Giải phương trình: \(x^3+3x+1=0\)
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Những câu hỏi liên quan
15 tháng 7 2023
1) \(\sqrt[]{3x+7}-5< 0\)
\(\Leftrightarrow\sqrt[]{3x+7}< 5\)
\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)
\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)
\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)
2
11 tháng 1 2023
Bài `1:`
`h)(3/4x-1)(5/3x+2)=0`
`=>[(3/4x-1=0),(5/3x+2=0):}=>[(x=4/3),(x=-6/5):}`
______________
Bài `2:`
`b)3x-15=2x(x-5)`
`<=>3(x-5)-2x(x-5)=0`
`<=>(x-5)(3-2x)=0<=>[(x=5),(x=3/2):}`
`d)x(x+6)-7x-42=0`
`<=>x(x+6)-7(x+6)=0`
`<=>(x+6)(x-7)=0<=>[(x=-6),(x=7):}`
`f)x^3-2x^2-(x-2)=0`
`<=>x^2(x-2)-(x-2)=0`
`<=>(x-2)(x^2-1)=0<=>[(x=2),(x^2=1<=>x=+-2):}`
`h)(3x-1)(6x+1)=(x+7)(3x-1)`
`<=>18x^2+3x-6x-1=3x^2-x+21x-7`
`<=>15x^2-23x+6=0<=>15x^2-5x-18x+6=0`
`<=>(3x-1)(5x-1)=0<=>[(x=1/3),(x=1/5):}`
`j)(2x-5)^2-(x+2)^2=0`
`<=>(2x-5-x-2)(2x-5+x+2)=0`
`<=>(x-7)(3x-3)=0<=>[(x=7),(x=1):}`
`w)x^2-x-12=0`
`<=>x^2-4x+3x-12=0`
`<=>(x-4)(x+3)=0<=>[(x=4),(x=-3):}`
2
11 tháng 1 2023
`m)(1-x)(5x+3)=(3x-7)(x-1)`
`<=>(1-x)(5x+3)+(1-x)(3x-7)=0`
`<=>(1-x)(5x+3+3x-7)=0`
`<=>(1-x)(8x-4)=0<=>[(x=1),(x=1/2):}`
`p)(2x-1)^2-4=0`
`<=>(2x-1-2)(2x-1+2)=0`
`<=>(2x-3)(2x+1)=0<=>[(x=3/2),(x=-1/2):}`
`r)(2x-1)^2=49`
`<=>(2x-1-7)(2x-1+7)=0`
`<=>(2x-8)(2x+6)=0<=>[(x=4),(x=-3):}`
`t)(5x-3)^2-(4x-7)^2=0`
`<=>(5x-3-4x+7)(5x-3+4x-7)=0`
`<=>(x+4)(9x-10)=0<=>[(x=-4),(x=10/9):}`
`u)x^2-10x+16=0`
`<=>x^2-8x-2x+16=0`
`<=>(x-2)(x-8)=0<=>[(x=2),(x=8):}`
Y
12 tháng 4 2022
\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)
\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)
\(\Leftrightarrow x^2-9-x^2+3x=0\)
\(\Leftrightarrow3x-9=0\)
\(\Leftrightarrow3x=9\)
\(\Leftrightarrow x=3\left(n\right)\)
Vậy \(S=\left\{3\right\}\)
Y
12 tháng 4 2022
\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)
\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)
\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)
\(\Leftrightarrow12x-9-12x+20+2x-7>0\)
\(\Leftrightarrow2x+4>0\)
\(\Leftrightarrow2x>-4\)
\(\Leftrightarrow x>-2\)
2N
1
TM
22 tháng 5 2022
\(\dfrac{3x}{x^2-x+3}-\dfrac{2x}{x^2-3x+3}+1=0\left(a\right)\)
Ta có : \(x^2-x+3=x^2-x+\dfrac{1}{4}+\dfrac{11}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{11}{4}>0\)
\(x^2-3x+3=x^2-3x+\dfrac{9}{4}+\dfrac{3}{4}=\left(x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}>0\)
\(\RightarrowĐKXĐ:x\in R\)
Đặt : \(t=x^2-x+3\)
\(\left(a\right)\Leftrightarrow\dfrac{3x}{t}-\dfrac{2x}{t-2x}+1=0\)
\(\Leftrightarrow3x\left(t-2x\right)-2xt+t\left(t-2x\right)=0\)
\(\Leftrightarrow t^2-xt-6x^2=0\)
\(\Leftrightarrow t^2+2xt-3xt-6x^2=0\)
\(\Leftrightarrow t\left(t+2x\right)-3x\left(t+2x\right)=0\)
\(\Leftrightarrow\left(t-3x\right)\left(t+2x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t-3x=0\\t+2x=0\end{matrix}\right.\left(b\right)\)
Thay \(t=x^2-x+3\) lại vào (b) được :
\(\left[{}\begin{matrix}x^2-x+3-3x=0\\x^2-x+3+2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2-4x+3=0\\x^2+x+3=0\end{matrix}\right.\left(c\right)\)
Mà : \(x^2-4x+3=x^2-x-3x+3\)
\(=x\left(x-1\right)-3\left(x-1\right)=\left(x-1\right)\left(x-3\right)\left(c'\right)\)
và : \(x^2+x+3=x^2+x+\dfrac{1}{4}+\dfrac{11}{4}\)
\(=\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}\left(c''\right)\)
Thay (c') và (c'') vào (c) được :
\(\left[{}\begin{matrix}\left(x-1\right)\left(x-3\right)=0\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{11}{4}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x-1=0\Leftrightarrow x=1\left(tmđk\right)\\x-3=0\Leftrightarrow x=3\left(tmđk\right)\end{matrix}\right.\\\left(x+\dfrac{1}{2}\right)^2=-\dfrac{11}{4}\Leftrightarrow x\in\varnothing\end{matrix}\right.\)
Vậy : Phương trình có tập nghiệm \(S=\left\{1;3\right\}\)
F
0
BC
1
16 tháng 1 2021
\(\left(x-1\right)^3-\left(x-1\right)\left(2x-3\right)\left(3x-5\right)+\left(2x-3\right)^3-\left(x-1\right)\left(2x-3\right)\left(3x-5\right)+\left(3x-5\right)^3-\left(x-1\right)\left(2x-3\right)\left(3x-5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(\left(x-1\right)^2-\left(2x-3\right)\left(3x-5\right)\right)+\left(2x-3\right)\left(\left(2x-3\right)^2-\left(x-1\right)\left(3x-5\right)\right)+\left(3x-5\right)\left(\left(3x-5\right)^2-\left(x-1\right)\left(2x-3\right)\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(7-5x\right)+\left(2x-3\right)\left(x-2\right)^2+\left(3x-5\right)\left(x-2\right)\left(7x-11\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(\left(x-1\right)\left(7-5x\right)+\left(2x-3\right)\left(x-2\right)+\left(3x-5\right)\left(7x-11\right)\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(18x^2-63x+54\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\18x^2-63x+54=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{2}\end{matrix}\right.\)
4 tháng 3 2022
a: 7x+35=0
=>7x=-35
=>x=-5
b: \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)
=>8-x-8(x-7)=1
=>8-x-8x+56=1
=>-9x+64=1
=>-9x=-63
hay x=7(loại)
4 tháng 3 2022
a, \(7x=-35\Leftrightarrow x=-5\)
b, đk : x khác 7
\(8-x-8x+56=1\Leftrightarrow-9x=-63\Leftrightarrow x=7\left(ktm\right)\)
vậy pt vô nghiệm
2, thiếu đề
Đây là phương trình bậc 3 nên cách giải sai, đáp số \(x=\sqrt[3]{\frac{\sqrt{5}-1}{2}}+\sqrt[3]{\frac{2}{\sqrt{5}-1}}\)
\(\Delta=3^2-4.1.1=5\)
\(\Delta>0,\text{phương trình có 2 nghiệm phân biệt:}\)
\(x_1=\frac{-3+\sqrt{\Delta}}{2.1}=\frac{-3+\sqrt{5}}{2};x_2=\frac{-3-\sqrt{\Delta}}{2.1}=\frac{-3-\sqrt{5}}{2}\)