Lưu ý: Giải chi tiết từng bước không bỏ bước
Mình đang cần gắp :((
Bài 1: Rút gọn:
\(\dfrac{x^2-6x+5}{5x^2-45}\)
\(\dfrac{\left(4x-3\right)^2-9x^2}{-7x^2+24x-9}\)
2/ Rút gọn rồi tính giá trị của biểu thức tại x = 2022
\(A=\dfrac{a^{16}-1}{\left(a^2+1\right)\left(a^4+1\right)\left(a^2+1\right)\left(a+1\right)}\)
3/ Chứng minh rằng biểu thức sau là giá trị hằng số:
\(B=\dfrac{a^3+b^3+c^3-\left(a+b+c\right)^3}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
3.
Áp dụng công thức: \(\left(x+y\right)^3=x^3+y^3+3xy\left(x+y\right)\) ta có:
\(B=\dfrac{a^3+b^3+c^3-\left(a+b+c\right)^3}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\dfrac{a^3+b^3+c^3-\left[\left(a+b\right)^3+c^3+3\left(a+b\right).c.\left(a+b+c\right)\right]}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\dfrac{a^3+b^3+c^3-\left[a^3+b^3+3ab\left(a+b\right)+c^3+3\left(a+b\right).c.\left(a+b+c\right)\right]}{ }\)
\(=\dfrac{-3ab\left(a+b\right)-3\left(a+b\right).c.\left(a+b+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\dfrac{-3\left(a+b\right)\left[ab+c\left(a+b+c\right)\right]}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\dfrac{-3\left(a+b\right)\left(ab+ca+bc+c^2\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\dfrac{-3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(=\dfrac{-3\left(a+b\right)\left(b+c\right)\left(c+a\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=-3\) là hằng số
Bài 1:
a: \(=\dfrac{\left(x-1\right)\left(x-5\right)}{5\left(x^2-9\right)}=\dfrac{\left(x-1\right)\left(x-5\right)}{5\left(x-3\right)\left(x+3\right)}\)
b: \(=\dfrac{\left(4x-3-3x\right)\left(4x-3+3x\right)}{-7x^2+21x+3x-9}\)
\(=\dfrac{\left(x-3\right)\left(7x-3\right)}{\left(x-3\right)\left(-7x+3\right)}=-1\)
