{2008}+\frac{2009-x}{2009}\)

\(\Leftrightarrow\frac{2009-x}{2007}-\frac{2009-x}{2008}-\frac{2009-x}{2009}=0\)

\(\Leftrightarrow\left(2009-x\right)\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)

Vì \(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\ne0\)

\(\Rightarrow2009-x=0\)

\(\Leftrightarrow x=2009\)

Vậy....

\(2\left(a^2+b^2\right)=5ab\)

\(\Leftrightarrow2a^2+2b^2-5ab=0\)

\(\Leftrightarrow\left(2a^2-4ab\right)+\left(2b^2-ab\right)=0\)

\(\Leftrightarrow2a\left(a-2b\right)-b\left(a-2b\right)=0\)

\(\Leftrightarrow\left(2a-b\right)\left(a-2b\right)=0\)

\(a>b\Rightarrow2a>b\Leftrightarrow2a-b>0\)

\(\Rightarrow a=2b\)

\(Q=\frac{3\left(2b-b\right)}{2\left(2b+b\right)}=\frac{3b}{6b}=\frac{1}{2}\)

thank bạn nha

4 ( x + y ) = xy + 11

\(\Leftrightarrow\)4x + 4y - xy = 11

\(\Leftrightarrow\)x ( 4 - y ) - 16 + 4y = -5

\(\Leftrightarrow\)x ( 4 - y ) - 4 ( 4 - y ) = -5

\(\Leftrightarrow\)( x - 4 ) ( 4 - y ) = -5

lập bảng giá trị, tìm được x,y