1) ta có: \(x:3=y.15\Rightarrow x\cdot\frac{1}{3}=y.15\Rightarrow\frac{x}{15}=\frac{y}{\frac{1}{3}}\)

ADTCDTSBN

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2) bn ghi thiếu đề r

3) ta có: \(3x=7y\Rightarrow\frac{x}{7}=\frac{y}{3}=k\Rightarrow\hept{\begin{cases}x=7k\\y=3k\end{cases}}\)

mà xy = 189 => 7k.3k = 189

                          21 k² = 189

                                 k² = 9 = 3² = (-3)² => k = 3 hoặc k  = - 3

TH1: k = 3

x = 7.3 => x  = 21

y = 3.3 => y = 9

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4) ta có: \(4x=5y\Rightarrow\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x^2}{25}=\frac{y^2}{16}\)

ADTCDTSBN

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X-x/3=5+2/4

3x/3-x/3=20/4+2/4

3x-x/3=22/4

2x/3=11/2

4x/6=33/6

4x=33

x=33/4

Vay...

x-2/4=5+x-3 =>x=33/4

X-2/4=5+X/3

X-X/3=2/4+5

2/3.X=11/2

x=33/4

NHO K DAY NHE!

Ta có :

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

\(B=1+\frac{1}{2}.\frac{2.3}{2}+\frac{1}{3}.\frac{3.4}{2}+...+\frac{1}{x}.\frac{x.\left(x+1\right)}{2}\)

\(B=1+\frac{3}{2}+\frac{4}{2}+...+\frac{x+1}{2}\)

\(B=\frac{2+3+4+...+\left(x+1\right)}{2}\)

để B = 115 thì \(\frac{2+3+4+...+\left(x+1\right)}{2}=115\)

\(\Rightarrow\)\(\left(x+3\right)x=115.2.2\)

\(\Rightarrow\)\(\left(x+3\right)x=23.20\)

\(\Rightarrow\)x = 20

1. x:(-12/5)=2/3=>x=2/3*(-12/5)=-8/5=-1,6

2.x=14/2*(3^2-2)+6

   x=7*7+6=55

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\(2\left(x-3\right)+2^3=3^2.2\\ 2\left(x-3\right)=18-8\\ 2\left(x-3\right)=10\\ x-3=\dfrac{10}{2}\\ x-3=5\\ x=5+3\\ x=8\\ ----\\ 2^2.3-3\left(x-4\right)=2^3-2\\ 12-3\left(x-4\right)=6\\ 3\left(x-4\right)=12-6\\ 3\left(x-4\right)=6\\ x-4=6:3\\ x-4=2\\ x=2+4\\ x=6\\ ----\\ 7.2^2-4\left(2x+1\right)=2^3\\ 28-4\left(2x+1\right)=8\\ 4\left(2x+1\right)=28-8\\ 4\left(2x+1\right)=20\\ 2x+1=20:4\\ 2x+1=5\\ 2x=5-1\\ 2x=4\\ x=\dfrac{4}{2}=2\\ ---\\ 600-2\left(x-3\right)^8=11.2^3\\ 600-2\left(x-3\right)^8=88\\ 2\left(x-3\right)^8=600-88\\ 2\left(x-3\right)^8=512\\ \left(x-3\right)^8=\dfrac{512}{2}\\ \left(x-3\right)^8=256=2^8=\left(-2\right)^8\\ x-3=2.hoặc.x-3=-2\\ x=5.hoặc.x=1\)