Giải:

\(A^2-\left(\dfrac{3}{5}\right)^2=\dfrac{1}{1.2}+\dfrac{1}{2.7}+\dfrac{1}{7.5}+\dfrac{1}{5.13}+\dfrac{1}{13.8}+\dfrac{1}{8.19}+\dfrac{1}{19.11}+\dfrac{1}{11.25}\) 

Gọi: \(B=\dfrac{1}{1.2}+\dfrac{1}{2.7}+\dfrac{1}{7.5}+\dfrac{1}{5.13}+\dfrac{1}{13.8}+\dfrac{1}{8.19}+\dfrac{1}{19.11}+\dfrac{1}{11.25}\) 

\(B=\left(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+\dfrac{1}{13.16}+\dfrac{1}{16.19}+\dfrac{1}{19.22}+\dfrac{1}{22.25}\right):\dfrac{1}{2}\) \(B=\left[\dfrac{1}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{19.22}+\dfrac{3}{22.25}\right)\right]:\dfrac{1}{2}\) 

\(B=\left[\dfrac{1}{3}.\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{25}\right)\right]:\dfrac{1}{2}\) 

\(B=\left[\dfrac{1}{3}.\left(\dfrac{1}{1}-\dfrac{1}{25}\right)\right]:\dfrac{1}{2}\) 

\(B=\left[\dfrac{1}{3}.\dfrac{24}{25}\right]:\dfrac{1}{2}\) 

\(B=\dfrac{8}{25}:\dfrac{1}{2}\) 

\(B=\dfrac{16}{25}\) 

\(\Rightarrow A^2-\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}\) 

                  \(A^2=\dfrac{16}{25}+\dfrac{9}{25}\) 

                  \(A^2=1\) 

\(\Rightarrow A^2=1^2\) hoặc \(A^2=\left(-1\right)^2\) 

      \(A=1\) hoặc \(A=-1\) 

Chúc bạn học tốt!

\(a^2-\frac{3}{5^2}=\frac{1}{1.2}+\frac{1}{2.7}+\frac{1}{7.5}+\frac{1}{5.13}+\frac{1}{13.8}+\frac{1}{8.19}+\frac{1}{19.11}+\frac{1}{11.25}\)

\(a^2-\frac{3}{5^2}=2.\left(\frac{1}{2.4}+\frac{1}{4.7}+\frac{1}{7.10}+\frac{1}{10.13}+\frac{1}{13.16}+\frac{1}{16.19}+\frac{1}{19.22}+\frac{1}{22.25}\right)\)

\(a^2-\frac{3}{5^2}=2.\frac{1}{3}\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+....+\frac{1}{22}-\frac{1}{25}\right)\)

\(a^2-\frac{3}{5^2}=\frac{2}{3}\left(\frac{1}{2}-\frac{1}{25}\right)\)

=> \(a^2-\frac{3}{25}=\frac{2}{3}.\frac{23}{50}=\frac{23}{75}\)

=> \(a^2=\frac{23}{75}+\frac{3}{25}=\frac{32}{75}\)

=> \(a=\sqrt{\frac{32}{75}}\)(Nếu thế thì đây phải là đề của lớp 7 chứ nhỉ)

 bÀI LÀM

a) x⁴+x³+2x²+x+1=(x⁴+x³+x²)+(x²+x+1)=x²(x²+x+1)+(x²+x+1)=(x²+x+1)(x²+1)

b)a³+b³+c³-3abc=a³+3ab(a+b)+b³+c³ -(3ab(a+b)+3abc)=(a+b)³+c³-3ab(a+b+c)

=(a+b+c)((a+b)²-(a+b)c+c²)-3ab(a+b+c)=(a+b+c)(a²+2ab+b²-ac-ab+c²-3ab)=(a+b+c)(a²+b²+c²-ab-ac-bc)

c)Đặt x-y=a;y-z=b;z-x=c

a+b+c=x-y-z+z-x=o

đưa về như bài b

d)nhóm 2 hạng tử đầu lại và 2hangj tử sau lại để 2 hạng tử sau ở trong ngoặc sau đó áp dụng hằng đẳng thức dề tính sau đó dặt nhân tử chung

e)x²(y-z)+y²(z-x)+z²(x-y)=x²(y-z)-y²((y-z)+(x-y))+z²(x-y)

=x²(y-z)-y²(y-z)-y²(x-y)+z²(x-y)=(y-z)(x²-y²)-(x-y)(y²-z²)=(y-z)(x²-2y²+xy+xz+yz)