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

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

1) \(\left(x+\dfrac{1}{3}\right)^3=x^3+3.x^2.\dfrac{1}{3}+3.x.\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3\)

\(=x^3+x^2+\dfrac{x}{3}+\dfrac{1}{27}\)

2) \(\left(2x+y^2\right)^3=\left(2x\right)^3+3.\left(2x\right)^2.y^2+3.2x.\left(y^2\right)^2+\left(y^2\right)^3\)

\(=8x^3+12x^2y^2+6xy^4+y^6\)

3) \(\left(\dfrac{1}{2}x^2+\dfrac{1}{3}y\right)^3=\left(\dfrac{1}{2}x^2\right)^3+3.\left(\dfrac{1}{2}x^2\right)^2.\dfrac{1}{3}y+3.\dfrac{1}{2}x^2.\left(\dfrac{1}{3}y\right)^2+\left(\dfrac{1}{3}y\right)^3\)

\(=\dfrac{1}{8}x^6+\dfrac{1}{4}x^4y+\dfrac{1}{6}x^2y^2+\dfrac{1}{27}y^3\)

4) \(\left(3x^2-2y\right)^3=\left(3x^2\right)^3-3.\left(3x^2\right)^2.2y+3.3x^2.\left(2y\right)^2-\left(2y\right)^3\)

\(=27x^6-54x^4y+36x^2y^2-8y^3\)

5) \(\left(\dfrac{2}{3}x^2-\dfrac{1}{2}y\right)^3=\left(\dfrac{2}{3}x^2\right)^3-3.\left(\dfrac{2}{3}x^2\right)^2.\dfrac{1}{2}y+3.\dfrac{2}{3}x^2.\left(\dfrac{1}{2}y\right)^2-\left(\dfrac{1}{2}y\right)^3\)

\(=\dfrac{8}{27}x^6-\dfrac{1}{3}x^4y+\dfrac{1}{2}x^2y^2-\dfrac{1}{8}y^3\)

6) \(\left(2x+\dfrac{1}{2}\right)^3=\left(2x\right)^3+3.\left(2x\right)^2.\dfrac{1}{2}+3.2x.\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3\)

\(=8x^3+6x^2+\dfrac{3}{2}x+\dfrac{1}{8}\)

7) \(\left(x-3\right)^3=x^3-3.x^2.3+3.x.3^2-3^3\)

\(=x^3-9x^2+27x-27\)

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

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

\(=x^3+1^3\)

\(=x+1\)

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

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

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

\(=x^3-27\)

10) \(\left(x-2\right)\left(x^2+2x+4\right)\)

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

\(=x^3-2^3\)

\(=x^3-8\)

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

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

\(=x^3+4^3\)

\(=x^3+64\)

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

\(=\left(x-3y\right)\left[x^2+x.3y+\left(3y\right)^2\right]\)

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

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

13) \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)\)

\(=\left(x^2-\dfrac{1}{3}\right)\left[\left(x^2\right)^2+x^2.\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^2\right]\)

\(=\left(x^2\right)^3-\left(\dfrac{1}{3}\right)^3\)

\(=x^6-\dfrac{1}{27}\)

14) \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)\)

\(=\left(\dfrac{1}{3}x+2y\right)\left[\left(\dfrac{1}{3}x\right)^2-\dfrac{1}{3}x.2y+\left(2y\right)^2\right]\)

\(=\left(\dfrac{1}{3}x\right)^3+\left(2y\right)^3\)

\(=\dfrac{1}{27}x^3+8y^3\)

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

\(=\left(x^3-6x^2y+9xy^2\right)+\left(y^3-6xy^2+9x^2y\right)\)

\(=x\left(x^2-6xy+9y^2\right)+y\left(y^2-6xy+9x^2\right)\)

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

\(\Rightarrow dpcm\)

\(VP=x\left(x^2-6xy+9y^2\right)+y\left(y^2-6xy+9x^2\right)=\)

\(=x^3-6x^2y+9xy^2+y^3-6xy^2+9x^2y=\)

\(=x^3+3x^2y+3xy^2+y^3=\left(x+y\right)^3=VT\)

a, F(\(x\)) = a\(x^2\) + b\(x\) + c  (a; b; c \(\in\) Q và a \(\ne\) 0)

 Vì F(\(x\)) có nghiệm là \(\sqrt{2}\) ta có F(\(\sqrt{2}\)) = 0

⇔ a.(\(\sqrt{2}\))² + b.(\(\sqrt{2}\)) + c = 0

    2a + \(\sqrt{2}\)b + c = 0 ⇒ c = - (2a + \(\sqrt{2}\)b) (1)

a\(x^2\) + b\(x\) + c = 0

a(\(x^2\) + 2. \(\dfrac{b}{2a}\)\(x\) + \(\dfrac{b^2}{4a^2}\)) - \(\dfrac{b^2-4ac}{4a}\)  = 0

a.(\(x\) + \(\dfrac{b}{2a}\))² = \(\dfrac{b^2-4ac}{4a}\)

   (\(x\) + \(\dfrac{b}{2a}\) )² = \(\dfrac{b^2-4ac}{4a^2}\)

    \(\left[{}\begin{matrix}x=\dfrac{-b+\sqrt{b^2-4ac}}{2a}\\x=\dfrac{-b-\sqrt{b^2-4ac}}{2a}\end{matrix}\right.\)

Thay (1) vào  \(x\) = \(\dfrac{-b-\sqrt{b^2-4ac}}{2a}\) ta có

 \(x\) = \(\dfrac{-b-\sqrt{b^2-4a\left(2a+\sqrt{2}b\right)}}{2a}\) 

a) \(f\left(x\right)=ax^2+bx+c=0\)

\(\Rightarrow f\left(x_1=\sqrt[]{2}\right)=2a+b\sqrt[]{2}+c=0\left(1\right)\)

\(S=x_1+x_2=-\dfrac{b}{a}\Rightarrow x_2=-\dfrac{b}{a}-x_1=-\dfrac{b}{a}-\sqrt[]{2}\)

\(P=x_1.x_2=\dfrac{c}{a}\Rightarrow x_2=\dfrac{c}{a.x_1}=\dfrac{c}{a.\sqrt[]{2}}\)

Vậy nghiệm còn lại là \(-\dfrac{b}{a}-\sqrt[]{2}\) hay \(\dfrac{c}{a.\sqrt[]{2}}\left(a,b,c\in Q;a\ne0\right)\)

b) \(P\left(x\right)=x^2-px+q\)

\(S=x_1+x_2=p;P=x_1.x_2=q\)

Để P(x) có nghiệm \(x_1;x_2\) đều là số nguyên

\(\Rightarrow S=p;P=q\) đều là số nguyên

mà \(p,q\) là số nguyên tố

\(\Rightarrow p;q⋮1\)

\(\Rightarrow\left(p;q\right)\in\left\{-1;1\right\}\Rightarrow p=\pm1;q=\pm1\)

Ta thay \(p=\pm1;q=\pm1\) vào \(P\left(x\right)=x^2-px+p=0\) ta được \(\Delta=5;\Delta=-4< 0\) \(\Rightarrow p,q\) không thỏa nghiệm đa thức nguyên

\(\Rightarrow\left(p;q\right)\in\varnothing\)