x=-1

x=3

x=-1

x=3

sk ưell

\(a,4\left(2-x\right)^2+xy-2y\)

\(=4\left(2-x\right)^2-y\left(2-x\right)\)

\(=4-y\left(2-x\right)^2\left(2-x\right)\)

\(=\left(2-x\right)\left[\left(2-x\right)4-y\right]\)

\(=\left(2-x\right)\left(4x-8+y\right)\)

\(c,x^3+y^3+z^3-3xyz\)

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

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

\(=\left(x+y\right)^3-3xy\left(x+1\right)+z^3-3xyz\)

\(=\left[\left(x+y\right)^3+z^3\right]-3xy\left(x+y\right)-3xyz\)

\(=\left[\left(x+y\right)+z\right]\left[\left(x+y\right)^2-\left(x+y\right)z+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-xz-yz+z^2\right)-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+y^2+z^2-xy-xz-yz\right)\)

a) 4(2 - x)² + xy - 2y = 4(x - 2)² + y(x - 2) = (4x - 8 + y)(x - 2)

b) 2(x - 1)³ - 5(x - 1)² - (x - 1) = (x - 1)[2(x - 1)² - 5(x - 1) - 1]

= (x - 1)(2x² - 4x + 2 - 5x + 5 - 1) = (x - 1)(2x² - 9x + 6)

c) x³ + y³ + z³ - 3xyz = (x + y)(x² - xy + y²) + z³ - 3xyz

= (x + y)³ + z³ - 3xy(x + y) - 3xyz = (x + y + z)(x² + 2xy + y² - xz - yz + z²) - 3xy(x + y + z)

= (x + y + z)(x² + y² + z² - xz - yz + 2xy - 3xy) = (x + y + z)(x² + y² + z² - xy - yz - xz)

\(2x\left(12x-5\right)-8x\left(3x-1\right)=30\)

\(\Leftrightarrow24x^2-10x-24x^2+8x=30\)

\(\Leftrightarrow-2x=30\)

\(\Leftrightarrow x=-15\)

<br class="Apple-interchange-newline"><div></div>2x(12x−5)−8x(3x−1)=30

<=>\(24x^2-10x-24x^2+8x=30\)

<=>\(-2x=30\)

<=>\(x=30:\left(-2\right)\)

<=>\(x=-15\)

\(\frac{a+b}{a-b}.\frac{b+c}{b-c}+\frac{b+c}{b-c}.\frac{c+a}{c-a}+\frac{c+a}{c-a}.\frac{a+b}{a-b}\)\(=\frac{\left(a+b\right)\left(b+c\right)\left(c-a\right)+\left(b+c\right)\left(c+a\right)\left(a-b\right)+\left(c+a\right)\left(a+b\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=\frac{\left(b^2+ab+bc+ca\right)\left(c-a\right)+\left(c^2+ab+bc+ca\right)\left(a-b\right)+\left(a^2+ab+bc+ca\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=\frac{\left(b^2c+bc^2+c^2a-ab^2-a^2b-ca^2\right)+\left(c^2a+a^2b+ca^2-bc^2-ab^2-b^2c\right)+\left(a^2b+ab^2+b^2c-ca^2-bc^2-c^2a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=\frac{\left(a^2b-ca^2\right)+\left(b^2c-bc^2\right)-\left(ab^2-c^2a\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(=\frac{a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b+c\right)\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{\left(b-c\right)\left(a^2+bc-ab-ac\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)\(=\frac{\left(b-c\right)\left[a\left(a-b\right)-c\left(a-b\right)\right]}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=\frac{\left(a-b\right)\left(b-c\right)\left(a-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=-1\)

abc  bnj k

    \(\left(\frac{75}{100}-\frac{60}{100}+\frac{3}{7}+\frac{3}{11}\right)+\left(\frac{275}{100}-\frac{220}{100}+\frac{11}{7}+\frac{11}{13}\right)\)\(=\frac{1311}{1540}+5\frac{331}{420}\)\(=6\frac{211}{330}\)

HỌC TỐT!!!

\(\left(\frac{75}{100}-\frac{60}{100}+\frac{3}{7}+\frac{3}{11}\right)+\left(\frac{275}{100}-\frac{220}{100}+\frac{11}{7}+\frac{11}{13}\right)\)

\(=\frac{3}{20}+\frac{3}{7}+\frac{3}{11}+\frac{11}{20}+\frac{11}{7}+\frac{11}{13}\)

\(=\frac{14}{20}+\frac{14}{7}+\frac{160}{143}=\frac{3231}{715}+\frac{14}{7}=\frac{4661}{715}\)

2x-3(x-1)-5(x-4(3-2x)+10)(-2x)

=2x-3x+3-5(x-12+8x+10)(-2x)

=-x+3-5(-2x²+24x-16x²-20x)

=-x+3-(-10x²+120x-80x²-100x)

=-x+3+10x²-120x+80x²+100x

=90x²-21x+3

2x - 3( x - 1 ) - 5[ x - 4( 3 - 2x ) + 10 ].(-2x)

= 2x - 3x + 3 + 10x[ x - 12 + 8x + 10 ]

= -x + 3 + 10x[ 9x - 2 ]

= -x + 3 + 90x² - 20x

= 90x² - 21x + 3 

Gọi ba số tự nhiên liên tiếp là n,n + 1,n + 2

Mà tích của hai số đầu nhỏ hơn tích của hai số sau là 18 đơn vị => (n + 1)(n + 2) - n(n + 1) = 18

=> n(n + 2)  + 1(n + 2) - n(n + 1) = 18

=> n² + 2n + n + 2 - n² - n = 18

=> (n² - n²) + (2n + n - n) + 2 = 18

=> 2n +2  = 18

=> 2n = 16

=> n = 8

+) n + 1 = 8 + 1 = 9

+) n + 2 = 8 + 2 = 10

Vậy ba số tự nhiên liên tiếp là 8,9,10

Gọi x - 1 là số thứ nhất 

x là số thứ hai 

x + 1 la số thứ ba 

Theo đề , ta có : 

\(\left(x-1\right)x+18=x\left(x+1\right)\) 

\(x^2-x+18=x^2+x\) 

\(2x=18\) 

\(x=9\) 

Vậy số thứ nhất là 9 - 1 = 8

Số thứ hai là 9 

Số thứ ba là 9 + 1 =10 

3xⁿ(4x^{n - 1} - 1) - 2x^{n + 1} (6x^{n - 2} - 1)

= 12x^{2n - 1} - 3xⁿ - 12x^{2n - 1} + 2x^{n + 1}

= 2x^{n + 1} - 3xⁿ

= xⁿ(2x - 3)

3xn(4xn-1-1)-2xn+1(6xn-2-1)
= 12x^2n^2 - 6xn - 2xn + 6xn - 3
= 12x^2n^2 - 2xn - 3

5ⁿ⁺¹-4.5ⁿ

=5n.5-4.5n

=5ⁿ

\(5^{n+1}-4.5^n=5^n.5-4.5^n=5^n.\left(5-4\right)=5^n.1=5^n\)