\(2^3x+5^2x=2.\left(5^2+2^3\right)-33\)

\(8x+25x=2.\left(25+8\right)-33\)

\(8x+25x=2.33-33\)

\(8x+25x=66-33\)

\(8x+25x=33\)

\(x+\left(8+25\right)=33\)

\(x+33=33\)

\(x=0\)

\(15:\left(x+2\right)=\left(3^3+3\right):10\)

\(15:\left(x+2\right)=\left(27+3\right):10\)

\(15:\left(x+2\right)=30:10\)

\(15:\left(x+2\right)=3\)

\(x+2=15:3\)

\(x+2=5\)

\(x=3\)

Ta có: \(15:\left(x+2\right)=\left(3^3+3\right):10\)

\(\Leftrightarrow15:\left(x+2\right)=3\)

\(\Leftrightarrow x+2=5\)

hay x=3

\(5^{10}:5^8+x=3^{20}:\left(-3\right)^{18}-2^{35}:\left(-2\right)^{33}\\ \Leftrightarrow5^2+x=3^2+\left(-2\right)^2\\ \Leftrightarrow25+x=13\\ \Leftrightarrow x=-12\)

5^10/5^8+x=3^20/(-3)^18-2^35/(-2)^33

5^2+x=(-3)^2+2^2

25+x= 9+4

25+x=13

      x=13-25

     x=-12

vậy x=-12

1: =>x^2+4x+3-x^2-2x=7

=>2x=4

hay x=2

x.3+1/2+3/2+5/2=33

x.3+9/2=33

x.3+4,5=33

x.3=33-4,5

x.3=28,5

x=28,5:3

x=9,5

Bài 1: 

a) Ta có: \(x\left(x^2-4\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{0;2;-2\right\}\)

b) Ta có: \(\left(2x-3\right)+\left(-3x\right)-\left(x-5\right)=40\)

\(\Leftrightarrow2x-3-3x-x+5=40\)

\(\Leftrightarrow-2x+2=40\)

\(\Leftrightarrow-2x=38\)

hay x=-19

Vậy: x=-19

Bài 2: 

a) Ta có: \(-45\cdot12+34\cdot\left(-45\right)-45\cdot54\)

\(=-45\cdot\left(12+34+54\right)\)

\(=-45\cdot100\)

\(=-4500\)

b) Ta có: \(43\cdot\left(57-33\right)+33\cdot\left(43-57\right)\)

\(=43\cdot57-43\cdot33+43\cdot33-33\cdot57\)

\(=43\cdot57-33\cdot57\)

\(=57\cdot\left(43-33\right)\)

\(=57\cdot10=570\)

\(x+\dfrac{1}{2}=\dfrac{33}{4}\\ \Rightarrow x=\dfrac{33}{4}-\dfrac{1}{2}\\ \Rightarrow x=\dfrac{31}{4}\\ \dfrac{5}{6}-x=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{5}{6}-\dfrac{1}{3}\\ \Rightarrow x=\dfrac{1}{2}\\ x+\dfrac{4}{5}=\dfrac{-2}{3}\\ \Rightarrow x=\dfrac{-2}{3}-\dfrac{4}{5}\\ \Rightarrow x=\dfrac{-22}{15}\)

\(6-2\left(x-1\right)=4\)

\(\Rightarrow2\left(x-1\right)=6-4\)

\(\Rightarrow2\left(x-1\right)=2\)

\(\Rightarrow x-1=1\)

\(\Rightarrow x=1+1=2\)

________________

\(2\cdot\left(x-2\right)+1=7\)

\(\Rightarrow2\cdot\left(x-2\right)=7-1\)

\(\Rightarrow2\cdot\left(x-2\right)=6\)

\(\Rightarrow x-2=3\)

\(\Rightarrow x=3+2=5\)

_______________

\(\left(2\cdot x-3\right)+4=9\)

\(\Rightarrow2\cdot x-3=5\)

\(\Rightarrow2\cdot x=3+5\)

\(\Rightarrow2\cdot x=8\)

\(\Rightarrow x=\dfrac{8}{2}=4\)

________________

\(\left(3\cdot x-2\right)-1=3\)

\(\Rightarrow3\cdot x-2=3+1\)

\(\Rightarrow3\cdot x-2=4\)

\(\Rightarrow3\cdot x=6\)

\(\Rightarrow x=\dfrac{6}{3}=2\)

a: =>2(x-1)=2

=>x-1=1

=>x=2

b: =>2(x-2)=6

=>x-2=3

=>x=5

c; =>2x-3=5

=>2x=8

=>x=4

d: =>3x-2=4

=>3x=6

=>x=2

e: =>2(6-x)=4

=>6-x=2

=>x=4

f: =>x-2=5

=>x=7

g: =>10-2x=4

=>2x=6

=>x=3

h: =>2x+4=3

=>2x=-1

=>x=-1/2

j: =>x+2=12

=>x=10

l: =>2x+3=3

=>2x=0

=>x=0

\(6-2\left(x-1\right)=4\\ 2\left(x-1\right)=6-4=2\\ x-1=\dfrac{2}{2}=1\\ x=1+1=2\)

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

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

\(2\left(x-2\right)+1=7\\ 2\left(x-2\right)=7-1=6\\ x-2=\dfrac{6}{2}=3\\ x=3+2=5\)

\(20+\left(x-2\right)=25\\ x-2=25-20=5\\ x=5+2=7\)

\(\left(x+2\right)-5=7\\ x+2=7+5=12\\ x=12-2=10\)

\(\left(2x-3\right)+4=9\\ 2x-3=9-4=5\\ 2x=5+3=8\\ x=\dfrac{8}{2}=4\)

\(\left(10-\dfrac{2}{x}\right)-1=3\\ 10-\dfrac{2}{x}=3+1=4\\ \dfrac{2}{x}=10-4=6\\ x=\dfrac{2}{6}=\dfrac{1}{3}\)

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

\(\left(3x-2\right)-1=3\\ 3x-2=3+1=4\\ 3x=4+2=6\\ x=\dfrac{6}{3}=2\)

\(a,\dfrac{12}{5}=\dfrac{x}{1,5}\Rightarrow x=\dfrac{12\cdot1,5}{5}=3,6\\ b,\dfrac{x}{5}=\dfrac{3}{20}\Rightarrow x=\dfrac{5\cdot3}{20}=\dfrac{3}{4}\\ c,\dfrac{4}{x}=\dfrac{10}{9}\Rightarrow x=\dfrac{4\cdot9}{10}=\dfrac{18}{5}\\ d,\Rightarrow\dfrac{x}{15}=\dfrac{60}{x}\Rightarrow x^2=60\cdot15=900\Rightarrow\left[{}\begin{matrix}x=30\\x=-30\end{matrix}\right.\\ 2,\)

a, Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x+y-z}{3+5-6}=\dfrac{8}{2}=4\\ \Rightarrow\left\{{}\begin{matrix}x=12\\y=20\\z=24\end{matrix}\right.\)

b, Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{x-y+z}{3-5+6}=\dfrac{-4}{4}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-6\end{matrix}\right.\)

c, Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=\dfrac{2y}{10}=\dfrac{3z}{18}=\dfrac{x-2y+3z}{3-10+18}=\dfrac{-33}{11}=-3\\ \Rightarrow\left\{{}\begin{matrix}x=-9\\y=-15\\z=-18\end{matrix}\right.\)

d, Đặt \(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{6}=k\Rightarrow x=3k;y=5k;z=6k\)

\(x^2-4y^2+2z^2=-475\\ \Rightarrow9k^2-100k^2+72z^2=-475\\ \Rightarrow-19k^2=-475\\ \Rightarrow k^2=25\Rightarrow\left[{}\begin{matrix}k=5\\k=-5\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=15;y=25;z=30\\x=-15;y=-25;z=-30\end{matrix}\right.\)