a)=1/2 . 8/15 - 3/4.47/9

=4/15 - 47/12

=-73/20

b)=2-1/3 . -21/20

=2+7/20

=47/20

đơn giản 

......dễ.....

Đặt \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{60^2}\)

\(\Rightarrow A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{59.60}\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{59}-\frac{1}{60}\)

\(\Rightarrow A< \frac{1}{2}-\frac{1}{60}=\frac{29}{60}>\frac{4}{9}\)

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{2}\\x-\frac{1}{3}=\frac{-1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{-1}{6}\end{cases}}}\)

Vậy x= 5/6 hoặc -1/6

b) - Nếu x=0 thì \(5^y=2^0+624=1+624=625=5^4\Rightarrow y=4\left(y\in N\right)\)

- Nếu x \(\ne\) 0 thì vế trái là số chẵn , vế phải là số lẻ \(\forall x;y\inℕ\) ( vô lí)

Vậy x=0, y=4

thank you bạn.

\(a,10.a^6+20a^5=10a^5\left(a+2\right)\)

\(b,5x^2-10xy+5y^2=5\left(x^2-2xy+y^2\right)=5\left(x-y\right)^2\)

\(c,3ab^3+6ab^2-18ab=3ab\left(b^2+2b-1\right)\)

\(d,15x^3y^2+10x^2y^2-20x^2y^3=5x^2y^2\left(3x+2-4y\right)\)

\(e,a^2\left(x-1\right)-b\left(1-x\right)=a^2\left(x-1\right)+b\left(x-1\right)=\left(x-1\right)\left(a^2+b\right)\)

\(f,x\left(x-5\right)-4\left(5-x\right)=x\left(x-5\right)+4\left(x-5\right)=\left(x-5\right)\left(x+4\right)\)

(mk sửa lại thứ tự là a,b,c,d,e,f nha)

chúc bn học tốt

\(1,10a^6+20a^5=10a^5\left(a+10\right)\)

\(2,5x^2-10xy+5y^2=5\left(x^2-2xy+y^2\right)\)

\(=5\left(x-y\right)^2\)

\(3,3ab^3+6ab^2-18ab\)

\(=3ab\left(b^2+2b-6\right)\)

\(4,15x^3y^2+10x^2y^2-20x^2y^3\)

\(=5x^2y^2\left(3x+2-4y\right)\)

\(5,a^2\left(x-1\right)-b\left(1-x\right)\)

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

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

\(6,x\left(x-5\right)-4\left(5-x\right)\)

\(=x\left(x-5\right)+4\left(x-5\right)\)

\(=\left(x+4\right)\left(x-5\right)\)

mk sửa lại câu 3 chút nha
\(3ab^3+6ab^2-18ab=3ab\left(b^2+2b-6\right)\)

xl vì sai sót
chúc bn học tốt nhé

\(B=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}x...x\frac{2002}{2003}x\frac{2003}{2004}\)

\(B=\frac{1x2x3x4x...x2002x2003}{2x3x4x5x...x2003x2004}\)

Rút gọn các thừa số ở tử và mẫu ta được:

\(B=\frac{1}{2004}\)

                                                                              Đ/S:\(\frac{1}{2004}\)

Ta có:

\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)....\left(1-\frac{1}{2003}\right).\left(1-\frac{1}{2004}\right)\)

\(=\frac{1}{2}.\frac{2}{3}....\frac{2002}{2003}.\frac{2003}{2004}\)

\(=\frac{1.2....2002.2003}{2.3....2003.2004}\)

Đơn giản hết sẽ là:

\(=\frac{1}{2004}\)

B = \(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{5}\right)x......x\left(1-\frac{1}{2003}\right)x\left(1-\frac{1}{2004}\right)\)

B = \(\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}x.........x\frac{2002}{2003}x\frac{2003}{2004}\)

B = \(\frac{1x2x3x4x..........x2002x2003}{2x3x4x5x.....x2003x2004}\)

Sau khi trực tiêu các số thì ta có:

B = \(\frac{1}{2004}\)