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

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

2(x-5)+3=-7

=>2x-10=-7+3=-4

=>2x=-4+10=6

=>x=6:2=3

4x+(27+2^3)=1

=>4x+(27+8)=1

=>4x+35=1

=>4x=1-35=-34

=>x=-34:4=8.5

lx-3l+5=-27

=>lx-3l=-27-5=-32

mà lx-3l>0 với mọi x

=>k co gt x t/m

-3lx-1l=-27

=>lx-1l=-27:(-3)=9

=>x-1=+9

+)x-1=9=>x=10

+)x-1=-9=>x=-8

27*x là j z bn?Tran Thuy vy

\(=27x^3+27-27x^3+1=28\)

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/

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

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

\(\Leftrightarrow x=0\)

\(1-27x^3\)

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

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

\(---\)

\(x-3^3+27\)

\(=x-27+27=x\) 

\(---\)

\(27x^3+27x^2+9x+1\)

\(=\left(3x\right)^3+3\cdot\left(3x\right)^2\cdot1+3\cdot3x\cdot1^2+1^3\)

\(=\left(3x+1\right)^3\)

\(---\)

\(\dfrac{x^6}{27}-\dfrac{x^4y}{3}+x^2y^2-y^3\) (sửa đề)

\(=\left(\dfrac{x^2}{3}\right)^3-3\cdot\left(\dfrac{x^2}{3}\right)^2\cdot y+3\cdot\dfrac{x^2}{3}\cdot y^2-y^3\)

 \(=\left(\dfrac{x^2}{3}-y\right)^3\)

#Ayumu

1-27x\(^3\)

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

Bạn ơi phần b sai đề rồi, phải là 27 chứ.

phần b nhé

(X+0,7)³=-27

(X+0,7)³=(-3)³

^X+0,7=-3

X=3-0,7=2,3

Vậy...

Ta có: 

\(C=1-x-\frac{x^2}{3}-\frac{x^3}{27}\)

\(C=1-\left(-27\right)-\frac{\left(-27\right)^2}{3}-\frac{\left(-27\right)^3}{27}\) tại x = -27

\(C=1+27-\frac{3^6}{3}+27^2\)

\(C=1+27-243+729\)

\(C=514\)

\(\left(3-x\right)^3=-\dfrac{27}{64}\)

\(\left(3-x\right)^3=\left(\dfrac{-3}{4}\right)^3\)

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

\(x=3-\dfrac{-3}{4}=\dfrac{12}{4}+\dfrac{3}{4}\)

\(x=\dfrac{15}{4}\)

________

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

\(\left(x-5\right)^3=\left(\dfrac{-1}{3}\right)^3\)

\(=>x-5=\dfrac{-1}{3}\)

\(x=\dfrac{-1}{3}+5=\dfrac{-1}{3}+\dfrac{15}{3}\)

\(x=\dfrac{14}{3}\)

_____________

\(\left(x-\dfrac{1}{2}\right)^3=\dfrac{27}{8}\)

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

\(=>x-\dfrac{1}{2}=\dfrac{3}{2}\)

\(x=\dfrac{3}{2}+\dfrac{1}{2}\)

\(x=2\)

________

\(\left(2x-1\right)^2=\dfrac{1}{4}\)            

\(\left(2x-1\right)^2=\left(\dfrac{1}{2}\right)^2\)           hoặc              \(\left(2x-1\right)^2=\left(\dfrac{-1}{2}\right)^2\)

\(=>2x-1=\dfrac{1}{2}\)                                       \(2x-1=\dfrac{-1}{2}\)

\(2x=\dfrac{1}{2}+1=\dfrac{1}{2}+\dfrac{2}{2}\)                               \(2x=\dfrac{-1}{2}+1=\dfrac{-1}{2}+\dfrac{2}{2}\)

\(2x=\dfrac{3}{2}\)                                                     \(2x=\dfrac{1}{2}\)

\(x=\dfrac{3}{2}:2=\dfrac{3}{2}.\dfrac{1}{2}\)                                     \(x=\dfrac{1}{2}:2=\dfrac{1}{2}.\dfrac{1}{2}\)

\(x=\dfrac{3}{4}\)                                                       \(x=\dfrac{1}{4}\)

____________

\(\left(2-3x\right)^2=\dfrac{9}{4}\)

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

\(=>2-3x=\dfrac{3}{2}\)                                               \(2-3x=\dfrac{-3}{2}\)

\(3x=2-\dfrac{3}{2}=\dfrac{4}{2}-\dfrac{3}{2}\)                                      \(3x=2-\dfrac{-3}{2}=\dfrac{4}{2}+\dfrac{3}{2}\)

\(3x=\dfrac{1}{2}\)                                                            \(3x=\dfrac{7}{2}\)

\(x=\dfrac{1}{2}.\dfrac{1}{3}\)                                                          \(x=\dfrac{7}{2}.\dfrac{1}{3}\)

\(x=\dfrac{1}{6}\)                                                               \(x=\dfrac{7}{6}\)

______________

\(\left(1-\dfrac{2}{3}\right)^2=\dfrac{4}{9}\) -> Kiểm tra đề câu này

(3-x)^{3_{=(-\(\dfrac{3}{4}\))}3}

^{3-x=-\(\dfrac{3}{4}\)}

  ^{x=3-(-\(\dfrac{3}{4}\))}

  ^{x=\(\dfrac{15}{4}\)}

1) \(\dfrac{1}{27}+a^3=\left(\dfrac{1}{3}+a\right)\left(\dfrac{1}{9}-\dfrac{a}{3}+a^2\right)\)

2) \(=\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)

3) \(=\left(\dfrac{1}{2}x+2y\right)\left(\dfrac{1}{4}x-xy+4y^2\right)\)

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

5) \(=\left(x^3+1\right)\left(x^6-x^3+1\right)\)

6) \(=\left(x-4\right)\left(x^2+4x+16\right)\)

7) \(=\left(x-5\right)\left(x^2+5x+25\right)\)

8) \(=\left(2x^2-3y\right)\left(4x^4+6x^2y+9y^2\right)\)

9) \(=\left(\dfrac{1}{4}x^2-5y\right)\left(\dfrac{1}{16}x^4+\dfrac{5}{4}x^2y+25y^2\right)\)

10) \(=\left(\dfrac{1}{2}x-2\right)\left(\dfrac{1}{4}x^2+x+4\right)\)

11) \(=\left(x+2\right)^3\)

12) \(=\left(x+3\right)^3\)