Tìm x
2x4-9x3-11x2+81x=63
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19 tháng 10 2021
Bài 3:
Ta có: \(2n^2+n-7⋮n-2\)
\(\Leftrightarrow2n^2-4n+5n-10+3⋮n-2\)
\(\Leftrightarrow n-2\in\left\{1;-1;3;-3\right\}\)
hay \(n\in\left\{3;1;5;-1\right\}\)
13 tháng 4 2023
`a,`
`P(x)=M(x)+N(x)`
`P(x)=`\(\left(5x^4+8x^2-9x^3-12x-6\right)+\left(-5x^2+9x^3-5x^4+12x-8\right)\)
`P(x)= 5x^4+8x^2-9x^3-12x-6-5x^2+9x^3-5x^4+12x-8`
`P(x)=(5x^4-5x^4)+(-9x^3+9x^3)+(8x^2-5x^2)+(-12x+12x)+(-6-8)`
`P(x)=3x^2-14`
`b,`
`M(x)=N(x)+Q(x)`
`-> Q(x)=M(x)-N(x)`
`-> Q(x)=(5x^4+8x^2-9x^3-12x-6)-(-5x^2+9x^3-5x^4+12x-8)`
`Q(x)=5x^4+8x^2-9x^3-12x-6+5x^2-9x^3+5x^4-12x+8`
`Q(x)=(5x^4+5x^4)+(-9x^3-9x^3)+(8x^2+5x^2)+(-12x-12x)+(-6+8)`
`Q(x)=10x^4-18x^3+13x^2-24x+2`
XZ
1
C
1
L
14 tháng 8 2021
a) \(x^2-x+x=4\)
\(x^2=4\)
\(x=\pm2\)
b) \(3x\left(x-5\right)-2\left(x-5\right)=0\)
\(\left(x-5\right)\left(3x-2\right)=0\)
\(\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)
c) Ta có: \(a+b+c=5-3-2=0\)
\(\left[{}\begin{matrix}x=1\\x=\dfrac{c}{a}=\dfrac{-2}{5}\end{matrix}\right.\)
d) Đặt \(x^2=t\left(t\ge0\right)\) . Lúc đó phương trình trở thành :
\(t^2-11t+18=0\)
\(\left[{}\begin{matrix}t=9\left(tmđk\right)\\t=2\left(tmđk\right)\end{matrix}\right.\)
\(t=9\rightarrow x^2=9\rightarrow x=\pm3\)
\(t=2\rightarrow x^2=2\rightarrow x=\pm\sqrt{2}\)
NH
3
20 tháng 8 2023
\(3^{8x+4}=81^{x+3}\)
\(3^{8x+4}=\left(3^4\right)^{x+3}\)
\(3^{8x+4}=3^{4x+12}\)
\(\Rightarrow8x+4=4x+12\)
\(\Rightarrow8x-4x=12-4\)
\(\Rightarrow4x=8\Rightarrow x=2\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
20 tháng 8 2023
38.x + 4 = 81x + 3
38.x + 4 = (34)x + 3
38.x + 4 = 34.x + 12
8.x + 4 = 4.x + 12
8.x - 4.x = 12 - 4
4.x = 8
x = 8 : 4
x = 2
KV
5 tháng 8 2023
3⁵ˣ⁺⁴ = 81ˣ⁺³
3⁵ˣ⁺⁴ = (3⁴)ˣ⁺³
3⁵ˣ⁺⁴ = 3⁴ˣ⁺¹²
5x + 4 = 4x +12
5x - 4x =12 - 4
x = 8
14 tháng 8 2021
a:Ta có: \(x\left(x-1\right)+x=4\)
\(\Leftrightarrow x^2-x+x=4\)
\(\Leftrightarrow x^2=4\)
hay \(x\in\left\{2;-2\right\}\)
b: Ta có: \(3x\left(x-5\right)-2x+10=0\)
\(\Leftrightarrow\left(x-5\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{2}{3}\end{matrix}\right.\)
c: Ta có: \(5x^2-3x-2=0\)
\(\Leftrightarrow5x^2-5x+2x-2=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)
d: Ta có: \(x^4-11x^2+18=0\)
\(\Leftrightarrow x^4-9x^2-2x^2+18=0\)
\(\Leftrightarrow x^2\left(x^2-9\right)-2\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)
14 tháng 8 2021
a) x(x-1)+x=4
⇔x2=4⇔\(x=\pm2\)
b)3x(x-5)-2x+10=0
⇔3x(x-5)-2(x-5)=0
⇔(x-5)(3x-1)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{3}\end{matrix}\right.\)
c)5x2-3x-2=0
⇔ 5x(x-1)+2(x-1)=0
⇔ (x-1)(5x+2)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{2}{5}\end{matrix}\right.\)
d)x4-11x2+18=0
⇔ x2(x2-2)-9(x2-2)=0
⇔ (x2-2)(x2-9)=0
\(\Leftrightarrow\left[{}\begin{matrix}x^2=2\\x^2=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{2}\\x=\pm3\end{matrix}\right.\)
\(2x^4-9x^3-11x^2+81x=63\)
\(\Rightarrow2x^4-9x^3-11x^2+81x-63=0\)
\(\Rightarrow2x^4-2x^3-7x^3+7x^2-18x^2+18x+63x-63=0\)
\(\Rightarrow2x^3\left(x-1\right)-7x^2\left(x-1\right)-18x\left(x-1\right)+63\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(2x^3-7x^2-18x+63\right)=0\)
\(\Rightarrow\left(x-1\right)\left[x^2\left(2x-7\right)-9\left(2x-7\right)\right]=0\)
\(\Rightarrow\left(x-1\right)\left(2x-7\right)\left(x^2-9\right)=0\)
\(\Rightarrow\left(x-1\right)\left(2x-7\right)\left(x-3\right)\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x-7=0\\x-3=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{2}\\x=3\\x=-3\end{matrix}\right.\)
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