A(x)=2X^2-4X+7:B(x)=5X^2+2X+3 A/TÍNH A(x)+b(x) B/TÍNH A(x)-b(x)
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1: A(x)=5x^4+4x^4+x^2+x^2-x+3
=9x^4+2x^2-x+3
B(x)=-8x^4-x^3-2x^2+3
2: A(x)+B(x)
=9x^4+2x^2-x+3-8x^4-x^3-2x^2+3
=x^4-x^3-x+6
A(x)-B(x)
=9x^4+2x^2-x+3+8x^4+x^3+2x^2-3
=17x^4+x^3+4x^2-x
bậc của A(x)-B(x) là 4
3: P(x)=x^4-x^3-x+6-x^4+x^3=-x+6
P(6)=-6+6=0
=>x=6 là nghiệm của P(x)
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\(1,\\ a,=-35x^5y^4z\\ b,=6x^2-30x-6x^2-3x=-33x\\ c,=x^3-9x^2-2x^2+18x-x+9=x^3-11x^2+17x+9\\ 2,\\ A\left(x\right)+B\left(x\right)=10-2x+4x^3-5x^2-10x^3-5x+6x^2-20\\ =-6x^3+x^2-7x-10\\ A\left(x\right)-B\left(x\right)=10-2x+4x^3-5x^2+10x^3+5x-6x^2+20\\ =14x^3-11x^2+3x+30\\ 3,\\ a,M\left(x\right)=5x+20=0\\ \Leftrightarrow x=-4\\ b,N\left(x\right)=100x^2-49=0\\ \Leftrightarrow\left(10x-7\right)\left(10x+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\\ c,P\left(x\right)=3x-15=0\\ \Leftrightarrow x=5\)
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Bài 1;
a)\(5x^3yz.\left(-7x^2y^3\right)=-35.x^5y^4z\)
b)\(6x\left(x-5\right)-x\left(6x+3\right)=6x^2-30x-6x^2-3x=-33x\)
c) \(\left(x-9\right)\left(x^2-2x-1\right)=x^3-2x^2-x-9x^2+18x+9=x^3-11x^2+17x+9\)
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Bài làm
a) ( 2x - 3 ).( x + 1 ) - 2x( 2 - x ) - 4x2 + 5x
= 2x2 + 2x - 3x - 3 - 4x + 2x2 - 4x2 + 5x
= -3
b) x3 - 6x2 + 9x + 14 : ( x - 7 )
Đặt cột chia ta được:
= x2 + x + 16 ( dư 126 )
Nhưng nếu đề thế này tính dễ hơn.
x3 - 6x2 - 9x + 14 : ( x - 7 )
= x3 - 8x2 + 7x + 2x2 - 16x + 14 : ( x - 7 )
= ( 7x + 14 ) + ( x3 - 8x2 + 2x2 - 16x ) : ( x - 7 )
= 7( x + 2 ) + x( x2 - 8x + 2x - 16 ) : ( x - 7 )
= 7( x + 2 ) + x[ ( x2 - 8x ) + ( 2x - 16 ) ] : ( x - 7 )
= 7( x + 2 ) + x[ x( x - 8 ) + 2( x - 8 ) : ( x - 7 )
= 7( x + 2 ) + x( x + 2 )( x - 8 ) : ( x - 7 )
= ( x + 2 )[ 7 + x( x - 8 )] : ( x - 7 )
= ( x + 2 )( x2 - 8x + 7 ) : ( x - 7 )
= ( x + 2 )( x2 - 7x - x + 7 ) : ( x - 7 )
= ( x + 2 )[ ( x2 - 7x ) - ( x - 7 ) ] : ( x - 7 )
= ( x + 2 )[ x( x - 7 ) - ( x - 7 ) ] : ( x - 7 )
= ( x - 2 )( x - 1 )( x - 7 ) : ( x - 7 )
= ( x - 2 )( x - 1 )
~ Bạn xem đề nào mới đunbgs nha ~
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a) \(A\left(x\right)=3x^3-4x^4-2x^3+4x^4-5x+3\)
\(\Rightarrow A\left(x\right)=-4x^4+4x^4+3x^3-2x^3-5x+3\)
\(\Rightarrow A\left(x\right)=x^3-5x+3\)
\(B\left(x\right)=5x^3-4x^2-5x^3-4x^2-5x-3\)
\(\Rightarrow B\left(x\right)=5x^3-5x^3-4x^2-4x^2-5x-3\)
\(\Rightarrow B\left(x\right)=-8x^2-5x-3\)
b) \(A\left(x\right)+B\left(x\right)=x^3-5x+3+\left(-8x^2-5x-3\right)\)
\(\Rightarrow A\left(x\right)+B\left(x\right)=x^3-5x+3-8x^2-5x-3\)
\(\Rightarrow A\left(x\right)+B\left(x\right)=x^3-8x^2-5x-5x+3-3\)
\(\Rightarrow A\left(x\right)+B\left(x\right)=x^3-8x^2-10x\)
\(A\left(x\right)-B\left(x\right)=x^3-5x+3-\left(-8x^2-5x-3\right)\)
\(\Rightarrow A\left(x\right)-B\left(x\right)=x^3-5x+3+8x^2+5x+3\)
\(\Rightarrow A\left(x\right)-B\left(x\right)=x^3+8x^2-5x+5x+3+3\)
\(\Rightarrow A\left(x\right)-B\left(x\right)=x^3+8x^2+6\)
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Bài 2:
a: \(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
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Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
\(A\left(x\right)=2x^2-4x+7\)
\(B\left(x\right)=5x^2+2x+3\)
a/ \(A\left(x\right)+B\left(x\right)=\left(2x^2-4x+7\right)+\left(5x^2+2x+3\right)\)
\(=2x^2-4x+7+5x^2+2x+3\)
\(=7x^2-2x+10\)
b/ \(A\left(x\right)-B\left(x\right)=\left(2x^2-4x+7\right)-\left(5x^2+2x+3\right)\)
\(=2x^2-4x+7-5x^2-2x-3\)
\(=-3x^2-6x+4\)
a, Ta có: \(A\left(x\right)+B\left(x\right)=2x^2-4x+7+5x^2+2x+3\\ \Rightarrow A\left(x\right)+B\left(x\right)=7x^2-2x+10\)
Vậy...
b, Ta có :\(A\left(x\right)-B\left(x\right)=2x^2-4x+7-5x^2-2x-3\\ \Rightarrow A\left(x\right)-B\left(x\right)=-3x^2-6x+4.\)
Vậy...