6x5+15x4+20x3+15x2+6x+1
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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
\(P\left(x\right)=5^6-6.5^5+6.5^4-6.5^3+6.5^2-6.5+1=5^6-6\left(5^5-5^4-5^3-5^2-5\right)+1=1556\)
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)
`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
hình như là hệ số đối xứng
phân tích thành nhân tử ak