Bài 3: Những hằng đẳng thức đáng nhớ

Bài 1 :

a) \(2.\left(x-1\right)^2-4.\left(3x+x\right)^2+2x.\left(x-5\right)\)

\(2.\left(\left(x-1\right)^2-2\left(3x+x\right)^2+x\left(x-5\right)\right)\)

\(2.\left(\left(x-1\right)^2-2.\left(4x\right)^2+x^2-5x\right)\)

\(2.\left(\left(x-1\right)^2-\left(2^2x\right)^2+x^2-5x\right)\)

\(2.\left(\left(x-1\right)^2-2.2^4x^2+x^2-5x\right)\)

\(2.\left(x-1\right)^2-2^5x^2+x^2-5x\)

\(2.\left(\left(x-1\right)^2-32x^2+x^2-5x\right)\)

\(2.\left(x^2-2x+1-31x^2-5x\right)\)

\(2.\left(-30x^2-7x+1\right)\)

b) \(2.\left(2x+5\right)^2-3\left(4x+1\right)\left(1-4x\right)\)

\(2.\left(4x^2+20x+25\right)-3\left(1+4x\right)\left(1-4x\right)\)

\(8x^2+40x+50-3\left(1-16x^2\right)\)

\(8x^2+40x+50-3+48x^2\)

\(56x^2+40x+47\)

c) \(4\left(x+1\right)^2+\left(2x-1\right)^2-8\left(x-1\right)\left(x+1\right)\)

\(4.\left(x^2+2x+1\right)+4x^2-4x+1-8\left(x^2-1\right)\)

\(4x^2+8x+4+4x^2-4x+1-8x^2+8\)

\(0+4x+13\)

\(4x+13\)

Thì bạn học thuộc rồi cứ thay số vào rồi tính thôi. Tui thấy cũng khá dễ mà.

a)9+6xy^2+x^2y^4

b)100-400m^2n+160m^4n^2

c)a^2-b^4

Có (a+b+c)² = 3(ab+bc+ac)

\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac=3ab+3bc+3ac\)

\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac-3ab-3bc-3ac\)\(=0\)

\(\Rightarrow a^2+b^2+c^2-ab-bc-ac\)\(=0\)

\(\Rightarrow a^2+b^2+c^2=ab+bc+ac\)

\(\Rightarrow2a^2+2b^2+2c^2=2ab+2bc+2ac\)

\(\Rightarrow a^2-2ab+b^2+b^2-2bc+c^2+a^2-2ac+c^2\)\(=0\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2=0\)

\(\Rightarrow a=b=c\)

\(P=9x^2-8x-5\)

\(P=9x^2-8x+\dfrac{16}{9}-\dfrac{61}{9}\)

\(P=\left(3x-\dfrac{4}{3}\right)^2-\dfrac{61}{9}\)

Xét :

\(\left(3x-\dfrac{4}{3}\right)^2\ge0\)

\(\Leftrightarrow\left(3x-\dfrac{4}{3}\right)^2-\dfrac{61}{9}\ge\dfrac{-61}{9}\)

Vậy GTNN của \(P=-\dfrac{61}{9}\)

Dấu \(=\) xảy ra khi \(x=\dfrac{4}{9}\)

Ta có : 9x² -8x - 5

= ( 3x )² - 2 . 3x . \(\dfrac{4}{3}\) + (\(\dfrac{4}{3}\))² - (\(\dfrac{4}{3}\))² - 5

= ( 3x - \(\dfrac{4}{3}\) )² - \(\dfrac{61}{9}\)

Vì ( 3x - \(\dfrac{4}{3}\) )² >= 0

Vậy GTNN = -61/9

Câu 1:

a,4a²b( 2ab² - 3a²b²)

= 8a³b³ - 12a⁴b³

b, ( x - 4 )( x² + 2x - 5)

= x( x² + 2x - 5) - 4(x² + 2x - 5)

= x³ + 2x² - 5x - 4x² - 8x + 20

= x³ - 2x² - 13x + 20

Câu 2 :

a, 4xy ( 2xy² - 3x²y)

= 8x²y³ - 12x³y²

b,( x + 2 )( 2x² - 3x + 4)

= x( 2x² - 3x + 4) + 2( 2x² - 3x + 4)

= 2x³ - 3x² + 4x + 4x² - 6x + 8

= 2x³ + x² - 2x + 8

Câu 3 :

a, ( x + y )² = x² + 2.x.y + y² = x² + 2xy + y²

b, ( 2m - n )³ = ( 2m)³ - 3.( 2m )².n + 3.2m.n² - n³

= 8m³ - 12m²n + 6mn² - n³

Chúc bạn học tốt

Vì ko có thời gian nên mình chỉ có thể giúp bạn câu 3 thôi nhé mong bạn thông cảm cho minh nha.

a, (x+y)^2=x^2+2*x*y+y^2=x^2+2xy+y^2

b, (2m-n)^3=2m^3-3*2m^2*n+3*2m*n^3-n^3=2m^3-6m^2n+6mn^3-n^3.

Mong bn thông cảm cho mình nha. Chúc bn luôn may mắn.

a: \(9x^2-6x+3\)

\(=\left(9x^2-6x+1\right)+2\)

\(=\left(3x-1\right)^2+2\ge2\)

b: \(6x-x^2+1\)

\(=-\left(x^2-6x-1\right)\)

\(=-\left(x^2-6x+9-10\right)\)

\(=-\left(x-3\right)^2+10\le10\)

a) (a+b)(a+b)² = (a+b)³

= a³+3a²b+3ab²+b³

b) (2x-y)²(2x-y) = (2x-y)³

= (2x)³-3.(2x)².y+3.2x.y²-y³

=8x³-12x²y+6xy²-y³

c) (3x+2y)(3x+2y)² = (3x+2y)³

= (3x)³+3.(3x)².2y+3.3x.(2y)²+(2y)³

= 27x³+54x²y+36xy²+8y³