\(6x-15x^2\Leftrightarrow3x\left(2-5x\right)\)

\(6x-15x^2=3x\left(2-5x\right)\)

`1)<=> -4x-3 + 5x+ 2 =0`

`<=> 5x-4x = -2+3`

`<=> x =1`

`2)<=> -5x +2-3x+6 =4`

`<=> -5x-3x = 4-6-2`

`<=> -8x=-4`

`<=> x=1/2`

`3) <=> -7x^2 +2 +7x^2 +14x =8`

`<=> 14x +2 =8`

`<=> 14x = 6`

`<=> x=3/7`

Đây nhé! Tích giúp mình nha

Ta có: \(\dfrac{4x^4+3x^3}{-x^3}+\dfrac{15x^2+6x}{3x}=0\)

\(\Leftrightarrow-4x-3+5x+2=0\)

\(\Leftrightarrow x-1=0\)

hay x=1

\(P\left(x\right)=5^6-6.5^5+6.5^4-6.5^3+6.5^2-6.5+1=5^6-6\left(5^5-5^4-5^3-5^2-5\right)+1=1556\)

mình quên là k dùng máy tính bỏ túi nha

\(P\left(x\right)=x^6-6x^5+6x^4-6x^3+6x^2-6x+1\)

\(=x^6-5x^5-x^5+5x^4+x^4-5x^3-x^3+5x^2+x^2-5x-x+1\)

\(=x^5\left(x-5\right)-x^4\left(x-5\right)+x^3\left(x-5\right)-x^2\left(x-5\right)+x\left(x-5\right)-x+1\)

-Thay \(x=5\) vào P(x) ta được:

\(P\left(5\right)=5^5\left(5-5\right)-5^4\left(5-5\right)+5^3\left(5-5\right)-5^2\left(5-5\right)+5\left(5-5\right)-5+1\)\(=-5+1=-4\)

1. \(=3x\left(2x+5\right)\)

2. \(=\left(3x-1\right)\left(3x+1\right)\)

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

1, = 3x.(2x + 5)

2. = (3x)² - 1² = (3x - 1).(3x +1 )

3, =(x² + 6x + 9) - y² = (x + 3)² - y² =(x + y -3 ). (x - y +3)

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

Chọn đáp án A

`1,(4x^3+3x^3):x^3+(15x^2+6x):(-3x)=0`

`<=> 4 + 3 + (-5x) + (-2)=0`

`<=> -5x+5=0`

`<=>-5x=-5`

`<=>x=1`

`2,(25x^2-10x):5x +3(x-2)=4`

`<=> 5x - 2 + 3x-6=4`

`<=> 8x -8=4`

`<=> 8x=12`

`<=>x=12/8`

`<=>x=3/2`

`3,(3x+1)^2-(2x+1/2)^2=0`

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

`<=>( 3x+1-2x-1/2)(3x+1+2x+1/2)=0`

`<=>( x+1/2) (5x+3/2)=0`

`@ TH1`

`x+1/2=0`

`<=>x=0-1/2`

`<=>x=-1/2`

` @TH2`

`5x+3/2=0`

`<=> 5x=-3/2`

`<=>x=-3/2 : 5`

`<=>x=-15/2`

`4, x^2+8x+16=0`

`<=>(x+4)^2=0`

`<=>x+4=0`

`<=>x=-4`

`5, 25-10x+x^2=0`

`<=> (5-x)^2=0`

`<=>5-x=0`

`<=>x=5`

Bài 2: Tìm x

a)ĐKXĐ: \(x\ne0\)

Ta có: \(\left(4x^4+3x^3\right):\left(-x^3\right)+\left(15x^2+6x\right):3x=0\)

\(\Leftrightarrow\frac{-x^3\left(4x+3\right)}{x^3}+\frac{3x\left(5x+2\right)}{3x}=0\)

\(\Leftrightarrow-4x-3+5x+2=0\)

\(\Leftrightarrow x-1=0\)

hay x=1(nhận)

Vậy: x=1

b) ĐKXĐ: \(x\notin\left\{0;\frac{1}{3}\right\}\)

Ta có: \(\left(x^2-12x\right):2x-\left(3x-1\right)^2:\left(3x-1\right)=0\)

\(\Leftrightarrow\frac{x\left(x-12\right)}{2x}-\frac{\left(3x-1\right)^2}{\left(3x-1\right)}=0\)

\(\Leftrightarrow\frac{x-12}{x}-3x+1=0\)

\(\Leftrightarrow\frac{x-12}{x}=3x-1\)

\(\Leftrightarrow x-12=x\left(3x-1\right)\)

\(\Leftrightarrow3x^2-x+x-12=0\)

\(\Leftrightarrow3x^2-12=0\)

\(\Leftrightarrow3x^2=12\)

\(\Leftrightarrow x^2=4\)

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

Vậy: \(x\in\left\{2;-2\right\}\)

a) 20x4+20x3+20x3

= 20 x ( 4+3 + 3)

= 20 x10

= 200

 b) (6x9-54)x(25 +26+27+.........41+42)

= 0 x (25 +26+27+.........41+42)

= 0 

20x4+20x3+20x3

= 20 x ( 4+3 + 3)

= 20 x10

= 200

 b) (6x9-54)x(25 +26+27+.........41+42)

= 0 x (25 +26+27+.........41+42)

= 0 

a) = 20 x ( 4 + 3 + 3 )

   = 20 x 10

  = 200

b) = 0 x ( 25+26+27+....41+42)

= x