Pt có nghiệm là 2 và -2 

`Answer:`

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

`<=>(x-1)(x+2)-(x+2)=0`

`<=>(x+2)(x-1-1)=0`

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

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

`<=>x^2=4`

`<=>x=2` hoặc `x=-2`

\(A=\left(100-99\right)\left(100+99\right)+\left(99-98\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\\ A=100+99+99+98+...+2+1\\ A=\left(100+1\right)\left(100-1+1\right):2=5050\)

\(B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^1-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)...\left(2^{64}+1\right)+1\\ B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\\ B=\left(2^{32}-1\right)\left(2^{32}+1\right)\left(2^{64}+1\right)+1\\ B=\left(2^{64}-1\right)\left(2^{64}+1\right)+1=2^{128}-1+1=2^{128}\)

\(C=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2+2ab-2ac-2bc-2a^2-4ab-2b^2\\ C=2c^2\)

Cũa mị:>>>

Tham khảo ạ !!!

A = 100² - 99² + 98² - 97² + ...... + 2² - 1²

= ( 100 - 99 ) ( 100 + 99 ) + ( 98 - 97 ) ( 98 + 97 ) + ......... + ( 2 - 1 ) ( 2 + 1 )

= 1 + 2 + 3 + ......... + 99 + 100

= ( 100 + 1 ) . 100 : 2 = 5050 

 B = 3 ( 2² + 1 ) ( 2⁴ + 1 ) ... ( 2⁶⁴ + 1 ) + 1²

= ( 2² - 1 ) ( 2² + 1 ) ( 2⁴ + 1 ) ... ( 2⁶⁴ + 1 ) + 1

= ( 2⁴ - 1 ) ( 2⁴ + 1 ) ... ( 2⁶⁴ + 1 ) + 1

= ( 2⁸ - 1 ) ( 2⁸ + 1 ) ... ( 2⁶⁴ + 1 ) + 1

= ( 2¹⁶ - 1 ) ( 2¹⁶ + 1 ) ... ( 2⁶⁴ + 1 ) + 1

= ( 2³² - 1 ) ( 2³² + 1 ) ( 2⁶⁴ + 1 ) + 1

= ( 2⁶⁴ - 1 ) ( 2⁶⁴ + 1 ) + 1

= 2¹²⁸ - 1 + 1 

= 2¹²⁸

C = ( a + b + c )² + ( a + b - c )² - 2 ( a + b )²

= a² + b² + c² + 2ab + 2bc + 2ca + a² + ab - ac + ab + b² - bc - ac - bc + c² - 2 ( a² + 2ab + b² )

= a² + b² + c² + 2ab + 2bc + 2ca + a² + ab - ac + ab + b² - bc - ac - bc + c² - 2a² - 4ab - 2b²

= 2c²

Theo hệ quả Ta lét \(\dfrac{AK}{AC}=\dfrac{NK}{BC}\Rightarrow NK=\dfrac{AK.BC}{AC}=4\)

có HK//BC

=> ak/ac = x/bc
=>3/(3+1,5) = x/6
=>3/4,5 = x/6

=> x = 4 or NK = 4

ta có 

\(x^2-2x+23=x^2-2x+1+22=\left(x-1\right)^2+22=0\text{ vô nghiệm vì}\)

\(\left(x-1\right)^2\ge0\Rightarrow VT=\left(x-1\right)^2+22\ge22>0\)

tổng điểm 4 môn ( đã nhân kèm hệ số) mà AN cần đạt tới là : 

\(8\times6=48\text{ điểm}\)

Tổng số điểm mà AN đang có là : \(8\times2+7+9=32\text{ điểm}\)

Tổng điểm môn toán ( đã nhân hệ số 2 ) cần đạt được là : \(48-32=16\text{ điểm}\)

vậy số điểm tối thiểu mà môn toán của An cần đạt được là : \(16:2=8\text{ điểm}\)

\(\left(\dfrac{7}{2}x-7\right)\left(21x-63\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{7}{2}x-7=0\\21x-63=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{7}{2}x=7\\21x=63\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

TH1: 2

TH2: 3

c) \(\dfrac{x-1}{2}+\dfrac{x-1}{3}+\dfrac{x-1}{2016}=0\\ \Leftrightarrow\dfrac{1008\left(x-1\right)}{2016}+\dfrac{672\left(x-1\right)}{2016}+\dfrac{x-1}{2016}=0\\ \Leftrightarrow1008x-1008+672x-672+x-1=0\\ \Leftrightarrow1008x+672x+x=0+1+672+1008\\ \Leftrightarrow1681x=1681\\ \Leftrightarrow x=1\)

d) \(\dfrac{x-4}{5}+\dfrac{3x-4}{10}-x=\dfrac{2x-5}{3}-\dfrac{7x+3}{6}\\ \Leftrightarrow\dfrac{6\left(x-4\right)}{30}+\dfrac{3\left(3x-4\right)}{30}-\dfrac{30x}{30}=\dfrac{10\left(2x-5\right)}{30}-\dfrac{5\left(7x+3\right)}{30}\\ \Leftrightarrow6x-24+9x-12-30x=20x-50-35x+15\\ \Leftrightarrow6x+9x-20x+35x-30x=-50+15+24+12\\ \Leftrightarrow0x=1\left(vô.lí\right)\)