Bài 1

a/ $\frac{-3}{7}-\left(\frac{2}{3}-\frac{3}{7}\right)$

$\backslash (=\backslash frac\{-3\}\{7\}-\backslash frac\{2\}\{3\}+\backslash frac\{3\}\{7\}\backslash )$

\(=\left(\frac{-3}{7}+\frac{3}{7}\right)-\frac{2}{3}\)

\(=0-\frac{2}{3}\)

\(=\frac{-2}{3}\)

Bài 1:

a) Ta có:

\(3,2\cdot x+\left(-1,2\right)\cdot x+2,7=-4,9\)

\(\Rightarrow\left[3,2+\left(-1,2\right)\right]\cdot x=\left(-4,9\right)-2,7\)

\(\Rightarrow2x=-7,6\)

\(\Rightarrow x=\left(-7,6\right):2\)

\(\Rightarrow x=-3,8\)

Vậy \(x=-3,8\)

b) Ta có:

-5,6.x+2,9.x-3,86=-9,8

=>[(-5,6)+2,9].x=(-9,8)+3,86

=>(-2,7).x=-5,94

=>x=(-5,94):(-2,7)

=>x=2,3

Vậy x=2,2

Dễ mà. ấn máy tính cũng ra

a) \(\frac{-3}{7}-\left(\frac{2}{3}-\frac{3}{7}\right)\)

\(=\frac{-3}{7}-\frac{2}{3}+\frac{3}{7}\)

\(=\left(\frac{3}{7}-\frac{3}{7}\right)-\frac{2}{3}\)

\(=0-\frac{2}{3}\)

\(=-\frac{2}{3}\)

b) \(\frac{2}{15}\div\left(\frac{1}{3}.\frac{4}{5}-\frac{1}{3}.\frac{6}{5}\right)\)

\(=\frac{2}{15}\div\left[\frac{1}{3}\left(\frac{4}{5}-\frac{6}{5}\right)\right]\)

\(=\frac{2}{15}\div\left[\frac{1}{3}.\frac{-2}{5}\right]\)

\(=\frac{2}{15}\div\frac{-2}{15}\)

\(=\frac{2}{15}\times\frac{15}{-2}\)

\(=\frac{2}{-2}=-1\)

\(\frac{3}{4}+\frac{1}{4}x=\frac{5}{8}\)

\(\Leftrightarrow\frac{1}{4}x=\frac{5}{8}-\frac{3}{4}\)

\(\Leftrightarrow\frac{1}{4}x=\frac{5}{8}-\frac{6}{8}\)

\(\Leftrightarrow\frac{1}{4}x=\frac{-1}{8}\)

\(\Leftrightarrow x=\frac{-1}{8}\div\frac{1}{4}\)

\(\Leftrightarrow x=\frac{-1}{8}\times\frac{4}{1}\)

​\(\Leftrightarrow x=\frac{-4}{8}\)​

\(\Leftrightarrow x=\frac{-1}{2}\)

\(25\%x+x=-1,25\)

\(\Leftrightarrow\frac{25}{100}x+x=\frac{-125}{100}\)

\(\Leftrightarrow\frac{1}{4}x+x=\frac{-25}{20}\)

\(\Leftrightarrow x\left(\frac{1}{4}+1\right)=\frac{-5}{4}\)

\(\Leftrightarrow x\left(\frac{1}{4}+\frac{4}{4}\right)=-\frac{5}{4}\)

\(\Leftrightarrow\frac{5}{4}x=\frac{-5}{4}\)

\(\Leftrightarrow x=\frac{-5}{4}\div\frac{5}{4}\)

\(\Leftrightarrow x=\frac{-5}{4}\times\frac{4}{5}\)

\(\Leftrightarrow x=\frac{-5}{5}=-1\)

Bài 1: Ta có:

a + b = a.b => a = a.b - b = b.(a - 1) (1)

=> a : b = a - 1 = a + b

=> b = -1

Thay b = -1 vào (1) ta có: a = -1.(a - 1) = -a + 1

=> a + a = 1 = 2a

\(\Rightarrow a=\frac{1}{2}\)

Vậy \(a=\frac{1}{2};b=-1\)

b) \(\frac{5}{x}+\frac{y}{4}=\frac{1}{8}\)

\(\Rightarrow\frac{5}{x}=\frac{1}{8}-\frac{y}{4}=\frac{1-2y}{8}\)

=> (1 - 2y).x = 40

\(\Rightarrow40⋮1-2y\)

Mà 1 - 2y là số lẻ \(\Rightarrow1-2y\in\left\{1;-1;5;-5\right\}\)

Ta có bảng sau:

Vậy các cặp giá trị (x;y) thỏa mãn đề bài là: (40;0) ; (-40;1) ; (8;-2) ; (-8;3)