Ta có:\(\dfrac{x^2}{4}=\dfrac{x}{2};\dfrac{y^2}{9}=\dfrac{y}{3};\dfrac{z^2}{25}=\dfrac{z}{5}\)

Aps dụng tính chất dãy tỉ số bằn nhau:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x-y+z}{2-3+5}=\dfrac{4}{4}=1\)

=>\(\dfrac{x}{2}=1=>x=2\)

  \(\dfrac{y}{3}=1=>y=3\)

\(\dfrac{z}{5}=1=>z=5\)

Vậy x=2, y=3, z=5

Ta có : \(\dfrac{x^2}{4}=\dfrac{y^2}{9}=\dfrac{z^2}{25}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}\)

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

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{5}=\dfrac{x-y+z}{2-3+5}=\dfrac{4}{4}=1\)

\(\Leftrightarrow x=2;y=3;z=5\)

bài 2:để Z là số nguyên thì 3n-5 \(⋮\)n+4

\(\Rightarrow[(3n-5)-3(n+4)]⋮(n+4)\)

\(\Rightarrow(3n-5-3n-12)⋮(n+4)\)

\(\Rightarrow-17⋮n+4\)

\(\Rightarrow n+4\inƯ(17)\)={1;-1;17;-17}

\(\Rightarrow\)n\(\in\){-3;-5;13;-21}

tick cho mk nha bn

2. Để phân số \(A=\dfrac{3n-5}{n+4}\in Z\)

\(\Leftrightarrow\left\{{}\begin{matrix}3n-5⋮n+4\\3n+12⋮n+4\end{matrix}\right.\)

\(\Leftrightarrow17⋮n+4\)

Vì \(n\in Z\Leftrightarrow n+4\in Z;n+4\inƯ\left(17\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}n+4=1\\n+4=17\\n+4=-1\\n+4=-17\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}n=-3\\n=13\\n=-5\\n=-21\end{matrix}\right.\)

Vậy ....

1.

\(\left(\dfrac{-2}{3}\right).0,75+1\dfrac{2}{3}:\left(\dfrac{-4}{9}\right)+\left(\dfrac{-1}{2}\right)^2\)

\(=\left(\dfrac{-2}{3}\right).\dfrac{3}{4}+\dfrac{5}{3}.\left(\dfrac{9}{-4}\right)+\dfrac{1}{4}\)

\(=-\dfrac{1}{2}+\dfrac{45}{-12}+\dfrac{1}{4}\)

\(=-\dfrac{6}{12}+\dfrac{-45}{12}+\dfrac{3}{4}\)

\(=\dfrac{-48}{12}\)

\(=-4\)

2.

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

\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{3}{4}-\dfrac{4}{5}\)

\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{-1}{20}\)

\(\Leftrightarrow x=\dfrac{-1}{20}-\dfrac{1}{2}\)

\(\Leftrightarrow x=\dfrac{-1}{20}-\dfrac{10}{20}\)

\(\Leftrightarrow x=\dfrac{-11}{20}\)

b) \(\left|x-\dfrac{2}{5}\right|+\dfrac{3}{4}=\dfrac{11}{4}\)

\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=\dfrac{11}{4}-\dfrac{3}{4}\)

\(\Leftrightarrow\left|x-\dfrac{2}{5}\right|=2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{2}{5}=-2\Rightarrow x=-2+\dfrac{2}{5}=\dfrac{-8}{5}\\x-\dfrac{2}{5}=2\Rightarrow x=2+\dfrac{2}{5}=\dfrac{12}{5}\end{matrix}\right.\)

3.

a) \(\dfrac{16}{2^n}=2\)

\(\Leftrightarrow2^n=16:2\)

\(\Leftrightarrow2^n=8\)

\(\Leftrightarrow2^n=2^3\)

\(\Leftrightarrow n=3\)

b) \(\dfrac{\left(-3\right)^n}{81}=-27\)

\(\Leftrightarrow\left(-3\right)^n=\left(-27\right).81\)

\(\Leftrightarrow\left(-3\right)^n=\left(-3\right)^3.\left(-3\right)^4\)

\(\Leftrightarrow\left(-3\right)^n=\left(-3\right)^7\)

\(\Leftrightarrow n=7\)

4. Ta có:

\(\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{x}{10}=\dfrac{y}{15}\) (1)

\(\dfrac{y}{5}=\dfrac{z}{4}\Rightarrow\dfrac{y}{15}=\dfrac{z}{12}\) (2)

Từ (1) và (2) suy ra \(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}\)

Vì \(x-y+x=-49\) ta có:

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-49}{7}=-7\)

Vậy \(\left\{{}\begin{matrix}x=\left(-7\right).10=-70\\y=\left(-7\right).15=-105\\z=\left(-7\right).12=-84\end{matrix}\right.\)

a, \(\frac{2}{3}x=\frac{3}{4}y=\frac{4}{5}z\)

\(\Rightarrow\frac{2x}{3.12}=\frac{3y}{4.12}=\frac{4z}{5.12}\)

\(\Rightarrow\frac{x}{18}=\frac{y}{16}=\frac{z}{15}=\frac{x+y+z}{18+16+15}=\frac{45}{49}\)

Đến đây tự làm tiếp nhé

b, \(2x=3y=5z\Rightarrow\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=\frac{95}{19}=5\)

=> x = 75, y = 50, z = 30

c, \(\frac{3}{4}x=\frac{5}{7}y=\frac{10}{11}z\)

\(\Rightarrow\frac{3x}{4.30}=\frac{5y}{7.30}=\frac{10z}{11.30}\)

\(\Rightarrow\frac{x}{40}=\frac{y}{42}=\frac{z}{33}\)

\(\Rightarrow\frac{2x}{80}=\frac{3y}{126}=\frac{4z}{132}=\frac{2x-3y+4z}{80-126+132}=\frac{8,6}{86}=\frac{1}{10}\)

=> x=... , y=... , z=...

d, Đặt \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow x=2k,y=5k\)

Ta có: xy = 90 => 2k.5k = 90 => 10k² = 90 => k² = 9 => k = 3 hoặc -3

Với k = 3 => x = 6, y = 15

Với k = -3 => x = -6, y = -15

Vậy...

e, Tương tự câu d

b) Ta có :\(\text{ 2x = 3y = 5z }=\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=\frac{x+y-z}{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}=\frac{95}{\frac{19}{30}}=\frac{1}{6}\)

=> \(2x=\frac{1}{6}\Rightarrow x=\frac{1}{12}\)

     \(3y=\frac{1}{6}\Rightarrow y=\frac{1}{18}\)

      \(5z=\frac{1}{6}\Rightarrow z=\frac{1}{30}\)

a: Áp dụng tính chất của DTSBN, ta được:

x/5=y/2=(x-y)/(5-2)=9/3=3

=>x=15; y=6

b: =>(x-3)/12=3/(x-3)

=>(x-3)^2=36

=>(x-9)(x+3)=0

=>x=9 hoặc x=-3

c; x/2=y/3

=>x/10=y/15

y/5=z/4

=>y/15=z/12

=>x/10=y/15=z/12=(x-y-z)/(10-15-12)=-49/-17=49/17

=>x=490/17; y=735/17; z=588/17

Câu 2

(xⁿ)^m=x^m.n

2:

(xⁿ)^m = x^{n . m}

Câu 1

Ta có: \(\dfrac{a}{3}=\dfrac{b}{2}=\dfrac{2c}{10}\) và a-b+2c=77

\(\dfrac{a-b+2c}{3-2+10}=\dfrac{77}{11}=7\)

\(\dfrac{a}{3}=7\) ⇒ a=21

\(\dfrac{b}{2}=7\) ⇒ b=14

\(\dfrac{c}{5}=7\) ⇒ c=35

Bài 2: 

a) Ta có: \(A=\dfrac{4}{n-1}+\dfrac{6}{n-1}-\dfrac{3}{n-1}\)

\(=\dfrac{4+6-3}{n-1}\)

\(=\dfrac{7}{n-1}\)

Để A là số tự nhiên thì \(7⋮n-1\)

\(\Leftrightarrow n-1\inƯ\left(7\right)\)

\(\Leftrightarrow n-1\in\left\{1;7\right\}\)

hay \(n\in\left\{2;8\right\}\)

Vậy: \(n\in\left\{2;8\right\}\)

ta có B=2n+9/n+2-3n+5n+1/n+2=4n+10/n+2                                                   Để B là STN thì 4n+10⋮n+2                          4n+8+2⋮n+2                                  4n+8⋮n+2                                                      ⇒2⋮n+2                                     n+2∈Ư(2)                                                        Ư(2)={1;2}                                  Vậy n=0                                                                                  

Ta có: =4+6−3n−1=4+6−3�−1