/3x-4/+4=3x tìm x
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6 tháng 7 2023
`@` `\text {Ans}`
`\downarrow`
`a,`
`P(x)+Q(x) = (3x^4-2x^3+3x+11)+(3x^2- x^3-5x+3x+4-x+2x^4)`
`= 3x^4-2x^3+3x+11+3x^2- x^3-5x+3x+4-x+2x^4`
`= (3x^4 + 2x^4) + (-2x^3 - x^3) + 3x^2 + (3x + 3x - 5x - x) + (11+4)`
`= 5x^4 - 3x^3 + 3x^2 + 15`
`b,`
` A(x) = P(x) + B(x)`
Thay `B(x) = 2x^3 - 3x^4 - 2`
`A(x) = P(x) + B (x)`
`=> A (x) = (2x^3 - 3x^4 - 2)+(3x^4 - 2x^3 + 3x + 11)`
`= 2x^3 - 3x^4 - 2+ 3x^4 - 2x^3 + 3x + 11`
`= (2x^3 - 2x^3) + (-3x^4 + 3x^4) + 3x + (-2+11) `
`= 3x + 9`
`A(x) = 3x+9 = 0`
`=> 3x = 0-9`
`=> 3x = -9`
`=> x = -9 \div 3`
`=> x = -3`
Vậy, nghiệm của đa thức là `x = -3.`
16 tháng 9 2021
1) \(\left(x+1\right)\left(x+2\right)-3x\left(x-4\right)=x^2+3x+2-3x^2+12x=-2x^2+15x+2\)
2) \(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)\)
\(\Leftrightarrow3x^2-10x+8=3x^2-27x\)
\(\Leftrightarrow17x=-8\Leftrightarrow x=-\dfrac{8}{17}\)
3) \(-3\left(x-4\right)\left(x-2\right)-x^2\left(-3x+18\right)+24x-25\)
\(=-3x^3+6x^2+12x^2-24x+3x^3-18x^2+24x-25=-25\)
H9
HT.Phong (9A5)
CTVHS
3 tháng 8 2023
a) \(\left(2x-3\right)\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=3\\2x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b) \(\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)
c) \(2x\left(3x-1\right)-3x\left(5+2x\right)=0\)
\(\Rightarrow x\left[2\left(3x-1\right)-3\left(5+2x\right)\right]=0\)
\(\Rightarrow x\left(6x-2-15-6x\right)\)
\(\Rightarrow-16x=0\)
\(\Rightarrow x=0\)
d) \(\left(3x-2\right)\left(3x+2\right)-4\left(x-1\right)=0\)
\(\Rightarrow9x^2-4-4x+4=0\)
\(\Rightarrow9x^2-4x=0\)
\(\Rightarrow x\left(9x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\9x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\end{matrix}\right.\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
3 tháng 8 2023
\(a,\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ b,\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)
24 tháng 10 2021
d: ta có: \(x^2-4x+4=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)
15 tháng 10 2021
a) \(\Rightarrow9x^2+24x+16-9x^2+1=49\)
\(\Rightarrow24x=32\Rightarrow x=\dfrac{4}{3}\)
b) \(\Rightarrow x^2-13x+22=0\)
\(\Rightarrow\left(x-11\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=11\\x=2\end{matrix}\right.\)
c) \(\Rightarrow x^2-3x-10=0\)
\(\Rightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
17 tháng 8 2016
Ta có x^4-3x^3+3x^2+ax+b= (x2 -3x + 4)( x2 - 1) + (ax - 3x) + (b - 4)
Để đây là phép chia hết thì (ax - 3x) = 0 và (b - 4) = 0
Hay a=3 và b =4
NN
1
NN
0
9 tháng 7 2019
b, \(\left(x-5\right)\left(x-4\right)-\left(x+1\right)\left(x-2\right)=7\)
\(\Rightarrow x^2-9x+20-x^2+x+2=7\)
\(\Rightarrow-8x+22=7\)
\(\Rightarrow-8x=-15\)
\(\Rightarrow x=\frac{15}{8}\)
c, \(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)-3\)
\(\Rightarrow3x^2-10x+8=3x^2-27x-3\)
\(\Rightarrow3x^2-10x-3x^2+27x=\left(-3\right)+\left(-8\right)\)
\(\Rightarrow17x=-11\)
\(\Rightarrow x=-\frac{11}{17}\)
d, \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(5-x^2\right)=6x\)
\(\Rightarrow x^3+3x^2+9x-3x^2-9x-27+5x-x^3=6x\)
\(\Rightarrow6x=-27\)
\(\Rightarrow x=-\frac{27}{6}\)
\(\Rightarrow x=-\frac{9}{2}\)
e, \(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)
\(\Rightarrow3x^2-2x-5-3x^2-2x+1=x-4\)
\(\Rightarrow-4=x-4\)
\(\Rightarrow x=0\)
9 tháng 7 2019
b) (x - 5)(x - 4) - (x + 1)(x - 2) = 7
<=> x2 - 9x + 20 - x2 + x + 2 - 7 = 0
<=> 8x - 15 = 0 <=> x = 15/8
c) (3x - 4)(x - 2) = 3x(x - 9) - 3
<=> 3x2 - 10x + 8 = 3x2 - 27x - 3
<=> 17x = -11 <=> x = -11/17
d) (x - 3)(x2 + 3x + 9) + x(5 - x2) = 6x
<=> x3 - 27 - x3 + 5x - 6x = 0
<=> x = -27
e) (3x - 5)(x + 1) - (3x - 1)(x + 1) = x - 4
<=> (x + 1)(3x - 5 - 3x + 1) - x + 4 = 0
<=> -4x - 4 - x + 4 = 0 <=> x = 0
|3x - 4| + 4 = 3x
=> |3x - 4| = 3x - 4
TH1: 3x - 4 = 3x - 4 với mọi x
TH2: 3x - 4 = -(3x - 4)
=> 3x - 4 = -3x + 4
=> 6x - 4 = 4
=> 6x = 8
=> x = \(\frac{4}{3}\)