Vào trang cá nhân của mình đi, có cái này hay lắm, nhớ kb vs mình nha

323 =17.19.

Ta có:  \(20^n+16^n-3^n-1=\left(20^n-3^n\right)+\left(16^n-1\right)\)

\(20^n-3^n⋮17,16^n-1⋮17\)(vì n chẵn)

\(\Rightarrow20^n+16^n-3^n-1⋮17\)(1)

Tương tự:

\(20^n+16^n-3^n-1=\left(20^n-1\right)+\left(16^n-3^n\right)\)

\(20^n-1⋮19,16^n-3^n⋮19\)(vì n chẵn)

\(\Rightarrow20^n+16^n-3^n-1⋮19\)(2)

Từ (1) và (2) \(\Rightarrow20^n+16^n-3^n-1⋮\left(17,19\right)=323\)(đpcm)

(3x - 4)(x - 2) = 3x(x - 9) - 3

=> 3x² - 10x + 8 = 3x² - 27x - 3

=> 27x - 10x = -3 - 8

=> 17x = -11

=> x = -11/17 

\(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)-3\)

\(\Leftrightarrow3x^2-6x-4x+8=3x^2-27x-3\)

\(\Leftrightarrow3x^2-10x+8=3x^2-27x-3\)

\(\Leftrightarrow17x+11=0\)

\(\Leftrightarrow17x=-11\)

\(\Leftrightarrow x=\frac{-11}{17}\)

a, \(A=\left(-5x+4\right)\left(3x-2\right)+\left(-2x+3\right)\left(x-2\right)\)

\(=-15x^2+10x+12x-8=-15x^2+22x-8\)

Thay x = -2 vào biểu thức ta có : \(-15\left(-2\right)^2+22\left(-2\right)-8\)

\(=-15.4-44-8=-112\)

b, \(B=\left(x-9\right)\left(2x+3\right)-2\left(x+7\right)\left(x-5\right)\)

\(=2x^2+3x-18x-27=2x^2-15x-27\)

Thay x = -1/2 vào biểu thức ta có : \(2\left(-\frac{1}{2}\right)^2-15\left(-\frac{1}{2}\right)-27\)

\(=2.\frac{1}{4}+\frac{15}{2}-27=\frac{11}{2}+\frac{15}{2}+27=40\)

Bài làm:

a) \(A=\left(-5x+4\right)\left(3x-2\right)+\left(-2x+3\right)\left(x-2\right)\)

\(A=-15x^2+22x-8-2x^2+7x-6\)

\(A=-17x^2+29x-14\)

Thay x = -2 vào ta được:

\(A=-17.\left(-2\right)^2+29.\left(-2\right)-14\)

\(A=-68-58-14\)

\(A=-140\)

b) \(B=\left(x-9\right)\left(2x+3\right)-2\left(x+7\right)\left(x-5\right)\)

\(B=2x^2-15x-27-2\left(x^2+2x-35\right)\)

\(B=2x^2-15x-27-2x^2-4x+70\)

\(B=-19x+43\)

Thay x = -1/2 vào B ta được:

\(B=-19.\left(-\frac{1}{2}\right)+43=\frac{19}{2}+43=\frac{105}{2}\)

Em lớp 5 thôi nhưng em nghĩ bài này nhân vế bình thường thôi ạ 

Ví dụ câu a : \(\left(2x^2+x-1\right)\left(-3x^2-7x-5\right)\)

\(=-6x^4-17x^3-14x^2+2x+5\)

Tương tự chị nhé 

Câu b : 

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

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

\(=\)\(2x^3-3x^2y+xy^2+2x^2y-3xy^2+y^3\) 

\(\left(5a-3b+8c\right)\left(5a-3b-8c\right)=\left(3a-5b\right)^2\)

\(\Leftrightarrow\left(5a-3b\right)^2-64c^2-\left(3a-5b\right)^2=0\)

\(\Leftrightarrow\left(5a-3b-3a+5b\right)\left(5a-3b+3a-5b\right)=64c^2\)

\(\Leftrightarrow\left(2a+2b\right)\left(8a-8b\right)=16\left(a^2-b^2\right)\)

\(\Leftrightarrow16\left(a^2-b^2\right)=16\left(a^2-b^2\right)\left(true\right)\)

Vậy \(\left(5a-3b+8c\right)\left(5a-3b-8c\right)=\left(3a-5b\right)^2\)khi \(a^2-b^2=4c^2\)

\(x^2+2y^2+4x-4y-2xy+5=0\)

\(\Leftrightarrow\left(x^2-2xy+y^2\right)+4\left(x-y\right)+4+y^2+1=0\)

\(\Leftrightarrow\left(x-y\right)^2+4\left(x-y\right)+4+y^2+1=0\)

\(\Leftrightarrow\left(x-y+2\right)^2+y^2+1=0\)

Đến đây thấy pt vô nghiệm ._.

\(1-2\left(x+1\right)\ge5\left(x-2\right)+2\)

\(\Leftrightarrow1-2x-2\ge5x-10+2\)

\(\Leftrightarrow-2x-5x\ge-10+2-1+2\)

\(\Leftrightarrow-7x\ge-7\)

\(\Leftrightarrow x\le1\)

\(\frac{3x+3}{3x-2}< 1\)

\(\Leftrightarrow\frac{3x-2+5}{3x-2}< 1\)

\(\Leftrightarrow1+\frac{5}{3x-2}< 1\)

\(\Leftrightarrow\frac{5}{3x-2}< 0\)

\(\Leftrightarrow3x-2< 0\)

\(\Leftrightarrow3x< 2\)

\(\Leftrightarrow x< \frac{2}{3}\)

1. -4x( x + 3 )( x - 4 ) - 3x( x² - x + 1 )

= -4x( x² - x - 12 ) - 3x( x² - x + 1 )

= -4x³ + 4x² + 48x - 3x³ + 3x² - 3x

= -7x³ + 7x² + 45x

2. a) 4x( x - 5 ) - ( x - 1 )( 4x - 3 ) = 5

<=> 4x² - 20x - ( 4x² - 7x + 3 ) = 5

<=> 4x² - 20x - 4x² + 7x - 3 = 5

<=> -13x - 3 = 5

<=> -13x = 8

<=> x = -8/13

b) 6( x - 3 )( x - 4 ) - 6x( x - 2 ) = 4

<=> 6( x² - 7x + 12 ) - 6x² + 12x = 4

<=> 6x² - 42x + 72 - 6x² + 12x = 4

<=> -30x + 72 = 4

<=> -30x = -68

<=> x = 34/15

Bài 1 : 

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

\(=-7x^3+7x^2+45x\)

Bài 2 : 

a, \(4x\left(x-5\right)-\left(x-1\right)\left(4x-3\right)=5\)

\(\Leftrightarrow4x^2-20x-\left[4x^2-7x+3\right]=5\)

\(\Leftrightarrow4x^2-20x-4x^2+7x-3=5\)

\(\Leftrightarrow-13x-8=0\Leftrightarrow x=-\frac{8}{13}\)

b, \(6\left(x-3\right)\left(x-4\right)-6x\left(x-2\right)=4\)

\(\Leftrightarrow6x^2-42x+72-6x^2+12x=4\)

\(\Leftrightarrow-30x+68=0\Leftrightarrow x=\frac{34}{15}\)