\(\text{phân tích đa thức sau thành nhân tử: (a-b)(b-c)(a-c)+(a+b)(b+c)(a-c)+(a+b)(a+c)(c-b)}\)
mong các bn giúp đỡ ạ cảm ơn nhìu
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: \(\left(a-b\right)\left(b-c\right)\left(a-c\right)+\left(a+b\right)\left(b+c\right)\left(a-c\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a-c\right).\left[\left(a-b\right)\left(b-c\right)+\left(a+b\right)\left(b+c\right)\right]+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a-c\right).\left(ab-ac-b^2+bc+ab+ac+b^2+bc\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a-c\right).\left(2ab+2bc\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=2b.\left(a-c\right).\left(a+c\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a+c\right)\left[2b\left(a-c\right)+\left(a+b\right)\left(c-b\right)\right]\)
\(=\left(a+c\right)\left(2ab-2bc+ac-ab+bc-b^2\right)\)
\(=\left(a+c\right)\left(ab-bc+ac-b^2\right)\)
\(=\left(a+c\right)\left[a.\left(b+c\right)-b.\left(b+c\right)\right]\)
\(=\left(a+c\right)\left(a-b\right)\left(b+c\right)\)
A = a( b + 2 ) + b( 2 + b )
= a( b + 2 ) + b( b + 2 )
= ( a + b )( b + 2 )
Với a = 2 ; b = 3
A = ( 2 + 3 )( 3 + 2 ) = 5.5 = 25
B = b2 + b + c( b + 1 )
= b( b + 1 ) + c( b + 1 )
= ( b + c )( b + 1 )
Với b = 1 ; c = 2
B = ( 1 + 2 )( 1 + 1 ) = 6
C = xy( x - y ) - 2x + 2y
= xy( x - y ) - 2( x - y )
= ( x - y )( xy - 2 )
Với xy = 8 ; x - y = 5
C = 5.( 8 - 2 ) = 30
D = x( x + y ) - xy( x + y )
= ( x + y )( x - xy )
= ( x + y )x( 1 - y )
Với x = 1 ; y = -5
D = ( 1 - 5 ).1.[ 1 - ( -5 ) ] = -24
Bài làm:
a) Ta có: \(x\left(x-3\right)-2x+6=0\)
\(\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
b) \(\frac{x}{3}+\frac{x^2}{2}=0\)
\(\Leftrightarrow\frac{3x^2+2x}{6}=0\)
\(\Leftrightarrow x\left(3x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\3x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-\frac{3}{2}\end{cases}}\)
c) \(x-2=\left(x-2\right)^2\)
\(\Leftrightarrow x^2-4x+4+2-x=0\)
\(\Leftrightarrow x^2-5x+6=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
d) \(\left(x^2+3\right)\left(x+1\right)+x=-1\)
\(\Leftrightarrow\left(x^2+3\right)\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^2+4\right)\left(x+1\right)=0\)
Vì \(x^2+4>\left(\forall x\right)\) => \(x=-1\)
a. \(x\left(x-3\right)-2x+6=0\)
\(\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
b. \(\frac{x}{3}+\frac{x^2}{2}=0\)
\(\Leftrightarrow\frac{2x+3x^2}{6}=0\)
\(\Leftrightarrow x\left(2+3x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\2+3x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{2}{3}\end{cases}}\)
c. \(x-2=\left(x-2\right)^2\)
\(\Leftrightarrow x-2-x^2+4x-4=0\)
\(\Leftrightarrow-\left(x-3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}\)
d. \(\left(x^2+3\right)\left(x+1\right)+x=-1\)
\(\Leftrightarrow x^3+x^2+3x+3+x+1=0\)
\(\Leftrightarrow x^3+x^2+4x+4=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^2+4=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x^2=-4\left(vo-ly\right)\end{cases}}\)
<=> x = - 1
a) x( x + 2018 ) - 2x - 4036 = 0
<=> x( x + 2018 ) - 2( x + 2018 ) = 0
<=> ( x + 2018 )( x - 2 ) = 0
<=> \(\orbr{\begin{cases}x+2018=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2018\\x=2\end{cases}}\)
b) x + 5 = 2( x + 5 )2
<=> x + 5 = 2( x2 + 10x + 25 )
<=> x + 5 = 2x2 + 20x + 50
<=> 2x2 + 20x + 50 - x - 5 = 0
<=> 2x2 + 19x + 45 = 0
<=> 2x2 + 10x + 9x + 45 = 0
<=> 2x( x + 5 ) + 9( x + 5 ) = 0
<=> ( x + 5 )( 2x + 9 ) = 0
<=> \(\orbr{\begin{cases}x+5=0\\2x+9=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=-\frac{9}{2}\end{cases}}\)
c) ( x2 + 1 )( 2x - 1 ) + 2x = 1
<=> 2x3 - x2 + 4x - 1 - 1 = 0
<=> 2x3 - x2 + 4x - 2 = 0
<=> x2( 2x - 1 ) + 2( 2x - 1 ) = 0
<=> ( 2x - 1 )( x2 + 2 ) = 0
<=> \(\orbr{\begin{cases}2x-1=0\\x^2+2=0\end{cases}\Leftrightarrow}x=\frac{1}{2}\)( vì x2 + 2 ≥ 2 > 0 ∀ x )
d) \(\frac{x}{3}-\frac{x^2}{4}=0\)
\(\Leftrightarrow\frac{4x}{12}-\frac{3x^2}{12}=0\)
\(\Leftrightarrow\frac{4x-3x^2}{12}=0\)
\(\Leftrightarrow4x-3x^2=0\)
\(\Leftrightarrow x\left(4-3x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\4-3x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)
Bài này đề sửa thành: \(H=a+4b+1\) mk ms lm được ạ
Ta có: \(a=111...1\) (2020 chữ số 1)
\(a=111...1\cdot100...0+111...1\)
\(a=b.\left(9b+1\right)+b\)
Thay vào:
\(H=a+4b+1=b\left(9b+1\right)+b+4b+1=9b^2+6b+1=\left(3b+1\right)^2\)
=> đpcm
\(8x\left(x-2017\right)-2x+4034=0\)\(\Leftrightarrow8x\left(x-2017\right)-2\left(x-2017\right)=0\)
\(\Leftrightarrow2\left(x-2017\right)\cdot\left(4x-1\right)=0\)\(\Leftrightarrow\hept{\begin{cases}x-2017=0\\4x-1=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=2017\\x=\frac{1}{4}\end{cases}}\)
Vậy \(x=2017\)hoặc \(x=\frac{1}{4}\)
8x( x - 2017 ) - 2x + 4034 = 0
<=> 8x( x - 2017 ) - 2( x - 2017 ) = 0
<=> ( 8x - 2 )( x - 2017 ) = 0
<=> \(\orbr{\begin{cases}8x-2=0\\x-2017=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{4}\\x=2017\end{cases}}\)
3x4 - 8x3 + 16
Thử với x = 2 ta được :
3.24 - 8.23 + 16 = 0
Vậy x = 2 là nghiệm của đa thức . Theo hệ quả của định lí Bézout thì đa thức trên chia hết cho x - 2
Thực hiện phép chia 3x4 - 8x3 + 16 cho x - 2 ta được 3x3 - 2x2 - 4x - 8
=> 3x4 - 8x3 + 16 = ( x - 2 )( 3x3 - 2x2 - 4x - 8 )
Ta có : 3x3 - 2x2 - 4x - 8
= 3x3 + 4x2 + 4x - 6x2 - 8x - 8
= x( 3x2 + 4x + 4 ) - 2( 3x2 + 4x + 4 )
= ( x - 2 )( 3x2 + 4x + 4 )
Tổng kết : 3x4 - 8x3 + 16 = ( x - 2 )( x - 2 )( 3x2 + 4x + 4 ) = ( x - 2 )2( 3x2 + 4x + 4 )
Ta có: \(3x^4-8x^3+16=\left(3x^4-12x^3+12x^2\right)+\left(4x^3-16x^2+16x\right)+\left(4x^2-16x+16\right)\)
\(=3x^2.\left(x^2-4x+4\right)+4x.\left(x^2-4x+4\right)+4.\left(x^2-4x+4\right)\)
\(=\left(3x^3+4x+4\right)\left(x-2\right)^2\)
Ta có: \(\left(a-b\right)\left(b-c\right)\left(a-c\right)+\left(a+b\right)\left(b+c\right)\left(a-c\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a-c\right).\left[\left(a-b\right)\left(b-c\right)+\left(a+b\right)\left(b+c\right)\right]+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a-c\right).\left(ab-ac-b^2+bc+ab+ac+b^2+bc\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a-c\right).\left(2ab+2bc\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=2b.\left(a-c\right).\left(a+c\right)+\left(a+b\right)\left(a+c\right)\left(c-b\right)\)
\(=\left(a+c\right)\left[2b\left(a-c\right)+\left(a+b\right)\left(c-b\right)\right]\)
\(=\left(a+c\right)\left(2ab-2bc+ac-ab+bc-b^2\right)\)
\(=\left(a+c\right)\left(ab-bc+ac-b^2\right)\)
\(=\left(a+c\right)\left[a.\left(b+c\right)-b.\left(b+c\right)\right]\)
\(=\left(a+c\right)\left(a-b\right)\left(b+c\right)\)