\(\frac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}=\frac{\left(3x^3-3x^2\right)-\left(4x^2-4x\right)+\left(x-1\right)}{\left(2x^3-2x^2\right)+\left(x^2-x\right)-\left(3x-3\right)}=\frac{\left(x-1\right).\left(3x^2-4x+1\right)}{\left(x-1\right).\left(2x^2+x-3\right)}\\ \)

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

\(\frac{3x^3-7x^2+5x-1}{2x^3-x^2-4x+3}\)

\(=\frac{3x^3-3x^2-4x^2+4x+x-1}{2x^3-2x^2+x^2-x-3x+3}\)

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

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

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

\(=\frac{3x^2-4x+1}{2x^2+x-3}\)

\(=\frac{3x^2-3x-x+1}{2x^2-2x+3x-3}\)

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

\(=\frac{3x\left(x-1\right)-\left(x-1\right)}{2x\left(x-1\right)-3\left(x-1\right)}\)

\(=\frac{3x-1}{2x-3}\)

\(\frac{a^3-125}{3a^2+15a+75}\)

\(=\frac{\left(a-5\right)\left(a^2+5a+25\right)}{3\left(a^2+5a+25\right)}\)

\(=\frac{a-5}{3}\)

Ủng hộ nhé~~~~!!! :3

\(\frac{a^3-125}{3a^2+15a+75}=\frac{\left(a-5\right).\left(a^2+5a+25\right)}{3.\left(a^2+5a+25\right)}=\frac{a-5}{3}\\ \)

2/

a/ \(25x^2-1=0\)

<=> \(\left(5x\right)^2-1=0\)

<=> \(\left(5x-1\right)\left(5x+1\right)=0\)

<=> \(\orbr{\begin{cases}5x-1=0\\5x+1=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{1}{5}\\x=-\frac{1}{5}\end{cases}}\)

b/ \(4\left(x-1\right)^2-9=0\)

<=> \(\left[2\left(x-1\right)\right]^2-3^2=0\)

<=> \(\left(2x-2\right)^2-3^2=0\)

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

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

<=> \(\orbr{\begin{cases}2x-5=0\\2x+1=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{1}{2}\end{cases}}\)

c/ \(\frac{1}{4}-9\left(x+1\right)^2=0\)

<=> \(\left(\frac{1}{2}\right)^2-\left[3\left(x-1\right)\right]^2=0\)

<=> \(\left(\frac{1}{2}\right)^2-\left(3x-3\right)^2=0\)

<=> \(\left(\frac{1}{2}-3x+3\right)\left(\frac{1}{2}+3x-3\right)=0\)

<=> \(\left(\frac{7}{2}-3x\right)\left(-\frac{5}{2}+3x\right)=0\)

<=> \(\orbr{\begin{cases}\frac{7}{2}-3x=0\\-\frac{5}{2}+3x=0\end{cases}}\)<=> \(\orbr{\begin{cases}3x=\frac{7}{2}\\3x=\frac{5}{2}\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{7}{6}\\x=\frac{5}{6}\end{cases}}\)

d/ \(\frac{1}{16}-\left(2x+\frac{3}{4}\right)^2=0\)

<=> \(\left(\frac{1}{4}\right)^2-\left(2x+\frac{3}{4}\right)^2=0\)

<=> \(\left(\frac{1}{4}-2x-\frac{3}{4}\right)\left(\frac{1}{4}+2x+\frac{3}{4}\right)=0\)

<=> \(\left(-\frac{1}{2}-2x\right)\left(1+2x\right)=0\)

<=> \(2\left(-\frac{1}{4}-x\right)\left(1+2x\right)=0\)

<=> \(\orbr{\begin{cases}-\frac{1}{4}-x=0\\1+2x=0\end{cases}}\)<=> \(\orbr{\begin{cases}x=-\frac{1}{4}\\x=-\frac{1}{2}\end{cases}}\)

\(\left(a+b\right)^6=a^6+6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6 \)
Học dỏi nha
~ Good luck ~

(a+b)^6 

= (a+b)^2 . (a+b)^2 . (a+b)^2

= a² + 2ab + b²   . a² + 2ab + b²  . a² + 2ab + b²  

=...

Ý kiến riêng , ko chắc

hok tốt

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

\(=x^2-2.x.1+1^2+y^2-2.y.1+1^2+1\)

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

(x+2)(x+3)-(x-2)(x+5)

= (x+2) [(x+3) - (x+5)]

=(x +2) [ x+3 - x - 5 }

= (x+2) [ x - x + 3 - 5 }

= (x +2 ) . ( -2)

hok tốt

Đặt \(t=x-1\)

Thế vào:\(t\left(t-1\right)+5=t^2-t+5\)

\(=t^2-2.\frac{1}{2}.t+\left(\frac{1}{2}\right)^2+5-\frac{1}{4}\)

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

Ta có : 

\(VT=\left(x-1\right)\left(x-2\right)+5=x^2-x-2x+2+5=x^2-3x+7\)

\(VT=\left(x^2-3x+\frac{9}{4}\right)+\frac{19}{4}=\left[x^2-2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2\right]+\frac{19}{4}=\left(x-\frac{3}{2}\right)^2+\frac{19}{4}\ge\frac{19}{4}>0\)

Vậy \(\left(x-1\right)\left(x-2\right)+5>0\) với mọi x 

Chúc bạn học tốt ~ 

a/ Ta có \(A=\frac{x-2}{x+2}\)

\(A=\frac{x+2-4}{x+2}\)

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

Để A > 1

<=> \(1-\frac{4}{x+2}>1\)

<=> \(\frac{4}{x+2}>0\)

<=> \(4>x+2\)

<=> \(2>x\)

<=> \(x< 2\)

Bạn coi lại đáp án câu a/ nha bạn. Mình ra là \(x< 2\).

b/ Để \(A\inℤ\)

<=> \(1-\frac{4}{x+2}\inℤ\)

Mà \(1\inℤ\)

<=> \(-\frac{4}{x+2}\inℤ\)

<=> \(\left(-4\right)⋮\left(x+2\right)\)

<=> \(x+2\in\)Ư (4)

Đến đây bạn giải quyết phần còn lại nhen. Mình lười lắm.

b) Để A có giá trị là số nguyên 

Thì (x—2) chia hết cho (x+2)

==> (x+2–4) chia hết cho (x+2)

Vì (x+2) chia hết cho (x+2)

Nên (—4) chia hết cho (x+2)

==> x+2 € Ư(4)

==> x+2 €{1;—1;2;—2;4;—4}

TH1: x+2=1

x=1–2

x=—1

TH2: x+2=—1

x=—1–2

x=—3

TH3: x+2=2

x=2–2

x=0

TH4: x+2=—2

x=—2–2

Xa=—4

TH5: x+2=4

x=4–2

x=2

TH6: x+2=—4

x=—4–2

x=—6

Vậy x€{—1;—3;0;—4;2;—6}