`x.(-3/7)=5/21`

`x=5/21:(-3/7)`

`x=5/21 . (-7/3)`

`=-5/9`

a. \(\dfrac{1}{7}=\dfrac{2}{14}=\dfrac{\left(-2\right)\left(-1\right)}{7.2}=\dfrac{-2}{7}.\dfrac{-1}{2}\)

b. \(\dfrac{1}{7}=\dfrac{-2}{7}.\dfrac{-1}{2}=\dfrac{-2}{7}:\dfrac{2}{-1}=\dfrac{-2}{7}:\left(-2\right)\)

a. \(\dfrac{-11}{84}=\dfrac{\left(-1\right).11}{42.2}=\dfrac{-1}{42}.\dfrac{11}{2}\)

b. \(\dfrac{-11}{84}=\dfrac{-1}{42}:\dfrac{2}{11}\)

Ta có :

\(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{99.101}\)

\(\Rightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)

\(\Rightarrow2A=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{101-99}{99.101}\)

\(\Rightarrow2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

\(\Rightarrow2A=1-\dfrac{1}{101}< 1\Rightarrow A< \dfrac{1}{2}=B\)

A=1.31​+3.51​+5.71​+...+99.1011​

\Rightarrow2A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}⇒2A=1.32​+3.52​+5.72​+...+99.1012​

\Rightarrow2A=\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{101-99}{99.101}⇒2A=1.33−1​+3.55−3​+...+99.101101−99​

\Rightarrow2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}⇒2A=1−31​+31​−51​+...+991​−1011​

\Rightarrow2A=1-\dfrac{1}{101}< 1\Rightarrow A< \dfrac{1}{2}=B⇒2A=1−1011​<1⇒A<21​=B

Nhân chia trước cộng trừ sau, số thập phân đổi ra phân số hoặc ngược lại, cái nào bạn thấy thuận tiện thì đổi. Sau thực hiện các phép tính theo bình thường

a, \(\dfrac{x+2}{x-1}\) ϵ Z ⇔ \(\dfrac{x+2}{x-1}\) = 1 + \(\dfrac{3}{x-1}\) ϵ Z ⇔ x - 1  ϵ Ư(3) = {-3;-1;1;3}

⇔ x ϵ { -2; 0; 2; 4}

b, \(\dfrac{x+2}{2x-1}\) ϵ Z ⇔ x + 2 = 2x - 1 hoặc x + 2 = -2x + 1

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

c, \(\dfrac{x-2}{x^{2^{ }}+5}\) ϵ Z ⇔ x - 2 = x² + 5 hoặc x-2 = -x² - 5

x - 2 = x² + 5 ⇔ x² - x + 7 = 0 ⇔ (x-1/2)² + 27/4 = 0 (vô nghiệm)

x - 2 = -x² - 5 ⇔x² + x + 3 = 0 ⇔ (x+1/2)² + 11/4 = 0 (vô nghiệm)

x ϵ \(\varnothing\)

d, \(\dfrac{x^2+3}{x-1}\) = x + 1 + \(\dfrac{4}{x-1}\) ϵ Z ⇔ x - 1 ϵƯ(4) ={-4;-2;-1;1;2;4}

⇔ x ϵ{ -3;-1;0;2; 3; 5}

e, \(\dfrac{x^2+2x-1}{x+1}\) = x + 1 - \(\dfrac{2}{x+1}\) ϵ Z ⇔ x+1 ϵƯ(2) ={-2;-1;1;2}

⇔ x ϵ { -3;-2;0;1}

f, \(\dfrac{2x-5}{3x+2}\) ϵ Z ⇔ 2x - 5 = 3x + 2 hoặc 2x -5 = -3x - 2

2x - 5 = 3x + 2 ⇔ x = -7

2x - 5 = -3x -2 ⇔ x =3/5 (loại)

`|5x-4|=|x+4|`

`=>5x-4=x+4` hoặc `5x-4=-(x+4)`

`=>5x-x=4+4` hoặc `5x+x=4-4`

`=>4x=8` hoặc `6x=0`

`=>x=2` hoặc `x=0`

0

\(\left(d\right):\dfrac{x-2}{7}=\dfrac{7}{21}\\ =>21\left(x-2\right)=49\\ =>x-2=\dfrac{49}{21}=\dfrac{7}{3}\\ =>x=\dfrac{7}{3}+2=\dfrac{13}{3}\\ \\ \left(d\right):x-\dfrac{2}{7}=\dfrac{7}{21}\\ =>x=\dfrac{2}{7}+\dfrac{7}{21}=\dfrac{6}{21}+\dfrac{7}{21}\\ =>x=\dfrac{13}{21}\)

\(\left(f\right):\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\\ =>\dfrac{11}{12}-\dfrac{2}{5}-x=\dfrac{2}{3}\\ =>\dfrac{31}{60}-x=\dfrac{2}{3}\\ =>x=\dfrac{31}{60}-\dfrac{2}{3}=-\dfrac{3}{20}\)

Em chú ý gõ công thức trực quan để rõ đề bài nhé. Các bạn trợ giúp sẽ dễ dàng hơn

\(\left(x^2-4\right)\left(x^2-3\right)=0\)

\(\left(x-2\right)\left(x+2\right)\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=0\)

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

\(\left(x^2-4\right).\left(x^2-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-4=0\\x^2-3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=4\\x^2=3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm2\\x=\pm\sqrt{3}\end{matrix}\right.\)

\(x^2-64=0\Leftrightarrow x^2-8^2=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+8\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-8=0\\x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

Đs:...

thanks