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9 tháng 10 2021
1)
a) \(=15x^3-20x^2+10x\)
b) \(=3x^4-x^3+4x^2-9x^3+3x-12x=3x^4-10x^3+4x^2-9x\)
2)
a) \(\Rightarrow x\left(x^2-6x+12\right)=0\)
\(\Rightarrow x=0\)(do \(x^2-6x+12=\left(x^2-6x+\dfrac{36}{4}\right)+3=\left(x-\dfrac{6}{2}\right)^2+3\ge3>0\))
b) \(\Rightarrow\left(x+3\right)^3=0\Rightarrow x=-3\)
23 tháng 3 2022
(3x²-5x+2)+(3x²+5x)= bao nhiêu ạ
Giúp em vs ạ . Em cảm ơn
17 tháng 9 2021
\(1,A⋮B\Leftrightarrow x^3-3x^2-ax+3=\left(x-1\right)\cdot a\left(x\right)\)
Thay \(x=1\)
\(\Leftrightarrow1-3-a+3=0\\ \Leftrightarrow a=1\)
\(2,A⋮B\Leftrightarrow3x^3-16x^2+25x+a=\left(x^2-4x+3\right)\cdot b\left(x\right)\\ \Leftrightarrow3x^3-16x^2+25x+a=\left(x-3\right)\left(x-1\right)\cdot b\left(x\right)\)
Thay \(x=1\)
\(\Leftrightarrow3-16+25+a=0\\ \Leftrightarrow a=-12\)
Thay \(x=3\)
\(\Leftrightarrow3\cdot27-16\cdot9+25\cdot3+a=0\\ \Leftrightarrow81-144+75+a=0\\ \Leftrightarrow12+a=0\Leftrightarrow a=-12\)
Vậy \(a=-12\)
17 tháng 11 2021
\(a,=2x^3y+2x^2y^2-6xy^3\\ b,=3x^3+6x^2-4x-8\\ c,=\left(4x^2+16x-20x-80+76\right):\left(x+4\right)\\ =\left[\left(x+4\right)\left(4x-20\right)+76\right]:\left(x+4\right)\\ =4x-20\left(dư.76\right)\\ d,=\left(x^4-x^2-x^3+x-2x^2+2\right):\left(x^2-1\right)\\ =\left(x^2-1\right)\left(x^2-x-2\right):\left(x^2-1\right)\\ =x^2-x-2\)
2 tháng 9 2021
Bài 2:
a) \(3x^2-7x-10=\left(x+1\right)\left(3x-10\right)\)
b) \(x^2+6x+9-4y^2=\left(x+3\right)^2-\left(2y\right)^2=\left(x+3-2y\right)\left(x+3+2y\right)\)
c) \(x^2-2xy+y^2-5x+5y=\left(x-y\right)^2-5\left(x-y\right)=\left(x-y\right)\left(x-y-5\right)\)
d) \(4x^2-y^2-6x+3y=\left(2x-y\right)\left(2x+y\right)-3\left(2x-y\right)=\left(2x-y\right)\left(2x+y-3\right)\)
e) \(1-2a+2bc+a^2-b^2-c^2=\left(a-1\right)^2-\left(b-c\right)^2=\left(a-1-b+c\right)\left(a-1+b-c\right)\)
f) \(x^3-3x^2-4x+12=\left(x+2\right)\left(x-3\right)\left(x-2\right)\)
g) \(x^4+64=\left(x^2+8\right)^2-16x^2=\left(x^2+8-4x\right)\left(x^2+6+4x\right)\)h) \(x^4-5x^2+4=\left(x+2\right)\left(x+1\right)\left(x-1\right)\left(x-2\right)\)
i) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+16=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+16=\left(x^2+8x+7\right)^2+8\left(x^2+8x+7\right)+16=\left(x^2+8x+11\right)^2\)
2 tháng 9 2021
a: \(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(=\left(x+1\right)\left(3x-10\right)\)
b: \(x^2+6x+9-4y^2\)
\(=\left(x+3\right)^2-4y^2\)
\(=\left(x+3-2y\right)\left(x+3+2y\right)\)
c: \(x^2-2xy+y^2-5x+5y\)
\(=\left(x-y\right)^2-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x-y-5\right)\)
17 tháng 12 2023
Bài 1:
a: \(M=x^2-10x+3\)
\(=x^2-10x+25-22\)
\(=\left(x^2-10x+25\right)-22\)
\(=\left(x-5\right)^2-22>=-22\forall x\)
Dấu '=' xảy ra khi x-5=0
=>x=5
b: \(N=x^2-x+2\)
\(=x^2-x+\dfrac{1}{4}+\dfrac{7}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\forall x\)
Dấu '=' xảy ra khi x-1/2=0
=>x=1/2
c: \(P=3x^2-12x\)
\(=3\left(x^2-4x\right)\)
\(=3\left(x^2-4x+4-4\right)\)
\(=3\left(x-2\right)^2-12>=-12\forall x\)
Dấu '=' xảy ra khi x-2=0
=>x=2
AH
Akai Haruma
Giáo viên
25 tháng 10 2021
a.
$x^2-y^2-2x+2y=(x^2-y^2)-(2x-2y)=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)$
b.
$x^2(x-1)+16(1-x)=x^2(x-1)-16(x-1)=(x-1)(x^2-16)=(x-1)(x-4)(x+4)$
c.
$x^2+4x-y^2+4=(x^2+4x+4)-y^2=(x+2)^2-y^2=(x+2-y)(x+2+y)$
d.
$x^3-3x^2-3x+1=(x^3+1)-(3x^2+3x)=(x+1)(x^2-x+1)-3x(x+1)$
$=(x+1)(x^2-4x+1)$
AH
Akai Haruma
Giáo viên
25 tháng 10 2021
e.
$x^4+4y^4=(x^2)^2+(2y^2)^2+2.x^2.2y^2-4x^2y^2$
$=(x^2+2y^2)^2-(2xy)^2=(x^2+2y^2-2xy)(x^2+2y^2+2xy)$
f.
$x^4-13x^2+36=(x^4-4x^2)-(9x^2-36)$
$=x^2(x^2-4)-9(x^2-4)=(x^2-9)(x^2-4)=(x-3)(x+3)(x-2)(x+2)$
g.
$(x^2+x)^2+4x^2+4x-12=(x^2+x)^2+4(x^2+x)-12$
$=(x^2+x)^2-2(x^2+x)+6(x^2+x)-12$
$=(x^2+x)(x^2+x-2)+6(x^2+x-2)=(x^2+x-2)(x^2+x+6)$
$=[x(x-1)+2(x-1)](x^2+x+6)=(x-1)(x+2)(x^2+x+6)$
h.
$x^6+2x^5+x^4-2x^3-2x^2+1$
$=(x^6+2x^5+x^4)-(2x^3+2x^2)+1$
$=(x^3+x^2)^2-2(x^3+x^2)+1=(x^3+x^2-1)^2$
9 tháng 9 2023
1: \(=\dfrac{-\left[\left(x+5\right)^2-9\right]}{\left(x+2\right)^2}=\dfrac{-\left(x+5-3\right)\left(x+5+3\right)}{\left(x+2\right)^2}\)
\(=\dfrac{-\left(x+2\right)\left(x+8\right)}{\left(x+2\right)^2}=\dfrac{-\left(x+8\right)}{x+2}\)
2: \(=\dfrac{2x\left(x^2-4x+16\right)}{\left(x+4\right)\left(x^2-4x+16\right)}=\dfrac{2x}{x+4}\)
3: \(=\dfrac{5x\left(x^2+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}=\dfrac{5x}{x^2-1}\)
4: \(=\dfrac{3\left(x^2-4x+4\right)}{x\left(x^3-8\right)}=\dfrac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}\)
\(=\dfrac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)
5: \(=\dfrac{2a\left(a-b\right)}{a\left(c+d\right)-b\left(c+d\right)}=\dfrac{2a\left(a-b\right)}{\left(c+d\right)\left(a-b\right)}=\dfrac{2a}{c+d}\)
6: \(=\dfrac{x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}\cdot\left(-1\right)=\dfrac{-x}{x+y}\)
7: \(=\dfrac{2\left(1-a\right)}{-\left(1-a^3\right)}=\dfrac{-2\left(1-a\right)}{\left(1-a\right)\left(1+a+a^2\right)}=-\dfrac{2}{1+a+a^2}\)
8: \(=\dfrac{x^4\left(x^3-1\right)}{\left(x^3-1\right)\left(x^3+1\right)}=\dfrac{x^4}{x^3+1}\)
9: \(=\dfrac{\left(x+2-x+2\right)\left(x+2+x-2\right)}{16x}=\dfrac{4\cdot2x}{16x}=\dfrac{1}{2}\)
10: \(=\dfrac{0.5\left(49x^2-y^2\right)}{0.5x\left(7x-y\right)}=\dfrac{1}{x}\cdot\dfrac{\left(7x-y\right)\left(7x+y\right)}{7x-y}\)
\(=\dfrac{7x+y}{x}\)
\(x^3+x^2+x=3\)
\(\Leftrightarrow x^3+x^2+x-3=0\)
\(\Leftrightarrow x^3+2x^2+3x-x^2-2x-3=0\)
\(\Leftrightarrow x\left(x^2+2x+3\right)-\left(x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+2x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x^2+2x+3=0\end{cases}}\)
+) x - 1 = 0 <=> x = 1
+) x2 + 2x + 3 = 0
Mà \(x^2+2x+3=\left(x+1\right)^2+2\ge2\)
=> Không có x tm trong th này
Vậy pt có nghiệm là x = 1
x3 + x2 + x = 3
<=> x3 + x2 + x - 3 = 0
<=> x3 + 2x2 - x2 + 3x - 2x - 3 = 0
<=> ( x3 + 2x2 + 3x ) - ( x2 + 2x + 3 ) = 0
<=> x( x2 + 2x + 3 ) - 1( x2 + 2x + 3 ) = 0
<=> ( x - 1 )( x2 + 2x + 3 ) = 0
<=> \(\orbr{\begin{cases}x-1=0\\x^2+2x+3=0\end{cases}}\)
+) x - 1 = 0 => x = 1
+) x2 + 2x + 3 = ( x2 + 2x + 1 ) + 2 = ( x + 1 )2 + 2 ≥ 2 > 0 ∀ x
Vậy phương trình có nghiệm duy nhất là x = 1