`a) x^3 - 9x^2 + 14x = 0`

\(\Rightarrow\) `x^3 - 7x^2 - 2x^2 + 14x = 0`

\(\Rightarrow\) `(x^3 - 2x^2) - (7x^2 - 14x) =0`

\(\Rightarrow\) `x^2.(x - 2) - 7x.(x - 2) =0`

\(\Rightarrow\) `(x^2 - 7x)(x-2)=0`

\(\Rightarrow\) `x.(x-7)(x-2)=0`

\(\Rightarrow\left[\begin{array}{l}x=0\\ x-7=0\\ x-2=0\end{array}\right.\) \(\Rightarrow\left[\begin{array}{l}x=0\\ x=0+7\\ x=0+2\end{array}\right.\) \(\Rightarrow\left[\begin{array}{l}x=0\\ x=7\\ x=2\end{array}\right.\)

Vậy \(x\in\left\lbrace0;7;2\right\rbrace\)

`3.x^3 - 5x^2 + 8x - 4 = 0`

\(\Rightarrow\) `x^3 - x^2 - 4x^2 + 4x + 4x - 4 =0`

\(\Rightarrow\) `x^2 . (x-1) - 4x(x-1) + 4.(x-1) =0`

\(\Rightarrow\) `(x^2 - 4x + 4)(x-1)=0`

\(\Rightarrow\) `(x-2)^2(x-1)=0`

\(\Rightarrow\left[\begin{array}{l}x-2=0\\ x-1=0\end{array}\right.\) \(\Rightarrow\left[\begin{array}{l}x=2\\ x=1\end{array}\right.\)

Vậy \(x\in\left\lbrace2;1\right\rbrace\)

giúp tôi với

1) 2x⁴ - 9x³ + 14x² - 9x + 2 = 0

<=> (2x⁴ - 4x³) - (5x³ - 10x²) + (4x² - 8x) - (x - 2) = 0

<=> 2x³(x - 2) - 5x²(x - 2) + 4x(x - 2) - (x - 2) = 0

<=> (2x³ - 5x² + 4x - 1)(x - 2) = 0

<=> [(2x³ - 2x²) - (3x² - 3x) + (x - 1)](x - 2) = 0

<=> [2x²(x - 1) - 3x(x - 1) + (x - 1)](x - 2) = 0

<=> (2x² - 2x - x + 1)(x - 1)(x - 2) = 0

<=> (2x - 1)(x - 1)²(x - 2) = 0

<=> 2x - 1=0

hoặc x - 1 = 0

hoặc x - 2 = 0

<=> x = 1/2

hoặc x = 1

hoặc x = 2

Vậy S = {1/2; 1; 2}

bỏ số 1 ở đầu thì giải dc á, còn có số 1 thì chịu

\(\dfrac{1}{x+2x}+\dfrac{1}{x^2+6x+8}+\dfrac{1}{x^2+10x+24}+\dfrac{1}{x^2+14x+48}=\dfrac{4}{105}\)

\(\dfrac{1}{x\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+8\right)}=\dfrac{4}{105}\)

\(\dfrac{2}{x\left(x+2\right)}+\dfrac{2}{\left(x+2\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}+\dfrac{2}{\left(x+6\right)\left(x+8\right)}=\dfrac{8}{105}\)

\(\dfrac{1}{x}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+8}=\dfrac{8}{105}\)

\(\dfrac{1}{x}-\dfrac{1}{x+8}=\dfrac{8}{105}\)

\(\dfrac{x+8-x}{x\left(x+8\right)}=\dfrac{8}{105}\)

\(\dfrac{8}{x.\left(x+8\right)}=\dfrac{8}{105}\)

\(\Rightarrow x\left(x+8\right)=105\)

\(x^2+8x-105=0\)

\(\left(x-7\right)\left(x+15\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=7\\x=-15\end{matrix}\right.\)

\(=\left(x+2\right)\left(x+4\right)\left(x+6\right)\left(x+8\right)+16\)

\(=\left(x+6\right)\left(x+4\right)\left(x+2\right)\left(x+8\right)+16\)

\(=\left(x^2+10x+24\right)\left(x^2+10x+16\right)+16\)

\(=\left(t+8\right)t+16=\left(t+4\right)^2=\left(x^2+10x+20\right)^2\)

1) x² - 4 = 0

=> x² = 4

=> x = \(\pm\)2

2) 2x² - 8 = 0

=> 2x² = 8

=> x² = 4

=> x = \(\pm2\)

3) (x + 3)² = 4 => (x + 3)² = 2²

=> \(\orbr{\begin{cases}x+3=2\\x+3=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=-5\end{cases}}\)

4) (x - 7)² = 36

=> (x - 7)² = 6²

=> \(\orbr{\begin{cases}x-7=6\\x-7=-6\end{cases}}\Rightarrow\orbr{\begin{cases}x=13\\x=1\end{cases}}\)

5) x² - 14x = -49

=> x² - 14x + 49 = 0

=> x² - 7x - 7x + 49 = 0

=> x(x - 7) - 7(x - 7) = 0

=> (x - 7)² = 0

=> x = 7

6) x² + 6x + 5 = 0

=> x² + x + 5x + 5 = 0

=> x(x + 1) + 5(x + 1) = 0

=> (x + 1)(x + 5) = 0

=> \(\orbr{\begin{cases}x=-1\\x=-5\end{cases}}\)

7) x² - 14x + 13 = 0

=> x² - x - 13x + 13 = 0

=> x(x - 1) - 13(x - 1) = 0

=> (x - 1)(x - 13) = 0

=> \(\orbr{\begin{cases}x=1\\x=13\end{cases}}\)

8) x² + 10x +16 = 0

=> x² + 2x + 8x + 16 = 0

=> x(x + 2) + 8(x + 2) = 0

=> (x + 2)(x + 8) = 0

=> \(\orbr{\begin{cases}x=-2\\x=-8\end{cases}}\)

Bài 4:

a: \(2x^4+18x^2=0\)

=>\(2x^2\left(x^2+9\right)=0\)

=>\(x^2=0\) (Vì \(2\left(x^2+9\right)=2x^2+18\ge18>0\forall x\) )

=>x=0

b: (x-5)(x+5)-15x+75=0

=>(x-5)(x+5)-15(x-5)=0

=>(x-5)(x+5-15)=0

=>(x-5)(x-10)=0

=>\(\left[\begin{array}{l}x-5=0\\ x-10=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=10\end{array}\right.\)

c: \(x^4=x^2\)

=>\(x^4-x^2=0\)

=>\(x^2\left(x^2-1\right)=0\)

=>\(\left[\begin{array}{l}x^2=0\\ x^2-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x^2=0\\ x^2=1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=1\\ x=-1\end{array}\right.\)

d: \(12x\left(6x-1\right)-24x^2=0\)

=>12x(6x-1-2x)=0

=>x(4x-1)=0

=>\(\left[\begin{array}{l}x=0\\ 4x-1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=\frac14\end{array}\right.\)

Bài 2:

a: 4x-16+3y(4-x)

=4(x-4)-3y(x-4)

=(x-4)(4-3y)

b: \(9y^2-6y+1=\left(3y\right)^2-2\cdot3y\cdot1+1^2=\left(3y-1\right)^2\)

c: \(25x^2-4=\left(5x\right)^2-2^2=\left(5x-2\right)\left(5x+2\right)\)

d: \(x^2-12x+36=x^2-2\cdot x\cdot6+6^2=\left(x-6\right)^2\)

e: \(8x^3+36x^2+54x+27\)

\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot3+3\cdot2x\cdot3^2+3^3\)

\(=\left(2x+3\right)^3\)

f: \(\left(2x-5\right)^2-\left(2x+y\right)^2\)

=(2x-5-2x-y)(2x-5+2x+y)

=(-y-5)(4x+y-5)

g: \(\left(2x-y\right)^3+\left(2x+y\right)^3\)

\(=8x^3-12x^2y+6xy^2-y^3+8x^3+12x^2y+6xy^2+y^3\)

\(=16x^3+12xy^2=4x\left(4x^2+3y^2\right)\)

Câu 1:

a: \(6x^2-72x=0\)

=>\(6\left(x^2-12x\right)=0\)

=>\(x^2-12x=0\)

=>x(x-12)=0

=>\(\left[\begin{array}{l}x=0\\ x-12=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=12\end{array}\right.\)

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

=>\(-2x\left(x^3-8\right)=0\)

=>\(x\left(x^3-8\right)=0\)

=>\(\left[\begin{array}{l}x=0\\ x^3-8=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x^3=8\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=2\end{array}\right.\)

c: \(\left(2x-1\right)^3-8x\left(x-3\right)\cdot\left(x+3\right)=-1\)

=>\(8x^3-12x^2+6x-1-8x\cdot\left(x^2-9\right)=-1\)

=>\(8x^3-12x^2+6x-1-8x^3+72x=-1\)

=>\(-12x^2+78x=0\)

=>-6x(2x-13)=0

=>x(2x-13)=0

=>\(\left[\begin{array}{l}x=0\\ 2x-13=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=\frac{13}{2}\end{array}\right.\)

d: \(x\left(x-5\right)-\left(x-3\right)^2=0\)

=>\(x^2-5x-\left(x^2-6x+9\right)=0\)

=>\(x^2-5x-x^2+6x-9=0\)

=>x-9=0

=>x=9

e: \(x\left(x-5\right)+3\left(x-5\right)=0\)

=>(x-5)(x+3)=0

=>\(\left[\begin{array}{l}x-5=0\\ x+3=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5\\ x=-3\end{array}\right.\)

f: 2x(x-8)-5(8-x)=0

=>2x(x-8)+5(x-8)=0

=>(x-8)(2x+5)=0

=>\(\left[\begin{array}{l}x-8=0\\ 2x+5=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=-\frac52\end{array}\right.\)

g: \(30x-15x^2=0\)

=>15x(2-x)=0

=>x(2-x)=0

=>\(\left[\begin{array}{l}x=0\\ 2-x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=0\\ x=2\end{array}\right.\)

h: \(-4x^3-12x=0\)

=>\(-4x\left(x^2+3\right)=0\)

=>x=0

Tham khảo

\(ĐKXĐ:x\ne0;-2;-4;-6;-8\)\(\frac{1}{x\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+8\right)}=\frac{4}{105}\)

\(\Leftrightarrow\frac{2}{x\left(x+2\right)}+\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{2}{\left(x+4\right)\left(x+6\right)}+\frac{2}{\left(x+6\right)\left(x+8\right)}=\frac{8}{105}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+4}+...+\frac{1}{x+6}-\frac{1}{x+8}=\frac{8}{105}\)

\(\Leftrightarrow\frac{1}{x}-\frac{1}{x+8}=\frac{8}{105}\)

Quy đồng làm nốt

Đặt \(f\left(x\right)=ax^3+bx^2+cx+d\)

\(\Rightarrow f\left(x+1\right)=a\left(x+1\right)^2+b\left(x+1\right)^2+c\left(x+1\right)+d\)

\(\Rightarrow f\left(x+1\right)=ax^3+\left(3a+b\right)x^2+\left(3a+2b+c\right)x+a+b+c+d\)

\(\Rightarrow f\left(x\right)+f\left(x+1\right)=2ax^3+\left(3a+2b\right)x^2+\left(3a+2b+2c\right)x+a+b+c+2d\)

Đồng nhất hệ số ta được:

\(\left\{{}\begin{matrix}2a=4\\3a+2b=14\\3a+2b+2c=16\\a+b+c+2d=17\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a=2\\b=4\\c=1\\d=5\end{matrix}\right.\)

Vậy \(f\left(x\right)=2x^3+4x^2+x+5\)