B1: Tính:
\(B=\dfrac{4.\left(x+3\right)^2}{\left(3x+5\right)^2-4x^2}-\dfrac{x^2-25}{9x^2-\left(2x+5\right)^2}-\dfrac{\left(2x+3\right)^2-x^2}{\left(4x+15\right)^2-x^2}\)
B2: Xác định a, b, c:
a, \(\dfrac{10x-4}{x^3-4x}=\dfrac{a}{x}+\dfrac{b}{1-2}+\dfrac{c}{n+2}\) với mọi x khác 0, x khác \(\pm2\)
b, \(\dfrac{1}{x^3-1}=\dfrac{a}{x-1}+\dfrac{bx+c}{x^2+x+1}\)
Help me!!!
Bài 1:
\(B=\dfrac{4\left(x+3\right)^2}{\left(3x+5\right)^2-4x^2}-\dfrac{\left(x^2-25\right)}{9x^2-\left(2x+5\right)^2}-\dfrac{\left(2x+3\right)^2-x^2}{\left(4x+15\right)^2-x^2}\)
\(=\dfrac{4\left(x+3\right)^2}{\left(3x+5-2x\right)\left(3x+5+2x\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(2x+3-x\right)\left(2x+3+x\right)}{\left(4x+15-x\right)\left(4x+15+x\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{5\left(x-5\right)\left(x+1\right)}-\dfrac{3\left(x+3\right)\left(x+1\right)}{15\left(x+5\right)\left(x+3\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{x+5}{5\left(x+1\right)}-\dfrac{x+1}{5\left(x+5\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x+5\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x+1\right)^2}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{4\left(x^2+6x+9\right)-\left(x^2+10x+25\right)-\left(x^2+2x+1\right)}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{4x^2+24x+36-x^2-10x-25-x^2-2x-1}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{2x^2+12x+10}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{2\left(x^2+6x+5\right)}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{2\left(x^2+5x+x+5\right)}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{2\left(x+5\right)\left(x+1\right)}{5\left(x+5\right)\left(x+1\right)}=\dfrac{2}{5}\)
Bài 2.
Sửa đề
a) \(\dfrac{10x-4}{x^3-4x}=\dfrac{a}{x}+\dfrac{b}{x-2}+\dfrac{c}{x+2}\)
Giải
Ta sẽ phân tích vế phải
VP = \(\dfrac{a}{x}+\dfrac{b}{x-2}+\dfrac{c}{x+2}\)
VP = \(\dfrac{a\left(x^2-4\right)+bx\left(x+2\right)+cx\left(x-2\right)}{x\left(x^2-4\right)}\)
VP = \(\dfrac{ax^2-4a+bx^2+2bx+cx^2-2cx}{x\left(x^2-4\right)}\)
VP = \(\dfrac{x^2\left(a+b+c\right)+2x\left(b-c\right)-4a}{x\left(x^2-4\right)}\)
Tương tự , ta cũng sẽ phân tích VT
VT = \(\dfrac{2x.5-4}{x\left(x^2-4\right)}\)
Đồng nhất hai VT và VP , ta có :
\(x^2\left(a+b+c\right)+2x\left(b-c\right)-4a=2.5x-4\)
* a + b + c = 0 => 1 + c + 5 + c = 0 => 2c = - 6 => c = - 3
* b - c = 5 => b = c + 5 => b = - 3 + 5 => b = 2
* a = 1
Vậy , a = 1 ; b = 2 ; c = -3
b) Ta sẽ phân tích VP
VP = \(\dfrac{a}{x-1}+\dfrac{bx+c}{x^2+x+1}\)
VP = \(\dfrac{a\left(x^2+x+1\right)+\left(bx+c\right)\left(x-1\right)}{x^3-1}\)
VP = \(\dfrac{ax^2+ax+a+bx^2-bx+cx-c}{x^3-1}\)
VP = \(\dfrac{x^2\left(a+b\right)+x\left(a-b+c\right)+a-c}{x^3-1}\)
Đồng nhất VP và VT , ta được :
\(x^2\left(a+b\right)+x\left(a-b+c\right)+a-c=1\)
* a + b = 0 => a = - b => b = \(-\dfrac{1}{3}\)
* a - b + c = 0 => a + a + a - 1 = 0 => 3a = 1 => a = \(\dfrac{1}{3}\)
* a - c = 1 => c = a - 1 => c = \(\dfrac{1}{3}\) - 1 = \(-\dfrac{2}{3}\)
Vậy , a = \(\dfrac{1}{3}\) ; b = \(-\dfrac{1}{3}\); c = \(-\dfrac{2}{3}\)
Bài 1 bạn Giang làm rồi thì thôi nhé
Kiểm tra giùm mk câu a bài 2 nha!!! ĐỀ BÀI!!!
Nguyễn Nam, Akai Haruma, Ribi Nkok Ngok, Trần Ngọc Bích, lê thị hương giang, Nguyễn Phương Trâm, Phạm Hoàng Giang, Ngân Hải, @Phùng Khánh Linh, @Ngô Thanh Sang, ...
