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(x-1)3+(2x+3)3=27x3+8
=> (x - 1 + 2x + 3)[(x - 1)2 - (x - 1)(2x + 3) + (2x + 3)2] = (3x)3 + 23
=> (3x + 2)[x2-2x+1-(2x2+x-3)+4x2+12x+9] = (3x + 2)[(3x)2 - 3x.2 + 22]
=> (3x + 2)(3x2 + 9x + 13) = (3x + 2)(9x2 - 6x + 4)
=> (3x + 2)(3x2 + 9x + 13) - (3x + 2)(9x2 - 6x + 4) = 0
=> (3x + 2)(3x2 + 9x + 13 - 9x2 + 6x - 4) = 0
=> (3x + 2)(-6x2 + 15x + 9) = 0
=>\(\left[{}\begin{matrix}3x+2=0\\-6x^2+15x+9=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}3x=-2\\-3\left(2x^2+5x\right)=-9\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\2x^2+5x=3\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\2x^2+6x-x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\2x\left(x+3\right)-\left(x+3\right)=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\\left(2x-1\right)\left(x+3\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\\x=\dfrac{1}{2}\\x=-3\end{matrix}\right.\)
Vậy phương trình (x-1)3+(2x+3)3=27x3+8 có nghiệm là {-2/3;1/2;-3}
=>x^3-3x^2+3x-1+8x^3+36x^2+54x+27=27x^3+8
=>37x^3+51x^2+57x+26-27x^3-8=0
=>10x^3+51x^2+57x+18=0
=>(5x+3)(2x^2+9x+6)=0
=>x=-3/5 hoặc \(x=\dfrac{-9\pm\sqrt{33}}{4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`x(x+3) - (2x-1) . (x+3) = 0`
`<=>(x+3)(x-2x+1)=0`
`<=>(x+3)(-x+1)=0`
`** x+3=0`
`<=>x=-3`
`** -x+1=0`
`<=>x=1`
`x(x-3) - 5 (x-3) = 0`
`<=>(x-3)(x-5)=0`
`** x-3=0`
`<=>x=3`
`** x-5=0`
`<=>x=5`
`3x + 12 = 0`
`<=>3x=-12`
`<=> x=-4`
`2x (x-2) + 5 (x-2) = 0`
`<=>(x-2)(2x+5)=0`
`** x-2=0`
`<=>x=2`
`** 2x+5=0`
`<=> x= -5/2`
Giải pt : a) 2/-x2+6x-8 - x-1/x-2 = x+3/x-4
b) 2/x3-x2-x+1 = 3/1-x2 - 1/x+1
c) x+2/x-2 - 2/x2-2x = 1/x
![](https://rs.olm.vn/images/avt/0.png?1311)
a,\(\frac{2}{-x^2+6x-8}-\frac{x-1}{x-2}=\frac{x+3}{x-4}\left(đkxđ:x\ne2;4\right)\)
\(< =>\frac{-2}{\left(x-2\right)\left(x-4\right)}-\frac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}=\frac{\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}\)
\(< =>-2-\left(x^2-5x+4\right)=x^2+x-5\)
\(< =>-x^2+5x-6-x^2-x+5=0\)
\(< =>-2x^2+4x-1=0\)
\(< =>2x^2-4x+1=0\)
đến đây thì pt bậc 2 dể rồi
\(\frac{2}{x^3-x^2-x+1}=\frac{3}{1-x^2}-\frac{1}{x+1}\left(đkxđ:x\ne\pm1\right)\)
\(< =>\frac{2}{x^2\left(x-1\right)-\left(x-1\right)}=\frac{3}{1-x^2}-\frac{1}{x+1}\)
\(< =>\frac{2}{\left(x^2-1\right)\left(x-1\right)}=-\frac{3}{x^2-1}-\frac{1}{x+1}\)
\(< =>\frac{2}{\left(x+1\right)\left(x-1\right)^2}=\frac{-3\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}\)
\(< =>2+3x-3+x^2-2x+1=0\)
\(< =>x^2+x=0< =>x\left(x+1\right)=0< =>\orbr{\begin{cases}x=-1\left(loai\right)\\x=0\left(tm\right)\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 2:
a: \(\Leftrightarrow3x^2+2x-1=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{3}\end{matrix}\right.\)
b: \(\Leftrightarrow x^3-4x-x^3-8=4\)
hay x=-3
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a) \(\left(x-1\right)^3\)
\(=x^3-3x^2+3x-1\)
b) \(\left(2x-3y\right)^3\)
\(=\left(2x\right)^3-3\left(2x\right)^23y+3.2x\left(3y\right)^3+\left(3y\right)^3\)
\(=8x^3-36x^2y+54xy^2-27y^3\)
Bài 3:
a: Ta có: \(\left(x-2\right)^3-x^2\left(x-6\right)=5\)
\(\Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2=5\)
\(\Leftrightarrow12x=13\)
hay \(x=\dfrac{13}{12}\)
b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=4\)
\(\Leftrightarrow x^3-1-x^3+4x=4\)
\(\Leftrightarrow4x=5\)
hay \(x=\dfrac{5}{4}\)
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a: \(=\dfrac{35x^3-14x^2+55x^2-22x+35x-14+9}{5x-2}\)
\(=7x^2-11x+7+\dfrac{9}{5x-2}\)
b: \(=\dfrac{\left(2x-3\right)\left(4x^2+6x+9\right)}{2x-3}=4x^2+6x+9\)
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1) tôi giải theo kt lớp 9 nhé nếu theo lp 8 thì phần tích theo đk trong gttđ
lập bảng xét dấu
x | 1 |
lx2-1l | 1-x2 0 x2-1 |
lx-1l | 1-x 0 x-1 |
lx2-1l+lx-1l | -x2-x+2 x2+x-2 |
với x <1 => x=1 x=-2
với x>1 >x=1 x=-2
vậy pt có 2 ng phân bịt x =1 và x=-2
các câu còn lại lm tương tự w nhé
chúc bn hc giỏi !!
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Ta có: \(\dfrac{2x}{x^2-x+1}-\dfrac{x}{x^2+x+1}=\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{2x\left(x^2+x+1\right)-x\left(x^2-x+1\right)}{\left(x^2-x+1\right)\left(x^2+x+1\right)}=\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{2x^3+2x^2+2x-x^3+x^2-x}{\left(x^2-x+1\right)\left(x^2+x+1\right)}=\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{x^3+3x^2+x}{\left(x^2+1\right)^2-x^2}=\dfrac{5}{3}\)
\(\Leftrightarrow3x^3+9x^2+3x=5\left(x^4+2x^2+1-x^2\right)\)
\(\Leftrightarrow3x^3+9x^2+3x=5x^4+5x^2+5\)
\(\Leftrightarrow5x^4+5x^2+5-3x^3-9x^2-3x=0\)
\(\Leftrightarrow5x^4-3x^3-4x^2-3x+5=0\)
\(\Leftrightarrow5x^4-5x^3+2x^3-2x^2-2x^2+2x-5x+5=0\)
\(\Leftrightarrow5x^3\left(x-1\right)+2x^2\left(x-1\right)-2x\left(x-1\right)-5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x^3+2x^2-2x-5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x^3-5x^2+7x^2-7x+5x-5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[5x^2\left(x-1\right)+7x\left(x-1\right)+5\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(5x^2+7x+5\right)=0\)
mà \(5x^2+7x+5>0\forall x\)
nên x-1=0
hay x=1
a,\(x\left(x+1\right)\left(x^2+x+2\right)\)
\(=\left(x^2+x\right)\left(x^2+x+2\right)\)
ĐẶT X^2+X=A\(\Rightarrow\left(x^2+x\right)\left(x^2+x+2\right)=a\left(a+2\right)=42\)
\(\Rightarrow a=\pm1,\pm2,\pm3,\pm6,\pm7,\pm42\)
SUY RA TÌM ĐC X
b,
a) \(x\left(x+1\right)\left(x^2+x-2\right)=48\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=48\)
Đặt \(x^2+x=t\Rightarrow t\left(t-2\right)=48\Leftrightarrow t^2-2t-48=0\Leftrightarrow\orbr{\begin{cases}x=-8\\x=6\end{cases}}\)
Với x = -8, ta có: \(x^2+x=-8\Leftrightarrow x^2+x+8=0\) (Vô nghiệm)
Với x = 6, ta có: \(x^2+x=6\Leftrightarrow x^2+x-6=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}\)
Vậy phương trình có tập nghiệm \(S=\left\{-3;2\right\}\)
b) \(\left(x-1\right)^3+\left(2x+3\right)^3=27x^3+8\)
\(\Leftrightarrow\left(x-1+2x+3\right)\left[\left(x-1\right)^2-\left(x-1\right)\left(2x+3\right)+\left(2x+3\right)^2\right]=\left(3x+2\right)\left(9x^2-6x+4\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(3x^2+9x+13\right)=\left(3x+2\right)\left(9x^2-6x+4\right)\)
\(\Leftrightarrow\left(3x+2\right)\left(3x^2+9x+13-9x^2+6x-4\right)=0\)
\(\Leftrightarrow\left(3x+2\right)\left(-6x^2+15x+9\right)=0\)
TH1: \(3x+2=0\Leftrightarrow x=-\frac{2}{3}\)
TH2: \(-6x^2+15x+9=0\Leftrightarrow\left(x-3\right)\left(-6x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-\frac{1}{2}\end{cases}}\)