1, \(x^2\left(x-3\right)-4x+12=x^2\left(x-3\right)-4\left(x-3\right)\)

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

2, \(2a\left(x+y\right)-x-y=2a\left(x+y\right)-\left(x+y\right)=\left(2a-1\right)\left(x+y\right)\)

3, \(2x-4+5x^2-10x=2\left(x-2\right)+5x\left(x-2\right)=\left(2+5x\right)\left(x-2\right)\)

4, sửa đề : 

 \(6x^2-12x-7x+14=6x\left(x-2\right)-7\left(x-2\right)=\left(6x-7\right)\left(x-2\right)\)

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

a) x²(x-3)-4x+12

=x²(x-3)-4(x-3)

=(x-3)(x²-4)

=(x-3)(x-2)(x+2)

b) 2a(x+y)-x-y

=2a(x+y)-(x+y)

=(x+y)(2a-1)

c) 2x-4+5x²-10x

=2(x-2)+5x(x-2)

=(x-2)(2+5x)

d) 5x²-12x-7x+14

=5x²-19x+14

e) xy-y²-3x+3y

=y(x-y)-3(x-y)

=(x-y)(y-3)

#H

\(3\left(1-4x\right)\left(x-1\right)+4\left(3x-2\right)\left(x+3\right)=-27\)

\(\Leftrightarrow3\left(-1-4x^2+5x\right)+4\left(3x^2+7x-6\right)=-27\)

\(\Leftrightarrow-12x^2+15x-3+12x^2+28x-24=-27\)

\(\Leftrightarrow43x-27=-27\Leftrightarrow x=0\)

\(3\left(1-4x\right)\left(x-1\right)+4\left(3x-2\right)\left(x+3\right)=-27\)

\(\Leftrightarrow\left(3-12x\right)\left(x-1\right)+\left(12x-8\right)\left(x+3\right)=-27\)

\(\Leftrightarrow3x-3-12x^2+12x+12x^2+36x-8x-24+27=0\)

\(\Leftrightarrow43x=0\)

\(\Leftrightarrow x=0\)

#H

Trả lời:

21, ( a + 1 )³ = a³ + 3a² + 3a + 1

22, ( a + 2 )³ = a³ + 3.a².2 + 3.a.2² + 2³ = a³ + 6a² + 12a + 8

23, ( x + 1 )³ = x³ + 3x² + 3x + 1

24, ( x + 2 )³ = x³ + 3.x².2 + 3.x.2² + 2³ = x³ + 6x² + 12x + 8

25, ( x + 3 )³ = x³ + 3.x².3 + 3.x.3² + 3³ = x³+ 9x² + 27x + 27

26, ( x + 4 )³ = x³ + 3.x².4 + 3.x.4² + 4³ = x³ + 12x² + 48x + 64

27, ( 2x + 3y )³ = (2x)³ + 3.(2x)².3y + 3.2x.(3y)² + (3y)³ = 8x³ + 36x²y + 54xy² + 27x³

28, ( 3x + 1 )³ = (3x)³ + 3.(3x)².1 + 3.3x.1² + 1³ = 27x³ + 27x² + 9x + 1

29, ( 5x + 1 )³ = (5x)³ + 3.(5x)².1 + 3.5x.1² + 1³ = 125x³ + 75x² + 15x + 1

30, ( 4x + 2y )³ = (4x)³ + 3.(4x)².2y + 3.4x.(2y)² + (2y)³ = 64x³ + 96x²y + 48xy² + 8y³

31, ( a - 1 )³ = a³ - 3a² + 3a - 1

32, ( a - 2 )³ = a³ - 3.a².2 + 3.a.2² - 2³ = a³ - 6a² + 12a - 8

33, ( x - 3 )³ = x³ - 3.x².3 + 3.x.3² - 3³ = x³ - 9x² + 27x - 27

34, ( x - 4 )³ = x³ - 3.x².4 + 3.x.4² - 4³ = x³ - 12x² + 48x - 64

35, ( x - 5 )³ = x³ - 3.x².5 + 3.x.5² - 5³ = x³ - 15x² + 75x - 125

36, ( 2x - 3y )³ = (2x)³ - 3.(2x)².3y + 3.2x.(3y)² - (3y)³ = 8x³ - 36x² + 54x - 27x³

37, ( x² - 2y )³ = (x²)³ - 3.(x²)².2y + 3.x².(2y)² - (2y)³ = x⁶ - 6x⁴y + 12x²y² - 8y³

38, ( 2x² - 3y )³ = (2x²)³ - 3.(2x²)².3y + 3.2x².(3y)² - (3y)³ = 8x⁶ - 36x⁴y + 54x²y² - 27y³

39, ( 4 - 4y² )³ = 4³ - 3.4².4y² + 3.4.(4y²)² - (4y²)³ = 64 - 48y² + 192y⁴ - 64y⁶

40, ( 3x² - 2y² )³ = (3x²)³ - 3.(3x²)².2y² + 3.3x².(2y²)² - (2y²)³ = 27x⁶ - 54x⁴y² + 36x²y⁴ - 8y⁶ 

a, \(a^4+64=\left(a^2\right)^2+8^2=\left(a^2\right)^2+16a^2+8^2-16a^2\)

\(=\left(a^2+8\right)^2-\left(4a\right)^2=\left(a^2-4a+8\right)\left(a^2+4a+8\right)\)

b, \(a^4+4b^2=\left(a^2\right)^2+\left(2b\right)^2=\left(a^2\right)^2+4a^2b+4b^2-4a^2b\)ĐK : b >= 0 

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

p/s : mình nghĩ phần b bạn nên để đề là \(a^4+4b^4\)sẽ hợp lí hơn 

Anh nhận bú lồn hoàn toàn miễn phí nha mấy em

nhiều vậy,chờ mình xíu

câu nào ạ

sửa đề : bài này chỉ có max thôi nhé 

\(B=-x^2+6x-7=-\left(x^2-3.2x+9-2\right)\)

\(=-\left(x-3\right)^2+2\le2\)

Dấu ''='' xảy ra khi x = 3 

Vậy GTLN B là 2 khi x = 3 

Trả lời:

B = - x² + 6x - 7 = - ( x² - 6x + 7 ) = - ( x² - 2.x.3 + 9 - 2 ) = - [ ( x - 3 )² - 2 ] = - ( x - 3 )² + 2 \(\le2\forall x\)

Dấu "=" xảy ra khi x - 3 = 0 <=> x = 3

Vậy GTLN của B = 2 khi x = 3

Bài 4 : 

a, ĐK : \(x\ne\pm\sqrt{2}\)

\(\frac{3x+6}{x^2-2}=\frac{3\left(x+2\right)}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}\)

b, Ta có : \(\frac{3\left(x+2\right)}{x^2-2}=-1\Rightarrow3x+6=-x^2+2x\Leftrightarrow x^2+x+6=0\)

\(\Leftrightarrow x^2+x+\frac{1}{4}+\frac{23}{4}=\left(x+\frac{1}{2}\right)^2+\frac{23}{4}>0\)

Vậy pt vô nghiệm 

Bài 5 : 

a, Xét tam giác HBA và tam giác ABC ta có : 

^B _ chung 

^AHB = ^CAB = 90⁰

Vậy tam giác HBA ~ tam giác ABC ( g.g ) 

\(\frac{AB}{BC}=\frac{AH}{AC}\)(*) ; \(\frac{BH}{AB}=\frac{AB}{BC}\)(**)

Áp dụng định lí Pytago tam giác ABC vuông tại A

\(BC^2=AB^2+AC^2=256+144=400\Rightarrow BC=20\)cm 

Từ (*) suy ra : \(AH=\frac{AB.AC}{BC}=\frac{12.16}{20}=\frac{48}{5}\)cm 

Từ (**) suy ra : \(BH=\frac{AB^2}{BC}=\frac{256}{20}=\frac{64}{5}\)cm 

b, Vì AD là đường phân giác : \(\frac{AB}{AC}=\frac{BD}{CD}\Rightarrow\frac{CD}{AC}=\frac{BD}{AB}\)

Áp dụng tính chất dãy tỉ số bằng nhau ta có : 

\(\frac{CD}{AC}=\frac{BD}{AB}=\frac{BD+CD}{AC+AB}=\frac{BC}{AC+AB}=\frac{20}{12+16}=\frac{20}{28}=\frac{5}{7}\)

\(CD=\frac{5}{7}.12=\frac{60}{7}cm;BD=\frac{5}{7}.16=\frac{80}{7}cm\)

Ta có 2x³ + 21x²+ 67x + 60 

= 2x³ + 10x² + 11x² + 55x + 12x + 60

= 2x²(x + 5) + 11x(x + 5) + 12(x + 5) 

 = (x + 5)(2x² + 11x + 12) 

Khi đó : ( 2x³ + 21x²+ 67x + 60) : (x + 5)

\(\Leftrightarrow\)(x + 5)(2x² + 11x + 12) : (x + 5) = 2x² + 11x + 12