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a) x4 + 6x3 + 11x2 + 6x + 1 = 0 <=> ( x2 + 3x + 1 ) 2 = 0 <=> x2 + 3x + 1 = 0
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a) \(x^2+4x+3\)
\(=x^2+3x+x+3\)
\(=x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
1. \(x^4+6x^3+11x^2+6x+1=0\)
\(\Leftrightarrow x^4+6x^3+9x^2+2x^2+6x+1=0\)
\(\Leftrightarrow\left(x^2+3x+1\right)^2=0\)
\(\Leftrightarrow x^2+3x+1=0\)
\(\Leftrightarrow\left(x+\frac{3}{2}\right)^2-\frac{5}{4}=0\)
\(\Leftrightarrow\left(x+\frac{3}{2}\right)^2=\frac{5}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{3}{2}=\frac{\sqrt{5}}{2}\\x+\frac{3}{2}=-\frac{\sqrt{5}}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-3+\sqrt{5}}{2}\\x=-\frac{3+\sqrt{5}}{2}\end{cases}}\)
2. \(x^4+x^3-4x^2+x+1=0\)
\(\Leftrightarrow\left(x^4+2x^2+1\right)+2.\frac{x}{2}\left(x^2+1\right)+\left(\frac{x}{2}\right)^2-\left(\frac{5}{2}x\right)^2=0\)
\(\Leftrightarrow\left(x^2+1+\frac{x}{2}\right)^2-\left(\frac{5}{2}x\right)^2=0\)
\(\Leftrightarrow\left(x^2-1\right)^2\left(x^2+3x+1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}\left(x-1\right)^2=0\\x^2+3x+1=0\end{cases}}\)
+) ( x - 1 )2 = 0
<=> x - 1 = 0
<=> x = 1
+) x2 + 3x + 1 = 0
<=> ( x + 3/2 )2 - 5/4 = 0
<=> ( x + 3/2 )2 = 5/4
<=> \(\hept{\begin{cases}x+\frac{3}{2}=\frac{\sqrt{5}}{2}\\x+\frac{3}{2}=-\frac{\sqrt{5}}{2}\end{cases}}\)
<=> \(\hept{\begin{cases}x=\frac{-3+\sqrt{5}}{2}\\x=-\frac{3+\sqrt{5}}{2}\end{cases}}\)
Vậy pt có tập nghiệm \(S=\left\{1;\frac{-3+\sqrt{5}}{2};-\frac{3+\sqrt{5}}{2}\right\}\)
Bài 1:
a) \(x^3-5x^2+8x-4\)
\(=x^3-4x^2+4x-x^2+4x-4\) \(=x\left(x^2-4x+4\right)-\left(x^2-4x+4\right)\)\(=\left(x-1\right)\left(x-2\right)^2\)
b) Ta có: \(\frac{A}{M}=\frac{10x^2-7x-5}{2x-3}=5x+4+\frac{7}{2x-3}\)
Với \(x\in Z\)thì \(A⋮M\)khi \(\frac{7}{2x-3}\in Z\)\(\Rightarrow7⋮\left(2x-3\right)\)\(\Rightarrow2x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
\(\Rightarrow=\left\{1;5;\pm2\right\}\)thì khi đó \(A⋮M\)
Các bài làm này có đúng ko ạ, ai đó duyệt giúp em, em cảm ơn.
Bài 1:
a)x3-5x2+8x-4=x3-4x2+4x-x2+4x-4
=x(x2-4x-4)-(x2-4x+4)
=(x-1) (x-2)2
b)Xét:
\(\frac{a}{b}-\frac{10x^2-7x-5}{2x-3}\)
=\(5x+4+\frac{7}{2x-3}\)
Với x thuộc Z thì A /\ B khi \(\frac{7}{2x-3}\) thuộc Z => 7 /\ (2x-3)
Mà Ư(7)={-1;1;-7;7} => x=5;-2;2;1 thì A /\ B
c)Biến đổi \(\frac{x}{y^3-1}-\frac{x}{x^3-1}=\frac{x^4-x-y^4+y}{\left(y^3-1\right)\left(x^3-1\right)}\)
=\(\frac{\left(x^4-y^4\right)\left(x-y\right)}{xy\left(y^2+y+1\right)\left(x^2+x+1\right)}\)(do x+y=1=>y-1=-x và x-1=-y)
=\(\frac{\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)-\left(x-y\right)}{xy\left[x^2y^2+y^2x+y^2+xy^2+xy+y+x^2+x+1\right]}\)
=\(\frac{\left(x-y\right)\left(x^2+y^2-1\right)}{xy\left[x^2y^2+xy\left(x+y\right)+x^2+y^2+xy+2\right]}\)
=\(\frac{\left(x-y\right)\left(x^2-x+y^2-y\right)}{xy\left[x^2y^2+\left(x+y\right)^2+2\right]}=\frac{\left(x-y\right)\left[x\left(x-1\right)+y\left(y-1\right)\right]}{xy\left(x^2y^2+3\right)}\)
=\(\frac{\left(x-y\right)\left[x\left(-y\right)+y\left(-x\right)\right]}{xy\left(x^2y^2+3\right)}=\frac{\left(x-y\right)\left(-2xy\right)}{xy\left(x^2y^2+3\right)}\)
=\(\frac{-2\left(x-y\right)}{x^2y^2+3}\)Suy ra điều phải chứng minh
Bài 2 )
a)(x2+x)2+4(x2+x)=12 đặt y=x2+x
y2+4y-12=0 <=>y2+6y-2y-12=0
<=>(y+6)(y-2)=0 <=> y=-6;y=2
>x2+x=-6 vô nghiệm vì x2+x+6 > 0 với mọi x
>x2+x=2 <=> x2+x-2=0 <=> x2+2x-x-2=0
<=>x(x+2)-(x+2)=0 <=>(x+2)(x-1) <=> x=-2;x-1
Vậy nghiệm của phương trình x=-2;x=1
b)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}+\frac{x+4}{2005}+\frac{x+5}{2004}\)\(+\frac{x+6}{2003}\)
=\(\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)+\left(\frac{x+4}{2005}+1\right)\)\(+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
<=>\(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}-\frac{x+2009}{2005}\)\(+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
<=>\(\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}\)\(-\frac{x+2009}{2005}-\frac{x+2009}{2004}-\frac{x+2009}{2003}=0\)
Nhờ OLM xét giùm em vs ạ !
mk giải từng nha == tại vì mk sợ nhiều qus bị troll
\(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(27x^3+18x^2+12x-18x^2-12x-8-3x\left(9x^2-3x+1\right)+\left(9x^2-3x+1\right)=x-4\)
\(27x^3-8-3\left(9x^2-3x+1\right)+9x^2-3x+1=x-4\)
\(27x^3-7-3x\left(9x^2-3x+1\right)+9x^2-3x=x-4\)
\(27x^3-7-27x^3+9x^2-3x+9x^2-3x=x-4\)
\(-7+18x^2-6x=x-4\)
\(3-18x^2+7x=0\)
\(x=\frac{-7+\sqrt{265}}{-36};\frac{-7-\sqrt{265}}{-36}\)
\(9\left(2x+1\right)=4\left(x-5\right)^2\)
\(18x+9=4x^2-40x+100\)
\(18x+9-4x^2+40x-100=0\)
\(58x-91-4x^2=0\)
\(x=\frac{29-3\sqrt{53}}{4};\frac{29+3\sqrt{53}}{4}\)
Câu hỏi của Trịnh Minh Châu - Toán lớp 8 - Học toán với OnlineMath
a) ( x - 3 )3 - ( x - 3 )( x2 + 3x + 9 ) + 9( x + 1 )2 = 4
<=> x3 - 9x2 + 27x - 27 - ( x3 - 27 ) + 9( x2 + 2x + 1 ) = 4
<=> x3 - 9x2 + 27x - 27 - x3 + 27 + 9x2 + 18x + 9 = 4
<=> 45x + 9 = 4
<=> 45x = -5
<=> x = -5/45 = -1/9
b) x( x - 5 )( x + 5 ) - ( x + 2 )( x2 - 2x + 4 ) = 17
<=> x( x2 - 25 ) - ( x3 + 8 ) = 17
<=> x3 - 25x - x3 - 8 = 17
<=> -25x - 8 = 17
<=> -25x = 25
<=> x = -1
a: \(8x\left(x-2017\right)-2x+4034=0\)
\(\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)