A=49x^2+a×x+b B=7x-1
Những câu hỏi liên quan
Tìm a để A chia hết cho B:A=49x^2+a×x+b ;B=7x-1
A chia hết cho B
=>\(49x^2+ax+b⋮7x-1\)
=>\(49x^2-7x+\left(a+7\right)x-\dfrac{1}{7}\left(a+7\right)+b+\dfrac{1}{7}\left(a+7\right)⋮7x-1\)
=>\(7x\left(7x-1\right)+\dfrac{1}{7}\left(a+7\right)\left(7x-1\right)+b+\dfrac{1}{7}\left(a+7\right)=0\)
b+1/7(a+7)=0
=>(a+7)+7b=0
=>a=-7b-7
Vậy: Với a,b là các số nguyên sao cho a=-7b-7 thì A chia hết cho B
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1) Làm phép chia:
a) (x^4 + 2x^2y^2 + y^4) : ( x^2+y^2)
b) ( 49x^2 + 81y^2) : (7x + 9x)
Thu gọn biểu thức
a) (1+2y)^2 + (1-2y)^2 + 2(1+2y) (1-2y)
b) (7x+2y)^2 + (7x-2y)^2 - 2(49x^2-4y^2)
Làm giúp mk bằng hằng đẳng thức số 3 nhé
a) Ta có (1 + 2y)2 + (1 - 2y)2 + 2(1 + 2y)(1 - 2y)
= (1 + 2y + 1 - 2y)2 = 22 = 4
b) Ta có (7x + 2y)2 + (7x - 2y)2 - 2(49x2 - 4y2)
= (7x + 2y)2 + (7x - 2y)2 - 2[(7x)2 - (2y)2]
= (7x + 2y)2 + (7x - 2y)2 - 2(7x - 2y)(7x + 2y)
= (7x + 2y - 7x + 2y)2
= (4y)2 = 16y2
1.Phân tích đa thức thành nhân tử
a)(4x^2-7x-50)^2-16x^4-56x^3-49x^2
b)(x^2+y^2-5)^2-4.x^2.y^2-16xy-16
c)x^4+x^3+3x^2+2x+12
Phân tích thành nhân tử :
a, \(x^4+3x^3-7x^2-27x-18\)
b, \(x^4+5x^3-7x^2-41x-30\)
c, \(x^6-14x^4+49x^2-36\)
a) \(x^4+3x^3-7x^2-27x-18\)
\(=\left(x^4+3x^3+2x^2\right)-\left(9x^2-27x-18\right)\)
\(=x^2\left(x^2+3x+2\right)-9\left(x^2+3x+2\right)=\left(x^2+x+2x+2\right)\left(x^2-9\right)\)
\(=\left(x+1\right)\left(x+2\right)\left(x-3\right)\left(x+3\right)\)
b) \(x^4+5x^3-7x^2-41x-30\)
\(=\left(x^4+2x^3-15x^2\right)+\left(3x^3+6x^2-45x\right)+\left(2x^2+4x-30\right)\)
\(=x^2\left(x^2+2x-15\right)+3x\left(x^2+2x-15\right)+2\left(x^2+2x-15\right)\)
\(=\left(x^2+2x-15\right)\left(x^2+3x+2\right)=\left(x^2+5x-3x-15\right)\left(x^2+x+2x+2\right)\)
\(=\left(x+5\right)\left(x-3\right)\left(x+1\right)\left(x+2\right)\)
c) \(x^6-14x^4+49x^2-36\)
\(=\left(x^6-9x^4\right)+\left(-5x^4+45x^2\right)+\left(4x^2-36\right)\)
\(=x^4\left(x^2-9\right)-5x^2\left(x^2-9\right)+4\left(x^2-9\right)\)
\(=\left(x^2-9\right)\left(x^4-5x^2+4\right)=\left(x^2-9\right)\left(x^4-4x^2-x^2+4\right)\)
\(=\left(x^2-1\right)\left(x^2-4\right)\left(x^2-9\right)=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x-3\right)\left(x+3\right)\)
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giúp mình với nha
phân tích đa thức thành nhân tử
a)14(x-y)^2+21(y-x)
b)7x^5(y-3)-49x^4(3-y)^3
c)(x^2-9)^2-x^2(x-3)^2
d)(4x^2-1)^2-9(2x-1)^2
a) \(14\left(x-y\right)^2+21\left(y-x\right)\)
\(=14\left(x-y\right)^2-21\left(x-y\right)\)
\(=7\left(x-y\right)\left[2\left(x-y\right)-3\right]\)
\(=7\left(x-y\right)\left(2x-2y-3\right)\)
b) \(7x^5\left(y-3\right)-49x^4\left(3-y\right)^3\)
\(=7x^4\left(y-3\right)\left[x+7\left(y-3\right)^2\right]\)
\(=7x^4\left(y-3\right)\left(x+7y^2-42y+63\right)\)
c) \(\left(x^2-9\right)^2-x^2\left(x-3\right)^2\)
\(=\left(x-3\right)^2\left(x+3\right)^2-x^2\left(x-3\right)^2\)
\(=\left(x-3\right)^2\left[\left(x+3\right)^2-x^2\right]\)
\(=\left(x-3\right)^2\left(x^2+6x+9-x^2\right)\)
\(=3\left(x-3\right)^2\left(x+3\right)\)
d) \(\left(4x^2-1\right)^2-9\left(2x-1\right)^2\)
\(=\left(2x-1\right)^2\left(2x+1\right)^2-9\left(2x-1\right)^2\)
\(=\left(2x-1\right)^2\left[\left(2x+1\right)^2-9\right]\)
\(=\left(2x-1\right)^2\left(4x^2+4x+1-9\right)\)
\(=4\left(2x-1\right)^2\left(x^2+x-2\right)\)
\(=4\left(2x-1\right)^2\left(x-1\right)\left(x+2\right)\)
a) 14( x - y )2 + 21( y - x )
= 14( x - y )2 - 21( x - y )
= ( x - y )[ 14( x - y ) - 21 ]
= ( x - y )( 14x - 14y - 21 )
= 7( x - y )( 2x - 2y - 3 )
b) 7x5( y - 3 ) - 49x4( 3 - y )3
= 7x5( y - 3 ) + 49x4( y - 3 )3
= 7x4( y - 3 )[ x + 7( y - 3 )2 ]
= 7x4( y - 3 )( x +7( y2 - 6y + 9 ) ]
= 7x4( y - 3 )( x + 7y2 - 42y + 63 )
c) ( x2 - 9 )2 - x2( x - 3 )2
= [ ( x - 3 )( x + 3 ) ]2 - x2( x - 3 )2
= ( x - 3 )2( x + 3 )2 - x2( x - 3 )2
= ( x - 3 )2[ ( x + 3 )2 - x2 ]
= ( x - 3 )2[ x2 + 6x + 9 - x2 ]
= ( x - 3 )2( 6x + 9 )
= 3( x - 3 )2( 2x + 3 )
d) ( 4x2 - 1 )2 - 9( 2x - 1 )2
= [ ( 2x - 1 )( 2x + 1 ) ]2 - 9( 2x - 1 )2
= ( 2x - 1 )2( 2x + 1 )2 - 9( 2x - 1 )2
= ( 2x - 1 )2[ ( 2x + 1 )2 - 9 ]
= ( 2x - 1 )2( 4x2 + 4x + 1 - 9 )
= ( 2x - 1 )2( 4x2 + 4x - 8 )
= 4( 2x - 1 )2( x2 + x - 2 )
= 4( 2x - 1 )2( x2 - x + 2x - 2 )
= 4( 2x - 1 )2[ x( x - 1 ) + 2( x - 1 ) ]
= 4( 2x - 1 )2( x - 1 )( x + 2 )
giải hệ pt \(\left\{{}\begin{matrix}x^3+xy^2+x^2+xy+2y^2-2y^2=0\\\left(7x+1\right)\sqrt{5x+2y}+\left(7x+6\right)\sqrt{7y}=49x^2+49x+12\end{matrix}\right.\)
Dùng điều kiện a + b + c = 0 hoặc a - b + c = 0 để tính nhẩm nghiệm của các phương trình sau.
a. 7x2 - 9x + 2 = 0 b. x2 + 49x - 50 = 0
c.1975x2 + 4x - 1979 = 0 c.\(\frac{1}{3}\)x2 - \(\frac{3}{2}\)x - \(\frac{11}{6}\) = 0
1, giải các pt
a, (2x+3)2+(5+x)(2x+3)=0
b,(2x+5)2-(2x-5)2=6x+8
c,(4x+3)2 =4(x-1)2
d,(7x-1)(3x-2)-49x2+14x=1
gipus mình nhanh nhanh nha ,cảm ơn ạ
a) Ta có: \(\left(2x+3\right)^2-\left(5+x\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(2x+3+5+x\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(3x+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\3x+8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-3\\3x=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-3}{2}\\x=\frac{-8}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-3}{2};\frac{-8}{3}\right\}\)
b) Ta có: \(\left(2x+5\right)^2-\left(2x-5\right)^2=6x+8\)
\(\Leftrightarrow\left(2x+5+2x-5\right)\left(2x+5-2x+5\right)-6x-8=0\)
\(\Leftrightarrow40x-6x-8=0\)
\(\Leftrightarrow34x=8\)
\(\Leftrightarrow x=\frac{8}{34}=\frac{4}{17}\)
Vậy: \(x=\frac{4}{17}\)
c) Ta có: \(\left(4x+3\right)^2=4\left(x-1\right)^2\)
\(\Leftrightarrow16x^2+24x+9=4\left(x^2-2x+1\right)\)
\(\Leftrightarrow16x^2+24x+9-4x^2+8x-4=0\)
\(\Leftrightarrow12x^2+32x+5=0\)
\(\Leftrightarrow12x^2+2x+30x+5=0\)
\(\Leftrightarrow2x\left(6x+1\right)+5\left(6x+1\right)=0\)
\(\Leftrightarrow\left(6x+1\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}6x+1=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}6x=-1\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{6}\\x=\frac{-5}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{-1}{6};\frac{-5}{2}\right\}\)
d) Ta có: \(\left(7x-1\right)\left(3x-2\right)-49x^2+14x=1\)
\(\Leftrightarrow\left(7x-1\right)\left(3x-2\right)-\left(49x^2-14x+1\right)=0\)
\(\Leftrightarrow\left(7x-1\right)\left(3x-2\right)-\left(7x-1\right)^2=0\)
\(\Leftrightarrow\left(7x-1\right)\left[3x-2-\left(7x-1\right)\right]=0\)
\(\Leftrightarrow\left(7x-1\right)\left(3x-2-7x+1\right)=0\)
\(\Leftrightarrow\left(7x-1\right)\left(-4x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7x-1=0\\-4x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}7x=1\\-4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{7}\\x=\frac{-1}{4}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{1}{7};\frac{-1}{4}\right\}\)