Bài 8 : Tính chất cơ bản của phép cộng phân số

a: \(=\dfrac{-3}{7}-\dfrac{4}{7}+\dfrac{5}{13}=\dfrac{5}{13}-1=-\dfrac{8}{13}\)

b: \(=\dfrac{-5}{21}-\dfrac{2}{21}+\dfrac{1}{3}=\dfrac{-7}{21}+\dfrac{1}{3}=0\)

c: \(=\dfrac{-1}{2}+\dfrac{4}{5}+\dfrac{2}{3}-\dfrac{2}{3}+\dfrac{7}{9}\)

\(=\dfrac{-45}{90}+\dfrac{72}{90}+\dfrac{70}{90}=\dfrac{97}{90}\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

Vậy: \(x\in\left\{5;-3\right\}\)

b) Ta có: \(\dfrac{x-4}{6}=\dfrac{-1}{3}\)

\(\Leftrightarrow x-4=-2\)

hay x=2

Vậy: x=2

a/ 

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

\(x+\dfrac{1}{4}+\dfrac{4}{x-1}=0\)

\(\dfrac{x\left(x-1\right)4}{4\left(x-1\right)}+\dfrac{16}{4\left(x-1\right)}=0\)

\(4x\left(x-1\right)+16=0\)(quy tắc khử mẫu lớp 8)

\(4x^2-4x+16=0\)

\(4x^2-2x-2x+16=0\)

\(\left(4x^2-2x\right)-\left(2x-16\right)=0\)

\(2x\left(2x-1\right)-2\left(x-16\right)=0\)

a: \(VT=\dfrac{1}{a+1}+\dfrac{1}{a}-\dfrac{1}{a+1}=\dfrac{1}{a}\)=VP

b: \(VP=\dfrac{a+1-a}{a\left(a+1\right)}=\dfrac{1}{a\left(a+1\right)}=VP\)

\(\dfrac{7}{8}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}\)

\(\dfrac{11}{12}=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}\)

\(\dfrac{14}{24}=\dfrac{7}{12}=\dfrac{1}{6}+\dfrac{1}{3}+\dfrac{1}{12}\)

Viết phân số ¾ dưới dạng tổng có tử số bằng 1 mẫu số khác nhau

\(A=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{99.100}\\ =\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{100}\\ =\dfrac{1}{2}-\dfrac{1}{100}=\dfrac{49}{100.}\)

\(\Rightarrow A=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2019}-\dfrac{1}{2020}\)

\(=1-\dfrac{1}{2020}=\dfrac{2019}{2020}\)

Vậy \(A=\dfrac{2019}{2020}\)

\(a)\dfrac{-5}{13}+\left(-\dfrac{8}{13}+1\right)\\ =\dfrac{-5}{13}+\dfrac{-8}{13}+1\\ =0+1=1\)

\(b)\dfrac{2}{3}+\left(\dfrac{3}{8}+\dfrac{-2}{3}\right)\\ =\dfrac{2}{3}-\dfrac{2}{3}+\dfrac{3}{8}\\ =\dfrac{3}{8}\)

\(c)\left(\dfrac{-3}{4}+\dfrac{5}{8}\right)+\dfrac{-1}{8}=\dfrac{-3}{4}+\dfrac{4}{8}\\=\dfrac{-6}{8}+\dfrac{4}{8}=\dfrac{-2}{8}=\dfrac{-1}{4}\)

Lời giải:

Gọi 2 số dương đề cập ở đề là $a$ và $b$. Theo bài ra ta có:

$\frac{a}{b}=\frac{7}{2}$ và $ab=21$

$\Rightarrow a=\frac{7}{2}b$ và $ab=21$

$\Rightarrow \frac{7}{2}b.b=21$

$\Rightarrow b^2=6$

$a^2=(\frac{7}{2}b)^2=\frac{49}{4}.b^2=\frac{147}{2}$

Tổng bình phương 2 số:

$a^2+b^2=\frac{147}{2}+6=\frac{159}{2}$