\(\left(x-1\right)^{x+2}=\left(x-1\right)^{x+4}\)
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a) \(=x^3-\dfrac{1}{27}-x^2+\dfrac{2}{3}x-\dfrac{1}{9}=x^3-x^2+\dfrac{2}{3}x-\dfrac{2}{27}\)
b) \(=x^6-6x^4+12x^2-8-x^3+x+x^2-3x=x^6-6x^4-x^3+13x^2-2x-8\)
\(b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow-2x=15-8=7\)
\(\Leftrightarrow x=\frac{-7}{2}\)
Vậy \(x=\frac{-7}{2}\)
\(=x^6-6x^4+12x^2-8-x^3+x+6x^2-18x\\ =x^6-6x^4-x^3+18x^2-17x-8\)
\(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+3\left(x-1\right)\left(x+1\right).\)
\(=x^3-3x^2+3x-1-\left(x^3-2^3\right)+3\left(x^2-1\right)\)
\(=x^3-3x^2+3x-1-x^3+8+3x^2-3\)
\(=3x+4\)
\(\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+3\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)^3-\left(x^3+8\right)+3\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)^3-x^3-8+3\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1-x\right)+3x\left(x-1\right)\left(x-1-x\right)-8+3\left(x-1\right)\left(x+1\right)\)(1)
\(=-1-3x\left(x-1\right)-8+3\left(x-1\right)\left(x+1\right)\)
\(=3\left(x-1\right)\left(-x+x+1\right)-9=3\left(x-1\right)-9=3\left(x-4\right)=3x-12\)
(1) là hằng đẳng thức \(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)\)
\(a,\Delta=4\left(m-1\right)^2-4\left(-2m-3\right)=4m^2-8m+4+8m+12\\ \Delta=4m^2+16>0\left(đpcm\right)\\ b,\Delta=\left(2m-1\right)^2-4\left(2m-2\right)=4m^2-4m+1-8m+8\\ \Delta=4m^2-12m+9=\left(2m-3\right)^2\ge0\left(đpcm\right)\\ c,Sửa:x^2-2\left(m+1\right)x+2m-2=0\\ \Delta=4\left(m+1\right)^2-4\left(2m-2\right)=4m^2+8m+4-8m+8\\ \Delta=4m^2+12>0\left(đpcm\right)\\ d,\Delta=4\left(m+1\right)^2-4\cdot2m=4m^2+8m+4-8m\\ \Delta=4m^2+4>0\left(đpcm\right)\\ e,\Delta=4m^2-4\left(m+7\right)=4m^2-4m+7=\left(2m-1\right)^2+6>0\left(đpcm\right)\\ f,\Delta=4\left(m-1\right)^2-4\left(-3-m\right)=4m^2-8m+4+12+4m\\ \Delta=4m^2-4m+16=\left(2m-1\right)^2+15>0\left(đpcm\right)\)
1) = x^3 + 3x^2 + 3x + 1 - x^3 + 3x^2 - 3x - 1
= 6x^2
2) = x^3 + 1 - ( x^3 - 1 )
= x^3 + 1 - x^3 + 1
= 2
3) dài lắm thôi ko viết ( Bạn áp dụng cái NHÂN ĐA THỨC VỚI ĐA THỨC nhé )
Học tốt ~
Em coi lại đề bài, \(8\left(x+\dfrac{1}{x}\right)\) hay \(8\left(x+\dfrac{1}{x}\right)^2\) nhỉ?
\(\left(x-\frac{1}{2}\right)^4=\left(x-\frac{1}{2}\right)^2\)
<=> \(\left(x-\frac{1}{2}\right)^4-\left(x-\frac{1}{2}\right)^2=0\)
<=> \(\left(x-\frac{1}{2}\right)^2\left[\left(x-\frac{1}{2}\right)^2-1\right]=0\)
<=> \(\left[\begin{array}{nghiempt}\left(x-\frac{1}{2}\right)^2=0\\\left(x-\frac{1}{2}\right)^2-1=0\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x=\frac{1}{2}\\\left[\begin{array}{nghiempt}x-\frac{1}{2}=1\\x-\frac{1}{2}=-1\end{array}\right.\end{array}\right.\)
<=> \(\left[\begin{array}{nghiempt}x=\frac{1}{2}\\\left[\begin{array}{nghiempt}x=\frac{3}{2}\\x=-\frac{1}{2}\end{array}\right.\end{array}\right.\)
Vậy x \(\in\left\{\frac{1}{2};\frac{3}{2};-\frac{1}{2}\right\}\)
mik chịu