\(\dfrac{16}{27}xX=\dfrac{10}{27}-\dfrac{6}{27}\)

\(\dfrac{16}{27}xX=\dfrac{4}{27}\)

\(X=\dfrac{4}{27}:\dfrac{16}{27}\)

X=\(\dfrac{1}{4}\)

16/27x X+2/9=10/27

16/27x X=10/27-2/9

16/27x X=4/27

X=4/27:16/27

X=1/4

x = 1/4

a)x⁷+x⁵+1=x⁷+x⁶-x⁶+2x⁵-x⁵+x⁴-x⁴+x³-x³+x²-x²+1

=x⁷-x⁶+x⁵-x³+x²+x⁶-x⁵+x⁴-x²+x+x⁵-x⁴+x³-x+1

=x²(x⁵-x⁴+x³-x+1)+x(x⁵-x⁴+x³-x+1)+1(x⁵-x⁴+x³-x+1)

=(x²+x+1)(x⁵-x⁴+x³-x+1)

b)4x⁴-32x²+1=4x⁴+12x³+2x²-12x³-36x²-6x+2x²+6x+1

=2x²(2x²+6x+1)-6x(2x²+6x+1)+1(2x²+6x+1)

=(2x²-6x+1)(2x²+6x+1)

c)x⁶+27=(x²+3)(x²-3x+3)(x²+3x+3)

d)3(x⁴+x²+1)-(x²+x+1)

=3x⁴-3x³+2x²+3x³-3x²+2x+3x²-3x+2

=x²(3x²-3x+2)+x(3x²-3x+2)+1(3x²-3x+2)

=(x²+x+1)(3x²-3x+2)

e)bạn tự làm nhé

= \(\dfrac{19}{27}\)

\(\dfrac{1}{4}\)+\(\dfrac{9}{20}\)

=\(\dfrac{5}{20}\)+\(\dfrac{9}{20}\)=\(\dfrac{5+9}{20}\)=\(\dfrac{14}{20}\)=\(\dfrac{7}{5}\)

\(\dfrac{1}{4}+\dfrac{9}{20}=\dfrac{5}{20}+\dfrac{9}{20}=\dfrac{14}{20}=\dfrac{7}{10}\)

1/4 + 9/20 = 5/20 + 9/20 = 14/20

hình như đề sai vì x + (-2/3) không thể bằng x + 1/4 đúng k nhỉ :)

Ko nhưng đề được giao nó là như thế còn mình chịu

2x³ - 32x = 0

=> 2x.(x² - 16) = 0

=> 2x = 0 hoặc x² - 16 = 0

=> x = 0 hoặc x² = 16

=> x = 0 hoặc x thuộc {4 ; -4}

Vậy x thuộc {0 ; 4 ; -4}

2x³-32x=0

2x(x²-16)=0

2x=0 hoặc x²-16=0

x=0 hoặc x²=16

x=0 hoặc x=-4 hoặc x=4

\(2x^3-32x=0\)

=>\(2x\left(x^2-16\right)=0\)

=>\(2x\left(x-4\right)\left(x+4\right)=0\)

=>x=0 hoặc x-4=0 hoặc x+4=0

=>x=0 hoặc x=4 hoặc x=-4

32x+9+7=11

<=> 32x=11-9-7=-5

<=> x=-5/32

32x + 9 + 7 = 11

32x = 11 - 9 - 7

32x = -5

=> x = -5/32

\(32x+9+7=11\)

\(\Leftrightarrow32x=11-9-7=-5\)

\(\Leftrightarrow x=-\frac{5}{32}\)

-1 . \(\dfrac{5}{27}\) - \(\left(3x.\dfrac{7}{9^{ }}\right)^3\) = -\(\dfrac{24}{27}\)

\(\Rightarrow\) \(\dfrac{-5}{27}\) - \(\left(3x.\dfrac{7}{9^{ }}\right)^3\) = \(\dfrac{-24}{27}\)

\(\Rightarrow\) \(\left(3x.\dfrac{7}{9^{ }}\right)^3\) = \(\dfrac{-5}{27}\) - \(\dfrac{-24}{27}\) = \(\dfrac{19}{27}\)

\(\Rightarrow\) \(3^3\) . \(\dfrac{7^3}{3^6}\) = \(\dfrac{19}{3^3}\)

\(\Rightarrow\) \(\dfrac{x^3.7^3}{3^3}\) = \(\dfrac{19}{3^3}\)

\(\Rightarrow\) \(x^3\) . \(7^3\) = 19

( 1 + 2 + 3 + ... + 8 + 9 + 10 ) x ( 6 x 8 - 48 )

= ( 1 + 2 + 3 + ... + 8 + 9 + 10 ) x ( 48 - 48 )

= ( 1 + 2 + 3 + ... + 8 + 9 + 10 ) x 0 = 0

=(1+2+3+.....+8+9+10) x (48-48)

=(1+2+3+.....+8+9+10) x 0

= 0

  (1 + 2 + 3 + ... + 8 + 9 + 10) x (6 x 8 - 48)

= (1 + 2 + 3 + ... + 8 + 9 + 10) x (48 - 48)

=(1 + 2 + 3 + ... + 8 + 9 + 10) x 0

= 0