\(C=\left(x+5\right)^3-x^3-125=x^3+15x^2+75x+125-x^3-125=15x^2+75x=15x\left(x+3\right)\)

\(D=\left(\frac{1}{2}a+b\right)^3+\left(\frac{1}{2}a-b\right)^3\)

\(=\frac{1}{8}a^3+\frac{3}{4}a^2b+3ab^2+b^3+\frac{1}{8}a^3-\frac{3}{4}a^2b+3ab^2-b^3=\frac{1}{4}a^3+6a^2b\)

Viết rõ đề bài ra đc không ạ

đấy là phân số

a/ \(16a^4+48a^3+4a^2-120a-72\)

b.\(2b^2+2a^2=4a\)

a/ 36

b/ 2a² + 2 b²

a: Ta có: \(\frac{1}{2a-b}-\frac{a^2-1}{2a^3-b+2a-a^2b}\)

\(=\frac{1}{2a-b}-\frac{a^2-1}{a^2\left(2a-b\right)+\left(2a-b\right)}\)

\(=\frac{1}{2a-b}-\frac{a^2-1}{\left(2a-b\right)\left(a^2+1\right)}=\frac{a^2+1-a^2+1}{\left(2a-b\right)\left(a^2+1\right)}=\frac{2}{\left(2a-b\right)\left(a^2+1\right)}\)

\(\frac{4a+2b}{a^3b+ab}-\frac{2}{a}\)

\(=\frac{4a+2b}{ab\left(a^2+1\right)}-\frac{2}{a}=\frac{4a+2b-2b\left(a^2+1\right)}{ab\left(a^2+1\right)}\)

\(=\frac{4a-2a^2b}{ab\left(a^2+1\right)}=\frac{2a\left(2-ab\right)}{ab\cdot\left(a^2+1\right)}=\frac{2\left(2-ab\right)}{b\left(a^2+1\right)}\)

Ta có: \(A=\left(\frac{1}{2a-b}-\frac{a^2-1}{2a^3-b+2a-a^2b}\right):\left(\frac{4a+2b}{a^3b+ab}-\frac{2}{a}\right)\)

\(=\frac{2}{\left(2a-b\right)\left(a^2+1\right)}:\frac{2\left(2-ab\right)}{b\left(a^2+1\right)}=\frac{2b\left(a^2+1\right)}{2\left(2-ab\right)\left(2a-b\right)\left(a^2+1\right)}=\frac{b}{\left(2-ab\right)\left(2a-b\right)}\)

b:

Sửa đề: b>a>0

\(4a^2+b^2=5ab\)

=>\(4a^2-5ab+b^2=0\)

=>\(4a^2-4ab-ab+b^2=0\)

=>(a-b)(4a-b)=0

TH1: a-b=0

=>a=b

mà a>b

nên Loại

TH2: 4a-b=0

=>b=4a(nhận)

\(A=\frac{b}{\left(2-ab\right)\left(2a-b\right)}\)

\(=\frac{4a}{\left(2-a\cdot4a\right)\left(2a-4a\right)}=\frac{4a}{\left(2-4a^2\right)\left(-2a\right)}\)

\(=\frac{4a}{-2a\cdot\left(-2\right)\left(2a^2-1\right)}=\frac{1}{2a^2-1}\)

\(7a\left(3a-5\right)+\left(2a-3\right)\left(4a+1\right)-\left(6a-2\right)^2\)

\(=21a^2-35a+8a^2+2a-12a-3-36a^2+24a-4\)

\(=-7a^2+4a-7\)