a. \(3\times2^{2x-1}=24\)

\(\Rightarrow2^{2x-1}=8=2^3\)

\(\Rightarrow2x-1=3\Rightarrow x=2\)

8: =>6x^2-9x+2x-3-6x^2-12x=16

=>-19x=19

=>x=-1

a: 3(x+7)-2x+5>0

=>3x+21-2x+5>0

=>x+26>0

=>x>-26

Sửa đề: \(\dfrac{x+2}{18}-\dfrac{x+3}{8}< \dfrac{x-1}{9}-\dfrac{x-4}{24}\)

=>\(\dfrac{4\left(x+2\right)}{72}-\dfrac{9\left(x+3\right)}{72}< \dfrac{8\left(x-1\right)}{72}< \dfrac{3\left(x-4\right)}{72}\)

=>\(4\left(x+2\right)-9\left(x+3\right)< 8\left(x-1\right)-3\left(x-4\right)\)

=>\(4x+8-9x-27< 8x-8-3x+12\)

=>-5x-19<5x+4

=>-10x<23

=>\(x>-\dfrac{23}{10}\)

b: \(3x+2+\left|x+5\right|=0\left(1\right)\)

TH1: x>=-5

(1) trở thành: 3x+2+x+5=0

=>4x+7=0

=>\(x=-\dfrac{7}{4}\left(nhận\right)\)

TH2: x<-5

=>x+5<0

=>|x+5|=-x-5

Phương trình (1) sẽ trở thành:

\(3x+2-x-5=0\)

=>2x-3=0

=>2x=3

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

\(3.2^{2x-1}=24\)

\(2^{2x-1}=8\)

\(\Rightarrow2^{2x-1}=2^3\)

\(\Rightarrow2x-1=3\)

\(2x=4\Rightarrow x=2\)

a) (x-2)³+6(x+1)²-x³+12=0

\(\Rightarrow\)x³-6x²+12x-8+6(x²+2x+1)-x³+12=0

\(\Rightarrow\)x³-6x²+12x-8+6x²+12x+6-x³+12=0

\(\Rightarrow\)24x+10=0

\(\Rightarrow\)24x=-10

\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)

b)(x-5)(x+5)-(x+3)²+3(x-2)²=(x+1)²-(x-4)(x+4)+3x²

\(\Rightarrow\)x²-25-(x²+6x+9)+3(x²-4x+4)=x²+2x+1-(x²-16)+3x²

\(\Rightarrow\)x²​-25-x²-6x-9+3x²-12x+12=x²+2x+1-x²+16+3x²

\(\Rightarrow\)3x²-18x-22=3x²+2x+17

\(\Rightarrow\)3x²-18x-22-3x²-2x-17=0

\(\Rightarrow\)-20x-39=0

\(\Rightarrow\)-20x=39

\(\Rightarrow\)x=\(-\dfrac{39}{20}\)

c) (2x+3)² +(x-1)(x+1)=5(x+2)²-(x-5)(x+1)+(x+4)

⇒4x²+12x+9+x²-1=5(x²+4x+4)-(x²+x-5x-5)+x+4

⇒5x²+12x+8=5x²+20x+20-x²-x+5x+5+x+4

⇒5x²+12x+8-5x²-20x-20+x²+x-5x-5-x-4=0

⇒x²-13x-21=0

theo bài ra ta có 

 \(\left(x+1\right)+2\left(\frac{3x}{2}+2\right)+3\left(2x+3\right)=26\)

\(\Leftrightarrow x+1+3x+4+6x+9=26\)

\(\Leftrightarrow10x+14=26\)

\(\Leftrightarrow10x=12\)

\(\Leftrightarrow x=\frac{12}{10}\)\(\Rightarrow y=\frac{9}{5}\)\(\Rightarrow z=\frac{12}{5}\)

+)Theo bài:2x=z;3x=2y(1)

+)Theo bài ta lại có:(x+1)+2.(y+2)+3.(z+3)=26

                    =>x+1+2y+4+3z+9=26

                   =>x+2y+3z+1+4+9=26

                   =>x+2y+3z+14=26

                   =>x+2y+3z      =26-14

                   =>x+2y+3z       =12(2)

+)Thay (1) vào (2) được:x+3x+6x=12

                                 =>(1+3+6)x=12

                                =>10x      =12

                                   =>x         =\(\frac{12}{10}=\frac{6}{5}\)

Vậy x=\(\frac{6}{5}\)

Chúc bn học tốt

Giải:

ta có:

2x=z => 6x=3z và 3x=2y *

Ta Lại có:

(x+1)+2(y+2)+3(z+3)=26

=x+1+2y+4+3z+9=26

=x+2y+3z+1+4+9=26

thay * vào biểu thức x+2y+3z+1+4+9 =26 ta được :

x+3x+6x+1+4+9=26

10x+14=26

10x=26-14=12

x=12:10=1.2 **

thay ** vào 2x=z ta được :

2*1.2=z

=> z=2.4

thay ** vào 3x=2y ta được:

3*1.2=2y

=> 2y=3.6

y=1.8

vậy x=1.2 ; y=1.8 ; x=2.4