4800:X+130/4+225=7,3x50\(\Rightarrow\)4800:x+130/4=110\(\Rightarrow\)x=4800:(110-130/4)\(\Rightarrow\)x=1920/31

\(4800:x+\frac{130}{4}+255=7,3\times50\)

\(4800:x+\frac{130}{4}+\frac{255\times4}{4}=365\)

\(4800:x+\frac{130}{4}+\frac{1020}{4}=365\)

\(4800:x+\frac{130+1020}{4}=365\)

\(4800:x+\frac{1150}{4}=365\)

\(4800:x+\frac{575}{2}=365\)

\(4800:x=365-\frac{575}{2}\)

\(4800:x=\frac{365\times2}{2}-\frac{575}{2}\)

\(4800:x=\frac{730}{2}-\frac{575}{2}\)

\(4800:x=\frac{730-575}{2}\)

\(4800:x=\frac{155}{2}\)

\(x=4800:\frac{155}{2}\)

\(x=\frac{1920}{31}\)

Chúc Em Học Tốt !

giải hộ em bài nữa nha :X/27=3/X

\(\frac{x}{27}=\frac{3}{x}\)

\(x\times x=27\times3\)

\(x^2=81\)

\(x=\sqrt{81}\)

\(x=9\)

\(1,\\ a,\Leftrightarrow4^{5-x}=4^2\Leftrightarrow5-x=2\Leftrightarrow x=3\\ b,\Leftrightarrow\left[{}\begin{matrix}x-1=5\\x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x+1=3\Leftrightarrow x=2\\ 2,\\ a,3^{100}=\left(3^2\right)^{50}=9^{50}\\ b,2^{98}=\left(2^2\right)^{49}=4^{49}< 9^{49}\\ c,5^{30}=5^{29}\cdot5< 6\cdot5^{29}\\ d,3^{30}=\left(3^3\right)^{10}=27^{10}>8^{10}\\ 4,\\ a,\Leftrightarrow5\left(x-10\right)=10\\ \Leftrightarrow x-10=2\Leftrightarrow x=12\\ b,\Leftrightarrow3\left(70-x\right)+5=92\\ \Leftrightarrow3\left(70-x\right)=87\\ \Leftrightarrow70-x=29\\ \Leftrightarrow x=41\\ c,\Leftrightarrow16+x-5=315-230=85\\ \Leftrightarrow x=74\\ d,\Leftrightarrow2^x-5+74=707:\left(16-9\right)=707:7=101\\ \Leftrightarrow2^x=32=2^5\\ \Leftrightarrow x=5\)

a) \(x\in\left\{15;30;45\right\}\)

b) \(x\in\left\{12;24;36\right\}\)

c) \(x\in\left\{60;120\right\}\)

1)\(\frac{252}{x}=\frac{84}{97}\Rightarrow\)\(\frac{84}{97}=\frac{252}{291}\Rightarrow x=291\)

6) \(\frac{y}{15}=\frac{2}{5}\Rightarrow\frac{2}{5}=\frac{6}{15}\Rightarrow x=6\)

\(a,\left(x-4\right).5=50\\ \Rightarrow x-4=10\\ \Rightarrow x=14\\ b,\left(-470-x\right)\left(-3\right)=-120\\ \Rightarrow-470-x=40\\ \Rightarrow x=-510\\ c,5x^3+\left(-170\right)=-210\\ \Rightarrow5x^3=-40\\ \Rightarrow x^3=-8\\ \Rightarrow x=-2\)

a,(x−4).5=50⇒x−4=10⇒x=14

b,(−470−x)(−3)=−120⇒−470−x=40⇒x=−510

c,5x3+(−170)=−210⇒5x3=−40⇒x3=−8⇒x=−2

x - 4 = 50. 5

x - 4 = 250

x = 250 + 4

x = 254

Vậy x = 254

b) -470 - x = -120. (-3)

-470 - x = 360

x = -470 - 360 

x = -830

Vậy x = -830

c) 5. x³ = -210 - (-170)

5. x³ = -40

x³ = -40: 5

x³ = -8

x = -2

Vậy x = -2

a,\(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)

\(\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\)

\(\dfrac{3}{2}x=-\dfrac{13}{10}\)

\(x=-\dfrac{13}{15}\)

b,\(\left(3x+1\right)^3=-27\)

\(3x+1=-3\)

\(3x=-4\)

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

a ) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\\\dfrac{1}{5}-\dfrac{3}{2}x=-\dfrac{3}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{3}{2}x=-\dfrac{13}{10}\\\dfrac{3}{2}x=\dfrac{17}{10}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{13}{15}\\x=\dfrac{17}{15}\end{matrix}\right.\)

Vậy .......

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

\(\Leftrightarrow3x+1=-3\)

\(\Leftrightarrow3x=-4\Leftrightarrow x=-\dfrac{4}{3}\)

Vậy ..........

c ) \(7.3^{x-1}-3^{x+2}=-540\)

\(\Leftrightarrow7.3^{x-1}-3^{x-1}.3^3=-540\)

\(\Leftrightarrow3^{x-1}\left(7-27\right)=-540\)

\(\Leftrightarrow-20.3^{x-1}=-540\)

\(\Leftrightarrow3^{x-1}=27\)

\(\Leftrightarrow x=4\).

Vậy ..........

a: ta có: \(\left(8x+2\right)\left(1-3x\right)+\left(6x-1\right)\left(4x-10\right)=-50\)

\(\Leftrightarrow8x-24x^2+2-6x+24x^2-60x-4x+40=-50\)

\(\Leftrightarrow-62x=-92\)

hay \(x=\dfrac{46}{31}\)

b: ta có: \(\left(1-4x\right)\left(x-1\right)+4\left(3x+2\right)\left(x+3\right)=38\)

\(\Leftrightarrow x-1-4x^2+4x+4\left(3x^2+9x+2x+6\right)=38\)

\(\Leftrightarrow-4x^2+5x-1+12x^2+44x+24-38=0\)

\(\Leftrightarrow8x^2+49x-15=0\)

\(\text{Δ}=49^2-4\cdot8\cdot\left(-15\right)=2881\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là: 

\(\left\{{}\begin{matrix}x_1=\dfrac{-49-\sqrt{2881}}{16}\\x_2=\dfrac{-49+\sqrt{2881}}{16}\end{matrix}\right.\)

a: x chia hết cho 4;5;10

nên \(x\in BC\left(4;5;10\right)\)

mà 10<=x<50

nên x=40

b: x=33