c) /x - 1/+/x - 2/+···+/x - 9/ = 10x -100
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19 tháng 7 2020
a, Ta có : \(\left|x-1\right|+\left|x-2\right|+...+\left|x-9\right|=10x-100\)
Ta thấy : \(\left\{{}\begin{matrix}\left|x-1\right|\ge0\\\left|...\right|\ge0\\\left|x-9\right|\ge0\end{matrix}\right.\)
=> \(\left|x-1\right|+\left|x-2\right|+...+\left|x-9\right|\ge0\)
=> \(10x-100\ge0\)
=> \(x\ge10\)
=> \(\left\{{}\begin{matrix}\left|x-1\right|=x-1\\\left|...\right|=...\end{matrix}\right.\)
=> \(x-1+x-2+...+x-9=10x-100\)
=> \(9x-45=10x-100\)
=> \(10x-9x=100-45\)
=> \(x=55\) ( TM )
Vậy ....
b, Ta có : \(\left|x-2\right|+...+\left|x-9\right|=1-x\)
Ta thấy : \(\left\{{}\begin{matrix}\left|x-2\right|\ge0\\\left|...\right|\ge0\end{matrix}\right.\)
=> \(\left|x-2\right|+...+\left|x-9\right|\ge0\)
=> \(1-x\ge0\)
=> \(x\le1\)
=> \(\left\{{}\begin{matrix}\left|x-2\right|=2-x\\\left|...\right|=-...\end{matrix}\right.\)
=> \(2-x+...+9-x=1-x\)
=> \(44-8x=1-x\)
=> \(8x-x=44-1\)
=> \(x=\frac{43}{7}\) ( KTM )
Vậy phương trình vô nghiệm .
16 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
`3x(4x-1) - 2x(6x-3) = 30`
`=> 12x^2 - 3x - 12x^2 + 6x = 30`
`=> 3x = 30`
`=> x = 30 \div 3`
`=> x=10`
Vậy, `x=10`
`b)`
`2x(3-2x) + 2x(2x-1) = 15`
`=> 6x- 4x^2 + 4x^2 - 2x = 15`
`=> 4x = 15`
`=> x = 15/4`
Vậy, `x=15/4`
`c)`
`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`
`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`
`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`
`=> 40x^2 -17x - 1 = 1`
`d)`
`(x+2)(x+2)-(x-3)(x+1)=9`
`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`
`=> 6x + 7 =9`
`=> 6x = 2`
`=> x=2/6 =1/3`
Vậy, `x=1/3`
`e)`
`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`
`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`
`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`
`=> 12x +8 = 0`
`=> 12x = -8`
`=> x= -8/12 = -2/3`
Vậy, `x=-2/3`
`g)`
`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`
`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`
`=> -3x + 4 =14`
`=> -3x = 10`
`=> x= - 10/3`
Vậy, `x=-10/3`
28 tháng 6 2017
a, \(25\left(x-2\right)^2-100\left(y-3\right)^2\)
\(=\left(5x-10\right)^2-\left(10y-30\right)^2\)
\(=\left(5x-10-10y+30\right)\left(5x-10+10y-30\right)\)
\(=\left(5x-10y+20\right)\left(5x+10y-40\right)\)
\(=25\left(x-2y+4\right)\left(x+2y-8\right)\)
b, \(4\left(x+3\right)^2-9\left(x+2\right)^2\)
\(=\left(2x+6\right)^2-\left(3x+6\right)^2\)
\(=\left(2x+6-3x-6\right)\left(2x+6+3x+6\right)\)
\(=-x\left(5x+12\right)\)
c, \(x^2-10x-y^2+25\)
\(=\left(x-5\right)^2-y^2\)
\(=\left(x-y-5\right)\left(x+y-5\right)\)
d, \(x^2-4x-y^2+4\)
\(=\left(x-2\right)^2-y^2\)
\(=\left(x-y-2\right)\left(x+y-2\right)\)
e, \(\left(x^2+2\right)^2-2\left(x^2+2\right)+1\)
\(=\left(x^2+2-1\right)^2\)
\(=\left(x^2+1\right)^2\)
HV
1
27 tháng 10 2023
x=9
=>x+1=10
\(A=x^{10}-10x^9+10x^8-...+10x^2-10x+1\)
\(=x^{10}-x^9\left(x+1\right)+x^8\left(x+1\right)-...+x^2\left(x+1\right)-x\left(x+1\right)+1\)
\(=x^{10}-x^{10}-x^9+x^8+...+x^3+x^2-x^2-x+1\)
=-x+1
=-9+1=-8
SN
1
2 tháng 11 2019
c) Đặt \(f\left(x\right)=x^{10}-10x+9\)
Giả sử \(f\left(x\right)⋮\left(x-1\right)^2\)
\(\Rightarrow f\left(x\right)=\left(x-1\right)^2Q\left(x\right)\)
\(\Leftrightarrow f\left(1\right)=\left(1-1\right)^2Q\left(1\right)\)
\(=0\)
\(\Leftrightarrow1^{10}-10.1+9=0\)
\(\Leftrightarrow0=0\)( đúng)
\(\Rightarrow\)điều giả sử đúng
\(\Rightarrow f\left(x\right)⋮\left(x-1\right)^2\left(đpcm\right)\)
13 tháng 7 2016
e) \(E=x^5-15x^4+16x^3-29x^2+13x\) tại x = 14
\(E=x^5-\left(x+1\right)x^4+\left(x+2\right)x^3-\left(2x+1\right)x^2+x\left(x-1\right)\)
\(E=x^5-x^5-x^4+x^4+2x^3-2x^3-x^2+x^2-x\)
\(E=-x\)
\(E=-14\)
13 tháng 7 2016
d) \(D=x^3-30x^2-31+1\) tại x = 31
\(D=31^3-30.31^2-31+1\)
\(D=31^2\left(31-30-1\right)+1\)
\(D=0+1\)
\(D=1\)
tks mn