\(A=4x^2+6x=2x\left(2x+3\right)\)

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

\(C=\left(9x^2-1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2=\left(3x-1\right)\left(3x+1-3x+1\right)=2\left(3x+1\right)\)

\(D=x^3-16x=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\)

\(E=4x^2-25y^2=\left(2x-5y\right)\left(2x+5y\right)\)

\(G=\left(2x+3\right)^2-\left(2x-3\right)^2=\left(2x+3-2x+3\right)\left(2x+3+3x-3\right)=6.4x=24x\)

\(A=2x\left(2x+3\right)\\ B=\left(2x+3\right)\left(2x+3-x\right)=\left(2x+3\right)\left(x+3\right)\\ C=\left(3x-1\right)\left(3x+1\right)-\left(3x-1\right)^2\\ =\left(3x-1\right)\left(3x+1-3x+1\right)\\ =2\left(3x-1\right)\\ D=x\left(x^2-16\right)=x\left(x-4\right)\left(x+4\right)\\ E=\left(2x-5y\right)\left(2x+5y\right)\\ G=\left(2x+3-2x+3\right)\left(2x+3+2x-3\right)\\ =24x\)

chào bê đê

Bài 1 : 

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

b, \(2a^3-54b^3=2\left(a^3-27b^3\right)=2\left(a-3b\right)\left(a^2+3ab+9b\right)\)

Bài 2 : tự kết luận nhé, ngại mà lười :( 

a, \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)

\(\Leftrightarrow\frac{4x-3}{5}-\frac{5x-4}{3}=\frac{6x-2}{7}+3\)

\(\Leftrightarrow\frac{12x-9-25x+20}{15}=\frac{6x-2+21}{7}\)

\(\Leftrightarrow\frac{-13x-29}{15}=\frac{6x+19}{7}\Rightarrow-91x-203=90x+285\)

\(\Leftrightarrow181x=-488\Leftrightarrow x=-\frac{488}{181}\)

b, \(\frac{x+2}{3}+\frac{3\left(2x-1\right)}{4}-\frac{5x-3}{6}=x+\frac{5}{12}\)

\(\Leftrightarrow\frac{4x+8+9\left(2x-1\right)}{12}-\frac{10x-6}{12}=\frac{12x+5}{12}\)

\(\Rightarrow4x+8+18x-9-10x+6=12x+5\)

\(\Leftrightarrow12x+5=12x+5\Leftrightarrow0x=0\)

Vậy phương trình có vô số nghiệm 

c, \(\left|2x-3\right|=4\)

Với \(x\ge\frac{3}{2}\)pt có dạng : \(2x-3=4\Leftrightarrow x=\frac{7}{2}\)

Với \(x< \frac{3}{2}\)pt có dạng : \(2x-3=-4\Leftrightarrow x=-\frac{1}{2}\)

d, \(\left|3x-1\right|-x=2\Leftrightarrow\left|3x-1\right|=x+2\)

Với \(x\ge\frac{1}{3}\)pt có dạng : \(3x-1=x+2\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)

Với \(x< \frac{1}{3}\)pt có dạng : \(3x-1=-x-2\Leftrightarrow4x=-1\Leftrightarrow x=-\frac{1}{4}\)

Câu 3 : 

a, \(A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)ĐK : \(x\ne\pm2\)

\(=\left(\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right):\left(\frac{x^2-4+10-x^2}{\left(x-2\right)\left(x+2\right)}\right)\)

\(=\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}.\frac{\left(x-2\right)\left(x+2\right)}{6}=\frac{-6}{6}=-1\)

b, :)) -1 > 0 ( vô lí ) Vậy ko có giá trị x để A > 0 

Bạn thử giải câu này xem

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

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

Đặt: \(x^2+2x=t\)

khi đó: \(\left(x^2+2x\right)\left(x^2+2x+2\right)+1=t\left(t+2\right)+1=\left(t+1\right)^2\)

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

b) Xét: \(\left(n+1\right)^2-n^2=\left(n+1+n\right)\left(n+1-n\right)=2n+1\)

Khi đó:

\(A=\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+\frac{7}{\left(3.4\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)

\(A=\frac{2^2-1^2}{1^2.2^2}+\frac{3^2-2^2}{2^2.3^2}+\frac{4^2-3^2}{3^2.4^2}+...+\frac{\left(n+1\right)^2-n^2}{n^2.\left(n+1\right)^2}\)

\(A=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{n^2}-\frac{1}{\left(n+1\right)^2}\)

\(A=1-\frac{1}{\left(n+1\right)^2}\)

Câu 1: =\(\frac{1}{4}x^2-\frac{1x}{2x}+\frac{1}{4x^2}\)

câu 2 dễ mà

a, \(x^4+2013x^2+2012x+2013\)

\(=x^4+2013x^2-x+2013x+2013\)

\(=\left(x^4-x\right)+\left(2013x^2+2013x+2013\right)\)

\(=x\left(x^3-1\right)+2013\left(x^2+x+1\right)\)

\(=x\left(x-1\right)\left(x^2+x+1\right)+2013\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left\{x\left(x-1\right)+2013\right\}\)

\(=\left(x^2+x+1\right)\left(x^2-x+2013\right)\)

\(1)\)

\(a,2x^3-6x^2=2x^2.\left(x-3\right)\)

\(b,6x-6y-x^2+xy=6.\left(x-y\right)-x.\left(x-y\right)=\left(x-y\right).\left(6-x\right)\)

\(2)\)

\(a,ĐKXĐ:x\ne0;x\ne1\)

\(b,B=\left(\frac{2}{x\left(1-x\right)}-\frac{1}{1-x}\right):\frac{2-x}{1-2x+x^2}\)

\(B=\left(\frac{2}{x\left(1-x\right)}-\frac{x}{x\left(1-x\right)}\right):\frac{2-x}{\left(1-x\right)^2}\)

\(B=\frac{2-x}{x\left(1-x\right)}.\frac{\left(1-x\right)^2}{2-x}\)

\(B=\frac{1-x}{x}\)

\(c,\)Thay x=-2014 vào B ta có :

\(B=\frac{-2014-1}{-2014}=\frac{2015}{2014}\)

Câu 1 ;

a) \(x^2-2x-15\)

= \(x^2-5x+3x-15\)

= \(x(x-5)+3(x-5)\)

= \((x+3).(x-5)\)

b) \(xy+\frac{1}{3}y-\frac{1}{4}x-\frac{1}{12}\)

= \((x+\frac{1}{3})y-\frac{1}{4}(x+\frac{1}{3})\)

= \((x-\frac{1}{4}).(x+\frac{1}{3})\)

Câu 2 : 

\(A=\left(x+1\right)\left(x^2-x+1\right)+x-\left(x-1\right)\left(x^2+x+1\right)+1994\)

=> \(A=x^3+1+x-x^3+1+1994\)

=> \(A=1+x+1+1994\)

=> \(A=x+1996=-1995+1996=1\)