[(x+2)(x-5)][(x+3)(x-6)]=180
(x2-3x-10)(x2-3x-18)=180
Đặt x2-3x-10=y
y(y-8)=180
y2-8y-180=0
y2+10y-18y-180=0
y(y+10)-18(y+10)=0
(y+10)(y-18)=0
y={-10;18}
TH1: x2-3x-10= -10
x(x-3)=0
x={0;3}
TH2: x2-3x-10=18
x2-3x-28=0
x2+4x-7x-28=0
x(x+4)-7(x+4)=0
(x-7)(x+4)=0
x={-4;7}
Vậy x={-4;0;3;7}
Giải các phương trình sau:
a)\(\frac{x-1}{x+2}-\frac{x}{x-2}=\frac{5x-2}{4-x^2}\)
b)x3- 5x2+ 6x = 0
c)x2+2x-15=0
d)/2x-3/= 4
e)/3x-1/ - x = 2
f)/x-7/ = 2x+3
g)(x+1)4 + (x-3)4 = 82
h)(x+2)(x+3)(x-5)(x-6)= 180
i)\(\frac{x^2-8}{92}+\frac{x^2-7}{93}=\frac{x^2-6}{94}+\frac{x^2-5}{95}\)
Giải phương trình:
b) \(\dfrac{7}{2}-\left(\dfrac{x}{5}-\dfrac{1}{4}\right)=\dfrac{9}{2}\)
c) (x+2) . (x-5). (x-6) (x+3) = 180
d) \(x-\dfrac{\dfrac{x}{2}-\dfrac{3+x}{4}}{2}=\dfrac{2x-\dfrac{10-7x}{3}}{2}-x-1\)
e) \(\left(\dfrac{1}{1.101}+\dfrac{1}{2.102}+........+\dfrac{1}{10.110}\right).\left(x-3\right)=\dfrac{1}{1.11}+\dfrac{1}{2.12}+.......+\dfrac{1}{100.110}\)
Bài 1: Cho biểu thức \(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}-\dfrac{1}{x-2}\)
a, Rút gọn biểu thức A
b, Tìm x biết A = -3
c, Tìm x nguyên để A đạt giá trị nguyên dương
Biết đa thức f(x) chia cho x-3 dư 7, chia cho x-2 dư 5. Tìm đa thức dư trong phép chia đa thức f(x) cho x^2-5x+6
Tìm x biết:
1. (x-2)^2-(x-3)(x+3)=6
2. 4(x-3)^2-(2x-1)(2x+1)=10
3. (x-4)^2-(x-2)(x+2)=6
4.9(x+1)^2-(3x-2)(3x+2)=10
5. 3x +2(5-x)=0
6.x(2x-1)(x+5)-(2x^2+1)(x+4,5)=3,5
7, 3x^2-3x(x-2)=36
8. (3x^2-x+1)(x-1) +x^2(4-3x)=5/2
Cho x+y=3, x.y=2
Tính x^2+y^2; x^3+y^3; x^4+y^4; x^5+y^5; x^6+y^6 ?
Bài 1
4) (x+3)(x-1)(x-5)(x+15)+64x^2
5) (x+2)(x-4)(x+6)(x-12)+36x^2
6) (x-2)(x-4)(x-5)(x-10)-54x^2
Giải các phương trình sau:
1. \(a,\dfrac{6}{x-1}-\dfrac{4}{x-3}=\dfrac{8}{2x-6}\)
\(b,\dfrac{1}{x-2}+\dfrac{5}{x+1}=\dfrac{3}{2-x}\)
\(c,\dfrac{3x}{x-2}-\dfrac{x}{x-5}=\dfrac{3x}{\left(x-2\right)\left(5-x\right)}\)
2. \(a,\left(x+2\right)\left(3-4x\right)=x^2+4x+4\)
\(b,2x^2-6x+1\)
a) x+2/x-2-1/x=2/x*(x-2)
b)2/2x-6+2/2x+2+2x/(x+1)*(3-x)=0
c) x+1/2017+x+2/2016=x+3/2015+x+4/2014
d) x-45/5+x-44/6+x-43/7+x-42/8=4
e) x-3/2011+x+2/2012=x-2012/2+x-2011/3
