Có thể dùng định lí Bezu nha
Tìm a,b sao cho
a) \(2x^3-x^2+ax+b\text{⋮}x^2-1\)
b) \(ax^3+bx^2+2x-1\text{⋮}x^2+5x-6\)
c) \(ax^{4\:}+bx^3+1\text{⋮}\left(x+1\right)^2\)
d) \(x^3-x-15\text{⋮}x^2+ax+b\)
e) \(x^3+ax+b\text{⋮}x^3+x-6\)
Tìm x, biết:
a) \(x\left(x-1\right)-x^2+2\text{x}=5\)
b) \(8\left(x-2\right)-2\left(3\text{x}-4\right)=2\)
c) \(\left(3\text{x}+2\right)\left(x-1\right)-3\left(x+1\right)\left(x-2\right)=4\)
d) \(\left(3\text{x}-5\right)\left(7-5\text{x}\right)-\left(5\text{x}+2\right)\left(2-3\text{x}\right)=4\)
Xác định a,b,c,d thỏa mãn đẳng thức với mọi x
a,\(\left(ax+b\right)\left(x^2+cx+1\right)=7x^3-3x+2\)
b, \(x^4+ax^2+b=\left(x^2-3x+2\right)\left(x^2+cx+d\right)\)
Tìm các số a,b,c thỏa mãn :
a) \(\frac{\text{x^2}-x+2}{\text{ }\left(x-1\right)^3}=\frac{A}{\left(x-1\right)^3}+\frac{B}{\left(x-1\right)^2}+\frac{C}{x-1}\) b)\(\frac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}\)
:| Giúp tớ với
tìm x, biết:
a) \(\left(3\text{x}+2\right)\left(x-1\right)-3\left(x+1\right)\left(x-2\right)=4\)
b) \(\left(3\text{x}-5\right)\left(7-5\text{x}\right)-\left(5\text{x}+2\right)\left(2-3\text{x}\right)=4\)
phan tích nhan tử thanh nhan tử:
a)\(3x^2-12y^2\)
b)\(5xy^2-10xyt+5xt^2\)
c)\(x^3+3x^2+3x+1-27x^3\)
d)\(\text{a}^3x-\text{a}b+b-x\)
e)\(3x^2\left(\text{a}+b+c\right)+36xy\left(\text{a}+b+c\right)+108y^2\left(\text{a}+b+c\right)\)
f)\(\text{a}b\left(\text{a}-b\right)+bc\left(b-c\right)+c\text{a}\left(c-\text{a}\right)\)
g)\(\left(\text{a}+b+c\right)^3-\text{a}^3-b^3-c^3\)
h)\(4\text{a}^2b^2-\left(\text{a}^2+b^2-c^2\right)^2\)
Giải và biện luận phương trình:
a)\(\frac{an}{a-x}+\frac{\left(a+n\right)\left(\text{anx}+nx^2+x^3\right)}{x^3+nx^2-a^2x-a^2n}=\frac{\text{ax}}{n+x}+\frac{nx^2}{x^2-a^2}\left(a\ne0\right)\)
b)\(\frac{a+x}{a^2+\text{ax}+x^2}-\frac{a-x}{\text{ax}-x^2-a^2}=\frac{3a}{2\left(a^4+a^2x^2+x^4\right)}\)
\(M=\left(1+\frac{a}{a^2+1}\right):\left(\frac{1}{a-1}-\frac{2a}{a^3-a^2+a-1}\right)\)
a)tìm điều kiện xác định
b)rút gọn M
Bài 2:
Cho f(x)=\(2x^2+\text{ax}+1v\text{à}g\left(x\right)=x-3\)
tìm a để f(x):g(x) dư 4
Tìm x:
a)\(\text{(x-5)(x+5)-(x+3)^2+3(x-2)^2=(x+1)^2-(x+4)(x-4)+3x^2}\)
\(\text{b)(2x+3)^2}+\left(x-1\right)\left(x+1\right)=5\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)+\left(x+4\right)^2\)