\(x^4+2x^3-5x^2-2x+1=0\)0
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2
10 tháng 1 2023
`a)2x^2+3(x-1)(x+1)=5x(x+1)`
`<=>2x^2+3x^2-3=5x^2+5x`
`<=>5x=-3`
`<=>x=-3/5`
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`b)(x-3)^3+3-x=0` nhỉ?
`<=>(x-3)^3-(x-3)=0`
`<=>(x-3)(x^2-1)=0`
`<=>[(x=3),(x^2=1<=>x=+-1):}`
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`c)5x(x-2000)-x+2000=0`
`<=>5x(x-2000)-(x-2000)=0`
`<=>(x-2000)(5x-1)=0`
`<=>[(x=2000),(x=1/5):}`
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`d)3(2x-3)+2(2-x)=-3`
`<=>6x-9+4-2x=-3`
`<=>4x=2`
`<=>x=1/2`
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`e)x+6x^2=0`
`<=>x(1+6x)=0`
`<=>[(x=0),(x=-1/6):}`
NV
14 tháng 7 2016
5x2 - 4(x2 - 2x + 1) - 5 = 0
=> 5x2 - 4x2 + 8x - 4 - 5 = 0
=> x2 + 8x - 9 = 0
=> x2 + 9x - x - 9 = 0
=> x(x + 9) - (x + 9) = 0
=> (x + 9)(x - 1) = 0
=> x + 9 = 0 => x = -9
hoặc x - 1 = 0 = > x = 1
Vậy x = -9, x = 1
14 tháng 7 2016
\(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\left(5x^2-5\right)-4\left(x^2-2.1.x+1^2\right)=0\)
\(5\left(x^2-1\right)-4\left(x-1\right)^2=0\)
\(5\left(x-1\right)\left(x+1\right)-4\left(x-1\right)\left(x-1\right)=0\)
\(\left[5\left(x+1\right)-4\left(x-1\right)\right]\left(x-1\right)=0\)
\(\left(5x+5-4x+4\right)\left(x-1\right)=0\)
\(\left(x+9\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+9=0\\x-1=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=-9\\x=1\end{cases}}\)
Vậy \(\orbr{\begin{cases}x=-9\\x=1\end{cases}}.\)
DT
13
VN
Võ Ngọc Phương
VIP
26 tháng 8 2023
a) \(\dfrac{1}{4}+\dfrac{3}{4}:x=-2\)
\(\dfrac{3}{4}:x=-2-\dfrac{1}{4}=\dfrac{-8}{4}-\dfrac{1}{4}\)
\(\dfrac{3}{4}:x=\dfrac{-9}{4}\)
\(x=\dfrac{3}{4}:\dfrac{-9}{4}=\dfrac{3}{4}.\dfrac{-4}{9}\)
\(x=\dfrac{-1}{3}\)
b) \(\dfrac{3}{4}+2.\left(2x-\dfrac{2}{3}\right)=-2\)
\(2.\left(2x-\dfrac{2}{3}\right)=-2-\dfrac{3}{4}=\dfrac{-8}{4}-\dfrac{3}{4}\)
\(2.\left(2x-\dfrac{2}{3}\right)=\dfrac{-11}{4}\)
\(2x-\dfrac{2}{3}=\dfrac{-11}{4}:2=\dfrac{-11}{4}.\dfrac{1}{2}\)
\(2x-\dfrac{2}{3}=\dfrac{-11}{8}\)
\(2x=\dfrac{-11}{8}+\dfrac{2}{3}=\dfrac{-33}{24}+\dfrac{16}{24}\)
\(2x=\dfrac{-17}{24}\)
\(x=\dfrac{-17}{24}:2=\dfrac{-17}{24}.\dfrac{1}{2}\)
\(x=\dfrac{-17}{48}\)
c) \(\left(\dfrac{1}{2}+5x\right).\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}+5x=0\\2x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{-1}{2}\\2x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{10}\\x=\dfrac{3}{2}\end{matrix}\right.\)
26 tháng 8 2023
a, 1/4 + 3/4 : x = -2
3/4 : x = -2 - 1/4
3/4 : x = -9/4
x = 3/4 : -9/4
x = -1/3
NY
4
20 tháng 10 2018
\(\left(2x-3\right)^4-\left(2x-3\right)^2=0\)
\(\Rightarrow\left(2x-3\right)^2\left[\left(2x-3\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x-3\right)^2=0\\\left(2x-3\right)^2=1\end{cases}}\)
Từ đó tìm được \(x=\frac{3}{2},x=2,x=1\)
20 tháng 10 2018
(2x-3)4-(2x-3)2=0
suy ra có 2 TH
TH1 (2x-3)4=0
(2x-3)4=04
2x-3=0
2x=0+3
2x=3
x=3:2
x=1,5
tH 2
(2X-3)2=0
(2X-3)2=02
2X-3=0
2x=0+3
2x=3
x=3:2
x=1.5
vậy x \(\in\){1,5}
HQ
4
29 tháng 6 2019
\(a,\frac{1}{2}x+\frac{5}{2}=\frac{7}{2}x-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x+\frac{5}{2}-\frac{7}{2}x=-\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{7}{2}x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x+\frac{5}{2}=-\frac{3}{4}\)
\(\Leftrightarrow-3x=-\frac{13}{4}\)
\(\Leftrightarrow x=-\frac{13}{4}:(-3)=-\frac{13}{4}:\frac{-3}{1}=-\frac{13}{4}\cdot\frac{-1}{3}=\frac{13}{12}\)
29 tháng 6 2019
\(b,\frac{2}{3}x-\frac{2}{5}=\frac{1}{2}x-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{2}{5}-\frac{1}{2}x=-\frac{1}{3}\)
\(\Leftrightarrow\frac{2}{3}x-\frac{1}{2}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x-\frac{2}{5}=-\frac{1}{3}\)
\(\Leftrightarrow\frac{1}{6}x=\frac{1}{15}\)
\(\Leftrightarrow x=\frac{1}{15}:\frac{1}{6}=\frac{1}{15}\cdot6=\frac{6}{15}=\frac{2}{5}\)
\(c,\frac{1}{3}x+\frac{2}{5}(x+1)=0\)
\(\Leftrightarrow\frac{1}{3}x+\frac{2}{5}x+\frac{2}{5}=0\)
\(\Leftrightarrow\frac{11}{15}x=-\frac{2}{5}\)
\(\Leftrightarrow x=-\frac{6}{11}\)
d,e,f Tương tự
=> \(x^4+x^4-\left(x^5+x^2\right)-2x=1\)
=> \(x^5-x^5-x^2-2x=1\)
=> \(0-x.\left(x+2\right)=1\)
=> \(x.\left(x+2\right)=-1\)
Ta có bảng:
=>
Vậy x = 1;-1;-3
\(x^4+3x^3-x^2-x^3-3x^2+x-x^2-3x+1.\)
\(\left(x^4-x^3-x^2\right)+3\left(x^3-x^2-x\right)-\left(x^2-x-1\right)=0\)
\(x^2\left(x^2-x-1\right)+3x\left(x^2-x-1\right)-\left(x^2-x-1\right)=0\)
\(\left(x^2-x-1\right)\left(x^2+3x-1\right)=0\)
đến đây dùng denta
\(x^2-x-1=0\Leftrightarrow\Delta=b^2-4ac=1+4=5>0\)
vậy pt có 2 nghiệm phân biệt
\(x_1=\frac{-b+\sqrt{\Delta}}{2a}=\frac{1+\sqrt{5}}{2}\) " 1)
\(x_2=\frac{1-\sqrt{5}}{2}\) (2)
\(x^2+3x-1=0\)
áp dụng denta ta có \(\Delta=b^2-4ac=9+4=13>0\)
vậy pt có 2 nghiệm phân biệt
\(x_3=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-3+\sqrt{13}}{2}\) (3)
\(x_4=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-3-\sqrt{13}}{2}\) (4)
gom hết lại rồi kl nghiệm của pt là ....................