\(A=x\left(2x+3\right)-4\left(x+1\right)-2x\left(x-\frac{1}{2}\right)\)

\(=2x^2+3x-4x-4-2x^2+x\)

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

\(=-4\)

\(\left(2x^3-3xy+12x\right)\left(-\frac{1}{6}xy\right)\)

\(=-\frac{2}{6}x^3.xy+\frac{3}{6}xy.xy-\frac{12}{6}x.xy\)

\(=-\frac{1}{3}x^4y+\frac{1}{2}x^2y^2-2x^2y\)

BÀi 1

Ta có A = x( 2x + 3 ) - 4( x + 1 ) - 2x( x - 1/2 ) = ( 2x.x + 3.x ) - ( 4.x + 4.1 ) - ( 2x.x - 1/2.2x )

= 2x² + 3x - 4x - 4 - 2x² + x

= - 4.

Chọn đáp án C

Bài 2

Ta có: ( 2x³ - 3xy + 12x )( - 1/6xy ) = ( - 1/6xy ).2x³ - 3xy( - 1/6xy ) + 12x( - 1/6xy )

= - 1/3x⁴y + 1/2x²y² - 2x²y

Chọn đáp án D

Hok tốt

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 \(a^2-b^2=4c^2\)

$\mathrm{(5}a-3b+8c)(5a-3b-8c)$

$={(5a-3b)}^2-{\left(8c\right)}^2$

$={(5a-3b)}^2-16.4c^2$

Thay $a^2-b^2=4c^2$ ta có :

$=25a^2-30ab+9b^2-16(a^2-b^2)$

$=25a^2-30ab+9b^2-16a^2+16b^2$

$=9a^2-30ab+25b^2$

$={(3a-5b)}^2\left(\mathrm{đp}cm\right)$

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

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

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

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

Đến đây thấy pt vô nghiệm ._.

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

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

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

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

\(\Leftrightarrow x\le1\)

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

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

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

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

\(\Leftrightarrow3x-2< 0\)

\(\Leftrightarrow3x< 2\)

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