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\(x^2-\left(y-3\right)^2-4x+4\)
\(=x^2-\left(y^2-6y+9\right)-4x+4\)
\(=x^2-y^2+6y-9-4x+4\)
\(=\left(x^2-4x+4\right)-\left(y^2-6y+9\right)\)
\(=\left(x-2\right)^2-\left(y-3\right)^2\)
\(=\left[\left(x-2\right)-\left(y-3\right)\right]\left[\left(x-2\right)+\left(y-3\right)\right]\)
\(=\left(x-y+5\right)\left(x+y-5\right)\)
1.
x2 - ( y - 3 )2 - 4x + 4
= ( x2 - 4x + 4 ) - ( y - 3 )2
= ( x - 2 )2 - ( y - 3 )2
= [ ( x - 2 ) - ( y - 3 ) ][ ( x - 2 ) + ( y - 3 ) ]
= ( x - 2 - y + 3 )( x - 2 + y - 3 )
= ( x - y + 1 )( x + y - 5 )
2.
a) Ta có : 2x4 + 8x3 + 9x2 - 4x - 5
= 2x4 + 10x2 - x2 + 8x3 - 4x - 5
= ( 2x4 - x2 ) + ( 8x3 - 4x ) + ( 10x2 - 5 )
= x2( 2x2 - 1 ) + 4x( 2x2 - 1 ) + 5( 2x2 - 1 )
= ( 2x2 - 1 )( x2 + 4x + 5 )
=>(2x4 + 8x3 + 9x2 - 4x - 5) : ( 2x2 - 1 ) = x2 + 4x + 5
b) Ta có : x2 + 4x + 5 = ( x2 + 4x + 4 ) + 1 = ( x + 2 )2 + 1 ≥ 1 > 0 ∀ x
=> đpcm
a) 3x3-2x2+2 chia x+1= 3x2-5x+5 dư -3 b) -3 chia hết x+1 vậy chon x =2
1)
a) \(-7x\left(3x-2\right)\)
\(=-21x^2+14x\)
b) \(87^2+26.87+13^2\)
\(=87^2+2.87.13+13^2\)
\(=\left(87+13\right)^2\)
\(=100^2\)
\(=10000\)
2)
a) \(x^2-25\)
\(=x^2-5^2\)
\(=\left(x-5\right)\left(x+5\right)\)
b) \(3x\left(x+5\right)-2x-10=0\)
\(\Leftrightarrow3x\left(x+5\right)-\left(2x-10\right)=0\)
\(\Leftrightarrow3x\left(x+5\right)-2\left(x-5\right)=0\)
\(\Leftrightarrow\left(x+5\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=0\\3x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\3x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=\dfrac{2}{3}\end{matrix}\right.\)
Vậy..........
3)
a) \(A:B=\left(3x^3-2x^2+2\right):\left(x+1\right)\)
Vậy \(\left(3x^3-2x^2+2\right):\left(x+1\right)=\left(3x^2-5x-5\right)+7\)
b)
Để \(A⋮B\Rightarrow7⋮\left(x+1\right)\)
\(\Rightarrow\left(x+1\right)\in U\left(7\right)=\left\{-1;1-7;7\right\}\)
Vì x là số nguyên nên x=0 ; x=6 thì \(A⋮B\)
Tuy mk không biết làm nhưng mình sẽ đánh dấu bài này mk không cần bạn k nhưng bạn k trong các câu khác nha.
Chưa có ai trả lời câu hỏi này, hãy gửi một câu trả lời để giúp Trang Nhung giải bài toán này.
\(C1:=3+1-3y\)
\(=4-3y\)
\(C2:\)
\(a.=3x\left(2y-1\right)\)
\(b.=\left(x-y\right)\left(x+y\right)+4\left(x+y\right)\)
\(=\left(x-y+4\right)\left(x+y\right)\)
\(C3:\)
\(a.6x^2+2x+12x-6x^2=7\)
\(14x=7\)
\(x=\frac{1}{2}\)
\(b.\frac{1}{5}x-2x^2+2x^2+5x=-\frac{13}{2}\)
\(\frac{26}{5}x=-\frac{13}{2}\)
\(x=-\frac{13}{2}\times\frac{5}{26}\)
\(x=-\frac{5}{4}\)
Bạn Moon làm kiểu gì vậy ?
1) \(\left(3x^2y^2+x^2y^2\right):\left(x^2y^2\right)-3y\)
\(=\left[\left(x^2y^2\right)\left(3+1\right)\right]:\left(x^2y^2\right)-3y\)
\(=4-3y\)
2) a, \(6xy-3x=\left(3x\right)\left(2y-1\right)\)
b, \(x^2-y^2+4x+4y=\left(x+y\right)\left(x-y\right)+4\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y+4\right)\)
3) a, \(2x\left(3x+1\right)+\left(4-2x\right)3x=7\)
\(< =>6x^2+2x+12x-6x^2=7\)
\(< =>14x=7< =>x=\frac{7}{14}\)
b, \(\frac{1}{2}x\left(\frac{2}{5}-4x\right)+\left(2x+5\right)x=-6\frac{1}{2}\)
\(< =>\frac{x}{2}.\frac{2}{5}-\frac{x}{2}.4x+2x^2+5x=-\frac{13}{2}\)
\(< =>\frac{x}{5}-2x^2+2x^2+5x=-\frac{13}{2}\)
\(< =>\frac{26x}{5}=\frac{-13}{2}\)
\(< =>26x.2=\left(-13\right).5\)
\(< =>52x=-65< =>x=-\frac{65}{52}=-\frac{5}{4}\)
Bài 1:
a) \(x.\left(x^2-2xy+1\right)=x^3-2x^2y+x\)
b) \(\left(2x-3\right).\left(x+2\right)=2x^2+4x-3x-6=2x^2-x-6\)
Bài 2:
a) \(x^3-2x^2+x=x.\left(x^2-2x+1\right)=x.\left(x-1\right)^2\)
b) \(x^2-xy+2x-2y=\left(x^2-xy\right)+\left(2x-2y\right)=x.\left(x-y\right)+2.\left(x-y\right)=\left(x-y\right).\left(x+2\right)\)
c) Đề sai.
b)\(\frac{9x^4-6x^3+15x^2+2x+1}{3x^2-2x+5}=\frac{3x^2.\left(3x^2-2x+5\right)+2x+1}{3x^2-2x+5}=3x^2+\frac{2x+1}{3x^2-2x+5}\)
=> đa thức dư trong phép chia là 2x+1
\(\frac{x^3+2x^2-3x+9}{x+3}=\frac{x^3+9x^2+27x+27-7x^2-30x-18}{x+3}=\frac{\left(x+3\right)^3-7x^2-30x-18}{x+3}\)
\(\left(x+3\right)^2-\frac{7x^2+21x+9x+18}{x+3}=\left(x+3\right)^2-\frac{7x.\left(x+3\right)+9.\left(x+3\right)-9}{x+3}\)
\(=\left(x+3\right)^2-\frac{\left(7x+9\right).\left(x+3\right)-9}{x+3}=\left(x+3\right)^2-\left(7x+9\right)-\frac{9}{x+3}\)
=> đa thức dư trong phép chia là 9
p/s: t mới lớp 7_sai sót mong bỏ qua :>
a: \(=\dfrac{x\left(x^2+x-2\right)}{x+2}=\dfrac{x\left(x+2\right)\left(x-1\right)}{x+2}=x^2-x\)
b: \(=\dfrac{x^3-3x^2+2x+24}{x+2}=\dfrac{x^3+2x^2-5x^2-10x+12x+24}{x+2}=x^2-5x+12\)