\(\Leftrightarrow3^x=27\)

hay x=3

\(\Leftrightarrow3^x+3^x.3^2=270\Leftrightarrow3^x.10=270\\\Leftrightarrow3^x=3^3\Leftrightarrow x=3\)

Sửa đề bài 1 : k => x  P/s : đề sai r :)) 

\(A=\left(3-2x\right)3x^2-8+\left(2x+5\right)\left(3x-2\right)-20x\)

\(=9x^2-6x^3-8+6x^2-4x+15x-10-20x=15x^2-6x^3-18-9x\)

Vậy biểu thức phụ thuộc biến x 

\(B=\left(3-5x\right)\left(2x+11\right)-\left(2x+3\right)\left(3x+7\right)\)

\(=6x+33-10x^2-55x-6x^2-14x-9x-21=-72x+12-16x^2\)

Vậy biểu thức phụ thuộc biến x 

Bài 2 : 

a, \(2x\left(x-1\right)-x^2+6=0\Leftrightarrow2x^2-2x-x^2+6=0\)

\(\Leftrightarrow x^2-2x+6=0\)( vô nghiệm )

b, \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)

\(\Leftrightarrow\left(x+3\right)\left(x-3\right)-x\left(x-2\right)\left(x+2\right)=15\)

\(\Leftrightarrow x^2-9-x\left(x^2-4\right)=15\Leftrightarrow x^2-9-x^3+12=15\)

\(\Leftrightarrow-x^3+x^2-12=0\Leftrightarrow x=2\)

Lời giải:

a. 

ƯCLN $ =5^2=25$
BCNN $=3.5^2.7=525$

b.

ƯCLN $=3$

BCLNN $=2^2.3^2.5.7.11=13860$

=3x^2-15x-3x^2+7x

=-15x+7x

=-8x

1: \(\frac{3x-2}{3}-2=\frac{4x+1}{4}\)

=>\(\frac{3x-2-6}{3}=\frac{4x+1}{4}\)

=>\(\frac{3x-8}{3}=\frac{4x+1}{4}\)

=>3(4x+1)=4(3x-8)

=>12x+3=12x-32

=>3=-32(vô lý)

=>Phương trình vô nghiệm

2: \(\frac{x-3}{4}+\frac{2x-1}{3}=\frac{2-x}{6}\)

=>\(\frac{3\left(x-3\right)+4\left(2x-1\right)}{12}=\frac{2\left(2-x\right)}{12}\)

=>3(x-3)+4(2x-1)=2(2-x)

=>3x-9+8x-4=4-2x

=>11x-13=4-2x

=>13x=17

=>\(x=\frac{17}{13}\)

3: \(\frac12\left(x+1\right)+\frac14\left(x+3\right)=3-\frac13\left(x+2\right)\)

=>\(\frac12x+\frac12+\frac14x+\frac34+\frac13x+\frac23=3\)

=>\(x\left(\frac12+\frac14+\frac13\right)+\frac{6}{12}+\frac{9}{12}+\frac{8}{12}=3\)

=>\(x\left(\frac{6}{12}+\frac{3}{12}+\frac{4}{12}\right)=3-\frac{23}{12}=\frac{36}{12}-\frac{23}{12}=\frac{13}{12}\)

=>\(x\cdot\frac{13}{12}=\frac{13}{12}\)

=>x=1

4: \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)

=>\(\frac{x+4}{5}+\frac{5\left(-x+4\right)}{5}=\frac{2x-3\left(x-2\right)}{6}\)

=>\(\frac{x+4-5x+20}{5}=\frac{2x-3x+6}{6}\)

=>\(\frac{-4x+24}{5}=\frac{-x+6}{6}\)

=>6(-4x+24)=5(-x+6)

=>-24x+144=-5x+30

=>-19x=-114

=>x=6

5: \(\frac{4-5x}{6}=\frac{2\left(-x+1\right)}{2}\)

=>\(\frac{4-5x}{6}=-x+1\)

=>6(-x+1)=-5x+4

=>-6x+6=-5x+4

=>-6x+5x=4-6

=>-x=-2

=>x=2

6: \(-\left(\frac{x-3}{2}-2\right)=\frac{5\left(x+2\right)}{4}\)

=>\(-\frac{x-3-4}{2}=\frac{5\left(x+2\right)}{4}\)

=>\(\frac{-2\left(x-7\right)}{4}=\frac{5\left(x+2\right)}{4}\)

=>5(x+2)=-2(x-7)

=>5x+10=-2x+14

=>7x=4

=>x=4/7

7: \(\frac{2\left(2x+1\right)}{5}-\frac{6+x}{3}=\frac{5-4x}{15}\)

=>\(\frac{6\left(2x+1\right)-5\left(x+6\right)}{15}=\frac{5-4x}{15}\)

=>6(2x+1)-5(x+6)=-4x+5

=>12x+6-5x-30=-4x+5

=>7x-24=-4x+5

=>7x+4x=5+24

=>11x=29

=>\(x=\frac{29}{11}\)

8: \(\frac{7-3x}{2}-\frac{5+x}{5}=1\)

=>\(\frac{5\left(7-3x\right)-2\left(x+5\right)}{10}=1\)

=>5(7-3x)-2(x+5)=10

=>35-15x-2x-10=10

=>-17x+25=10

=>-17x=-15

=>x=15/17

a) x= chịu

b) x=2

Trình bày cách giải ra >:(

a, 7\(x\) - \(x\) = 5²¹ : 5¹⁹ + 3.2².7

     6\(x\)    = 5³ + 3.4.7

    6\(x\)    = 125 + 12.7

    6\(x\)  = 125 + 84

    6\(x\) = 209

     \(x\)  = 209 : 6

    \(x\) = \(\dfrac{209}{6}\)

b; 11\(x\) - 7\(x\) + 3⁴ : 3³ = 5⁴ + 2\(x\)

    4\(x\) + 3 = 625 + 2\(x\)

   4\(x\) - 2\(x\) = 625 - 3

   2\(x\)        = 622

     \(x\)        = 622 : 2

    \(x\)        = 311

c; 75 - 5.(\(x-3\))³ = 700

          5.(\(x\) - 3)³ = 700 - 75

         5.(\(x\) - 3)³ = - 625

           (\(x\) - 30)³ = - 625 : 5

           (\(x\) - 30)³ = - 125

           (\(x-3\))³ =  (-5)³

           \(x\) - 3 = - 5 

           \(x\)         = - 5 + 3

            \(x\)       = -2

d, 3.(2\(x\) - 1)² = 75

       (2\(x\) - 1)² = 75 : 3

       (2\(x\) - 1)² = 25

       \(\left[{}\begin{matrix}2x-1=-5\\2x-1=5\end{matrix}\right.\)

       \(\left[{}\begin{matrix}2x=-5+1\\2x=5+1\end{matrix}\right.\)

       \(\left[{}\begin{matrix}2x=-4\\2x=6\end{matrix}\right.\)

       \(\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1