tất cả các số bé kia là mũ nha các bạn(số 2,3 ấy)

1. biến đổi vế trái 

= a²x² + a²y² + b²x² + b²y² 

= (ax -by)² + (bx+ ay)² - 2abxy + 2abxy 

= (ax -by)² + ( bx + ay)² = vế phải( dpcm)

=a, a(b²+c²)+b(a²+c²)+c(a²+b²)+2abc

= ab²+ac²+ba²+bc²+ca²+cb²+2abc

= c²(a+b)+ab(a+b)+c(a²+b²+2ab)

= c²(a+b)+ab(a+b)+c(a+b)²

= (a+b)\(\left[c^2+ab+c\left(a+b\right)\right]\)

= (a+b)(c²+ab+ca+cb)

= (a+b)\(\left[c\left(a+c\right)+b\left(a+c\right)\right]\)

=(a+b)(a+c)(b+c)

b, a(b-c)³+b(c-a)³+c(a-b)³

= a(b-c)³-b\(\left[\left(b-c\right)+\left(a-b\right)\right]\)³+c(a-b)³

= a(b-c)³-b(b-c)³-3b(b-c)²(a-b)-3b(b-c)(a-b)²-b(a-b)³+c(a-b)³

= a(b-c)³-b(b-c)³-3b(b-c)(a-b)(b-c+a-b)-b(a-b)³+c(a-b)³

= a(b-c)³-b(b-c)³-3b(b-c)(a-b)(a-c)-b(a-b)³+c(a-b)³

= (b-c)³(a-b)-3b(b-c)(a-b)(a-c)-(a-b)³(b-c)

= (b-c)(a-b)\(\left[\left(b-c\right)^2-3b\left(a-c\right)-\left(a-b\right)^2\right]\)

=(b-c)(a-b)(b²-2bc+c²-3ab+3bc-a²+2ab-b²)

= (b-c)(a-b)(c²-a²+bc-ab)

= (b-c)(a-b)\(\left[\left(c-a\right)\left(c+a\right)+b\left(c-a\right)\right]\)

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

c, a²b²(a-b)+b²c²(b-c)+c²a²(c-a)

= a²b²(a-b)-b²c²\(\left[\left(a-b\right)+\left(c-a\right)\right]\)+c²a²(c-a)

= a²b²(a-b)-b²c²(a-b)-b²c²(c-a)+c²a²(c-a)

= b²(a-b)(a²-c²)+c²(c-a)(a²-b²)

= b²(a-b)(a-c)(a+c)-c²(a-c)(a-b)(a+b)

= (a-c)(a-b)\(\left[b^2\left(a+c\right)-c^2\left(a+b\right)\right]\)

= (a-c)(a-b)(b²a+b²c-c²a-c²b)

= (a-c)(a-b)\(\left[a\left(b^2-c^2\right)+bc\left(b-c\right)\right]\)

= (a-c)(a-b)\(\left[a\left(b-c\right)\left(b+c\right)+bc\left(b-c\right)\right]\)

= (a-c)(a-b)(b-c)\(\left[a\left(b+c\right)+bc\right]\)

= (a-c)(a-b)(b-c)(ab+ac+bc)

d, a⁴(b-c)+b⁴(c-a)+c⁴(a-b)

= a⁴(b-c)-b⁴[(b-c)+(a-b)]+c⁴(a-b)
= (b-c)(a⁴-b⁴)+(a-b)(c⁴-b⁴)
= (b-c)(a²-b²)(a²+b²)+(a-b)(c²-b²)(c²+b²)
= (b-c)(a-b)(a+b)(a^2+b^2)-(a-b)(b-c)(b+c)(b²+c²)
= (b-c)(a-b)(a³+ab²+ba²+b³-bc²-b³-cb²-c³)

= (b-c)(a-b)(a³+ab²+ba²-bc²-c³-cb²)
= (b-c)(a-b)(a³-c³)+b²(a-c)+b(a²-c²)
= (b-c)(a-b)(a-c)(a²+ac+c²)+b²(a-c)+b(a-c)(a+c)
= (b-c)(a-b)(a-c)(a²+ac+c²+b²+ab+ac)

= (a-b)(b-c)(c-a)(a²+b²+c²+ab+bc+ca)

bạn làm giỏi thế có phương pháo nào ko mách mk

a) VP = a³ + 3a²b + 3ab² + b³ - 3a²b - 3ab² = a³ + b³ ( đpcm )

b) VP = a³ - 3a²b + 3ab² - b³ + 3a²b - 3ab² = a³ - b³ ( đpcm )

Áp dụng

a³ - b³ = a³ - 3a²b + 3ab² - b³ + 3a²b - 3ab² 

            = ( a - b )³ + 3ab( a - b )

Thế ab = 8 ; a - b = 12 ta được

( 12 )³ + 3.8.12 = 1728 + 288 = 2016

Được cái khai triển ... 

a,  \(a^3+b^3=\left(a+b\right)^3-3ab\left(a+b\right)\)

\(VP=a^3+3a^2b+3b^2a+b^3-3a^2b-3ab^2\)

Ta có : \(VP=a^3+b^3\left(đpcm\right)\)

b, \(a^3+b^3=\left(a-b\right)^3+3ab\left(a-b\right)\)

Cách khác : \(\left(a-b\right)^3+3ab\left(a-b\right)=\left(a-b\right)^3+3ab\left(a-b\right)\)

Ta có đpcm 

Ta có : \(a^3-b^3=\left(a-b\right)^3+3ab\left(a-b\right)\)

Thay ab = 8 và a - b = 12 :

\(12^3+3.8.12=2016\)

1a)\(a^2+b^2+1\ge ab+a+b\)

\(\Leftrightarrow2\left(a^2+b^2+1\right)\ge2\left(ab+b+a\right)\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)(luôn đúng)

Dấu "=" xảy ra khi x=y=1

b)\(a^2+b^2+c^2\ge a\left(b+c\right)\)

\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2ac\)

\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+b^2+c^2\ge0\)

\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+b^2+c^2\ge0\)(luôn đúng)

Dấu "=" xảy ra khi a=b=c=0

a.

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

= \(a^3+b^3+a^3-b^3\)

= \(2a^3\)

b.

\(\left(a+b\right)\left[\left(a-b\right)^2+ab\right]\)

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

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

= \(a^3+b^3\)

a.=(a³+b³)+(a³-b³)

=a³+b³+a³-b³

=2a³

\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\frac{a^2}{ab+ac}+\frac{b^2}{bc+ab}+\frac{c^2}{ac+bc}\ge\frac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}\ge\frac{3\left(ab+bc+ac\right)}{2\left(ab+bc+ac\right)}=\frac{3}{2}\)

\("="\Leftrightarrow a=b=c\)

\(\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=\left(\frac{b}{a}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\ge2\sqrt{\frac{ab}{ab}}+2\sqrt{\frac{ac}{ac}}+2\sqrt{\frac{bc}{bc}}=2+2+2=6\)

\("="\Leftrightarrow a=b=c\)

giải sao ra hay vậy bạn ?

a) \(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\dfrac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\dfrac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\dfrac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\dfrac{1}{2}\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)

\(=\dfrac{1}{2}\left(3^{32}-1\right)\left(3^{32}+1\right)\)

\(=\dfrac{1}{2}\left(3^{64}-1\right)\)

\(=\dfrac{3^{64}-1}{2}\)

b) \(\left(a+b+c\right)2+\left(a-b-c\right)2+\left(b-c-a\right)2+\left(c-a-b\right)2\)

\(=2\left[\left(a+b+c\right)+\left(a-b-c\right)+\left(b-c-a\right)+\left(c-a-b\right)\right]\)

\(=2\left(a+b+c+a-b-c+b-c-a+c-a-b\right)\)

\(=2.0\)

\(=0\)

c)\(\left(a+b+c+d\right)2+\left(a+b-c-d\right)2+\left(a+c-b-d\right)2+\left(a+d-b-c\right)2\)

\(=2\left(a+b+c+d+a+b-c-d+a+c-b-d+a+d-b-c\right)\)

\(=2.4a\)

\(=8a\)

Bài 1:
x² + 2xy + 4y² = ( x + 2y )²
\(\Rightarrow\)Đúng

Bài 2
( a + b )² = ( a - b )² + 4ab
Xét VP : ( a - b)² - 4ab = a² - 2ab + b² + 4ab
= a² + 2ab + b² = ( a + b )²
= VT
\(\Rightarrow\)đpcm
( a - b)² = ( a + b )² - 4ab
Xét VP: a² + 2ab + b² -4ab
= a² - 2ab + b² = ( a - b)²
= VT
\(\Rightarrow\)đpcm
Áp dụng:
a) Ta có: ( a - b)² = ( a + b)² - 4ab
Thay a + b = 7 ; ab = 12
\(\Rightarrow\)7² - 4.12 = 49 - 48 = 1
b) Ta có : ( a + b )² = ( a - b)² + 4ab
Thay a - b = 20 ; ab = 3
\(\Rightarrow\) 20² + 4.3 = 400 + 12 = 412

Bài 3:
Ta có: 49x² - 70x + 25
= ( 7x)² - 2.7x.5 + 5²
= (7x - 5 )²
a) Thay x = 5
\(\Rightarrow\) ( 7.5 - 5)² = 30² = 900
b) Thay x = 7
\(\Rightarrow\)( 7 . \(\dfrac{1}{7}\)- 5 )² = 16

Bài 4: Tính
a) ( a + b + c )²
= [ ( a + b ) + c ] ²
= ( a+ b)² + 2.( a + b).c + c²
= a² + 2ab + b² + 2ac + 2bc + c²

b) ( a + b - c)²
= [ a + ( b - c)]²
= a² + 2.a.( b - c) + ( b - c )²
= a² + 2ab - 2ac + b² - 2bc + c²

c) ( a - b - c)²
= [( a - b)-c ]²
= ( a- b)² - 2. ( a - b ).c + c²
= a² - 2ab + b² - 2ac + 2bc + c²