a: \(27^{2-x}< =9\)

=>\(\left(3^3\right)^{2-x}< =3^2\)

=>\(3^{6-3x}< =3^2\)

=>6-3x<=2

=>-3x<=-4

=>\(x>=\dfrac{4}{3}\)

b: \(7^{3-x}< 49\)

=>\(7^{3-x}< 7^2\)

=>3-x<2

=>-x<2-3=-1

=>x>1

c: \(27^{3-x}>9\)

=>\(\left(3^3\right)^{3-x}>3^2\)

=>\(3^{9-3x}>3^2\)

=>9-3x>2

=>-3x>-7

=>\(x< \dfrac{7}{3}\)

d: \(2^{3-x}< 2^3\)

=>3-x<3

=>-x<0

=>x>0

e: \(27^{3-x^2}< 27^{x+1}\)

=>\(3-x^2< x+1\)

=>\(-x^2-x+2< 0\)

=>\(x^2+x-2>0\)

=>(x+2)(x-1)>0

=>\(\left[{}\begin{matrix}x>1\\x< -2\end{matrix}\right.\)

j mà hjhj thế nguyenthuy

j mà j mà hjhj thế nguyenthuy thế trieu dang

1/

\(3\left(-1-4x^2+5x\right)+4\left(3x^2+7x-6\right)=-27\)

\(\Leftrightarrow-3-12x^2+15x+12x^2+28x-24=-27\)

\(\Leftrightarrow43x=0\Rightarrow x=0\)

2/

\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-1\right)=27\)

\(\Leftrightarrow x^3+27-x^3+x=27\)

\(\Leftrightarrow x=0\)

Hướng dẫn giải:

a) 5 x (10 + 9)

Cách 1:

5 x (10 + 9) = 5 x 19 = 95

Cách 2:

5 x (10 + 9) = 5 x 10 + 5 x 9 = 50 + 45 = 95

b) 10 x (30 + 5)

Cách 1:

10 x (30 + 5) = 10 x 35 = 350

Cách 2:

10 x (30 + 5) = 10 x 30 + 10 x 5 = 300 + 50 = 350

Hướng dẫn giải nè :

a) 5 x (10 + 9)

Cách 1:

5 x (10 + 9) = 5 x 19 = 95

Cách 2:

5 x (10 + 9) = 5 x 10 + 5 x 9 = 50 + 45 = 95

b) 10 x (30 + 5)

Cách 1:

10 x (30 + 5) = 10 x 35 = 350

Cách 2:

10 x (30 + 5) = 10 x 30 + 10 x 5 = 300 + 50 = 350

1).( 27,56 x 35 ) + ( 27,56 x 67 ) - ( 27,56 x 2)

= (964 + 1846,52) - 55,12

=2810,52 - 55,12

= 2755,4

2).( 4x 35 ) x ( 25 x 5 ) x 2

= ( 140 x 125 ) x2

= 17500 x 2

=35000

4). 3/10

5). 1188

6).  61/6

1)2756   2)35000   3)0   4)3/10   5)10/1/6

\(\frac{2}{x-3}\)-\(\frac{27}{x^3-27}\)=\(\frac{3}{x^2+3x+9}\)

\(\frac{2}{x-3}\)-\(\frac{27}{\left(x-3\right)\left(x^2+3x+9\right)}\)=\(\frac{3}{x^2+3x+9}\)

\(\frac{2\left(x^2+3x+9\right)}{\left(x-3\right)\left(x^2+3x+9\right)}\)-\(\frac{27}{\left(x-3\right)\left(x^2+3x+9\right)}\)=\(\frac{3\left(x-3\right)}{\left(x-3\right)\left(x^2+3x+9\right)}\)

2x²+6x+18-27=3x-9

2x²+6x-3x=27-18-9

2x²+3x=0

x(2x+3)=0

x=0 hoặc 2x+3=0

x=0 hoặc x=

x^3+27=27

x=0

=>27=9-0^2

=>27=9

=>x không tồn tại

\(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2-9\right)=27\\ x.x^2-x.3x+x.9-x.x^2+x.9=27\\ x^3-3x^2+9x-x^3+9x=27\\ 3x^2+18x=27\\ 21x^2=27\\ x^2=\dfrac{9}{7}\\ \Rightarrow x=\sqrt{\dfrac{9}{7}}\)

Sai rồi