Bài này khó dữ chị ơi! Em chỉ mới học lớp 4! Sorry chị nha!

em bó tay.com. vn

em mới lớp 5 thui chị ơi

a: ĐKXĐ: \(x\notin\left\{-1;3\right\}\)

b: \(A=\dfrac{3x\left(x+1\right)}{\left(x+1\right)\left(2x-6\right)}=\dfrac{3x}{2x-6}\)

Để A=0 thì 3x=0

hay x=0

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\(a,\)\(\left(4x-2\right)\left(10x+4\right)\left(5x+7\right)\left(2x+1\right)+17\)

\(=\left(4x-2\right)\left(5x+7\right)\left(10x+4\right)\left(2x+1\right)+17\)

\(=\left(20x^2+18x-5\right)\left(20x^2+18x+4\right)+17\)

Đặt ....

\(5x^4y+10x^3y+10x^2y^3+5xy^4\)

\(=5xy.x^3+5xy.2x^3+5xy.2xy^3+5xy.y^3\)

\(=5xy\left(x^3+2x^3+2xy^3+y^3\right)\)

Ht pt

m-2m-n+1 = (m-2m+1)-n = (m-1)² -n

\(15x^3y^2+10x^2y^2-2x^2y^3\)

\(=x^2y^2\left(15x+10-2y\right)\)

`15x^3 y^2 + 10x^2 y^2 - 2x^2 y^3`

`=x^2y^2(15x+10-2y)`

1) đặt 2x+1 = a => \(a^4-3a^2+2=\left(a^2-1\right)\left(a^2-2\right)=\)\(\left(a-1\right)\left(a+1\right)\left(a-\sqrt{2}\right)\left(a+\sqrt{2}\right)\)

=(2x+1-1)(2x+1+1)(2x+1-\(\sqrt{2}\))(2x+1+\(\sqrt{2}\)) = 4x(x+1)(2x+1-\(\sqrt{2}\))(2x+1+\(\sqrt{2}\))

2) =(x²-x)(x²-x-2)-3

đặt x²-x = b => b(b-2)-3 = b²-2b-3 = (b+1)(b-3) = (x²-x+1)(x²-x-3)

3) đặt x²+2x-1 = c => c²-3xc+2x² = (c-x)(c-2x) = (x²+2x-1-x)(x²+2x-1-2x) = (x²+x-1)(x²-1) = (x²+x-1)(x-1)(x+1)

tìm x

x³-8 +(x-2)(x+1)=0 <=> (x-2)(x²+2x+4)+(x-2)(x+1)=0 <=>(x-2)(x²+2x+4+x+1)=0 <=> x=2 (vì x²+3x+5= (x+\(\frac{3}{2}\))² +\(\frac{11}{4}\)>0)

vậy x=2 

2) \(x\left(x-1\right)\left(x+1\right)\left(x-2\right)-3\)

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

Đặt \(x^2-x=t\)

\(\Rightarrow\left(1\right)=t\left(t-2\right)-3=t^2-2t+1-4\)

\(=\left(t-1\right)^2-4\)

\(=\left(t+3\right)\left(t-5\right)\)

Thay \(x^2-x=t\), ta được:

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

\(x^3-8+\left(x-2\right)\left(x+1\right)=0\)

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

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

\(TH1:x-2=0\Leftrightarrow x=2\)

\(TH2:x^2-x+5=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2+\frac{19}{4}=0\)

Mà \(\left(x-\frac{1}{2}\right)^2+\frac{19}{4}>0\)nên loại TH2

Vậy x = 2

bài này 1h rùi,chắc chờ tui ngủ dậy làm;

= (x+y)³ - (x+y) + xy(x+y) =

= (x+y)((x+y)² -1 +xy)) = (x+y)(x² +3xy +y² -1)

\(1,3x^2-6xy+3y^2-12z^2\)

\(=3\left(x^2-2xy+y^2-4z^2\right)\)

\(=3[\left(x^2-2xy+y^2\right)-4z^2]\)

\(=3\left[\left(x-y\right)^2-4z^2\right]\)

\(=3\left(x-y-2x\right)\left(x-y+2z\right)\)

\(2,x^2-2x-15\)

\(=x^2-5x+3x-15\)

\(=\left(x^2-5x\right)+\left(3x-15\right)\)

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

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