3x-5>9x+1
tìm x
1) (3x-2)(9x^2+6x+4)-(2x-5)(2x+5)=(3x-1)^3-(2x+3)^2+9x(3x-1)
2) (2x+1)^3-(3x+2)^2=(2x-5)(4x^2+10x+25)+6x(2x+1)-9x^2
1) (3x-2)(9x^2+6x+4)-(2x-5)(2x+5)=(3x-1)^3-(2x+3)^2+9x(3x-1)
Tìm x
( 3x - 2 )( 9x2 + 6x + 4 ) - ( 2x - 5 )( 2x + 5 ) = ( 3x - 1 )3 - ( 2x + 3 )2 + 9x( 3x - 1 )
⇔ 27x3 - 8 - ( 4x2 - 25 ) = 27x3 - 27x2 + 9x - 1 - ( 4x2 + 12x + 9 ) + 27x2 - 9x
⇔ 27x3 - 8 - 4x2 + 25 = 27x3 - 1 - 4x2 - 12x - 9
⇔ 27x3 - 4x2 + 17 - 27x3 + 4x2 + 12x + 10 = 0
⇔ 12x + 27 = 0
⇔ 12x = -27
⇔ x = -27/12 = -9/4
Bài 2: Tìm x, biết:
a/ 12x(x – 5) – 3x(4x - 10) = 120
b/ 9x(x + 4) – 5x(3x + 2) = 112 - 2x(3x + 1)
c/ 3x(1 – x) - 5x(3x + 7) = 154 + 9x(5 – 2x)
$ a/ 12x(x – 5) – 3x(4x - 10) = 120$
`<=>12x^2-60x-12x^2+30x=120`
`<=>-30x=120`
`<=>x=-4`
Vậy `x=-4`
$b/ 9x(x + 4) – 5x(3x + 2) = 112 - 2x(3x + 1)$
`<=>9x^2+36x-15x^2-10x=112-6x^2-2x`
`<=>-6x^2+26x=112-6x^2-2x`
`<=>28x=112`
`<=>x=4`
Vậy `x=4`
$c/ 3x(1 – x) - 5x(3x + 7) = 154 + 9x(5 – 2x)$
`<=>3x-3x^2-15x^2-35x=154+45x-18x^2`
`<=>-32x-18x^2=154+45x-18x^2`
`<=>77x=-154`
`<=>x=-2`
Vậy `x=-2`
1 -4( -x+5) + 9x - 1= 4x + 7 - 2x +3
2) 9x -3( 2x +1) +3x - 3 = 3x
1) \(-4\left(-x+5\right)+9x-1=4x+7-2x+3\)
\(\Leftrightarrow4x-20+9x-1=2x+10\)
\(\Leftrightarrow9x=31\)
\(\Leftrightarrow x=\frac{31}{9}\)
2) \(9x-3\left(2x+1\right)+3x-3=3x\)
\(\Leftrightarrow9x-6x-3-3=0\)
\(\Leftrightarrow3x=6\)
\(\Leftrightarrow x=2\)
3x^2(x^3-3x^2+1)
=3x^2.x^3+3x^2.(-3x^)+3x^2.1
=3x^5-9x^4+3x^2
mấy bạn ơi sao ra 3x^5-9x^4 vậy chỉ mk với mình cần gấp ạ
giải phương trình 1)\(\dfrac{1-6x}{x-2}+\dfrac{9x+4}{x+2}=\dfrac{x\left(3x-2\right)+1}{x^2-4}\)2) \(\dfrac{3x+2}{3x-2}-\dfrac{6}{2+3x}=\dfrac{9x^2}{9x^2-4}\)3) \(\dfrac{x+5}{3x-6}-\dfrac{1}{2}=\dfrac{2x-3}{2x-4}\)4) \(\dfrac{x-1}{x}+\dfrac{1}{x+1}=\dfrac{2x-1}{2x^2+2}\)5) \(\dfrac{2}{x+1}+\dfrac{3x+1}{x+1}=\dfrac{1}{\left(x+1\right)\left(x-2\right)}\)
giúp mình với ạ câu nào cũng được
9x^2-5-(4-3x).(-3x)=1
\(9x^2-5-\left(4-3x\right)\left(-3x\right)=1\)
\(\Rightarrow9x^2-5-\left[4\left(-3x\right)-3x\left(-3x\right)\right]=1\)
\(\Rightarrow9x^2-5-\left[-12x-\left(-9x^2\right)\right]=1\)
\(\Rightarrow9x^2-5-\left[-12x+9x^2\right]=1\)
\(\Rightarrow9x^2-5+12x-9x^2=1\)
\(\Rightarrow12x-5=1\)
\(\Rightarrow12x=6\)
\(\Rightarrow x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
\(9x^2-5-\left(4-3x\right)\left(-3x\right)=1\)
\(\Leftrightarrow9x^2-5-\left(-12x+9x^2\right)=1\)
\(\Leftrightarrow9x^2-5+12x-9x^2=1\)
\(\Leftrightarrow-5+12x=1\)
\(\Leftrightarrow12x=6\)
\(\Rightarrow x=\frac{6}{12}=\frac{1}{2}\)
giải phương trình :
( 3x - 5 ) ( 3x+5 ) - x (9x-1) = 4
Thực hiện phép chia:
1. (-3x3 + 5x2 - 9x + 15) : ( 3x + 5)
2. ( 5x4 + 9x3 - 2x2 - 4x - 8) : ( x-1)
3. ( 5x3 + 14x2 + 12x + 8 ) : (x + 2)
4. ( x4 - 2x3 + 2x -1 ) : ( x2 - 1)
5. ( 5x2 - 3x3 + 15 - 9x ) : ( 5 - 3x)
6. ( -x2 + 6x3 - 26x + 21) : ( 3 -2x )
1: Sửa đề: 3x-5
\(=\dfrac{-x^2\left(3x-5\right)-3\left(3x-5\right)}{3x-5}=-x^2-3\)
2: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)
=5x^2+14x^2+12x+8
3: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)
4: \(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}=x^2+1-2x\)
5: \(=\dfrac{x^2\left(5-3x\right)+3\left(5-3x\right)}{5-3x}=x^2+3\)