\(\left(2^3.9^4+9^3.45\right):\left(9^2.10-9^2\right)=\left(2^3.9^4+9^3.9.5\right):\left(9^2.9\right)=\left(9^4\left(2^3+5\right)\right):9^3=9^4.13:9^3=9.13=117\)

đúng cho mk nha Nguyễn Ngọc Thuỳ Vy

Đặt a/b = c/d = k

=> a = bk; c = dk

Thay vào đk đề bài ta đc:

(bk)² + b²/ (dk)² + d² ​ = b² (k² + 1)/d²(k² + 1) = b/d (2)

ab/cd = bk.b/dk.d = b².k/d².k = b²/d² = b/d (1)

Từ (1) và (2) suy ra a² + b²/c² + d² = ab/cd → ĐPCM.

Ta co \(\frac{1}{a}-\frac{1}{a+1}=\frac{a+1}{a\left(a+1\right)}-\frac{a}{a+1}=\frac{a+1-a}{a\left(a+1\right)}=\frac{1}{a\left(a+1\right)}\)

Vay \(\frac{1}{a\left(a+1\right)}=\frac{1}{a}-\frac{1}{a+1}\)

Ap dung cong thuc tren:

=> A = \(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{132}\)

     A = \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{11.12}\)

     A = \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{11}-\frac{1}{12}\)

     A = \(\frac{1}{2}-\frac{1}{12}\)

     A = \(\frac{5}{12}\)

C : \(0,2.\sqrt{\left(-10\right)^2.3}+2\sqrt{\left(\sqrt{3}-\sqrt{5}\right)^2}=10.0,2\sqrt{3}+2l\sqrt{3}-\sqrt{5}l=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}=2\sqrt{5}\)

Mk nè