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

Đơn giản mà em cô sẽ chứng minh cho em thấy ngay bây giờ ha.

320 : ( 16 \(\times\) 2 ) = 320 : 16 : 2

Ta có: 

   320 : ( 16 \(\times\) 2)

= 320 : ( \(\dfrac{16\times2}{1}\) )

= 320 \(\times\) \(\dfrac{1}{16\times2}\) 

= \(\dfrac{320}{16\times2}\)

= \(\dfrac{320}{16}\) \(\times\) \(\dfrac{1}{2}\)

= \(\dfrac{320}{16}\) : 2

= 320 : 16 : 2 ( đpcm)

`4-x=2(x-4)^2`

`<=>4-x=2(x^2-8x+16)`

`<=> 4-x=2x^2 - 16x+32`

`<=> 4-x-2x^2+16x-32=0`

`<=> -2x^2 +15x-28=0`

`<=> -(2x^2-15x+28)=0`

`<=>-(2x^2-7x-8x+28)=0`

`<=> - [x(2x-7) - 4(2x-7)]=0`

`<=> -(2x-7)(x-4)=0`

\(\Leftrightarrow\left[{}\begin{matrix}-2x+7=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-2x=-7\\x=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\)

__

`(x^2 +1) (x-2)+2x=4`

`<=> x^3 -2x^2 +x-2+2x-4=0`

`<=> x^3 -2x^2 +3x-6=0`

`<=> (x^3+3x)-(2x^2+6)=0`

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x^2+3=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=2\end{matrix}\right.\)

__

`x^4 -16x^2=0`

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

`<=>x^2(x-4)(x+4)=0`

\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)

\(4-x=2\left(x-4\right)^2\)

\(\Leftrightarrow4-x=2\left(x^2-8x+16\right)\)

\(\Leftrightarrow4-x=2x^2-16x+32\)

\(\Leftrightarrow2x^2-15x+28=0\)

\(\Leftrightarrow2x^2-7x-8x+28=0\)

\(\Leftrightarrow x\left(2x-7\right)-4\left(2x-7\right)=0\)

\(\Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-7\\x=4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\)

___________

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

\(\Leftrightarrow x^3-2x^2+x-2+2x=4\)

\(\Leftrightarrow x^3-2x^2+3x-2-4=0\)

\(\Leftrightarrow x^3-2x^2+3x-6=0\)

\(\Leftrightarrow x^2\left(x-2\right)+3\left(x-2\right)=0\)

\(\Leftrightarrow\left(x^2+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=-3\left(\text{vô lý}\right)\\x=2\left(tm\right)\end{matrix}\right.\)

\(\Leftrightarrow x=2\)

________________

\(x^4-16x^2=0\)

\(\Leftrightarrow\left(x^2\right)^2-\left(4x\right)^2=0\)

\(\Leftrightarrow\left(x^2-4x\right)\left(x^2+4x\right)=0\)

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

\(\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x-4=0\\x+4=0\end{matrix}\right.\)

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

\(a,=\left(5x-1\right)^2\\ b,=\left(x+4\right)^2\\ c,=\left(4x+3y\right)^2\\ d,=\left(\dfrac{x}{4}+2y\right)^2\)

Đáp án là 6 

TL:

đáp án là:

6

-HT-

\(\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\)

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

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

\((X-y)^2-4=(x-y-2)(x-y+2)\)\((X^2+4)^2-16x^2=(x^2+4)^2-(4x)^2=(x^2+4-4x)(x^2+4+4x)\)

\(9-(x-y)^2=(3-x+y)(3+x-y)\)

a)(x-y)²-4=(x-y)²-2²=(x-y-2)(x-y+2)

b)9-(x-y)²=3²-(x-y)²=(3-x+y)(3+x-y)

c)(x²+4)²-16x²=(x²+4)²-(4x)²=(x²+4-4x)(x²+4+4x)=(x-2)²(x+2)²=(x²-2)²

Chọn D

\(6x^3-9x^2=3x^2\left(2x-3\right)\\ 25x^2-0,09=\left(5x-0,3\right)\left(5x+0,3\right)\\ x^2-x-y^2-y=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\\ \left(x^2+4\right)^2-16x^2=\left(x^2-4x+4\right)\left(x^2+4x+4\right)=\left(x-2\right)^2\left(x+2\right)^2\)

c: \(=2\sqrt{3}+3\sqrt{3}-\sqrt{3}=4\sqrt{3}\)