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3 tháng 4 2018
\(b,\left(x^2+1\right)^2+3x\left(X^2+1\right)+2x^2=0\)
đặt x^2+1 là y ta đc
\(y^2+3xy+2x^2=0< =>y^2+2xy+xy+2x^2=0< =>y\left(y+2x\right)+x\left(y+2x\right)=0< =>\left(y+x\right)\left(y+2x\right)=0< =>\left[{}\begin{matrix}y=-x\left(1\right)\\y=-2x\left(2\right)\end{matrix}\right.\)
giải 1 ta có;\(x^2+1=-x< =>x^2+x+1=0< =>x^2+2.\dfrac{1}{2}x+\left(\dfrac{1}{2}\right)^2+\dfrac{3}{4}=0< =>\left(x+\dfrac{1}{2}\right)^2=-\dfrac{3}{4}\left(vônghiemej\right)\)
giải 2:\(x^2+1=-2x< =>x^2+2x+1=0< =>\left(x+1\right)^2=0< =>x+1=0< =>x=-1\left(nhận\right)\)
vậy......
3 tháng 4 2018
b)Cách khác:\(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)
\(\Leftrightarrow\left(x^2+1\right)^2+x\left(x^2+1\right)+2x\left(x^2+1\right)+2x^2=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x^2+x+1\right)+2x\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+1=0\left(loai\right)\\x^2+2x+1=0\end{matrix}\right.\)
\(\Leftrightarrow x=-1\)
4 tháng 11 2017
1.
2x3-x2+5x+3=2x3+x2-2x2-x+6x+3=(2x+1)(x2-x+3)
2.
(x+2)(x+3)(x+4)(x+5)-24=(x2+7x+10)(x2+7x+12)-24
Đặt x2+7x+11=a
Ta cso:
(a-1)(a+1)-24=a2-1-24=a2-25=(a-5)(a+5)=(x2+7x+6)(x2+7x+16)
3.
27x3-27x2+18x-4=27x3-9x2-18x2+6x+12x-4=(3x-1)(9x2-6x+4)
5 tháng 11 2017
\(\text{1) }2x^3-x^2+5x+3\\ =2x^3-2x^2+x^2+6x-x+3\\ =\left(2x^3-2x^2+6x\right)+\left(x^2-x+3\right)\\ =2x\left(x^2-x+3\right)+\left(x^2-x+3\right)\\ =\left(2x+1\right)\left(x^2-x+3\right)\\ \)
\(\text{2) }\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\\ =\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\\ =\left(x^2+2x+5x+10\right)\left(x^2+3x+4x+12\right)-24\\ =\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\left(1\right)\\ \text{Đặt }x^2+7x+11=y\text{ }\text{ }\left(\text{*}\right)\\ Thay\text{ }\: \left(\text{*}\right)\text{ vào }\left(1\right)\\ \text{ }Ta\text{ }đư\text{ợc }:\\ \left(1\right)=\left(y-1\right)\left(y+1\right)-24\\ =y^2-1-24\\ =y^2-25\\ =\left(y-5\right)\left(y+5\right)\text{ }\text{ }\left(2\right)\\ Thay\text{ }\left(\text{*}\right)\text{ vào }\left(2\right)\\ \text{Ta lại được: }\left(2\right)=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\\ =\left(x^2+7x+6\right)\left(x^2+7x+17\right)\\ =\left(x^2+6x+x+6\right)\left(x^2+7x+17\right)\\ =\left[\left(x^2+6x\right)+\left(x+6\right)\right]\left(x^2+7x+17\right)\\ =\left[x\left(x+6\right)+\left(x+6\right)\right]\left(x^2+7x+17\right)\\ =\left(x+1\right)\left(x+6\right)\left(x^2+7x+17\right)\\ \)
\(\text{3) }27x^3-27x^2+18x-4\\ =27x^3-18x^2-9x^2+12x+6x-4\\ =\left(27x^3-18x^2+12x\right)-\left(9x^2-6x+4\right)\\ =3x\left(9x^2-6x+4\right)-\left(9x^2-6x+4\right)\\ =\left(3x-1\right)\left(9x^2-6x+4\right)\\ \)
CD
4
24 tháng 5 2020
cách 1: khai triển ra rồi làm
cách 2: đặt \(x+1=a\)và \(2x-1=b\)\(\Rightarrow a+b=3x\)
phương đã cho \(\Leftrightarrow a^3+b^3=\left(a+b\right)^3\)
phần còn lại tự giải nhé
TH
29 tháng 8 2018
a) \(27x^3+27x^2+9x+1=64\)
\(\Rightarrow27x^3+27x^2+9x-63=0\)
\(\Rightarrow27x^3-27x^2+54x^2-54x+63x-63=0\)
\(\Rightarrow27x^2\left(x-1\right)+54x\left(x-1\right)+63\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(27x^2+54x+63\right)=0\)
\(\Rightarrow\left(x-1\right).9\left(3x^2+6x+7\right)=0\)
\(\Rightarrow\left(x-1\right)\left(3x^2+6x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\3x^2+6x+7=0\end{matrix}\right.\)
Mà ta có:
\(3x^2+6x+7\)
\(=3\left(x^2+2x+\dfrac{7}{3}\right)\)
\(=3\left(x^2+2x+1-1+\dfrac{7}{3}\right)\)
\(=3\left(x+1\right)^2+4\)
Vì \(3\left(x+1\right)^2\ge0\) với mọi x
\(\Rightarrow3\left(x+1\right)^2+4\ge4\)
\(\Rightarrow3x^2+6x+7\) vô nghiệm
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
b) \(\left(x-2\right)^3-x^2\left(x-6\right)=4\)
\(\Rightarrow x^3-6x^2+12x-8-x^3+6x^2=4\)
\(\Rightarrow12x-8=4\)
\(\Rightarrow12x=12\)
\(\Rightarrow x=1\)
c) \(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+3\left(x-2\right)\left(x+2\right)=2\)
\(\Rightarrow x^3-3x^2+3x-1-\left(x^3+3^3\right)+3\left(x^2-2^2\right)=2\)
\(\Rightarrow x^3-3x^2+3x-1-x^3-9+3x^2-12=2\)
\(\Rightarrow3x-22=2\)
\(\Rightarrow3x=24\)
\(\Rightarrow x=8\)
\(\left(x-1\right)^3+\left(2x+3\right)^3=27x^3+8\)
\(\left(3x+2\right)\left(3x^2+9x+13\right)=\left(3x+2\right)\left(9x^2-6x+4\right)\)
\(\left(3x+2\right)\left[3x^2+9x+13-\left(9x^2-6x+4\right)\right]=0\)
\(\left(3x+2\right)\left[3x^2+9x+13-9x^2+6x-4\right]=0\)
TH1: \(3x+2=0\Leftrightarrow3x=-2\Leftrightarrow\frac{-2}{3}\)
TH2: \(3x^2+9x+13+9x^2+6x-4=0\)
\(\Leftrightarrow-6x^2+15x+9=0\)
\(\Leftrightarrow\left(-6x-3\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-6x-3=0\\x-3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}-6x=3\\x=3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\x=3\end{cases}}}\)