,(3x-1) mũ 2=9/16
<=> (3x-1)^2 = ( ±3/4)^2
<=> l3x-1l  = 3/4
Hoặc 3x-1 = 3/4
 <=> 3x= 3/4 + 1
<=> x = 7/4 : 3
<=> x= 7/1

a,676 phần 9

b,Ko bt,sao lại có y?

\(a)\)

\(21\left(x+3\right)^3:\left(3x+9\right)^2\)

\(=[21\left(x+3\right)^3]:[3^2\left(x+3\right)^2]\)

\(=7\left(x+3\right):3\)

Thay vào ta được: \(7.\frac{\left(-6+3\right)}{3}=7.\left(-3\right):3=-7\)

\(b)\)

Thay vào ta được:

\(\left(2.2^2-5.2+3\right)^4:[\left(2.2-3\right)^3:\left(2-1\right)^2]\)

\(=\left(2.4-10+3\right)^4:[\left(4-3\right)^31^2]\)

\(=1^4:\left(1^3.1\right)\)

\(=1:1\)

\(=1\)

\(c)\)

Thay vào ta được:

\(36.10^4.7^3:\left(-6.10^3.7^2\right)\)

\(=-6.10.7\)

\(=-420\)

a, \(\left(2x-1\right)^2-\left(3x-1\right)^2=\left(2x-1-3x+1\right)\left(2x-1+3x-1\right)=-x\left(5x-2\right)\)

b, \(\left(x+1\right)^2-9=\left(x+1-3\right)\left(x+1+3\right)=\left(x-2\right)\left(x+4\right)\)

c, \(\left(4x-1\right)^2-9x^2=\left(4x-1-3x\right)\left(4x-1+3x\right)=\left(x-1\right)\left(7x-1\right)\)

d, \(x^2-9=\left(x-3\right)\left(x+3\right)\); e, \(x^2-25=\left(x-5\right)\left(x+5\right)\)

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

i, \(x^6-y^4=\left(x^3\right)^2-\left(y^2\right)^2=\left(x^3-y^2\right)\left(x^3+y^2\right)\)

có cậu đấy Đỗ Thùy Dương

Đỗ Thuỳ Dương có cần mik cho bn một vé báo cáo ko

(dak bủh bủn lmao lmao ) đoạn này dành cho Đào Phúc Khánh

a, \(x^2-2.\frac{1}{3}x+\frac{1}{9}=\left(x-\frac{1}{3}\right)^2\)

Thay x = 9 vào ta được : \(=\left(9-\frac{1}{3}\right)^2=\left(\frac{26}{3}\right)^2=\frac{676}{9}\)

\(x^2-\frac{2}{3}x+\frac{1}{9}\)

Thay \(x=9\) và ta được:

\(9^2-\frac{2}{3}9+\frac{1}{9}\)\(=81-6+\frac{1}{9}\)\(=\frac{676}{9}\)

a) ( 5x - y )( 25x² + 5xy + y² ) = ( 5x )³ - y³ = 125x³ - y³

b) ( x - 3 )( x² + 3x + 9 ) - ( 54 + x³ ) = x³ - 3³ - 54 - x³ = -27 - 54 = -81

c) ( 2x + y )( 4x² - 2xy + y² ) - ( 2x - y )( 4x² + 2xy + y² ) = ( 2x )³ + y³ - [ ( 2x )³ - y³ ]= 8x³ + y³ - 8x³ + y³ = 2y³

d) ( x + y )² + ( x - y )² + ( x + y )( x - y ) - 3x² = x² + 2xy + y² + x² - 2xy + y² + x² - y² - 3x² = y²

e) ( x - 3 )³ - ( x - 3 )( x² + 3x + 9 ) + 6( x + 1 )²

= x³ - 9x² + 27x - 27 - ( x³ - 3³ ) + 6( x² + 2x + 1 )

= x³ - 9x² + 27x - 27 - x³ + 27 + 6x² + 12x + 6

= -3x² + 39x + 6

= -3( x² - 13x - 2 )

f) ( x + y )( x² - xy + y² ) + ( x - y )( x² + xy + y² ) - 2x³

= x³ + y³ + x³ - y³ - 2x³

= 0

g) x² + 2x( y + 1 ) + y² + 2y + 1

= x² + 2x( y + 1 ) + ( y² + 2y + 1 )

= x² + 2x( y + 1 ) + ( y + 1 )²

= ( x + y + 1 )²

= [ ( x + y ) + 1 ]²

= ( x + y )² + 2( x + y ) + 1

= x² + 2xy + y² + 2x + 2y + 1

a)\(6x-9-x^2\)

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

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

b)\(x^2+4y^2+4xy\)

\(=\left(x+2y\right)^2\)

c)\(x^2+8x+16\)

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

d)\(9x^2-12xy+4y^2\)

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

e)\(-25x^2y^2+10xy-1\)

\(=-\left(25x^2y^2-10xy+1\right)\)

\(=-\left(5xy-1\right)^2\)

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

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

j)\(x^2+6x+9\)

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

h)\(9x^2-6x+1\)

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

#H

a, 6x - 9 - x² = - x² + 6x - 9 = - (x² - 6x + 9) = - (x - 3)²

b, x² + 4y² + 4xy = x² + 2. x . 2y + (2y)² = (x + 2y)²

c, x² + 8x + 16 = x² + 2 . x . 4 + 4² = (x + 4)²

d, 9x² - 12xy + 4y² = (3x)² - 2 . 3x . 2y + (2y)² = (3x - 2y)²

e, - 25x²y² + 10xy - 1 = - (25x²y² - 10xy + 1) = - [(5xy)² - 2 . 5xy + 1] = - (5xy - 1)²

f, 4x² - 4x + 1 = (2x)² - 2 . 2x + 1 = (2x - 1)²

j, x² + 6x + 9 = x² + 2 . x . 3 + 3² = (x + 3)²

h, 9x² - 6x + 1 = (3x)² - 2 . 3x + 1 = (3x - 1)²