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a) 4x3y - 12x2y3 - 8x4y3 = 4x2y( x - 3y2 - 2x2y2 )
b) 2x2 + 4x + 2 - 2y2 = 2( x2 + 2x + 1 - y2 ) = 2[ ( x2 + 2x + 1 ) - y2 ] = 2[ ( x + 1 )2 - y2 ] = 2( x - y + 1 )( x + y + 1 )
c) x3 - 2x2 + x - xy2 = x( x2 - 2x + 1 - y2 ) = x[ ( x2 - 2x + 1 ) - y2 ] = x[ ( x - 1 )2 - y2 ] = x( x - y - 1 )( x + y - 1 )
d) x( x - 2y ) + 3( 2y - x ) = x( x - 2y ) - 3( x - 2y ) = ( x - 2y )( x - 3 )
e) x4 + 4 = ( x4 + 4x2 + 4 ) - 4x2 = ( x2 + 2 )2 - ( 2x )2 = ( x2 - 2x + 2 )( x2 + 2x + 2 )
f) 5x2 - 7x - 6 = 5x2 - 10x + 3x - 6 = 5x( x - 2 ) + 3( x - 2 ) = ( x - 2 )( 5x + 3 )
\(4x^4-21x^2y^2+y^4\)
\(=\left(4x^4+4x^2y^2+y^4\right)-25x^2y^2\)
\(=\left(2x^2+y^2\right)^2-\left(5xy\right)^2\)
\(=\left(2x^2+y^2-5xy\right)\left(2x^2+y^2+5xy\right)\)
\(a,4x^4-21x^2y^2+y^4=\left(2x^2\right)^2+4x^2y^2+y^4-4x^2y^2-21x^2y^2\)
\(=\left(2x^2+y^2\right)^2-25x^2y^2\)
\(=\left(2x^2+y^2-5xy\right)\left(2x^2+y^2+5xy\right)\)
\(b,x^5-5x^3+4x=x\left(x^4-5x^2+4\right)\)
\(=x\left(x^4-4x^2-x^2+4\right)\)
\(=x\left[x^2\left(x^2-4\right)-\left(x^2-4\right)\right]\)
\(=x\left(x^2-4\right)\left(x^2-1\right)\)
\(=x\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)
\(c,x^3+5x^2+3x-9=x^3-x^2+6x^2-6x+9x-9\)
\(=x^2\left(x-1\right)+6x\left(x-1\right)+9\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+6x+9\right)\)
\(=\left(x-1\right)\left(x^2+3x+3x+9\right)\)
\(=\left(x-1\right)\left[x\left(x+3\right)+3\left(x+3\right)\right]\)
\(=\left(x-1\right)\left(x+3\right)\left(x+3\right)\)
\(=\left(x-1\right)\left(x+3\right)^2\)
\(d,x^{16}+x^8-2=x^{16}+2x^8-x^8-2\)
\(=x^8\left(x^8-1\right)+2\left(x^8-1\right)\)
\(=\left(x^8-1\right)\left(x^8+2\right)\)
\(x^4+4=x^4+4x^2+4-4x^2=\left(x^2+2\right)^2-4x^2=\left(x^2+2x+2\right)\left(x^2-2x+2\right)\)
\(4x^8+1=\left(2x^4\right)^2+1=\left(2x^4\right)^2-2.2x^4+1+2.2.x^4=\left(2x^4+1\right)^2-4x^4\)
\(=\left(2x^4+2x^2+1\right)\left(4x^4-2x^2+1\right)\)
\(x^2-8x-9==x^2+x-9x-9=x\left(x+1\right)-9\left(x+1\right)=\left(x+1\right)\left(x-9\right)\)
\(x^2+14x+48=x^2+6x+8x+48=x\left(x+6\right)+8\left(x+6\right)=\left(x+6\right)\left(x+8\right)\)
a) \(x^4+4=x^4+4x^2+4-4x^2=\left(x^2+2\right)^2-4x^2=\left(x^2+2x+2\right)\left(x^2-2x+2\right)\)
b) \(4x^8+1=\left(2x^4\right)^2+1=\left(2x^4\right)^2-2.2x^4+1+2.2.x^4=\left(2x^4+1\right)^2-4x^4\)
c) \(x^2-8x-9==x^2+x-9x-9=x\left(x+1\right)-9\left(x+1\right)=\left(x+1\right)\left(x-9\right)\)
d) \(x^2+14x+48=x^2+6x+8x+48=x\left(x+6\right)+8\left(x+6\right)=\left(x+6\right)\left(x+8\right)\)
khi x=1 thì f(1)=0
f(1)= 3-7+5-36-4+8-a-1=0
<=> -32-a=0
<=> a=-32
Căng, sự thật là nó rất căng
Nhg dù sao thì.....
1) \(A\left(x\right)=\left(x-4\right)^2-\left(2x+1\right)^2\)
Xét \(A\left(x\right)=0\)
\(\Rightarrow\left(x-4\right)^2-\left(2x+1\right)^2=0\)
\(\Rightarrow x^2-8x+16-4x^2-4x-1=0\)
\(\Rightarrow-3x^2-12x+15=0\)
\(\Rightarrow-3x^2+3x-15x+15=0\)
\(\Rightarrow-3x\left(x-1\right)-15\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(-3x-15\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\-3x-15=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
2)(Sửa đề nha, sai cmnr) \(B\left(x\right)=x^3+x^2-4x-4\)
Xét \(B\left(x\right)=0\)
\(\Rightarrow x^3+x^2-4x-4=0\)
\(\Rightarrow x^2\left(x+1\right)-4\left(x+1\right)=0\)
\(\Rightarrow\left(x^2-4\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-4=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\pm2\\x=-1\end{matrix}\right.\)
Đó là những j mình biết 
\(a,x^2-4x+6\)
\(=x^2-2.2.x+2^2+2\)
\(=\left(x-2\right)^2+2\)
\(\left(x-2\right)^2\ge0\Rightarrow\left(x-2\right)^2+2\ge2\)
\(\Rightarrow\)biểu thức nhận giá trị dương với mọi x
\(b,x^2+5x+10\)
\(=x^2+2.\frac{5}{2}.x+\left(\frac{5}{2}\right)^2+\frac{15}{4}\)
\(=\left(x+\frac{5}{2}\right)^2+\frac{15}{4}\)
\(\left(x+\frac{5}{2}\right)^2\ge0\Rightarrow\left(x+\frac{5}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}\)
\(\Rightarrow\)biểu thức luôn nhận giá trị dương với mọi x
\(c,4x^2-4xy+2y^2+3\)
\(=\left(2x\right)^2-2.2x.y+y^2+y^2+3\)
\(=\left(2x-y\right)^2+y^2+3\)
\(\hept{\begin{cases}\left(2x+y\right)^2\ge0\\y^2\ge0\end{cases}\Rightarrow\left(2x+y\right)^2+y^2\ge0}\)
\(\Rightarrow\left(2x+y\right)^2+y^2+3\ge3\)
\(\Rightarrow\)biểu thức luôn nhận giá trị dương với mọi x
\(a,x^2-4x+6=x^2-4x+4+2=\left(x-2\right)^2+2\)
\(\left(x-2\right)^2\ge0\Rightarrow\left(x-2\right)^2+2\ge2>0\)
\(b,x^2+5x+10=x^2+2\cdot\frac{5}{2}\cdot x+\frac{25}{4}+\frac{15}{4}=\left(x+\frac{5}{2}\right)^2+\frac{15}{4}\)
\(\left(x+\frac{5}{2}\right)^2\ge0\Rightarrow\left(x+\frac{5}{2}\right)^2+\frac{15}{4}\ge\frac{15}{4}>0\)
Bài làm
A = 5x( x - 2y ) + 2( 2y - x )2
A = 5x( x - 2y ) + 2( x - 2y )2 [ *chỗ này dùng tính chất ( a - b )2 = ( b - a )2 nha, nếu k hiểu, vào ib ]
A = ( x - 2y )( 5 + 2( x - 2y )
A = ( x - 2y )( 5 + 2x - 4y )
B = ( 4x - 8 )( x2 + 6 ) - ( 4x - 8 )( x + 7 ) + 9( 8 - 4x )
B = ( 4x - 8 )( x2 + 6 ) - ( 4x - 8 )( x + 7 ) - 9( 4x - 8 ) [* Chỗ này mik đổi dấu, nếu thắc mắc vào ib ]
B = ( 4x - 8 )( x2 + 6 - x - 7 - 4x + 8 )
B = ( 4x - 8 )( x2 + 7 - 5x )
B = 4( x - 2 )( x2 + 7 - 5x )
# Học tốt #