\(\hept{\begin{cases}a+b=9\\ab=20\end{cases}}\Leftrightarrow\hept{\begin{cases}a=9-b\\ab=20\end{cases}}\)

\(\Leftrightarrow\left(9-b\right)b=20\)

\(\Leftrightarrow9b-b^2-20=0\)

\(\Leftrightarrow5b-20+4b-b^2=0\)

\(\Leftrightarrow5\left(b-4\right)-b\left(b-4\right)=0\)

\(\Leftrightarrow\left(5-b\right)\left(b-4\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}5-b=0\\b-4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}b=5\\b=4\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}a=9-b=9-5=4\\a=9-b=9-4=5\end{cases}}\)

@ giải phức tạp thế ai bắt tính a, b đâu

(a+b)=9

(a+b)^2=9^2

(a-b)^2=(a+b)^2-4ab=1

Ia-bI=1 a<b=> (a-b)=-1

=> (a-b)^2015=-1

ta có: a+b = 9

=> (a+b)² = 81

a² + 2ab + b² = 81

=> a² - 2ab + b² + 4ab = 81

(a-b)² + 4ab = 81

(a-b)² + 80= 81

(a-b)² = 1 = 1² = (-1)²

=> a-b = 1 hoặc a-b = -1

=> (a-b)²⁰¹⁵ = 1²⁰¹⁵ = 1

(a-b)²⁰¹⁵ = (-1)²⁰¹⁵ = -1

KL:...

a + b = 9 => ( a + b )² = 81

=> a² + 2ab + b² = 81

=> a² + 2.20 + b² = 81

=> a² + b² + 40 = 81

=> a² + b² = 41

Xét ( a - b )² = a² - 2ab + b² = ( a² + b² ) - 2 . 20 = 41 - 40 = 1

=> ( a - b )² = 1

=> a - b = { 1; -1 }

mà a > b => a - b = 1

=> ( a - b )²⁰¹⁵ = 1²⁰¹⁵ = 1

Vậy,......

a<b 

mà

Ta có: a+b=9
=> (a+b)^2=81
=> (a-b)^2 + 4ab =81
=> (a-b)^2=81-4.20
=> (a-b)^2=80-81
=>(a-b)^2=1
=> a-b=1 hoặc a-b=-1
mà a<b nên a-b <0 => a-b=-1
Vậy (a-b)^2015=(-1)^2015=-1

a) ta có : \(\left(a+b\right)^2=a^2+2ab+b^2=a^2-2ab+b^2+4ab\)

\(=\left(a-b\right)^2+4ab\left(đpcm\right)\)

b) ta có : \(a+b=9\Rightarrow\left(a+b\right)^2=9^2=81\)

\(\Leftrightarrow a^2+2ab+b^2=81\Leftrightarrow a^2-2ab+b^2+4ab=81\)

\(\Leftrightarrow\left(a-b\right)^2+4ab=81\) ............................................(1)

thay \(ab=20\) vào (1)

ta có (1) \(\Leftrightarrow\left(a-b\right)^2+4\left(20\right)=81\)

\(\Leftrightarrow\left(a-b\right)^2+80=81\Leftrightarrow\left(a-b\right)^2=81-80=1\)

\(\Leftrightarrow\left\{{}\begin{matrix}a-b=1\\a-b=-1\end{matrix}\right.\)

th1: \(a-b=1\Rightarrow\left(a-b\right)^{2015}=\left(1\right)^{2015}=1\)

th2: \(a-b=-1\Rightarrow\left(a-b\right)^{2015}=\left(-1\right)^{2015}=-1\)

vậy \(\left(a-b\right)^{2005}=\pm1\)

a) \(\left(a-b\right)^2=\left(a+b\right)^2-4ab\)

\(=7^2-4.12=49-48=1\)

b(\(\left(a+b\right)^2=\left(a-b\right)^2+4ab\)

\(=20^2+4.3=400+12=412\)

Cm: a, Ta có:

(a+b)² = a² + 2ab +b² (1)

(a-b)² + 4ab = a² - 2ab +b² + 4ab = a² + 2ab +b² ( 2)

Từ (1), (2) => đpcm

b. Ta có 

(a-b)² = a² - 2ab +b²  (3)

(a+b)² - 4ab = a² + 2ab +b² - 4ab = a² - 2ab +b² (4)

Từ (3),(4)=> đpcm

Áp dụng tính chất:

a, (a-b)² = (a+b)² - 4ab = 7² -4.12 = 1

b,(a+b)² = (a-b)² + 4ab = 20² + 4.3 = 412

Chúc bn hc tốt!

4a) \(\left(a+b\right)^2=a^2+2ab+b^2\)

\(\left(a-b\right)^2+4ab=a^2-2ab+b^2+4ab=a^2+b^2+2ab\)

=> (a+b)^2=(a-b)^2+4ab

(x – 3)(2x + 1) = 0

x = 3 hay x = -1/2