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3) tìm m để x = -1 là nghiệm của đa thức M(x) = x^2 - mx +2
\(\Rightarrow M\left(x\right)=x^2-mx+2\)
\(\Leftrightarrow\left(-1\right)^2-m\left(-1\right)+2=0\)
\(\Leftrightarrow1-m\left(-1\right)=-2\)
\(\Leftrightarrow m\left(-1\right)=3\)
\(\Leftrightarrow m=-3\)
vậy với m = -3 thì x= -1 là nghiệm của đa thức M(x)
4) \(K\left(x\right)=a+b\left(x-1\right)+c\left(x-1\right)\left(x-2\right)\)
\(\Leftrightarrow K\left(1\right)=a+b\left(1-1\right)+c\left(1-1\right)\left(1-2\right)=1\)
\(\Leftrightarrow a=1\)
\(\Leftrightarrow K\left(2\right)=a+b\left(2-1\right)+c\left(2-1\right)\left(2-2\right)=3\)
\(\Leftrightarrow K\left(2\right)=a+b=3\)
\(\Leftrightarrow K\left(0\right)=a+b\left(0-1\right)+c\left(0-1\right)\left(0-2\right)=5\)
\(\Leftrightarrow a+\left(-b\right)+c2=5\)
ta có \(\hept{\begin{cases}a=1\\a+b=3\\a+\left(-b\right)+c2=5\end{cases}}\Leftrightarrow\hept{\begin{cases}a=1\\1+b=3\\1+\left(-b\right)+c2=5\end{cases}}\Leftrightarrow\hept{\begin{cases}a=1\\b=2\\-1+c2=5\end{cases}}\Leftrightarrow\hept{\begin{cases}a=1\\b=2\\c2=6\end{cases}}\Leftrightarrow\hept{\begin{cases}a=1\\b=2\\c=3\end{cases}}\)
vậy \(a=1;b=2;c=3\)
1. a) Sắp xếp :
f(x) = -x5 - 7x4 - 2x3 + x4 + 4x + 9
g(x) = x5 + 7x4 + 2x3 + 2z2 - 3x - 9
b) h(x) = f(x) + g(x)
= -x5 - 7x4 - 2x3 + x2 + 4x + 9 + x5 + 7x4 + 2x3 + 2x2 - 3x - 9
= ( x5 - x5 ) + ( 7x4 - 7x4 ) + ( 2x3 - 2x3 ) + ( 2x2 + x2 ) - 3x + ( 9 - 9 )
= 3x2- 3x
c) h(x) có nghiệm <=> 3x2 - 3x = 0
<=> 3x( x - 1 ) = 0
<=> 3x = 0 hoặc x - 1 = 0
<=> x = 0 hoặc x = 1
Vậy nghiệm của h(x) là x= 0 hoặc x = 1
2. D(x) = A(x) + B(x) - C(x)
= 6x3 + 5x2 + x3 - x2 - ( -2x3 + 4x2 )
= 6x3 + 5x2 + x3 - x2 + 2x3 - 4x2
= ( 6x3 + x3 + 2x3 ) + ( 5x2 - x2 - 4x2 )
= 9x3
b) D(x) có nghiệm <=> 9x3 = 0 => x = 0
Vậy nghiệm của D(x) là x = 0
3. M(x) = x2 - mx + 2
x = -1 là nghiệm của M(x)
=> M(-1) = (-1)2 - m(-1) + 2 = 0
=> 1 + m + 2 = 0
=> 3 + m = 0
=> m = -3
Vậy với m = -3 , M(x) có nghiệm x = -1
4. K(x) = a + b( x - 1 ) + c( x - 1 )( x - 2 )
K(1) = 1 => a + b( 1 - 1 ) + c( 1 - 1 )( 1 - 2 ) = 1
=> a + 0b + c.0.(-1) = 1
=> a + 0 = 1
=> a = 1
K(2) = 3 => 1 + b( 2 - 1 ) + c( 2 - 1 )( 2 - 2 ) = 3
=> 1 + 1b + c.1.0 = 3
=> 1 + b + 0 = 3
=> b + 1 = 3
=> b = 2
K(0) = 5 => 1 + 5( 0 - 1 ) + c( 0 - 1 )( 0 - 2 ) = 5
=> 1 + 5(-1) + c(-1)(-2) = 5
=> 1 - 5 + 2c = 5
=> 2c - 4 = 5
=> 2c = 9
=> c = 9/2
Vậy a = 1 ; b = 2 ; c = 9/2
Mk lm câu b bài 2 há!
b, ( 8x - 3 )( 3x + 2 ) - ( 4x + 7 )( x + 4 ) = ( 2x +1 )( 5x - 1) =- 33
Pt <=> 3x ( 8x - 3 ) + 2( 8x- 33) - ( x ( 4x + 7) ) + ( 2x + 1) - 5x ( 2x + 1) + 33 = 0
<=> 24x2 - 9x + 16x - 6 - ( 4x2 + 7x + 16x + 28) + 2x + 1 - 10x2 - 5x + 33 = 0
<=> 24x2 - 9x + 16x - 6 - 4x2 - 7x - 16x - 28 + 2x + 1 - 10x2 - 19x = 0 <=> x ( 10x - 19) = 0
=> \(\orbr{\begin{cases}x=0\\10x-19=0\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{19}{10}\end{cases}}\)
^^ Ok con tê tê!
Bài 1: Bài này tớ làm không đảm bảo đúng 100% nên nếu có gì sai sót mong bạn thông cảm 
a) Nếu F(x) = G(x)
\(\Rightarrow ax+b-mx-n=0\)
\(\Rightarrow x\left(a-m\right)+\left(b-n\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x\left(a-m\right)=0\\b-n=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-m=0\\b=n\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=m\\b=n\end{matrix}\right.\)
b) Nếu F(x) = G(x)
\(\Rightarrow ax^2+bx+c-mx^2-nx-p=0\)
\(\Rightarrow x^2\left(a-m\right)+x\left(b-n\right)+\left(c-p\right)=0\)
\(\Rightarrow\left\{{}\begin{matrix}x^2\left(a-m\right)=0\\x\left(b-n\right)=0\\c-p=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a-m=0\\b-n=0\\c-p=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a=m\\b=n\\c=p\end{matrix}\right.\)
Bài 2:
a) \(A\left(x\right)=0\)
\(\Leftrightarrow2\left(\dfrac{1}{3}x-\dfrac{1}{2}\right)-\dfrac{1}{2}\left(3-x\right)=0\)
\(\Leftrightarrow2.\dfrac{1}{3}x-2.\dfrac{1}{2}-\dfrac{1}{2}.3+\dfrac{1}{2}x=0\)
\(\Leftrightarrow\dfrac{2}{3}x-1-\dfrac{3}{2}+\dfrac{1}{2}x=0\)
\(\Leftrightarrow\dfrac{7}{6}x-\dfrac{5}{2}=0\)
\(\Leftrightarrow\dfrac{7}{6}x=\dfrac{5}{2}\)
\(\Leftrightarrow x=\dfrac{15}{7}\)
b) Nếu B (x) = 0
\(\Leftrightarrow\left(2x-5\right)\left(x^2-\dfrac{9}{16}\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x-5=0\\x^2-\dfrac{9}{16}=0\\x^2+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x=5\\x^2=\dfrac{9}{16}\\x^2=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{3}{4};x=-\dfrac{3}{4}\\x=1;x=-1\end{matrix}\right.\)
c) Nếu C(x) = 0
\(\Leftrightarrow x^3-2x=0\)
\(\Leftrightarrow x\left(x^2-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x^2=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{2};x=-\sqrt{2}\end{matrix}\right.\)
d) Nếu D(x) = 0
\(\Leftrightarrow9x^2+16=0\)
\(\Leftrightarrow9x^2=-16\)
\(\Leftrightarrow x^2=-\dfrac{16}{9}\)
Vậy không tồn tại x thỏa mãn
e) Nếu M(x) = 0
\(\Leftrightarrow x^2+4x+4=0\)
\(\Leftrightarrow\left(x+2\right)^2=0\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
\(\frac{1}{9}.27^n=3^n\)
\(\Rightarrow\frac{3^n}{27^n}=\frac{1}{9}\)
\(\Rightarrow\left(\frac{3}{27}\right)^n=\frac{1}{9}\)
\(\Rightarrow\left(\frac{1}{9}\right)^n=\frac{1}{9}\)
\(\Rightarrow n=1\)
1. Ta có :
f(x) = ( m - 1 ) . 12 - 3m . 1 + 2 = 0
f(x) = m - 1 - 3m + 2 = -2m + 1 = 0
\(\Rightarrow m=\frac{1}{2}\)
2.
a) M(x) = -2x2 + 5x = 0
\(\Rightarrow-2x^2+5x=x.\left(-2x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\-2x+5=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{2}\end{cases}}\)
b) N(x) = x . ( x - 1/2 ) + 2 . ( x - 1/2 ) = 0
N(x) = ( x + 2 ) . ( x - 1/2 ) = 0
\(\Rightarrow\orbr{\begin{cases}x+2=0\\x-\frac{1}{2}=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{1}{2}\end{cases}}\)
c) P(x) = x2 + 2x + 2015 = x2 + x + x + 1 + 2014 = x . ( x + 1 ) + ( x + 1 ) + 2014 = ( x + 1 ) . ( x + 1 ) + 2014 = ( x + 1 )2 + 2014
vì ( x + 1 )2 + 2014 > 0 nên P(x) không có nghiệm
trắc nghiệm
câu 1: c
câu 2: B
câu 3: D
câu 4: A
câu 5: C
câu 6: D
tự luận
câu 1:
a)M(x) = x4 + 2x2 + 1
b) M(x) + N(x) = -4x4 + x3 + 5x2 - 2
M(x) - N(x) = 6x4 - x3 - x2 + 4
c) \(M\left(-\dfrac{1}{2}\right)=\left(-\dfrac{1}{2}\right)^4+2\left(-\dfrac{1}{2}\right)^2+1=\dfrac{25}{16}\)
Bài 1:
a)
\(F+G+H=(x^3-2x^2+3x+1)+(x^3+x-1)+(2x^2-1)\)
\(=2x^3+4x-1\)
b)
\(F-G+H=0\)
\(\Leftrightarrow (x^3-2x^2+3x+1)-(x^3+x-1)+(2x^2-1)=0\)
\(\Leftrightarrow 2x+1=0\)
\(\Leftrightarrow x=-\frac{1}{2}\)
Bài 2:
a)
\(A=-4x^5-x^3+4x^2-5x+9+4x^5-6x^2-2\)
\(=(-4x^5+4x^5)-x^3+(4x^2-6x^2)-5x+(9-2)\)
\(=-x^3-2x^2-5x+7\)
\(B=-3x^4-2x^3+10x^2-8x+5x^3\)
\(=-3x^4+(5x^3-2x^3)+10x^2-8x\)
\(=-3x^4+3x^3+10x^2-8x\)
b)
\(P=A+B=(-x^3-2x^2-5x+7)+(-3x^4+3x^3+10x^2-8x)\)
\(=-3x^4+(3x^3-x^3)+(10x^2-2x^2)-(8x+5x)+7\)
\(=-3x^4+2x^3+8x^2-13x+7\)
\(P(-1)=-3.(-1)^4+2(-1)^3+8(-1)^2-12(-1)+7=23\)
\(Q=A-B=(-x^3-2x^2-5x+7)-(-3x^4+3x^3+10x^2-8x)\)
\(=3x^4-(x^3+3x^3)-(2x^2+10x^2)+(8x-5x)+7\)
\(=3x^4-4x^3-12x^2+3x+7\)
I . Trắc Nghiệm 1B . 2D . 3C . 5A II . Tự luận 2,a,Ta có: A+(x22y-2xy22+5xy+1)=-2x22y+xy22-xy-1 ⇔⇔ A=(-2x22y+xy22-xy-1) - (x22y-2xy22+5xy+1) =-2x22y+xy22-xy-1 - x22y+2xy22-5xy-1 =(-2x22y - x22y) + (xy22+ 2xy22) + (-xy - 5xy ) + (-1 - 1) = -3x22y + 3xy22 - 6xy - 2 b, thay x=1,y=2 vào đa thức A Ta có A= -3x22y + 3xy22 - 6xy - 2 = -3 . 122 . 2 + 3 .1 . 222 - 6 . 1 . 2 -2 = -6 + 12 - 12 - 2 = -8 3,Sắp xếp f(x) =9-x55+4x-2x33+x22-7x44 =9-x55-7x44-2x33+x22+4x g(x) = x55-9+2x22+7x44+2x33-3x =-9+x55+7x44+2x33+2x22-3x b,f(x) + g(x)=(9-x55-7x44-2x33+x22+4x) + (-9+x55+7x44+2x33+2x22-3x) =9-x55-7x44-2x33+x22+4x-9+x55+7x44+2x33+2x22-3x =(9-9)+(-x55+x55)+(-7x44+7x44)+(-2x33+2x33)+(x22+2x22)+(4x-3x) = 3x22 + x g(x)-f(x)=(-9+x55+7x44+2x33+2x22-3x) - (9-x55-7x44-2x33+x22+4x) =-9+x55+7x44+2x33+2x22-3x-9+x55+7x44+2x 33-x22-4x =(-9-9)+(x55+x55)+(7x44+7x44)+(2x33+2x33)+(2x22-x22)+(3x-4x) = -18 + 2x55 + 14x44 + 4x33 + x22 - x
\(A\left(x\right)⋮x-1\)
=>\(mx^2-mx+\left(m-n-1\right)x-m+n+1+m-n-1-3⋮x-1\)
=>m-n-4=0(2)
\(A\left(x\right)⋮x+1\)
=>\(mx^2+mx-\left(m+n+1\right)x-\left(m+n+1\right)+m+n-2⋮x+1\)
=>m+n-2=0(1)
Từ (1),(2) ta có hệ phương trình:
\(\left\{{}\begin{matrix}m+n=2\\m-n=4\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2m=6\\m+n=2\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m=3\\n=2-3=-1\end{matrix}\right.\)
b: \(B\left(x\right)⋮x-2\)
=>\(\left(m+1\right)x^2-\left(2m+2\right)x+\left(2m+6\right)x-2\left(2m+2\right)+2\left(2m+2\right)+3⋮x-2\)
=>2(2m+2)+3=0
=>4m+7=0
=>\(m=-\dfrac{7}{4}\)
c: \(C\left(x\right)⋮x-1\)
=>\(\left(2n-3\right)x^2-\left(n+2\right)x-9⋮x-1\)
=>\(\left(2n-3\right)x^2-\left(2n-3\right)+\left(2n-3-n-2\right)x-9⋮x-1\)
=>\(\left(n-5\right)x-9⋮x-1\)
=>\(x\left(n-5\right)-\left(n-5\right)+n-5-9⋮x-1\)
=>n-14=0
=>n=14
d: \(D\left(x\right)⋮x-3\)
=>\(5x^2-15x+\left(-3n-1+15\right)x+7⋮x-3\)
=>\(\left(-3n+14\right)x+7⋮x-3\)
=>\(\left(-3n+14\right)x-3\left(-3n+14\right)+3\left(-3n+14\right)+7⋮x-3\)
=>3(-3n+14)+7=0
=>-9n+49=0
=>-9n=-49
=>\(n=\dfrac{49}{9}\)