1) (x - 5)² - (x + 3)(x - 3) = 14

=> x² - 10x + 25 - x² + 9 = 14

=> -10x + 34 = 14

=> -10x = 14 - 34

=> -10x = -20

=> x = 2

2) (x + 7)² - 3x - 21 = 0

=> x² + 14x + 49 - 3x - 21 = 0

=> x² + 11x + 28 = 0

=> x² + 4x + 7x + 28 = 0

=> x(x + 4) + 7(x + 4) = 0

=> (x + 7)(x + 4) = 0

=> \(\orbr{\begin{cases}x+7=0\\x+4=0\end{cases}}\)

=> \(\orbr{\begin{cases}x=-7\\x=-4\end{cases}}\)

a

\(\left(x-5\right)^2-\left(x+3\right)\left(x-3\right)=14\)

\(\Leftrightarrow x^2-10x+25-x^2+9=14\)

\(\Leftrightarrow-10x=-20\)

\(\Leftrightarrow x=2\)

b

\(\left(x+7\right)^2-3x-21=0\)

\(\Leftrightarrow x^2+14x+49-3x-21=0\)

\(\Leftrightarrow x^2+11x+28=0\)

\(\Leftrightarrow\left(x^2+4x\right)+\left(7x+28\right)=0\)

\(\Leftrightarrow x\left(x+4\right)+7\left(x+4\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x+7\right)=0\)

\(\Leftrightarrow x=-4;x=-7\)

\(=4x^2+12x+9-4x^2-12x\)

\(=0+0+9\)

\(=9\)

\(=x^2-5x+6x-30\)

\(=x\left(x-5\right)+6\left(x-5\right)\)

\(=\left(x+6\right)\left(x-5\right)\)

\(x^2+x-30\)

\(=\left(x^2-5x\right)+\left(6x-30\right)\)

\(=x\left(x-5\right)+6\left(x-5\right)\)

\(=\left(x-5\right)\left(x+6\right)\)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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