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16 tháng 8 2021
a: Ta có: \(\left(x+1\right)^3-\left(x+2\right)\left(x-1\right)^2-3\left(x-3\right)\left(x+3\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1-\left(x+2\right)\left(x^2-2x+1\right)-3\left(x^2-9\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-2x^2+x+2x^2-4x+2\right)-3\left(x^2-9\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x-2-3x^2+9=5\)
\(\Leftrightarrow6x=-3\)
hay \(x=-\dfrac{1}{2}\)
b: Ta có: \(\left(x+1\right)^3+\left(x-1\right)^3=\left(x+2\right)^3+\left(x-2\right)^3\)
\(\Leftrightarrow x^3+3x^2+3x+1+x^3-3x^2+3x-1=x^3+6x^2+12x+8+x^3-6x^2+12x-8\)
\(\Leftrightarrow2x^3+6x=2x^3+24x\)
\(\Leftrightarrow x=0\)
c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-1=-10\)
\(\Leftrightarrow12x=-11\)
hay \(x=-\dfrac{11}{12}\)
NL
3
25 tháng 12 2022
\(3\left(x-2\right)+4\left(x-1\right)=25\)
\(\Leftrightarrow3x-6+4x-4=25\)
\(\Leftrightarrow7x=35\)
\(\Leftrightarrow x=5\)
25 tháng 12 2022
\(\left(5x-3\right)\left(x-2\right)=\left(x-1\right)\left(x-2\right)\)
\(\Leftrightarrow\left(5x-3\right)\left(x-2\right)-\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(5x-3-x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\4x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{-1}{2}\end{matrix}\right.\)
21 tháng 6 2021
\(C=\left(\dfrac{2x^2+1}{x^3-1}-\dfrac{1}{x-1}\right)\div\left(1-\dfrac{x^2-2}{x^2+x+1}\right)\)
ĐKXĐ: \(x\ne1\)
\(C=[\left(\dfrac{2x^2+1}{(x-1)\left(x^2+x+1\right)}-\dfrac{1}{x-1}\right)]\div\left(1-\dfrac{x^2-2}{x^2+x+1}\right)\)
\(\Leftrightarrow C=[\left(\dfrac{2x^2+1}{(x-1)\left(x^2+x+1\right)}-\dfrac{1\left(x^2+x+1\right)}{(x-1)\left(x^2+x+1\right)}\right)]\div[\dfrac{(x-1)\left(x^2+x+1\right)}{(x-1)\left(x^2+x+1\right)}-\dfrac{(x^2-2)(x-1)}{(x^2+x+1)\left(x-1\right)}]\)
\(\Rightarrow C=\left[2x^2+1-1\left(x^2+x+1\right)\right]\div\left[\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2\right)\right]\)
\(\Rightarrow C=(2x^2+1-x^2-x-1)\div\left[\left(x-1\right)\left(x^2+x+1-x^2+2\right)\right]\)
\(\Rightarrow C=\left(x^2-x\right)\div\left[\left(x-1\right)\left(x+3\right)\right]\)
10 tháng 11 2021
\(a,\Leftrightarrow\left(x+2\right)\left(x+2-x+3\right)=0\\ \Leftrightarrow5\left(x+2\right)=0\Leftrightarrow x=-2\\ b,\Leftrightarrow2x\left(x-1\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\\ c,\Leftrightarrow\left(x-1-2x-1\right)\left(x-1+2x+1\right)=0\\ \Leftrightarrow3x\left(-x-2\right)=0\Leftrightarrow-3x\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
KK
5 tháng 9 2021
d. (x - 3)(x2 + 3x + 9) + x(x + 2)(2 - x) = 1
<=> x3 - 9 + (x2 + 2x)(2 - x) = 1
<=> x3 - 9 + 2x2 - x3 + 4x - 2x2 = 1
<=> 4x = 10
<=> x = \(\dfrac{10}{4}=\dfrac{5}{2}\)
S
5 tháng 9 2021
d)(x - 3)(x^2 + 3x + 9) + x(x + 2)(2 - x) = 1
\(<=> x^3-27-x(x^2-4)=1\)
\(<=> x^3-27-x^3-4x=1<=>-4x=28<=> x=-7\)
=> ptrình có tập nghiệm S={-7}
e) (x + 1)^3 - (x - 1)^3 - 6(x - 1)^2 = -19
\(<=> x^3+3x^2+3x+1-(x^3-3x^2+3x-1)-6(x^2-2x+1)+19=0\)
\(<=>x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+19=0\)
\(<=>12x=15<=>x=12/15 \)
=> ptrình có tập nghiệm S={12/15}
LM
2
12 tháng 1 2020
\(\frac{1-x}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{\left[x\left(x^4+x^2+1\right)\right]}\)
\(\Leftrightarrow\frac{\left(1-x\right)x\left(x^2-x+1\right)\left(x^4+x^2+1\right)}{x\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^4+x^2+1\right)}\)\(-\)\(\frac{x\left(x-1\right)\left(x^2+x+1\right)\left(x^4+x^2+1\right)}{x\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^4+x^2+1\right)}\)\(=\)\(\frac{3\left(x^2-x+1\right)\left(x^2+x+1\right)}{x\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^4+x^2+1\right)}\)
\(\Rightarrow\left(1-x\right)x\left(x^2-x+1\right)\left(x^4+x^2+1\right)-x\left(x-1\right)\left(x^2+x+1\right)\left(x^4+x^2+1\right)=\)\(3\left(x^2-x+1\right)\left(x^2+x+1\right)\)
\(\Leftrightarrow\left(x-x^2\right)\left(x^2-x+1\right)\left(x^4+x^2+1\right)-\left(x^2-x\right)\left(x^2+x+1\right)\left(x^4+x^2+1\right)=\)\(\left(3x^2-3x+3\right)\left(x^2+x+1\right)\)
\(\Leftrightarrow\left(x^3-x^2+x-x^4+x^3-x^2\right)\left(x^4+x^2+1\right)-\left(x^4+x^3+x^2-x^3-x^2-x\right)\left(x^4+x^2+1\right)=\) \(3x^4+3x^3+3x^2-3x^3-3x^2-3x+3x^2+3x+3\)
\(\Leftrightarrow\left(2x^3-2x^2+x-x^4\right)\left(x^4+x^2+1\right)-\left(x^4-x\right)\left(x^4+x+1\right)=3x^4+3x^2+3\)
\(\Leftrightarrow\left(x^4+x^2+1\right)\left(2x^3-2x^2+x-x^4-x^4+x\right)=3x^4+3x^2+3\)
\(\Leftrightarrow\left(x^4+x^2+1\right)\left(2x^3-2x^2+2x-2x^4\right)=3x^4+3x^2+3\)
\(\Leftrightarrow2x^7-2x^6+2x^5-2x^8+2x^5-2x^4+2x^3-2x+2x^3-2x^2+2x-2x^4-3x^4-3x^2-3=0\)
\(\Leftrightarrow2x^7-2x^6+4x^5-2x^8-7x^4+x^2-3=0\)
Đến đây thì chịu òi :^ Sr nha
13 tháng 1 2020
\(\frac{1-x}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\)
Ta có \(x^4+x^2+1=\left(x^2+1\right)^2-x^2=\left(x^2-x+1\right)\left(x^2+x+1\right)\)
=> \(\left(1-x\right)\left(\frac{1}{x^2+x+1}+\frac{1}{x^2-x+1}\right)=\frac{3}{x\left(x^4+x^2+1\right)}\)
<=>\(\left(1-x\right)\left(2x^2+2\right).x=3\)
Do \(2x^2+2>0\)
=> \(\left(1-x\right).x>0\)
=> \(0< x< 1\)=> \(2x^2+2< 4\)
Pt<=> \(\left(x-x^2\right)\left(2x^2+2\right)=3\)
Mà \(x-x^2\le\frac{1}{4};2x^2+2< 4\)
=> \(VT< 1\)
=> PT vô nghiệm
1
1
24 tháng 10 2021
1) \(\Rightarrow x^2+4x+4-x^2+1=9\)
\(\Rightarrow4x=4\Rightarrow x=1\)
2) \(\Rightarrow x\left(2x+7\right)+2\left(2x+7\right)=0\)
\(\Rightarrow\left(2x+7\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=-2\end{matrix}\right.\)
3) \(\Rightarrow x^3+3x^2+3x+1-x^3-3x^2=2\)
\(\Rightarrow3x=1\Rightarrow x=\dfrac{1}{3}\)
TT
0
Answer:
\(\left(x-2\right).\left(x-3\right)-\left(x-1\right)^2=-1\)
\(\Rightarrow x^2-3x-2x+6-\left(x^2-2x+1\right)=-1\)
\(\Rightarrow x^2-5x+6-x^2+2x-1=-1\)
\(\Rightarrow-3x+5=-1\)
\(\Rightarrow-3x=-1-5\)
\(\Rightarrow x=2\)
x^2 - 2x -3x + 6 - x^2 + 2x - 1 + 1 = 0
-3x+6=0
x=-2