a: \(\dfrac{1}{8}\cdot\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{1}{10}\cdot\dfrac{5}{3}=\dfrac{1}{2\cdot3}=\dfrac{1}{6}\)

b: \(=\dfrac{8-3}{12}\cdot\dfrac{6}{5}=\dfrac{6}{12}=\dfrac{1}{2}\)

c: \(=\dfrac{24}{35}:\dfrac{32}{35}=\dfrac{3}{4}\)

d: =63/21+15/21-7/21=71/21

a \(\dfrac{1}{8}\times\dfrac{4}{5}\times\dfrac{10}{6}=\dfrac{1}{6}\)

b \(\left(\dfrac{8}{12}-\dfrac{3}{12}\right)\times\dfrac{6}{5}=\dfrac{5}{12}\times\dfrac{6}{5}=\dfrac{1}{2}\)

c \(\dfrac{24}{35}:\dfrac{32}{35}=\dfrac{24}{35}\times\dfrac{35}{32}=\dfrac{3}{4}\)

d \(\dfrac{63}{21}+\dfrac{15}{21}-\dfrac{7}{21}=\dfrac{71}{21}\)

a) \(\left(x+1\right)^3-\left(x-1\right)^3-6\cdot\left(x-1\right)^2=10\)

\(\Rightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\cdot\left(x^2-2x+1\right)=10\)

\(\Rightarrow6x^2+2-6x^2+12x-6=10\)

\(\Rightarrow12x-4=10\)

\(\Rightarrow12x=14\)

\(\Rightarrow x=\dfrac{7}{6}\)

b) \(x\left(x+5\right)\left(x-5\right)-\left(x+2\right)\left(x^2-2x+4\right)=42\)

\(\Rightarrow x\left(x^2-25\right)-\left(x^3+8\right)=42\)

\(\Rightarrow x^3-25x-x^3-8=42\)

\(\Rightarrow-25x-8=42\)

\(\Rightarrow-25x=50\)

\(\Rightarrow x=\dfrac{50}{-25}=-2\)

c) \(\left(x-2\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=49\)

\(\Rightarrow x^3-6x^2+12x-8-\left(x^3-27\right)+6\left(x^2+2x+1\right)=49\)

\(\Rightarrow x^3-6x^2+12x-8-x^3+27+6x^2+12x+6=49\)

\(\Rightarrow24x+25=49\)

\(\Rightarrow24x=24\)

\(\Rightarrow x=\dfrac{24}{24}=1\)

a: Ta có: \(5\left(4x-1\right)+2\left(1-3x\right)-6\left(x+5\right)=10\)

\(\Leftrightarrow20x-5+2-6x-6x-30=10\)

\(\Leftrightarrow8x=43\)

hay \(x=\dfrac{43}{8}\)

b: ta có: \(2x\left(x+1\right)+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)+6x^2=0\)

\(\Leftrightarrow2x^2+2x+3x^2-3-5x^2-5x+6x^2=0\)

\(\Leftrightarrow6x^2-3x-3=0\)

\(\Leftrightarrow2x^2-x-1=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)

câu c,d đâu 

a: \(\dfrac{4}{5}-\dfrac{5}{6}< =\dfrac{x}{30}< =\dfrac{1}{3}-\dfrac{3}{10}\)

=>\(\dfrac{24-25}{30}< =\dfrac{x}{30}< =\dfrac{10-9}{30}\)

=>\(\dfrac{-1}{30}< =\dfrac{x}{30}< =\dfrac{1}{30}\)

=>-1<=x<=1

mà x nguyên

nên \(x\in\left\{-1;0;1\right\}\)

b: \(\dfrac{a}{7}+\dfrac{1}{14}=\dfrac{-1}{b}\)

=>\(\dfrac{2a+1}{14}=\dfrac{-1}{b}\)

=>\(\left(2a+1\right)\cdot b=-14\)

mà 2a+1 lẻ (do a là số nguyên)

nên \(\left(2a+1\right)\cdot b=1\cdot\left(-14\right)=\left(-1\right)\cdot14=7\cdot\left(-2\right)=\left(-7\right)\cdot2\)

=>\(\left(2a+1;b\right)\in\left\{\left(1;-14\right);\left(-1;14\right);\left(7;-2\right);\left(-7;2\right)\right\}\)

=>\(\left(a;b\right)\in\left\{\left(0;-14\right);\left(-1;14\right);\left(3;-2\right);\left(-4;2\right)\right\}\)

.

Xin lỗi m.n nhé gửi nhầm tí

Bài 1:

a: 76-6(x-1)=10

\(\Leftrightarrow x-1=11\)

hay x=12

c: \(5x+15⋮x+2\)

\(\Leftrightarrow x+2=5\)

hay x=3

Bài 1:

a) 76 - 6 (x - 1) = 10

           6 (x - 1) = 76 - 10

           6 (x - 1) = 66

               x - 1 = 66 : 6

               x - 1 = 11

               x      = 11 + 1

               x = 12

b) 3 . 4³ - 7 - 185

= 3 . 64 - 7 - 185

= 192 - 7 - 185

= 185 - 185

= 0

\(\dfrac{-6}{x}=\dfrac{-5}{10}\\ x=\dfrac{10.\left(-6\right)}{5-}=12\\ x=\dfrac{1}{4}+\dfrac{3}{2}=\dfrac{7}{4}\)

a) \(\dfrac{-6}{x}=\dfrac{-5}{10}\)

\(x=12\)

b) \(x-\dfrac{3}{2}=\dfrac{1}{4}\)

   \(x=\dfrac{1}{4}+\dfrac{3}{2}\)

  \(x=\dfrac{7}{4}\)

\(a,PT\Leftrightarrow x^3-6x^2+12x-8-x^3+x+6x^2-18x-10=0\)

\(\Leftrightarrow-5x-18=0\)

\(\Leftrightarrow x=-\dfrac{18}{5}\)

Vậy ...

\(b,PT\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+10=0\)

\(\Leftrightarrow12x+6=0\)

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

Vậy ...

\(c,PT\Leftrightarrow\left(x+1\right)^3+3^3=0\)

\(\Leftrightarrow\left(x+1+3\right)\left(x^2+2x+1-3x-3+9\right)=0\)

\(\Leftrightarrow\left(x+4\right)\left(x^2-x+7\right)=0\)

Thấy : \(x^2-\dfrac{2.x.1}{2}+\dfrac{1}{4}+\dfrac{27}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}>0\)

\(\Rightarrow x+4=0\)

\(\Leftrightarrow x=-4\)

Vậy ...

\(d,PT\Leftrightarrow\left(x-2\right)^3+1^3=0\)

\(\Leftrightarrow\left(x-2+1\right)\left(x^2-4x+4-x+2+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+7\right)=0\)

Thấy : \(x^2-5x+7=x^2-\dfrac{5.x.2}{2}+\dfrac{25}{4}+\dfrac{3}{4}=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)

\(\Rightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy ...

sao lại trả lời lại nhỉ ??

a: =>4x^2-24x+36-4x^2+4x-1<10

=>-20x<10-35=-25

=>x>=5/4

b: =>x(x^2-25)-x^3-8<=3

=>x^3-25x-x^3-8<=3

=>-25x<=11

=>x>=-11/25