a) 3(x – 5)(x – 2)(x + 2) + 4 = 7 + 3x3 – 15x2
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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
`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`
a: \(5x^2\left(3x^3-2x^2+x+2\right)\)
\(=15x^5-10x^4+5x^3+10x^2\)
b: \(3x^4\left(-2x^3+5x^2-\dfrac{2}{3}x+\dfrac{1}{3}\right)\)
\(=-6x^7+15x^6-2x^5+x^4\)
Bài 1: Tính
a, 1/2 x 2/3 x 3/4 = 1/4
b, 7/8 x 8/9 x 9/10 = 7/10
c, 5/14 x 7/15 x 28/7 = 2/3
d, 2 x 1/2 x 3 x 1/3 x 4 x 1/4 x 5 x 1/5 = 1
Bài 2: Tính
a, 7/20 - ( 5/8 - 3/5 ) = 7/20 - 1/40 = 14/40 - 1/40 = 13/40
b, 5/6 + ( 5/9 - 1/4 ) = 5/6 + 11/36 = 30/36 + 11/36 = 41/36
c, 9/10 - ( 2/5 + 3/10 ) + 7/20 = 9/10 - 7/10 + 7/20 = 2/10 + 7/20 = 4/20 + 7/20 = 11/20.
Bài 3: Tìm x:
a, 1/2 + x = 5/6
x = 5/6 - 1/2
x = 1/3
b, 5/6 - x = 1/3
x = 5/6 - 1/3
x = 1/2
c, x - 1/mấy vậy bạn
d, 1/3 x x = 1/6
x = 1/6 : 1/3
x = 1/2
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)
Thương Q(x) = 2x2 – x + 5
Dư R(x) = 2x – 1
Ta có: F(x) = 3x2 . (2x2 – x + 5) + 2x – 1
b)
Thương Q(x) = 4x2 + 2x – 2
Dư R(x) = -x – 1
Ta có: F(x) = (3x2 + x + 1) . (4x2 + 2x – 2) – x – 1
\(3\left(x-5\right)\left(x-2\right)\left(x+2\right)+4=7+3x^3-15x^2\)
\(\Rightarrow\left(3x-15\right)\left(x^2-4\right)+4-7-3x^3+15x^2=0\)
\(\Rightarrow3x^3-12x-15x^2+6x-3-3x^3+15x^2=0\)
\(\Rightarrow57=12x\)
\(\Rightarrow x=\frac{57}{12}\)