a. \(3\times2^{2x-1}=24\)

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

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

a: \(280-\left(x-140\right):35=270\)

=>(x-140):35=280-270=10

=>x-140=350

=>x=350+140

=>x=490

b: \(\left(190-2x\right):35-32=16\)

=>\(\left(190-2x\right):35=32+16=48\)

=>\(190-2x=35\cdot48=1680\)

=>2x=190-1680=-1490

=>x=-745

c: \(720:\left\lbrack41-\left(2x-5\right)\right\rbrack=2^3\cdot5\)

=>\(720:\left\lbrack41-\left(2x-5\right)\right\rbrack=8\cdot5=40\)

=>41-(2x-5)=720:40=18

=>2x-5=41-18=23

=>2x=28

=>x=14

d: \(\left(x:23+45\right)\cdot37-22=2^4\cdot105\)

=>\(\left(\frac{x}{23}+45\right)\cdot37=16\cdot105+22=1702\)

=>\(\frac{x}{23}+45=46\)

=>\(\frac{x}{23}=1\)

=>x=23

e: \(\left(3x-4\right)\left(x-1\right)^3=0\)

=>\(\left[\begin{array}{l}3x-4=0\\ x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac43\\ x=1\end{array}\right.\)

f: \(2^{2x-1}:4=8^3\)

=>\(2^{2x-1-2}=2^9\)

=>2x-3=9

=>2x=12

=>x=6

g: \(x^{17}=x\)

=>\(x^{17}-x=0\)

=>\(x\left(x^{16}-1\right)=0\)

=>\(\left[\begin{array}{l}x=0\\ x^{16}-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^{16}=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\\ x=-1\end{array}\right.\)

h: \(\left(x-5\right)^4=\left(x-5\right)^6\)

=>\(\left(x-5\right)^6-\left(x-5\right)^4=0\)

=>\(\left(x-5\right)^4\cdot\left\lbrack\left(x-5\right)^2-1\right\rbrack=0\)

=>\(\left(x-5\right)^4\cdot\left(x-4\right)\left(x-6\right)=0\)

=>x∈{4;5;6}

i: \(\left(x+2\right)^5=2^{10}\)

=>\(\left(x+2\right)^5=\left(2^2\right)^5=4^5\)

=>x+2=4

=>x=2

k: 1+2+3+...+x=78

=>\(\frac{x\left(x+1\right)}{2}=78\)

=>x(x+1)=156

=>\(x^2+x-156=0\)

=>(x+13)(x-12)=0

=>x=-13(loại) hoặc x=12(nhận)

l: \(\left(3x-2^4\right)\cdot7^3=2\cdot7^4\)

=>\(3x-16=2\cdot\frac{7^4}{7^3}=2\cdot7=14\)

=>3x=16+14=30

=>\(x=\frac{30}{3}=10\)

n: \(5^{x}:5^2=125\)

=>\(5^{x-2}=5^3\)

=>x-2=3

=>x=5

m: \(\left(x+1\right)^2=\left(x+1\right)^0\)

=>\(\left(x+1\right)^2=1\)

=>\(\left[\begin{array}{l}x+1=1\\ x+1=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=-2\end{array}\right.\)

o: Số số hạng của dãy số 2;4;..;52 là:

(52-2):2+1=50:2+1=25+1=26(số)

Tổng của dãy số 2;4;...;52 là:

\(\left(52+2\right)\cdot\frac{26}{2}=54\cdot13=702\)

(2+x)+(4+x)+...+(52+x)=780

=>26x+702=780

=>26x=78

=>x=3

p: \(70=2\cdot5\cdot7;80=2^4\cdot5\)

=>ƯCLN(70;80)=\(2\cdot5=10\)

70⋮x; 80⋮x

=>x∈ƯC(70;80)

=>x∈Ư(10)

mà x>8

nên x=10

q: \(12=2^2\cdot3;25=5^2;30=2\cdot3\cdot5\)

=>BCNN(12;25;30)=\(2^2\cdot3\cdot5^2=300\)

x⋮12; x⋮25; x⋮30

=>x∈BC(12;25;30)

=>x∈B(300)

mà 0<x<500

nên x=300

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

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=\(\frac{6}{5}\)

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