\(25^x:5^4=125^2\)

\(\left(5^2\right)^x:5^4=\left(5^3\right)^2\)

\(5^{2x}:5^4=5^6\)

\(5^{2x-4}=5^6\)

\(2x-4=6\)

\(2x=4+6\)

\(2x=10\)

\(x=5\)

\(1,\dfrac{2}{3}+\dfrac{1}{3}:x=\dfrac{4}{3}\\ =>\dfrac{1}{3}:x=\dfrac{4}{3}-\dfrac{2}{3}\\ =>\dfrac{1}{3}:x=\dfrac{2}{3}\\ =>x=\dfrac{1}{3}:\dfrac{2}{3}=\dfrac{1}{2}\\ 2,\dfrac{4}{5}+\dfrac{1}{3}:x=\dfrac{2}{3}\\ =>\dfrac{1}{3}:x=\dfrac{2}{3}-\dfrac{4}{5}=-\dfrac{2}{15}\\ =>x=\dfrac{1}{3}:\dfrac{-2}{15}=\dfrac{-5}{2}\\ 3,\dfrac{2}{3}+\dfrac{5}{2}:x=\dfrac{3}{4}\\ =>\dfrac{5}{2}:x=\dfrac{3}{4}-\dfrac{2}{3}\\ =>\dfrac{5}{2}:x=\dfrac{1}{12}\\ =>x=\dfrac{5}{2}:\dfrac{1}{12}=30\)

b: \(\frac{x}{y}=\frac{1.2}{2.5}\)

=>\(\frac{x}{y}=\frac{12}{25}\)

=>\(\frac{x}{12}=\frac{y}{25}\)

mà y-x=26

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x}{12}=\frac{y}{25}=\frac{y-x}{25-12}=\frac{26}{13}=2\)

=>\(\begin{cases}x=2\cdot12=24\\ y=2\cdot25=50\end{cases}\)

c: Sửa đề: \(\frac{x}{4}=\frac{y}{3}=\frac{z}{-2}\)

Ta có: \(\frac{x}{4}=\frac{y}{3}=\frac{z}{-2}\)

mà x-y-z=45

nên Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x}{4}=\frac{y}{3}=\frac{z}{-2}=\frac{x-y-z}{4-3-\left(-2\right)}=\frac{45}{1+2}=\frac{45}{3}=15\)

=>\(\begin{cases}x=15\cdot4=60\\ y=15\cdot3=45\\ z=15\cdot\left(-2\right)=-30\end{cases}\)

Do \(\widehat{xOy}\) và \(\widehat{yOz}\) là 1 góc kề bù nên:

\(\widehat{xOy}+\widehat{yOz}=180^0\)

\(\Rightarrow2.\widehat{yOz}+\widehat{yOz}=180^0\) (do \(\widehat{xOy}=2.\widehat{yOz}\))

\(\Rightarrow3.\widehat{yOz}=180^0\)

\(\Rightarrow\widehat{yOz}=180^0:3=60^0\)

\(\Rightarrow\widehat{xOy}=2.\widehat{yOz}=2.60^0=120^0\)

1. energy

2. ride

3. calories

4.  spend

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6. fit

7. from

8. in

9. exercise

10. tired

1 energy

2 ride

3 calories

4 spend

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8 in

9 exercise

10 tired

Ta có: \(\widehat{xOz}+\widehat{xOy}=180^0\)(hai góc kề bù)

=>\(\widehat{xOz}=180^0-50^0=130^0\)

Ot là phân giác của góc xOz

=>\(\widehat{zOt}=\dfrac{\widehat{xOz}}{2}=65^0\)

Ta có: \(\widehat{zOt}+\widehat{yOt}=180^0\)(hai góc kề bù)

=>\(\widehat{yOt}=180^0-65^0=115^0\)

a: \(x=\left(x^3\right)^{\dfrac{1}{3}}\)

b: \(x=\left(x^5\right)^{\dfrac{1}{5}}\)

\(\dfrac{1}{3^6}=\dfrac{1}{3^4\cdot3^2}=\dfrac{1}{81\cdot9}=\dfrac{1}{729}\)

  \(\dfrac{1}{3^6}\) = \(\dfrac{1}{3^4.3^2}\) = \(\dfrac{1}{81.9}\) = \(\dfrac{1}{729}\) 

\(\left(-5\right)^5=\left(-5\right)^4\cdot\left(-5\right)=5^4\cdot\left(-5\right)=625\cdot\left(-5\right)=-3125\)