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

\(\Rightarrow\left(3x+1\right)\left(-x-5\right)=0\Rightarrow\left[{}\begin{matrix}3x+1=0\\-x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-5\end{matrix}\right.\)

b)\(9\left(2x+1\right)^2-4\left(x+1\right)^2=0\Rightarrow\left[3\left(2x+1\right)+2\left(x+1\right)\right]\left[3\left(2x+1\right)-2\left(x+1\right)\right]=0\)

\(\Rightarrow\left[8x+5\right]\left[4x+1\right]=0\Rightarrow\left[{}\begin{matrix}8x+5=0\\4x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{8}\\x=\dfrac{1}{4}\end{matrix}\right.\)

c)\(x^3-6x^2+9x=0\Rightarrow x\left(x^2-6x+9\right)=0\Rightarrow x\left(x-3\right)^2=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

d) \(x^2\left(x+1\right)-x\left(x+1\right)+x\left(x-1\right)=0\)

\(\Rightarrow x\left(x+1\right)\left(x^2-1\right)+x\left(x-1\right)=0\)

\(\Rightarrow x\left(x+1\right)\left(x-1\right)\left(x+1\right)+x\left(x-1\right)=0\)

\(\Rightarrow x\left(x-1\right)\left[\left(x+1\right)\left(x+1\right)+1\right]=0\)

\(\Rightarrow x\left(x-1\right)\left[\left(x+1\right)^2+1\right]=0\)

Do \(\left(x+1\right)^2+1>0\)

\(\Rightarrow x\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

A=2(x³-y³)-3(x+y)²

A=2(x-y)(x²+xy+y²)-3(x²+2xy+y²)

A=2.2(x²+xy+y²)-3(x²+2xy+y²)

A=4(x²+xy+y²)-3x²+6xy+3y²

A=4x²+4xy+y²-3x²-6xy+3y²

A=x²-2xy+y²

A=(x-y)²

A= 2²

A=4

?????

a)x²-6x+9

=x²-2.x.3+3²

=(x-3)²

b)4x²+4x+1

=(2x)²+2.2x.1+1²

=(2x+1)²

c)4x²+12xy+9y²

=(2x)²+2.2x.3y+(3y)²

=(2x+3y)²

d)4x⁴-4x²+4

=(2x²)²-2.2x².2+2²

=(2x²-2)²

y²(\(x\)  + y) - ( \(x\) - 7)² (đk \(x\) +y ≥ 0)

= (y\(\sqrt{\left(x+y\right)}\) )² - (\(x\) - 7)²

= (y\(\sqrt{x+y}\) - (\(x-7\)))( y\(\sqrt{x+y}\) + (\(x\) - 7))

= (y\(\sqrt{x+y}\) - \(x\) + 7)(y\(\sqrt{x+y}\) + \(x\) - 7)

a) Ta có:

VT = (x - y)² + 4xy

= x² - 2xy + y² + 4xy

= x² + 2xy + y²

= (x + y)²

= VP

b) Ta có:

(x + y)² = (x - y)² + 4xy

= 5² + 4.3

= 25 + 12

= 37

a : 7 dư 3 cm a² : 7 dư 2

Ta có:     a = 7k + 3

          ⇔ a² = (7k + 3)²

          ⇔ a² = 49k² + 42k + 9

          ⇔ a² = 7.(7k² + 6k + 1) + 2

                7 ⋮ 7 ⇔ 7.(7k² + 6k + 1) ⋮ 7

          ⇔ a² = 7.(7k² + 6k + 1) + 2 : 7 dư 2 (đpcm)

Cách 2 sử dụng đồng dư thức:

a \(\equiv\) 3 (mod 7) ⇔ a² \(\equiv\) 3² (mod 7)  3² : 7 dư 2 ⇔ a² : 7 dư 2 (đpcm)

\(\text{∘ Ans}\)

\(\downarrow\)

`1,`

`86.15 + 150. 1,4`

`= 86. 15 + 15. 14`

`= 15.(86 + 14)`

`= 15.100`

`= 1500`

`2,`

`93.32 + 14.16`

`= 93.32 + 2.7.16`

`= 93.32 + 32.7`

`= 32.(93 + 7)`

`= 32.100`

`= 3200`

`3,`

\(98,6\cdot199-990\cdot9,86\)

`=`\(98,6\cdot199-99\cdot98,6\)

`=`\(98,6\cdot\left(199-99\right)\)

`=`\(98,6\cdot100\)

`=`\(9860\)

`4,`

\(85\cdot12,7+5\cdot3\cdot12,7?\)

`=`\(85\cdot12,7+15+12,7\)

`=`\(12,7\cdot\left(85+15\right)\)

`=`\(12,7\cdot100\)

`= 1270`

`5,`

\(0,12\cdot90-110\cdot0,6+36-25\cdot6?\)

\(=6\cdot1,8-11\cdot6+6\cdot6-25\cdot6\)

\(=6\cdot\left(1,8-11+6-25\right)\)

\(=6\cdot\left(-28,2\right)=-169,2\)

a. \(8x\left(x-2007\right)-2x+4034=0\)

\(\Rightarrow\left(x-2017\right)\left(4x-1\right)\)

\(\Rightarrow\left[{}\begin{matrix}x-2017=0\\4x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2017\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)

Vậy x=2017 hoặc x=1/4

b.\(\dfrac{x}{2}+\dfrac{x^2}{8}=0\)

\(\Rightarrow\dfrac{x}{2}\left(1+\dfrac{x}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}=0\\1+\dfrac{x}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{x}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)

Vậy x=0 hoặc x=-4

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

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

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

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

Vậy x=4 hoặc x=7/2

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

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

Nxet: (x²+3)>0 với mọi x

=> x-2=0 <=>x=2

Vậy x=2

a, 8\(x\).(\(x-2007\)) - 2\(x\) + 4034 = 0

     4\(x\)(\(x\) - 2007) - \(x\) + 2017 = 0

     4\(x^2\) - 8028\(x\) - \(x\) + 2017 = 0

     4\(x^2\) - 8029\(x\) + 2017 = 0

     4(\(x^2\) - 2. \(\dfrac{8029}{8}\) \(x\) +( \(\dfrac{8029}{8}\))²) - (\(\dfrac{8029}{4}\))²  + 2017 = 0

    4.(\(x\) + \(\dfrac{8029}{8}\))² = (\(\dfrac{8029}{4}\))² - 2017

       \(\left[{}\begin{matrix}x=-\dfrac{8029}{8}+\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\\x=-\dfrac{8029}{8}-\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\end{matrix}\right.\)