c: C=125x^3+150x^2+60x+8+125x^3-150x^2+60x-8-2(x^2-4)

=250x^3+120x-2x^2+8

=250x^3-2x^2+120x+8

d: D=(4x)^3-3^3-(4x)^3-3^3

=64x^3-27-64x^3-27

=-54

c) \(C=\left(5x+2\right)^3+\left(5x-2\right)^3-2\left(x-2\right)\left(x+2\right)\)

\(=\left[\left(5x\right)^3+3\cdot\left(5x\right)^2\cdot2+3\cdot5x\cdot2^2+2^3\right]+\left[\left(5x\right)^3-3\cdot\left(5x\right)^2\cdot2+3\cdot5x\cdot2^2-2^3\right]-2\left(x^2-4\right)\)

\(=125x^3+150x^2+60x+8+125x^3-150x^2+60x-8-2x^2+8\)

\(=\left(125x^3+125x^3\right)+\left(150x^2-150x^2-2x^2\right)+\left(60x+60x\right)+\left(8-8+8\right)\)

\(=250x^3-2x^2+120x+8\)

d) \(D=\left(4x-3\right)\left(16x^2+12x+9\right)-\left(4x+3\right)\left(16x^2-12x+9\right)\)

\(=\left(4x\right)^3-3^3-\left[\left(4x\right)^3+3^3\right]\)

\(=64x^3-27-\left(64x^3+27\right)\)

\(=64x^3-27-64x^3-27\)

\(=-27-27\)

\(=-54\)

\(\left(4x+3\right)\left(x^2-9\right)=\left(x+3\right)\left(16x^2-9\right)\\ \left(4x+3\right)\left(x-3\right)\left(x+3\right)=\left(x+3\right)\left(4x-3\right)\left(4x+3\right)\\ \left(4x+3\right)\left(x-3\right)\left(x+3\right)-\left(x+3\right)\left(4x-3\right)\left(4x+3\right)=0\)

\(\left(4x+3\right)\left(x+3\right)\left(x-3-4x+3\right)=0\\ \left(4x+3\right)\left(x+3\right)\cdot\left(-3x\right)=0\\ \left[{}\begin{matrix}4x+3=0< =>x=-\dfrac{3}{4}\\x+3=0< =>x=-3\\-3x=0< =>x=0\end{matrix}\right.\)

=>(4x+3)(x-3)(x+3)-(x+3)(4x-3)(4x+3)=0

=>(4x+3)(x+3)(x-3-4x+3)=0

=>-3x(4x+3)(x+3)=0

=>\(x\in\left\{0;-\dfrac{3}{4};-3\right\}\)

\(\left(4x+3\right)\left(x^2-9\right)=\left(x+3\right)\left(16x^2-9\right)\)

\(\Leftrightarrow\left(4x+3\right)\left(x-3\right)\left(x+3\right)=\left(x+3\right)\left(4x-3\right)\left(4x+3\right)\)

\(\Leftrightarrow\left(4x+3\right)\left(x-3\right)\left(x+3\right)-\left(x+3\right)\left(4x-3\right)\left(4x+3\right)=0\)

\(\Leftrightarrow\left(4x+3\right)\left(x+3\right)\left(x-3-4x+3\right)=0\)

\(\Leftrightarrow\left(4x+3\right)\left(x+3\right)\left(-3x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x+3=0\\x+3=0\\-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=-3\\x=0\end{matrix}\right.\)

Chọn D

lim x → - 1   x   +   1 6 x 2   +   3   +   3 x   =   1

1)

x^3 -16x=0`

`<=>x(x^2 -16)=0`

\(< =>\left[{}\begin{matrix}x=0\\x^2-16=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x^2=16\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

b)

`x^4 -2x^3=0`

`<=>x^3 (x-2)=0`

\(< =>\left[{}\begin{matrix}x^3=0\\x-2=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

3)

`(2x-11)(x^2 -1)=0`

\(< =>\left[{}\begin{matrix}2x-11=0\\x^2-1=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}2x=11\\x^2=1\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{11}{2}\\x=1\\x=-1\end{matrix}\right.\)

4)

`x^3 -36x=0`

`<=>x(x^2 -36)=0`

\(< =>\left[{}\begin{matrix}x=0\\x^2-36=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x^2=36\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\\x=6\\x=-6\end{matrix}\right.\)

5)

`2x+19=0`

`<=>2x=-19`

`<=>x=-19/2`

bài về nghiệm của đa thức

a) Ta có: \(a^3\)

\(=\left(\sqrt{5}+\sqrt{3}\right)^3\)

\(=5\sqrt{5}+15\sqrt{3}+9\sqrt{5}+3\sqrt{3}\)

b) Ta có: \(a^4-16a^2+4=0\)

\(\Leftrightarrow\left(\sqrt{5}+\sqrt{3}\right)^4-16\left(\sqrt{5}+\sqrt{3}\right)^2+4=0\)

\(\Leftrightarrow\left(8+2\sqrt{15}\right)^2-16\left(8+2\sqrt{15}\right)+4=0\)

\(\Leftrightarrow64+32\sqrt{15}+60-128-32\sqrt{15}+4=0\)

\(\Leftrightarrow0=0\)(đúng)

Ta có

P   =   ( 4 x   +   1 ) 3   –   ( 4 x   +   3 ) ( 16 x 2   +   3 )     =   ( 4 x ) 3   +   3 . ( 4 x ) 2 . 1   +   3 . 4 x . 1 2   +   1 3   –   ( 64 x 3   +   12 x   +   48 x 2   +   9 )     =   64 x 3   +   48 x 2   +   12 x   +   1   –   64 x 3   –   12 x   –   48 x 2   –   9   =   - 8

Nên P = -8

Q   =   ( x   –   2 ) 3   –   x ( x   +   1 ) ( x   –   1 )   +   6 x ( x   –   3 )   +   5 x     =   x 3   –   3 . x 2 . 2   +   3 x . 2 2   –   2 3   –   x ( x 2   –   1 )   +   6 x 2   –   18 x   +   5 x     =   x 3   –   6 x 2   +   12 x   –   8   –   x 3   +   x   +   6 x 2   –   18 x   +   5 x   =   - 8

=> Q = -8

Vậy P = Q

Đáp án cần chọn là: A

\(16x^2-8x+1\\ =\left(16x^2-4x\right)-\left(4x-1\right)\\ =4x\left(4x-1\right)-\left(4x-1\right)\\ =\left(4x-1\right)\left(4x-1\right)\\ =\left(4x-1\right)^2\)

 (4x)^2 - 2.4x.1 + 1 = 0

 (4x - 1)^2 = 0

4x - 1 = 0

x = 1/4

\(16x^2-8x+1=\left(4x-1\right)^2\)

câu 1 B 

câu 2 D

câu 3 ko bt 

câu 4 x=-1/2; x = -(căn bậc hai(3)*i-1)/4;x = (căn bậc hai(3)*i+1)/4;

câu 5 x=-5/3, x=0, x=1

Câu 1:  x² + 2 xy + y²   bằng:

A. x² + y²                   B.(x + y)²                  C. y² – x²                  D. x² – y²

Câu 2:  (4x + 2)(4x – 2)  bằng:

A. 4x² + 4                  B. 4x² – 4                 C. 16x² + 4                D. 16x² – 4

Câu 3: 25a²  + 9b²  - 30ab  bằng:

A.(5a-9b)²                  B.(5a – 3b)²              C.(5a+3b)²                D.(5a)² – (3b)²

Câu 4: 8x³ +1 bằng

A.(2x+1).(4x²-2x+1)      B. (2x-1).(4x²+2x+1)       C.(2x+1)³            D.(2x)³-1³

Câu 5:Thực hiện phép nhân  x(3x² + 2x - 5) ta được:

A.3x³ - 2x² – 5x          B. 3x³ + 2x² – 5x      C. 3x³ - 2x² +5x         D. 3x³ + 2x² + 5x