Phân tích đa thức thành nhân tử : ( phương pháp đổi biến )
\(2\left(x^4+y^4+z^4\right)-\left(x^2+y^2+z^2\right)^2-2\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(x+y+z\right)^4\)
K
Khách
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4 tháng 12 2020
c) ( x2 + x + 1 )( x2 + x + 2 ) - 12
Đặt t = x2 + x + 1
<=> t( t + 1 ) - 12
= t2 + t - 12
= t2 - 3t + 4t - 12
= t( t - 3 ) + 4( t - 3 )
= ( t - 3 )( t + 4 )
= ( x2 + x + 1 - 3 )( x2 + x + 1 + 4 )
= ( x2 + x - 2 )( x2 + x + 5 )
= ( x2 - x + 2x - 2 )( x2 + x + 5 )
= [ x( x - 1 ) + 2( x - 1 ) ]( x2 + x + 5 )
= ( x - 1 )( x + 2 )( x2 + x + 5 )
d) ( x + 2 )( x + 3 )( x + 4 )( x + 5 ) - 24
= [ ( x + 2 )( x + 5 ) ][ ( x + 3 )( x + 4 ) ] - 24
= ( x2 + 7x + 10 )( x2 + 7x + 12 ) - 24
Đặt t = x2 + 7x + 10
<=> t( t + 2 ) - 24
= t2 + 2t - 24
= t2 - 4t + 6t - 24
= t( t - 4 ) + 6( t - 4 )
= ( t - 4 )( t + 6 )
= ( x2 + 7x + 10 - 4 )( x2 + 7x + 10 + 6 )
= ( x2 + 7x + 6 )( x2 + 7x + 16 )
= ( x2 + 6x + x + 6 )( x2 + 7x + 16 )
= [ x( x + 6 ) + ( x + 6 ) ]( x2 + 7x + 16 )
= ( x + 6 )( x + 1 )( x2 + 7x + 16 )
NC
4 tháng 12 2020
a, Sửa đề:\(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
Đặt \(t=x^2+x\)
\(\Rightarrow t^2-2t-15\)
\(=t^2-5t+3t-15\)
\(=t\left(t-5\right)+3\left(t-5\right)\)
\(=\left(t+3\right)\left(t-5\right)\)
\(=\left(x^2+x+3\right)\left(x^2+x-5\right)\)
b,\(x^2+2xy+y^2-x-y-12\)
\(=\left(x+y\right)^2-\left(x+y\right)-12\)
Đặt \(t=x+y\)
\(\Rightarrow t^2-t-12\)
\(=t^2-4t+3t-12\)
\(=t\left(t-4\right)+3\left(t-4\right)\)
\(=\left(t+3\right)\left(t-4\right)\)
\(=\left(x+y+3\right)\left(x+y-4\right)\)
c,\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
Đặt \(t=x^2+x+1\)
\(\Rightarrow t\left(t+1\right)-12\)
\(=t^2+t-12\)
\(=t^2-3t+4t-12\)
\(=t\left(t-3\right)+4\left(t-3\right)\)
\(=\left(t+4\right)\left(t-3\right)\)
\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+5\right)\left(x^2-x+2x-2\right)\)
\(=\left(x^2+x+5\right)\left[x\left(x-1\right)+2\left(x-1\right)\right]\)
\(=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)
d,\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(t=x^2+7x+10\)
\(\Rightarrow t\left(t+2\right)-24\)
\(=t^2+2t-24\)
\(=t^2-4t+6t-24\)
\(=t\left(t-4\right)+6\left(t-4\right)\)
\(=\left(t+6\right)\left(t-4\right)\)
\(=\left(x^2+7x+16\right)\left(x^2+7x+6\right)\)
\(=\left(x^2+7x+16\right)\left(x^2+x+6x+6\right)\)
\(=\left(x^2+7x+16\right)\left[x\left(x+1\right)+6\left(x+1\right)\right]\)
\(=\left(x^2+7x+16\right)\left(x+6\right)\left(x+1\right)\)
4 tháng 12 2020
a, \(x^2-\frac{x^4}{x^2}-1=x^2-x^2-1=-1\)
b, \(\frac{x+9}{x^2-9}-\frac{3}{x^2+3x}=\frac{x+9}{\left(x-3\right)\left(x+3\right)}-\frac{3}{x\left(x+3\right)}\)
\(=\frac{x\left(x+9\right)}{x\left(x-3\right)\left(x+3\right)}-\frac{3\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}=\frac{x^2+9x-3x+9}{x\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2+6x+9}{x\left(x-3\right)\left(x+3\right)}=\frac{\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}=\frac{x+3}{x\left(x-3\right)}\)
4 tháng 12 2020
c, \(\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x-4x^2}=\frac{1-3x}{2x}+\frac{3x-2}{2x-1}+\frac{3x-2}{2x\left(1-2x\right)}\)
\(=\frac{1-3x}{2x}-\frac{3x-2}{1-2x}+\frac{3x-2}{2x\left(1-2x\right)}=\frac{\left(1-3x\right)\left(1-2x\right)}{2x\left(1-2x\right)}-\frac{2x\left(3x-2\right)}{2x\left(1-2x\right)}+\frac{3x-2}{2x\left(1-2x\right)}\)
\(=\frac{1-2x-3x+6x^2-6x^2+4x+3x-2}{2x\left(1-2x\right)}=\frac{-1+2x}{2x\left(1-2x\right)}=\frac{-\left(1-2x\right)}{2x\left(1-2x\right)}=\frac{-1}{2x}\)
d, viết lại đề đy nhé
e, \(\frac{x+1}{x-3}-\frac{1-x}{x+3}-\frac{2x\left(1-x\right)}{9-x^2}=\frac{x+1}{x-3}-\frac{1-x}{x+3}-\frac{2x-2x^2}{\left(3-x\right)\left(x+3\right)}\)
\(=\frac{x+1}{x-3}-\frac{1-x}{x+3}+\frac{2x-2x^2}{\left(x-3\right)\left(x+3\right)}=\frac{\left(x+1\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(1-x\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{2x-2x^2}{\left(x-3\right)\left(x+3\right)}\)
\(=\frac{x^2+3x+x+3-x+3+x^2+3x+2x-2x^2}{\left(x-3\right)\left(x+3\right)}=\frac{8x+6}{\left(x-3\right)\left(x+3\right)}\)
ZT
Cho tam giác ABC , đường cao AH . Biết AB = 15cm , AC = 41cm , HB = 12cm.Tính diện tích tam giác ABC
1
4 tháng 12 2020
Xét tam giác ABH ta có : \(AB^2=BH^2+AH^2\Rightarrow15^2=12^2+AH^2\)
\(AH^2=AB^2-BH^2=15^2-12^2=81\Rightarrow AH=9\)cm
Xét tam giác ACH ta có : \(AC^2=AH^2+HC^2\Rightarrow AC^2=AH^2+HC^2\)
\(HC^2=AC^2-AH^2=41^2-9^2=1600\Rightarrow HC=40\)cm
Ta có : \(BC=CH+HB=40+12=52\)cm
\(\Rightarrow S_{ABC}=\frac{1}{2}AH.BC=\frac{1}{2}.9.52=234\)cm2
VN
2
4 tháng 12 2020
Bài làm
Ta có :\(\frac{x^3+x^2-x-1}{x^3+2x-3}\)( ĐKXĐ : x ≠ 1 )
Để phân thức = 0 thì x3 + x2 - x - 1 = 0
<=> x2( x + 1 ) - ( x + 1 ) = 0
<=> ( x + 1 )( x2 - 1 ) = 0
<=> ( x + 1 )( x - 1 )( x + 1 ) = 0
<=> ( x + 1 )2( x - 1 ) = 0
<=> \(\orbr{\begin{cases}\left(x+1\right)^2=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\left(nhan\right)\\x=1\left(loai\right)\end{cases}}\)
Vậy x = -1 thì phân thức = 0