Bài 2: Phân số bằng nhau

a) Ta có: \(x+\dfrac{1}{3}=\dfrac{2}{6}\)

\(\Leftrightarrow x+\dfrac{1}{3}=\dfrac{1}{3}\)

hay x=0

Vậy: x=0

b) Ta có: \(x-\dfrac{1}{4}=\dfrac{1}{-2}\)

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

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

Vậy: \(x=-\dfrac{1}{4}\)

c) Ta có: \(\dfrac{-1}{6}=\dfrac{3}{2}x\)

\(\Leftrightarrow x=\dfrac{-1}{6}:\dfrac{3}{2}=\dfrac{-1}{6}\cdot\dfrac{2}{3}\)

hay \(x=\dfrac{-1}{9}\)

Vậy: \(x=\dfrac{-1}{9}\)

\(a.x=\dfrac{1}{3}-\dfrac{1}{3}\)

\(x=0\)

\(b.x-\dfrac{1}{4}=\dfrac{-1}{2}\)

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

\(x=\dfrac{-1}{4}\)

c. \(\dfrac{-1}{6}=\dfrac{3}{2x}\)

\(-2x=18\)

\(x=-9\)

d. \(\dfrac{4}{5}=\dfrac{-12}{9-x}\)

\(4.\left(9-x\right)=-60\)

\(9-x=-15\)

\(x=24\)

\(e.\dfrac{x+1}{3}=\dfrac{3}{x+1}\)

\(\left(x+1\right)^2=9\)

\(\left[{}\begin{matrix}x+1=-3\\x+1=3\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=-4\\x=2\end{matrix}\right.\)

f.\(\dfrac{x-1}{-4}=\dfrac{-4}{x-1}\)

\(\left(x-1\right)^2=16\)

\(\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\)

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

`5/12=-x/72`

`=>30/72=-x/72`

`=>-x=30`

`=>x=-30`

Vậy `x=-30`

\(\dfrac{5}{12}=-\dfrac{x}{72}\)

\(=>-1x=\dfrac{5\cdot72}{12}=30\)

\(-1x=30\)

\(=>x=-30\)

Với \(x\in N\), ta có:

 \(\dfrac{5}{12}=\dfrac{-x}{72}\)

\(\Rightarrow-12x=360\)

\(\Leftrightarrow x=-30\) (Loại)

Vậy \(x\in\varnothing\)

Phần trắc nghiệm

Câu 1: D

Câu 2: C

Câu 3: B

Câu 4: D

\(x^2=\dfrac{4}{9} \)

Mà: \(\left(-\dfrac{2}{3}\right)^2=\left(\dfrac{2}{3}\right)^2=\dfrac{4}{9}\)

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

Vậy \(S=\left\{-\dfrac{2}{3};\dfrac{2}{3}\right\}\)

Đáp án D : \(\dfrac{2}{3}\)   hoặc  \(\dfrac{-2}{3}\)

\(x=+-\sqrt{\dfrac{4}{9}}=+-\dfrac{2}{3}\)

=> câu D

x=12

y=8

Ta có: \(\dfrac{x-3}{y-2}=\dfrac{3}{2}\)

nên \(\dfrac{x-3}{3}=\dfrac{y-2}{2}\)

mà x-y=4

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

\(\dfrac{x-3}{3}=\dfrac{y-2}{2}=\dfrac{x-y-3+2}{3-2}=3\)

Do đó:

\(\left\{{}\begin{matrix}\dfrac{x-3}{3}=3\\\dfrac{y-2}{2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-3=9\\y-2=6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=12\\y=8\end{matrix}\right.\)

Vậy: (x,y)=(12;8)

\(x-\dfrac{1}{81}=\dfrac{2}{3}\)

\(\Leftrightarrow x=\dfrac{2}{3}+\dfrac{1}{81}=\dfrac{55}{81}\)

\(x-\dfrac{1}{81}=\dfrac{2}{3}\)

\(x=\dfrac{2}{3}+\dfrac{1}{81}\)

\(x=\dfrac{55}{81}\)

Vậy....

x - \(\dfrac{1}{81}=\dfrac{2}{3}\)

x         = ​​​\(\dfrac{2}{3}+\dfrac{1}{81}\)​

x      = \(\dfrac{55}{81}\)

\(\dfrac{x}{2}=\dfrac{18}{x}\)

\(\Leftrightarrow x^2=18\cdot2=36\)

\(\Leftrightarrow x=\pm\sqrt{36}=\pm6\)

\(\dfrac{x}{2}=\dfrac{18}{x}\)

\(=>x^2=18\cdot2\)

\(x^2=36\)

\(x^2=\left(\pm6\right)^2\)

\(x=\pm6\)

Vậy x=.....

\(\dfrac{x}{2}=\dfrac{18}{x}\)

\(\Leftrightarrow x^2=18\cdot2=36\)

\(\Leftrightarrow x=\pm\sqrt{36}=\pm6\)

a)

\(\dfrac{13}{x-1}\in Z\\ \Rightarrow\left(x-1\right)\inƯ\left(13\right)\\ \Rightarrow\left(x-1\right)\in\left\{1;-1;13;-13\right\}\\ \Rightarrow x\in\left\{2;0;14;-12\right\}\)

b) 

\(\dfrac{x+3}{x-2}=\dfrac{x-2+5}{x-2}=\dfrac{x-2}{x-2}+\dfrac{5}{x-2}=1+\dfrac{5}{x-2}\\ 1+\dfrac{5}{x-2}\in Z\\ \Rightarrow\dfrac{5}{x-2}\in Z\\ \Rightarrow\left(x-2\right)\inƯ\left(5\right)\\ \Rightarrow\left(x-2\right)\in\left\{1;-1;5;-5\right\}\\ \Rightarrow x\in\left\{3;1;7;-3\right\}\)

tham khảo 

https://olm.vn/hoi-dap/detail/99049659825.html

a) Để phân số \(\dfrac{13}{x-1}\) là số nguyên thì \(13⋮x-1\)

\(\Leftrightarrow x-1\inƯ\left(13\right)\)

\(\Leftrightarrow x-1\in\left\{1;-1;13;-13\right\}\)

hay \(x\in\left\{2;0;14;-12\right\}\)

Vậy: Để phân số \(\dfrac{13}{x-1}\) là số nguyên thì \(x\in\left\{2;0;14;-12\right\}\)

b) Để phân số \(\dfrac{x+3}{x-2}\) là số nguyên thì \(x+3⋮x-2\)

\(\Leftrightarrow x-2+5⋮x-2\)

mà \(x-2⋮x-2\)

nên \(5⋮x-2\)

\(\Leftrightarrow x-2\inƯ\left(5\right)\)

\(\Leftrightarrow x-2\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{3;1;7;-3\right\}\)

Vậy: Để phân số \(\dfrac{x+3}{x-2}\) là số nguyên thì \(x\in\left\{3;1;7;-3\right\}\)

x = 30, y = 63, z = 168;

\(\dfrac{x}{42}=\dfrac{45}{y}=\dfrac{120}{z}=\dfrac{15}{21}\\ \cdot\dfrac{x}{42}=\dfrac{15}{21}\Rightarrow x=\dfrac{42.15}{21}=30\\ \cdot\dfrac{45}{y}=\dfrac{15}{21}\Rightarrow y=\dfrac{45.21}{15}=63\\ \cdot\dfrac{120}{z}=\dfrac{15}{21}\Rightarrow z=\dfrac{120.21}{15}=168\)

Vậy \(x=30;y=63;z=168\)

Ta có: \(\dfrac{x}{42}=\dfrac{45}{y}=\dfrac{120}{z}=\dfrac{15}{21}\)

\(\Leftrightarrow\dfrac{x}{42}=\dfrac{45}{y}=\dfrac{120}{z}=\dfrac{5}{7}\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x}{42}=\dfrac{5}{7}\\\dfrac{45}{y}=\dfrac{5}{7}\\\dfrac{120}{z}=\dfrac{5}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{42\cdot5}{7}=30\\y=\dfrac{45\cdot7}{5}=63\\z=\dfrac{120\cdot7}{5}=168\end{matrix}\right.\)

Vậy: (x,y,z)=(30;63;168)