2:

a: x=2,4-0,4=2

b: =>2x=-1,5+0,8=-0,7

=>x=-0,35

c: =>x-16=-15

=>x=1

Bài 2: 

\(=\dfrac{28}{25}\cdot\dfrac{15}{7}\cdot5=\dfrac{75}{25}\cdot4=12\)

Bài 1: 

a: \(x+\dfrac{7}{8}=\dfrac{13}{2}:4=\dfrac{13}{8}\)

nên x=13/8-7/8=6/8=3/4

b: \(x:\dfrac{5}{3}=\dfrac{6}{5}-\dfrac{2}{3}=\dfrac{18-10}{15}=\dfrac{8}{15}\)

nên \(x=\dfrac{8}{15}\cdot\dfrac{5}{3}=\dfrac{8}{9}\)

1)

a) \(1\dfrac{5}{6}=\dfrac{-x}{5}\)

\(\Rightarrow\dfrac{11}{6}=\dfrac{-x}{5}\)

\(\Rightarrow-x=\dfrac{5.11}{6}=\dfrac{55}{6}\)

\(\Rightarrow x=-\dfrac{55}{6}\)

b) 4,25 : 8 = -3,5 : x

\(\dfrac{4,25}{8}=\dfrac{-3,5}{x}\)

\(x=\dfrac{-3,5.8}{4,25}\)

\(x=\dfrac{-28}{4,25}\)

2.

\(-\dfrac{12}{1,6}=\dfrac{55}{-7\dfrac{1}{3}}\)

\(\Rightarrow-\dfrac{12}{1,6}=\dfrac{55}{-\dfrac{22}{3}}\)

Ta có thể lặp đc các tỉ lệ thức sau:

\(-\dfrac{12}{1,6}=\dfrac{55}{-\dfrac{22}{3}}\)

\(\dfrac{-\dfrac{22}{3}}{1,6}=\dfrac{55}{-12}\)

\(-\dfrac{12}{55}=\dfrac{1,6}{-\dfrac{22}{3}}\)

\(\dfrac{1,6}{-12}=\dfrac{-\dfrac{22}{3}}{55}\)

1: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)

\(\Leftrightarrow\dfrac{5x^2-12}{\left(x-1\right)\left(x+1\right)}+\dfrac{3x+3}{\left(x-1\right)\left(x+1\right)}=\dfrac{5x^2-5x}{\left(x+1\right)\left(x-1\right)}\)

Suy ra: \(5x^2+3x-9=5x^2-5x\)

\(\Leftrightarrow8x=9\)

hay \(x=\dfrac{9}{8}\left(tm\right)\)

2: Ta có: \(\dfrac{3}{x-5}-\dfrac{15-3x}{x^2-25}=\dfrac{3}{x+5}\)

\(\Leftrightarrow\dfrac{3x+15}{\left(x-5\right)\left(x+5\right)}+\dfrac{3x-15}{\left(x-5\right)\left(x+5\right)}=\dfrac{3x-15}{\left(x+5\right)\left(x-5\right)}\)

Suy ra: \(6x=3x-15\)

\(\Leftrightarrow3x=-15\)

hay \(x=-5\left(loại\right)\)

2. ĐKXĐ: $x\neq \pm 5$
PT \(\Leftrightarrow \frac{3}{x-5}+\frac{3x-15}{x^2-25}=\frac{3}{x+5}\)

\(\Leftrightarrow \frac{3}{x-5}+\frac{3(x-5)}{(x-5)(x+5)}=\frac{3}{x+5}\)

\(\Leftrightarrow \frac{3}{x-5}+\frac{3}{x+5}=\frac{3}{x+5}\Leftrightarrow \frac{3}{x-5}=0\) (vô lý)

Vậy pt vô nghiệm.

3. ĐKXĐ: $x\neq \pm 4$
PT \(\Leftrightarrow \frac{-3(x+4)}{(x-4)(x+4)}-\frac{3-5x}{(x-4)(x+4)}=\frac{x-4}{(x-4)(x+4)}\)

\(\Rightarrow -3(x+4)-(3-5x)=x-4\)

\(\Leftrightarrow 2x-15=x-4\Leftrightarrow x=11\) (thỏa mãn)

Bài 1 : Ta thấy

\(\dfrac{10}{15}=\dfrac{2}{3};\dfrac{14}{21}=\dfrac{2}{3}\Rightarrow10:15=14:21\Rightarrow\dfrac{10}{15}=\dfrac{14}{21}\)

\(\dfrac{16}{\left(-4\right)}=-4;\dfrac{12}{\left(-3\right)}=-4\Rightarrow16:\left(-4\right)=12:\left(-3\right)\Rightarrow\dfrac{16}{\left(-4\right)}=\dfrac{12}{\left(-3\right)}=-4\)

\(\dfrac{\left(-5\right)}{15}=\dfrac{\left(-1,2\right)}{3,6}=-\dfrac{1}{3}\Rightarrow\left(-5\right):15=\left(-1,2\right):3,6\)

\(\dfrac{2}{3}:\dfrac{1}{4}=\dfrac{2}{3}.4=\dfrac{8}{3};\dfrac{16}{9}:\dfrac{16}{24}=\dfrac{16}{9}.\dfrac{24}{16}=\dfrac{8}{3}\)

\(\Rightarrow\left(\dfrac{2}{3}:\dfrac{1}{4}\right)=\left(\dfrac{16}{9}:\dfrac{16}{24}\right)=\dfrac{8}{3}\)

Bài 2 :

a) \(14.15=10.21\Rightarrow\dfrac{14}{10}=\dfrac{21}{15}=\dfrac{7}{5}\)

b) \(0,2.4,5=0,6.1,5\Rightarrow\dfrac{0,2}{0,6}=\dfrac{1,5}{4,5}=\dfrac{1}{3}\)

1, \(x^2+2xy+y^2=\left(x+y\right)^2\)

2, \(4x^2+12x+9=\left(2x\right)^2+2\cdot3\cdot2x+3^2=\left(2x+3\right)^2\)

3, \(x^2+5x+\dfrac{25}{4}=x^2+2\cdot\dfrac{5}{2}\cdot x+\left(\dfrac{5}{2}\right)^2=\left(x+\dfrac{5}{2}\right)^2\)

4, \(16x^2-8x+1=\left(4x\right)^2-2\cdot4x\cdot1+1^2=\left(4x-1\right)^2\)

5, \(x^2+x+\dfrac{1}{4}=x^2+2\cdot\dfrac{1}{2}\cdot x+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)

1: =(x+y)^2

2: =(2x+3)^2

3: =(x+5/2)^2

4: =(4x-1)^2

5: =(x+1/2)^2

6: =(x-3/2)^2

7: =(x+1)^3

8: =(1/2x+1)^2

9: =(3y-1/3)^3

10: =(2x+y)^3

6, \(x^2-3x+\dfrac{9}{4}=x^2-2\cdot\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^2=\left(x-\dfrac{3}{2}\right)^2\)

7, \(x^3+3x^2+3x+1=x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=\left(x+1\right)^3\)

8, \(\dfrac{x^2}{4}+x+1=\left(\dfrac{x}{2}\right)^2+2\cdot\dfrac{x}{2}\cdot1+1^2=\left(\dfrac{x}{2}+1\right)^2\)

9, \(27y^3-9y^2+y-\dfrac{1}{27}=\left(3y\right)^3-3\cdot\left(3y\right)^2\cdot\dfrac{1}{3}+3\cdot3y\cdot\left(\dfrac{1}{3}\right)^2-\left(\dfrac{1}{3}\right)^3=\left(3y-\dfrac{1}{3}\right)^3\)

10, \(8x^3+12x^2y+6xy^2+y^3=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2+y^3=\left(2x+y\right)^3\)

a) \(\dfrac{-2}{5}=\dfrac{-14}{35}\)

b) \(\dfrac{9}{4}=\dfrac{\dfrac{1}{4}}{\dfrac{1}{9}}\)

c) \(\dfrac{\dfrac{7}{4}}{\dfrac{1}{2}}=\dfrac{7}{2}\)