Bài 12:

a) \(\left(\dfrac{1}{2}x+4\right)^2\)

\(=\left(\dfrac{1}{2}x\right)^2+2\cdot\dfrac{1}{2}x\cdot4+4^2\)

\(=\dfrac{1}{4}x^2+4x+16\)

b) \(\left(7x-5y\right)^2\)

\(=\left(7x\right)^2-2\cdot7x\cdot5y+\left(5y\right)^2\)

\(=49x^2-70xy+25y^2\)

c) \(\left(6x^2+y^2\right)\left(y^2-6x^2\right)\)

\(=\left(y^2+6x^2\right)\left(y^2-6x^2\right)\)

\(=y^4-36x^4\)

d) \(\left(x+2y\right)^2\)

\(=x^2+2\cdot x\cdot2y+\left(2y\right)^2\)

\(=x^2+4xy+4y^2\)

e) \(\left(x-3y\right)\left(x+3y\right)\)

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

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

f) \(\left(5-x\right)^2\)

\(=5^2-2\cdot5\cdot x+x^2\)

\(=25-10x+x^2\)

\(11,\)

\(a,\left(7x+4\right)^2-\left(7x+4\right)\left(7x-4\right)\)

\(=\left(7x+4\right)\left(7x+4-7x+4\right)\)

\(=\left(7x+4\right).8=56x+32\)

\(b,\left(x+2y\right)^2-6xy\left(x+2y\right)\)

\(=\left(x+2y\right)\left(x+2y-6xy\right)\)

Bài `12`

`(1/2x+4)^2`

`=(1/2x)^2 + 2 . 1/2x.4 + 4^2`

`= 1/4 x^2 +4x + 16`

__

`(7x-5y)^2`

`=(7x)^2-2.7x.5y+(5y)^2`

`= 49x^2 -  70xy   + 25y^2`

__

`(6x^2+y^2)(y^2-6x^2)`

`=(y^2+6x^2)(y^2-6x^2)`

`=(y^2)^2 - (6x^2)^2`

`=y^4-36x^4`

__

`(x+2y)^2`

`=x^2+ 2.x.2y+(2y)^2`

`= x^2 + 4xy +4y^2`

__

`(x-3y)(x+3y)`

`=x^2 - (3y)^2`

`=x^2 - 9y^2`

__

`(5-x)^2`

`=5^2 -2.5.x+x^2`

`=25 - 10x+x^2`

Bài `11`

`(7x+4)^2 -(7x+4)(7x-4)`

`= (7x+4)(7x+4) -(7x+4)(7x-4)`

`=(7x+4)(7x+4-7x+4)`

`=8(7x+4)`

`= 56x+32`

__

`(x+2y)^2-6xy (x+2y)`

`= (x+2y) (x+2y-6xy)`

tả về bộ phận của thân cây ( hoa, lá, quả,....)

( NHANH NHÉ, MÌNH ĐANG CẦN GẤP, GẤP LẮM )

tớ mới học lớp 1

lớp 1 chưa hk mấy cái này đâu 

mà nếu hk rùi thì phải bt chứ 

nhân chia trước ,cộng trừ sau

Đáp án: B hoặc C (Sao B và C lại giống nhau nhỉ ?!?)

\(3x^2-5x+2+3x^2+5x=\left(3x^2+3x^2\right)+\left(-5x+5x\right)+2=6x^2+2\)

(3x²-5x+2)+(3x²+5x)

=3x²-5x+2+3x²+5x

=6x²+2

=>Đáp án đúng là B và C

bruh

a,x(2x-1)-(x-1)^2-x^2=0

<=>x(2x-1-x)-(x-1)^2=0

<=>x(x-1)-(x-1)^2=0

<=>(x-x+1)(x-1)=0

<=>x-1=0

<=>x=1

b,(x+2)^3-x^3-6x^2=4

<=>x^3+6x^2+12x+8-x^3-6x^2=4

<=>12x+8=4

<=>x=-1/3

tick mik nha

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

`<=>2x^2-x-x^2+2x-1-x^2=0`

`<=>x-1=0`

`<=>x=1`

Vậy `x=1.`

`b)(x+2)^3-x^3-6x^2=4`

`<=>x^3+6x^2+12x+8-x^3-6x^2=4`

`<=>12x+8=4`

`<=>12x=-4`

`<=>x=-1/3`

Vậy `x=-1/3.`

a: Ta có: \(x\left(2x-1\right)-\left(x-1\right)^2-x^2=0\)

\(\Leftrightarrow2x^2-x-x^2+2x-1-x^2=0\)

\(\Leftrightarrow x=1\)

b: Ta có: \(\left(x+2\right)^3-x^3-6x^2=4\)

\(\Leftrightarrow x^3+6x^2+12x+8-x^3-6x^2=4\)

\(\Leftrightarrow12x=-4\)

hay \(x=-\dfrac{1}{3}\)

a)
=(x-2)³
b)\(\left(2-x\right)^3\)
c)\(\left(x+\dfrac{1}{3}\right)^3\)
d)\(\left(\dfrac{x}{2}+y\right)^3\)
e)
\(=\left(x-1\right)^2\left(x-1-15\right)+25\left[3\left(x-1\right)-5\right]\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-3-5\right)\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-8\right)\)
 

a) (a² - 1)² + 4a²

= a⁴ - 2a² + 1 + 4a²

= (a²)² + 2.a².1 + 1²)

=(a² + 1)²

b) (6x² + y²)(y² - 6x²)

= (y² + 6x²)(y² - 6x²)

= (y²)² - (6x²)²

= y⁴ - 36x⁴

\(\left(2x+1\right)2-4\left(x+2\right)2=9\)

\(4x+2-8x-16=9\)

\(4x-8x=9+16-2\)

\(-4x=23\)

\(x=-\frac{23}{4}\)

a, \(\left(2x+1\right)2-4\left(x+2\right)2=9\)

\(\Leftrightarrow4x+2-8x-16=0\Leftrightarrow-4x-14=0\Leftrightarrow x=-\frac{7}{2}\)

b, \(\left(x+1\right)3-2x\left(x+3\right)=2\)

\(\Leftrightarrow3x+3-2x^2-6x=2\Leftrightarrow-3x+1-2x^2=0\)

e: x=5; -5

f: x=4

giải giúp em này ik mng ơiiiii