\(25-a^2+2ab-b^2\)

\(=25-\left(a^2-2ab+b^2\right)\)

\(=5^2-\left(a-b\right)^2\)

\(=\left(5-a+b\right)\left(5+a-b\right)\)

\(=25-\left(a-b\right)^2\)

\(=\left(25-a+b\right)\left(25+a-b\right)\)

\(M=3x^2+6x+9\)

\(M=3\left(x^2+2x+3\right)\)

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

\(M=3\left[\left(x+1\right)^2+2\right]\)

\(M=3\left(x+1\right)^2+6\)

\(\left(x+1\right)^2\ge0\)

\(\Rightarrow3\left(x+1\right)^2\ge0\)

\(\Rightarrow3\left(x+1\right)^2+6\ge6\)

Vậy biểu thức M luôn luôn dương \(\forall x\)

\(M=3x^2+6x+9=3x^2+6x+3+6\)

    \(=3\left(x^2+2x+1\right)+6\)\(=3\left(x+1\right)^2+6\)

Vì \(\left(x+1\right)^2\ge0\forall x\)\(\Rightarrow3\left(x+1\right)^2\ge0\forall x\)

\(\Rightarrow M\ge6\forall x\)\(\Rightarrow\)M luôn dương ( đpcm )

\(\frac{1}{k\left(k+1\right)\left(k+2\right)}=\frac{1}{2}.\frac{k+2-k}{k\left(k+1\right)\left(k+2\right)}=\frac{1}{2}.\left(\frac{1}{k\left(k+1\right)}-\frac{1}{\left(k+1\right)\left(k+2\right)}\right)\)

\(=\frac{1}{2}\left[\frac{k+1-k}{k\left(k+1\right)}-\frac{\left(k+2\right)-\left(k+1\right)}{\left(k+1\right)\left(k+2\right)}\right]\)

\(=\frac{1}{2}\left(\frac{1}{k}-\frac{1}{k+1}-\frac{1}{k+1}+\frac{1}{k+2}\right)\)

\(=\frac{1}{2}\left(\frac{1}{k}+\frac{1}{k+2}\right)-\frac{1}{k+1}\)

Ko viết lại đề

Câu 1: chia ra làm 3 trường hợp

Câu 2: 

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

\(4\left(x+2\right)=0\)

\(\Rightarrow x+2=0\)

\(x=-2\)

câu 1:suy ra 

5x=0vậy x=0

x-3=0vậy x=3

-2x+6=0vậy x=3

\(5x^2+5y^2+8xy+2x-2y+2=0\)

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

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

\(\Rightarrow x=-1;y=1\)

Khi đó:

\(M=\left(1-1\right)^{2010}+\left(2-1\right)^{2011}+\left(1-1\right)^{2012}\)

\(=1\)

Phân tích này ra nhìn cho dễ bạn nhé, ko thì nhẩm nhanh ra :>

(16-4x)(x+3)-x(1-4x)=39

<=>16(x+3)-4x(x+3)-x+4x²=39

<=>16x+48-4x²-12x-x+4x²=39

<=>(16x-12x-x)-(4x²-4x²)+48=39

<=>3x+48=39

<=>3x=-9

<=>x=-3

\(2x\left(x^2-25\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x=0\\x^2-25=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)

\(2x\left(3x-5\right)+\left(3x-5\right)=0\)

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

\(\Rightarrow\orbr{\begin{cases}2x+1=0\\3x-5=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{5}{3}\end{cases}}\)

\(9\left(3x-2\right)-x\left(2-3x\right)=0\)

\(9\left(3x-2\right)+x\left(3x-2\right)=0\)

\(\left(9+x\right)\left(3x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}9+x=0\\3x-2=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-9\\x=\frac{2}{3}\end{cases}}\)

\(\left(2x-1\right)^2=25\)

\(\Rightarrow\orbr{\begin{cases}2x-1=5\\2x-1=-5\end{cases}}\)

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

(16 - 4x)(x + 3) - x(1 - 4x) = 39

16x + 48 - 4x² - 12x - x + 4x² = 39

3x = 39 - 48

3x = -9

x = -3

Thank you Lộc nhìu nha!  ^ - ^