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\(P:\frac{4x-2-16}{2x+1}=\frac{4x^2+4x+1}{x-2}\)
\(\Rightarrow P=\frac{4x^2+4x+1}{x-2}.\frac{4x^2-16}{2x+1}\)
= \(\frac{\left(2x+1\right)^2}{x-2}.\frac{4.\left(x-2\right)\left(x+2\right)}{2x+1}\)
\(\Rightarrow P=4.\left(2x+1\right).\left(x+2\right)\)
\(=4.\left(2x^2+x+4x+2\right)\)
= \(8x^2+40x+8\)
Chúc bạn học tốt !!!
Đề Phân tích đa thức thành nhân tử 1/(1 - x )+ 1/(1+x)+2/(1+x^2)+ 4/(1+x^4)+8/(1+x^8) - 16/(1+ x^16)
1: =(x+y-3x)(x+y+3x)
=(-2x+y)(4x+y)
2: =(3x-1-4)(3x-1+4)
=(3x+3)(3x-5)
=3(x+1)(3x-5)
3: =(2x)^2-(x^2+1)^2
=-[(x^2+1)^2-(2x)^2]
=-(x^2+1-2x)(x^2+1+2x)
=-(x-1)^2(x+1)^2
4: =(2x+1+x-1)(2x+1-x+1)
=3x(x+2)
5: =[(x+1)^2-(x-1)^2][(x+1)^2+(x-1)^2]
=(2x^2+2)*4x
=8x(x^2+1)
6: =(5x-5y)^2-(4x+4y)^2
=(5x-5y-4x-4y)(5x-5y+4x+4y)
=(x-9y)(9x-y)
7: =(x^2+xy+y^2+xy)(x^2+xy-y^2-xy)
=(x^2+2xy+y^2)(x^2-y^2)
=(x+y)^3*(x-y)
8: =(x^2+4y^2-20-4xy+16)(x^2+4y^2-20+4xy-16)
=[(x-2y)^2-4][(x+2y)^2-36]
=(x-2y-2)(x-2y+2)(x+2y-6)(x+2y+6)
(x - 4)(x2 + 4x + 16) - x(x2 - 6) = 2
x3 - 64 - x3 + 6x = 2
6x = 2 + 64
6x = 66
x = 66 : 6
x = 11
x3 - 27 + 3x(x - 3)
= (x - 3)(x2 + 3x + 9) + 3x(x - 3)
= (x - 3)(x2 + 3x + 9 + 3x)
= (x - 3)(x2 + 6x + 9)
= (x - 3)(x + 3)2
5x3 - 7x2 + 10x - 14
= 5x(x2 + 2) - 7(x2 + 2)
= (x2 + 2)(5x - 7)
ta có
\(5x=-3y=4z\)
\(\Rightarrow\frac{x}{12}=-\frac{y}{20}=\frac{z}{15}\)
\(\Rightarrow\frac{x}{12}=-\frac{y}{20}=\frac{3z}{45}=\frac{x-y+3z}{12+20+45}=\frac{7}{77}=\frac{1}{11}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{1}{11}.12=\frac{12}{11}\\-y=\frac{1}{11}.20=\frac{20}{11}\\3z=\frac{1}{11}.45=\frac{45}{11}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\frac{12}{11}\\y=-\frac{20}{11}\\z=\frac{45}{11}:3=\frac{15}{11}\end{cases}}\)
Vậy \(\hept{\begin{cases}x=\frac{12}{11}\\y=\frac{-20}{11}\\z=\frac{15}{11}\end{cases}}\)
\(\dfrac{2x+4}{x^3-1}-A=\dfrac{2}{x-1}-\dfrac{x+2}{x^2+x+1}\)
\(\Leftrightarrow\dfrac{2\left(x+2\right)}{\left(x-1\right)\left(x^2+x+1\right)}-A=\dfrac{2}{x-1}-\dfrac{x+2}{x^2+x+1}\)
\(\Leftrightarrow A=\dfrac{2\left(x+2\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{2}{x-1}+\dfrac{x+2}{x^2+x+1}\)
\(\Leftrightarrow A=\dfrac{2\left(x+2\right)}{\left(x-1\right)\left(x^2+x+1\right)}-\dfrac{2\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}+\dfrac{\left(x+2\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow A=\dfrac{2x+4-2x^2-2x-2+x^2-x+2x-2}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow A=\dfrac{-x^2+x}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow A=\dfrac{-x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Leftrightarrow A=\dfrac{-x}{x^2+x+1}\)
P : 4 x 2 - 16 2 x + 1 = 4 x 2 + 4 x + 1 x - 2