1)

a) \(=3x^2\left(x^2-1\right)-\left(x^3-1\right)+x^8-3x^4+3x^2-1\)

\(=3x^4-3x^2-x^3+1+x^8-3x^4+3x^2-1=x^8-x^3\)

2) 

\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)-6\left(x^2+5x\right)+45\)

\(=\left(x^2+5x\right)^2-6\left(x^2+5x\right)-36+45\)

\(=\left(x^2+5x\right)^2-6\left(x^2+5x\right)+9=\left(x^2+5x-3\right)^2\)

a) \(\dfrac{x}{-6}=\dfrac{-15}{45}\)

    \(\dfrac{-x}{6}=\dfrac{-15}{45}\)

    \(\dfrac{x}{6}=\dfrac{15}{45}\)

    \(x=\dfrac{\left(15\cdot6\right)}{45}\)

    \(x=2\)

b) \(\dfrac{x}{5}=\dfrac{16}{25}\)

    \(x=\dfrac{\left(16\cdot5\right)}{25}\)

    \(x=\dfrac{80}{25}\)

   \(x=\dfrac{16}{5}\)

c) \(\dfrac{5}{x-3}=\dfrac{20}{-12}\)

  \(x-3=\dfrac{\left(5\cdot-12\right)}{20}\)

 \(x-3=-3\)

 \(x=\left(-3\right)+3\)

\(x=0\)

d) \(\dfrac{2}{5}\cdot x=\dfrac{6}{35}\)

         \(x=\dfrac{6}{35}\div\dfrac{2}{5}\)

        \(x=\dfrac{3}{7}\)

a) x(x - 5) - 4x + 20 = 0

\(\Leftrightarrow\) x(x - 5) - (4x + 20)

\(\Leftrightarrow\) x(x - 5) - 4(x - 5) = 0

\(\Leftrightarrow\) (x - 5)(x - 4)

Khi x - 5 = 0 hoặc x - 4 = 0

 \(\Leftrightarrow\) x = 5           \(\Leftrightarrow\) x = 4

 Vậy S = \(\left\{5;4\right\}\)

b) x(x + 6) - 7x - 42 = 0

 \(\Leftrightarrow\) x(x + 6) - (7x - 42) = 0

 \(\Leftrightarrow\) x(x + 6) - 7(x + 6) = 0

 \(\Leftrightarrow\) (x + 6)(x - 7) = 0

Khi x - 6 = 0 hoặc x - 7 = 0

   \(\Leftrightarrow\) x = 6           \(\Leftrightarrow\) x = 7

 Vậy S = \(\left\{6;7\right\}\)

c) x³ - 5x² - x + 5 = 0

 \(\Leftrightarrow\) (x³ - 5x²) - (x + 5) = 0

 \(\Leftrightarrow\) x² (x - 5) - (x - 5) = 0

 \(\Leftrightarrow\) (x - 5)(x² - 1) = 0

 \(\Leftrightarrow\) (x - 5)(x - 1)(x + 1) = 0

 Khi x - 5 = 0 hoặc x - 1 = 0 hoặc x + 1 = 0

   \(\Leftrightarrow\) x = 5           \(\Leftrightarrow\) x = 1            \(\Leftrightarrow\) x = -1

 Vậy S = \(\left\{5;1;-1\right\}\)

d) 4x² - 25 - (2x - 5)(3x + 7) = 0

\(\Leftrightarrow\) (2x)² - 5² - (2x - 5)(3x + 7) = 0

\(\Leftrightarrow\) (2x - 5)(2x + 5) - (2x - 5)(3x + 7) = 0

\(\Leftrightarrow\) (2x - 5) \([\left(2x+5\right)-\left(3x+7\right)]\) = 0

\(\Leftrightarrow\) (2x - 5) ( 2x + 5 - 3x + 7) = 0

\(\Leftrightarrow\) (2x - 5)( -x + 12) = 0

Khi 2x - 5 = 0 hoặc -x + 12 = 0

  \(\Leftrightarrow\) 2x = 5             \(\Leftrightarrow\)   -x = -12

  \(\Leftrightarrow\) x = \(\dfrac{5}{2}\)              \(\Leftrightarrow\) x = 12

 Vậy S = \(\left\{\dfrac{5}{2};12\right\}\)

e) x³ + 27 + (x + 3)(x - 9) = 0

\(\Leftrightarrow\) x³ - 3³ + (x + 3)(x - 9) = 0

\(\Leftrightarrow\) (x - 3)(x² - 3x + 9) + (x + 3)(x - 9) = 0

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

\(\Leftrightarrow\) (x - 3) ( x² - 3x + 9 + x - 9) = 0

\(\Leftrightarrow\) (x - 3)(x² - 2x) = 0

\(\Leftrightarrow\) (x - 3)x(x - 2)

 Khi x - 3 = 0 hoặc x = 0 hoặc x - 2 = 0

    \(\Leftrightarrow\) x = 3                            \(\Leftrightarrow\) x = 2

 Vậy S = \(\left\{3;0;2\right\}\)

 Chúc bạn học tốt

a) Ta có: \(x\left(x-5\right)-4x+20=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)

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

b) Ta có: \(x\left(x+6\right)-7x-42=0\)

\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)

\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=7\end{matrix}\right.\)

c) Ta có: \(x^3-5x^2-x+5=0\)

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

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

d) Ta có: \(4x^2-25-\left(2x-5\right)\left(3x+7\right)=0\)

\(\Leftrightarrow\left(2x-5\right)\left(2x+5-3x-7\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-2\end{matrix}\right.\)

\(b,\left(x+2\right)^2-25\)

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

\(=\left(x-3\right)\left(x+7\right)\)

\(c,36\left(x-y\right)^2\)

\(=36\left(x^2-2xy+y^2\right)\)

\(=36x^2-72xy+36y^2\)

\(d,x^2+\dfrac{1}{2}x+\dfrac{1}{16}\)

\(=x^2+2.x.\dfrac{1}{4}+\dfrac{1}{4}^2\)

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

\(e,2x^4y^3-3x^2y^4+5x^3y^4\)

\(=x^2y^3\left(2x^2-3y+5xy\right)\)

Các câu còn lại làm tương tự, chú ý sd HĐT

\((x-6)(3x-9)>0\)
TH1:
\(\orbr{\begin{cases}x-6< 0\\3x-9< 0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x< 6\\x< 3\end{cases}}\)\(\Rightarrow x< 3\)
TH2:
\(\orbr{\begin{cases}x-6>0\\3x-9>0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x>6\\x>3\end{cases}}\)\(\Rightarrow x>6\)
Vậy \(x< 3\) hoặc \(x>6\)thì \((x-6)(3x-9)>0\)
Học tốt!

20.
\((2x-1)(6-x)>0\)
TH1:
\(\orbr{\begin{cases}2x-1>0\\6-x>0\end{cases}\Rightarrow\orbr{\begin{cases}x< \frac{1}{2}\\x< 6\end{cases}}\Rightarrow x< 6}\)
TH2
\(\orbr{\begin{cases}2x-1< 0\\6-x< 0\end{cases}\Rightarrow\orbr{\begin{cases}x>\frac{1}{2}\\x>6\end{cases}}\Rightarrow x>\frac{1}{2}}\)
Vậy \(x< 6\)hoặc \(x>\frac{1}{2}\)thì \((2x-1)(6-x)>0\)
 

21.
\((2-x)(x+7)< 0\)
TH1.
\(\orbr{\begin{cases}2-x>0\\x+7< 0\end{cases}\Rightarrow\orbr{\begin{cases}x< 2\\x>-7\end{cases}}\Rightarrow-7< x< 2}\)
TH2.
\(\orbr{\begin{cases}2-x< 0\\x+7>0\end{cases}\Rightarrow\orbr{\begin{cases}x>2\\x< -7\end{cases}}\Rightarrow2< x< -7}\)(vô lí)
Vậy \(-7< x< 2\) thì \((2-x)(x+7)< 0\)
 

a: =>x-5/14=6/14-1/14=5/14

=>x=10/14=5/7

b; =>2/9:x=3/6+4/6=7/6

=>x=2/9:7/6=2/9*6/7=4/21

c: =>32:x=8

=>x=4

a

\(x-\dfrac{5}{14}=\dfrac{3}{7}-\dfrac{1}{14}\\ x-\dfrac{5}{14}-\dfrac{3}{7}+\dfrac{1}{14}=0\\ x+\dfrac{1}{14}-\dfrac{5}{14}-\dfrac{6}{14}=0\\ x+\dfrac{1-5-6}{14}=0\\ x-\dfrac{5}{7}=0\\ x=0+\dfrac{5}{7}\\ x=\dfrac{5}{7}\)

b

\(\dfrac{2}{9}:x=\dfrac{1}{2}+\dfrac{2}{3}\\ \dfrac{2}{9}:x=\dfrac{3}{6}+\dfrac{4}{6}\\ \dfrac{2}{9}:x=\dfrac{3+4}{6}=\dfrac{7}{6}\\ x=\dfrac{2}{9}:\dfrac{7}{6}\\ x=\dfrac{2}{9}\times\dfrac{6}{7}=\dfrac{2.3.2}{3.3.7}=\dfrac{4}{21}\)

c

\(\dfrac{6}{32:x}=\dfrac{12}{16}\\ 32:x=6:\dfrac{12}{16}\\ 32:x=6\times\dfrac{16}{12}\\ 32:x=\dfrac{3\times2\times4\times4}{3\times4}\\ 32:x=8\\ x=\dfrac{32}{8}\\ x=4\)

`@` `\text {Answer}`

`\downarrow`

`a,`

`x - 5/14 = 3/7 - 1/14`

`x - 5/14 = 5/14`

`=> x = 5/14 + 5/14`

`=> x = 5/7`

Vậy, `x = 5/7`

`b,`

`2/9 \div x = 1/2 + 2/3`

`2/9 \div x = 7/6`

`x = 2/9 \div 7/6`

`x = 4/21`

Vậy, `x = 4/21`

`c,`

\(\dfrac{6}{32\div x}=\dfrac{12}{16}\)

`6/(32 \div x) = 3/4`

`32 \div x = 6 \div 3/4`

`32 \div x = 8`

` x = 32 \div 8`

`x = 4`

Vậy, `x = 4`

a: =>x=-7/6+5/8=-13/24

b: =>x=-14/25-3/4=-131/100

c: \(x=\dfrac{-33}{26}:\dfrac{-9}{13}=\dfrac{33}{26}\cdot\dfrac{13}{9}=\dfrac{11}{3}\cdot\dfrac{1}{2}=\dfrac{11}{6}\)

d: \(x=\dfrac{4}{9}:\dfrac{5}{3}=\dfrac{4}{9}\cdot\dfrac{3}{5}=\dfrac{12}{45}=\dfrac{4}{15}\)

1: Áp dụng tính chất của DTSBN, ta được:

\(\dfrac{x}{7}=\dfrac{y}{13}=\dfrac{x-y}{7-13}=\dfrac{42}{-6}=-7\)

=>x=-48; y=-91

2: x/y=3/4

=>4x=3y

=>4x-3y=0

mà 2x+y=10

nên x=3 và y=4

3: =>7x-3y=0 và x-y=-24

=>x=18 và y=42

4: =>7x-5y=0 và x+y=24

=>x=10 và y=14

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