6x\(^4\) - 9x\(^3\)
thực hiện phép tính
a: (6x^5y^2-9x^4y^3+15x^3y^4)/3x^3y^2
b: (27x^3-8)/(6x+9x^2+4)
a) (x^2+4)(x+2)(x-2)-(x^2+3)(x^2-3)
b)(3x-2)(9x^2+6x+4)-3(9x^3-2)
c)(3x+5)^2+(6x+10)(2-3x)+(2-3x)^2
a: \(=\left(x^2+4\right)\left(x^2-4\right)-\left(x^4-9\right)\)
\(=x^4-16-x^4+9=-7\)
b: \(=27x^3-8-27x^3+6=-2\)
c: \(=\left(3x+5+2-3x\right)^2=7^2=49\)
giải pt:
a,\(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
b,\(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
rút gọn
\(\frac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}\)
rút gọn
\(\frac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}\)
\(\frac{x^4+6x^3+9x^2-1}{x^4+6x^3+7x^2-6x+1}=\frac{\left(x^2\right)^2+2.x^2.3x+\left(3x\right)^2-1}{\left(x^2\right)^2+2.x^2.3x+\left(3x\right)^2-2x^2-6x+1}\)
\(=\frac{\left(x^2+3x\right)^2-1}{\left(x^2+3x\right)^2-2\left(x^2+3x\right)+1}\)
\(=\frac{\left(x^2+3x-1\right)\left(x^2+3x+1\right)}{\left(x^2+3x-1\right)^2}=\frac{x^2+3x+1}{x^2+3x-1}\)
tìm x
1) (3x-2)(9x^2+6x+4)-(2x-5)(2x+5)=(3x-1)^3-(2x+3)^2+9x(3x-1)
2) (2x+1)^3-(3x+2)^2=(2x-5)(4x^2+10x+25)+6x(2x+1)-9x^2
Tìm GTLN của biểu thức :
\(A=x^4-6x^3+9x^2+6x+2021\)
giải pt :
a,\(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
b, \(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
x^4 - 6x^3 + 9x^2
x^4 - 6x^3 + 9x^2
=𝑥2 ( 𝑥2 − 6 𝑥 + 9 )
(3-9x)(6x-4)(5-15x)=0
(3-9x)(6x-4)(5-15x)=0
⇔ 3-9x=0 -> x=1/3
6x-4=0 -> x= 2/3
5-15x=0 -> x= 1/3
Vậy tập nghiệm S={ 2/3;1/3}
(3-9x) (6x-4) (5-15x)=0
<=> 3-9x=0 hoặc 6x-4=0 hoặc 5-15x=0
<=> 9x=3-0 hoặc 6x=0+4 hoặc 15x=5-0
<=> 9x=3 hoặc 6x=4 hoặc 15x=5
<=> 9x:9=3:9 hoặc 6x:6=4:6 hoặc 15x:15=5:15
<=> x=1/3 hoặc x=2/3 hoặc x=1/3
Vậy S={1/3;2/3}