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a, Xét x=0 không phải nghiệm pt chia 2 vế cho x2 , đặt t= x+1/x từ đó suy ra phương trình ẩn t, giải ra ta được các phương trình ẩn x rồi ra x.
b, Tách đa thức thành tích của đơn thức (x+1) và 1 đa thức bậc 4 rồi làm như câu a,.
\(2x^4+3x^3-x^2+3x+2=0\)
\(\Leftrightarrow2x^4+4x^3-x^3-2x^2+x^2+2x+x+2=0\)
\(\Leftrightarrow2x^3.\left(x+2\right)-x^2.\left(x+2\right)+x.\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(2x^3-x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(2x^3+x^2-2x^2-x+2x+1\right)=0\)
\(\Leftrightarrow\left(x+2\right).\left(2x+1\right).\left(x^2-x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+2=0\\2x+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=-\frac{1}{2}\end{cases}}}\)
\(\text{Vì }x^2-x+1=x^2-x+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy phương trình có nghiệm \(S=\left\{-2,-\frac{1}{2}\right\}\)
a, Đặt pt trên là (1)
Nhận thấy : x = 0 không là nghiệm của (1)
Với x khác 0 , chia cả 2 vế của (1) cho \(x^2\) ta được :
\(2x^2+3x-1+\dfrac{3}{x}+\dfrac{2}{x^2}=0\)
\(\Leftrightarrow2\left(x^2+\dfrac{1}{x^2}\right)+3\left(x+\dfrac{1}{2}\right)-1=0\circledast\)
Đặt \(x+\dfrac{1}{x}=y\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=y^2\)
\(\Leftrightarrow x^2+2x.\dfrac{1}{2}+\dfrac{1}{x^2}=4x^2\)
\(\Leftrightarrow x^2+\dfrac{1}{x^2}=4^2-2\)
\(\Rightarrow\circledast\Leftrightarrow2\left(y^2-2\right)+3y-1=0\)
\(\Leftrightarrow2y^2+3y-5=0\)
\(\Leftrightarrow2y^2-2y+5y-5=0\)
\(\Leftrightarrow\left(2y+5\right)\left(y-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}y=\dfrac{-5}{2}\\y=1\end{matrix}\right.\)
\(\)+ Với \(y=\dfrac{-5}{2}\Rightarrow x+\dfrac{1}{x}=\dfrac{-5}{2}\)
\(\Leftrightarrow\dfrac{2x^2+2}{2x}=\dfrac{-5x}{2x}\)
\(\Leftrightarrow2x^2+5x+2=0\)
\(\Leftrightarrow2x^2+x+4x+2=0\)
\(\Leftrightarrow x\left(2x+1\right)+2\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=-2\end{matrix}\right.\)
+ Với \(y=1\Rightarrow x+\dfrac{1}{x}=1\)
\(\Leftrightarrow\dfrac{x^2+1}{x}=\dfrac{x}{x}\)
\(\Leftrightarrow x^2+1=x\)
\(\Leftrightarrow x^2-x=-1\)
\(\Leftrightarrow x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}=-1+\dfrac{1}{4}\)
\(\Leftrightarrow\left(x-\dfrac{1}{2}\right)^2=-\dfrac{3}{4}\)
=> Vô nghiệm
Vậy phương trình có tập nghiệm là \(S=\left\{-2;-\dfrac{1}{2}\right\}\)
a. (3x - 1)2 - (x + 3)2 = 0
\(\Leftrightarrow\left(3x-1+x+3\right)\left(3x-1-x-3\right)=0\)
\(\Leftrightarrow\left(4x+2\right)\left(2x-4\right)=0\)
\(\Leftrightarrow4x+2=0\) hoặc \(2x-4=0\)
1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)
2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)
S=\(\left\{-\dfrac{1}{2};2\right\}\)
b. \(x^3=\dfrac{x}{49}\)
\(\Leftrightarrow49x^3=x\)
\(\Leftrightarrow49x^3-x=0\)
\(\Leftrightarrow x\left(49x^2-1\right)=0\)
\(\Leftrightarrow x\left(7x+1\right)\left(7x-1\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(7x+1=0\) hoặc \(7x-1=0\)
1. x=0
2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)
3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)
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):}`
`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):}`
1,\(3x-1=0\Leftrightarrow3x=-1\Leftrightarrow x=-\dfrac{1}{3}\)
2,\(2-x=3x+1\Leftrightarrow2-1=3x+x\rightarrow1=4x\Rightarrow x=-\dfrac{1}{4}\)
3,\(2\left(x-2\right)-1=5x\Leftrightarrow2x-4-1=5x\Leftrightarrow2x-5x=4+1\Rightarrow3x=5\Rightarrow x=\dfrac{5}{3}\)
4,\(\dfrac{x}{3}-\dfrac{x}{5}=4\Leftrightarrow\dfrac{5x}{15}-\dfrac{3x}{15}=\dfrac{60}{15}\Rightarrow5x-3x=60\Rightarrow2x=60\Rightarrow x=\dfrac{60}{2}=30\)
b) 5x(x-2000)-x+2000=0
\(\Rightarrow5x\left(x-2000\right)-\left(x-2000\right)=0\\ \Rightarrow\left(x-2000\right)\left(5x-1\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x-2000=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0+2000\\5x=0+1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\5x=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2000\\x=\dfrac{1}{5}\end{matrix}\right.\)
a: Ta có: \(3x-\left(3x+2\right)=x+3\)
\(\Leftrightarrow x+3=-2\)
hay x=-5
b: Ta có: \(\dfrac{5x-1}{4}+\dfrac{2x-1}{3}=\dfrac{3x}{2}\)
\(\Leftrightarrow15x-3+8x-4=18x\)
\(\Leftrightarrow5x=7\)
hay \(x=\dfrac{7}{5}\)
\(3x^2-13x+4=0\)
\(\Leftrightarrow3x^2-12x-x+4=0\)
\(\Leftrightarrow\left(3x^2-12x\right)-\left(x-4\right)=0\)
\(\Leftrightarrow3x\left(x-4\right)-\left(x-4\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-4\right)=0\)
\(\Leftrightarrow x=\dfrac{1}{3}\) hoặc \(x=4\)
ĐKXĐ: x khác 4
\(\dfrac{2x^3+5x^2-3x}{x^2-x-12}=0\)
\(\Leftrightarrow\dfrac{2x^3+6x^2-x^2-3x}{x^2+3x-4x-12}=0\)
\(\Leftrightarrow\dfrac{\left(2x^3+6x^2\right)-\left(x^2+3x\right)}{\left(x^2+3x\right)-\left(4x+12\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2\left(x+3\right)-x\left(x+3\right)}{x\left(x+3\right)-4\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{\left(2x^2-x\right)\left(x+3\right)}{\left(x-4\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{2x^2-x}{x-4}=0\)
\(\Leftrightarrow\dfrac{x\left(2x-1\right)}{x-4}=0\)
\(\Leftrightarrow x\left(2x-1\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(x=\dfrac{1}{2}\)