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a)\(x^2+3x+2\)
\(=x^2+x+2x+2\)
\(=x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+2\right)\left(x+1\right)\)
k)\(4x^2+4x+1\)
\(=\left(2x\right)^2+2.2x+1^2\)
\(=\left(2x+1\right)^2\)
\(A=x^2-6x-4=x^2-6x+9-13=\left(x-3\right)^2-13\ge-13\)
Vậy \(A_{min}=-13\Leftrightarrow x=3\)
\(B=x^2-x+1=x^2-2.\frac{1}{2}x+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\)
Vậy \(B_{min}=\frac{3}{4}\Leftrightarrow x=\frac{1}{2}\)
\(=x^3+x^2-\left(4x+4\right)=x^2\left(x+1\right)-4\left(x+1\right)=\left(x^2-4\right)\left(x+1\right)\)
\(=\left(x-2\right)\left(x+1\right)\left(x+2\right)\)
\(x^4+x^3+x^2-1=x^3\left(x+1\right)+\left(x-1\right)\left(x+1\right)=\left(x+1\right)\left(x^3+x-1\right)\)
\(c,=\left(x+y\right)^2-2\left(x+y\right)+1=\left(x+y-1\right)^2\)
\(d,=x^2y^2-y^2-x^2+1=\left(x^2-1\right)\left(y^2-1\right)=\left(x-1\right)\left(y-1\right)\left(x+1\right)\left(y+1\right)\)
\(e,4x^2+4x-15=\left(4x^2+4x+1\right)-16=\left(2x+1\right)^2-4^2=\left(2x+5\right)\left(2x-3\right)\)
\(3x^2-7x+2=\left(3x^2-6x\right)-\left(x-2\right)=3x\left(x-2\right)-\left(x-2\right)=\left(3x-1\right)\left(x-2\right)\)
\(4x^2-5x+1=\left(4x^2-4x\right)-\left(x-1\right)=4x\left(x-1\right)-\left(x-1\right)=\left(4x-1\right)\left(x-1\right)\)
Phân tích à :v
a) x3 + x2 - 4x - 4 = x2( x + 1 ) - 4( x + 1 ) = ( x + 1 )( x2 - 4 ) = ( x + 1 )( x - 2 )( x + 2 )
b) x4 + x3 + x2 - 1 = x3( x + 1 ) + ( x - 1 )( x + 1 ) = ( x + 1 )( x3 + x - 1 )
c) x2 + 2xy + y2 - 2x - 2y + 1 = ( x2 + 2xy + y2 ) - ( 2x + 2y ) + 1 = ( x + y )2 - 2( x + y ) + 12 = ( x + y - 1 )2
d) x2y2 + 1 - x2 - y2 = ( x2y2 - x2 ) - ( y2 - 1 ) = x2( y2 - 1 ) - ( y2 - 1 ) = ( y2 - 1 )( x2 - 1 ) = ( y - 1 )( y + 1 )( x - 1 )( x + 1 )
e) 4x2 + 4x - 15 = ( 4x2 + 4x + 1 ) - 16 = ( 2x + 1 )2 - 42 = ( 2x + 1 - 4 )( 2x + 1 + 4 ) = ( 2x - 3 )( 2x + 5 )
g) 3x2 - 7x + 2 = 3x2 - 6x - x + 2 = 3x( x - 2 ) - ( x - 2 ) = ( x - 2 )( 3x - 1 )
h) 4x2 - 5x + 1 = 4x2 - 4x - x + 1 = 4x( x - 1 ) - ( x - 1 ) = ( x - 1 )( 4x - 1 )
a) Đặt t = x2
bthuc <=> t2 - 7t + 16
Từ đây ta không thể phân tích được :)
b) x3 - 2x2 + 5x - 4
= x3 - x2 - x2 + x + 4x - 4
= x2( x - 1 ) - x( x - 1 ) + 4( x - 1 )
= ( x - 1 )( x2 - x + 4 )
c) x3 - 2x2 + x - 3 ( phân tích hổng ra :)) )
d) 3x3 - 4x2 + 12x - 4 ( phân tích hổng ra p2 :)) )
e) 6x3 + x2 + x + 1
= 6x3 + 3x2 - 2x2 - x + 2x + 1
= 3x2( 2x + 1 ) - x( 2x - 1 ) + ( 2x + 1 )
= ( 2x + 1 )( 3x2 - x + 1 )
f) 4x3 + 6x2 + 4x + 1
= 4x3 + 2x2 + 4x2 + 2x + 2x + 1
= 2x2( 2x + 1 ) + 2x( 2x + 1 ) + ( 2x + 1 )
= ( 2x + 1 )( 2x2 + 2x + 1 )
a) 4x2 - 12x + 5
= 4x2 - 2x - 10x + 5
= (4x2 - 2x) + (-10x + 5)
= 2x(x - 2) - 5 ( x - 2)
= (x - 2) (2x - 5)
còn lại tương tự nha
ok mk nha!!! 5547565876876876345645645666575676575688768898778978234532344543
ý lộn mk làm lại:
4x2 - 12x + 5
= 4x2 - 2x - 10x + 5
= 2x(2x - 1) - 5(2x - 1)
= (2x - 1) (2x - 5)
ok mk nhé!!!! 546465765876876876769789785653645645234634634457756756
Bài làm :
a) x( 2x - 7 ) - 4x + 14 = 0
<=> x( 2x - 7 ) - 2( 2x - 7 ) = 0
<=> ( 2x - 7 )( x - 2 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}2x-7=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=2\end{cases}}\)
b) Sửa đề : 5x3 + x2 - 4x + 9 = 0
<=>( 5x3 + 5 ) + (x2 - 4x +4)=0
<=> 5(x3 + 1) + (x-2)2 = 0
<=> 5(x+1)(x2 - x +1) + (x+2)2 =0
\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-2\end{cases}}\)
c) 3x3 - 7x2 + 6x - 14 = 0
<=> 3x2( x - 7/3 ) + 6( x - 7/3 ) = 0
<=> ( x - 7/3 )( 3x2 + 6 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{7}{3}=0\\3x^2+6=0\end{cases}}\Leftrightarrow x=\frac{7}{3}\)
d) 5x2 - 5x = 3( x - 1 )
<=> 5x( x - 1 ) - 3( x - 1 ) = 0
<=> ( x - 1 )( 5x - 3 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\5x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{3}{5}\end{cases}}\)
e) 4x2 - 25 - ( 4x - 10 ) = 0
<=> ( 2x - 5 )( 2x + 5 ) - 2( 2x - 5 ) = 0
<=> ( 2x - 5 )( 2x + 5 - 2 ) = 0
<=> ( 2x - 5 )( 2x + 3 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}2x-5=0\\2x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{3}{2}\end{cases}}\)
f) x3 + 27 + ( x + 3 )( x - 9 ) = 0
<=> ( x + 3 )( x2 - 3x + 9 ) + ( x + 3 )( x - 9 ) = 0
<=> ( x + 3 )( x2 - 3x + 9 + x - 9 ) = 0
<=> ( x + 3 )( x2 - 2x ) = 0
<=> x( x + 3 )( x - 2 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}\\\end{cases}}\begin{cases}x=0\\x=-3\\x=2\end{cases}\)
a) x( 2x - 7 ) - 4x + 14 = 0
<=> x( 2x - 7 ) - 2( 2x - 7 ) = 0
<=> ( 2x - 7 )( x - 2 ) = 0
<=> \(\orbr{\begin{cases}2x-7=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{2}\\x=2\end{cases}}\)
b) 5x3 + x2 - 4x - 9 = 0 ( đề sai )
c) 3x3 - 7x2 + 6x - 14 = 0
<=> 3x2( x - 7/3 ) + 6( x - 7/3 ) = 0
<=> ( x - 7/3 )( 3x2 + 6 ) = 0
<=> \(\orbr{\begin{cases}x-\frac{7}{3}=0\\3x^2+6=0\end{cases}}\Leftrightarrow x=\frac{7}{3}\)( do 3x2 + 6 ≥ 6 > 0 với mọi x )
d) 5x2 - 5x = 3( x - 1 )
<=> 5x( x - 1 ) - 3( x - 1 ) = 0
<=> ( x - 1 )( 5x - 3 ) = 0
<=> \(\orbr{\begin{cases}x-1=0\\5x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=\frac{3}{5}\end{cases}}\)
e) 4x2 - 25 - ( 4x - 10 ) = 0
<=> ( 2x - 5 )( 2x + 5 ) - 2( 2x - 5 ) = 0
<=> ( 2x - 5 )( 2x + 5 - 2 ) = 0
<=> ( 2x - 5 )( 2x + 3 ) = 0
<=> \(\orbr{\begin{cases}2x-5=0\\2x+3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{3}{2}\end{cases}}\)
f) x3 + 27 + ( x + 3 )( x - 9 ) = 0
<=> ( x + 3 )( x2 - 3x + 9 ) + ( x + 3 )( x - 9 ) = 0
<=> ( x + 3 )( x2 - 3x + 9 + x - 9 ) = 0
<=> ( x + 3 )( x2 - 2x ) = 0
<=> x( x + 3 )( x - 2 ) = 0
<=> x = 0 hoặc x + 3 = 0 hoặc x - 2 = 0
<=> x = 0 hoặc x = -3 hoặc x = 2
a) Ta có: \(A=x^2-6x+11\)
\(=x^2-6x+9+2\)
\(=\left(x^2-6x+9\right)+2\)
\(=\left(x-3\right)^2+2\)
Ta có: \(\left(x-3\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-3\right)^2+2\ge2\forall x\)
Dấu '=' xảy ra khi
\(\left(x-3\right)^2=0\Leftrightarrow x-3=0\Leftrightarrow x=3\)
Vậy: GTNN của đa thức \(A=x^2-6x+11\) là 2 khi x=3
b) Ta có: \(B=x^2-4x+3\)
\(=x^2-4x+4-1\)
\(=\left(x^2-4x+4\right)-1\)
\(=\left(x-2\right)^2-1\)
Ta có: \(\left(x-2\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-2\right)^2-1\ge-1\forall x\)
Dấu '=' xảy ra khi
\(\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)
Vậy: GTNN của đa thức \(B=x^2-4x+3\) là -1 khi x=2
c) Ta có: \(C=x^2+5x\)
\(=x^2+2\cdot x\cdot\frac{5}{2}+\frac{25}{4}-\frac{25}{4}\)
\(=\left(x^2+2\cdot x\cdot\frac{5}{2}+\frac{25}{4}\right)-\frac{25}{4}\)
\(=\left(x+\frac{5}{2}\right)^2-\frac{25}{4}\)
Ta có: \(\left(x+\frac{5}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{5}{2}\right)^2-\frac{25}{4}\ge\frac{-25}{4}\forall x\)
Dấu '=' xảy ra khi
\(\left(x+\frac{5}{2}\right)^2=0\Leftrightarrow x+\frac{5}{2}=0\Leftrightarrow x=\frac{-5}{2}\)
Vậy: GTNN của đa thức \(C=x^2+5x\) là \(\frac{-25}{4}\) khi \(x=\frac{-5}{2}\)
d) Ta có: \(D=x^2+x+1\)
\(=x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}+\frac{3}{4}\)
\(=\left(x^2+2\cdot x\cdot\frac{1}{2}+\frac{1}{4}\right)+\frac{3}{4}\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\)
Ta có: \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
Dấu '=' xảy ra khi
\(\left(x+\frac{1}{2}\right)^2=0\Leftrightarrow x+\frac{1}{2}=0\Leftrightarrow x=\frac{-1}{2}\)
Vậy: GTNN của đa thức \(D=x^2+x+1\) là \(\frac{3}{4}\) khi \(x=\frac{-1}{2}\)
e) Ta có: \(E=4x^2+4x-2\)
\(=\left(2x\right)^2+2\cdot2x\cdot1+1-3\)
\(=\left[\left(2x\right)^2+2\cdot2x\cdot1+1\right]-3\)
\(=\left(2x+1\right)^2-3\)
Ta có: \(\left(2x+1\right)^2\ge0\forall x\)
\(\Rightarrow\left(2x+1\right)^2-3\ge-3\forall x\)
Dấu '='xảy ra khi
\(\left(2x+1\right)^2=0\Leftrightarrow2x+1=0\Leftrightarrow2x=-1\Leftrightarrow x=\frac{-1}{2}\)
Vậy: GTNN của đa thức \(E=4x^2+4x-2\) là -3 khi \(x=\frac{-1}{2}\)
g) Ta có: \(G=x^2-7x\)
\(=x^2-2\cdot x\cdot\frac{7}{2}+\frac{49}{14}-\frac{49}{14}\)
\(=\left(x^2-2\cdot x\cdot\frac{7}{2}+\frac{49}{4}\right)-\frac{49}{4}\)
\(=\left(x-\frac{7}{2}\right)^2-\frac{49}{4}\)
Ta có: \(\left(x-\frac{7}{2}\right)^2\ge0\forall x\)
\(\Rightarrow\left(x-\frac{7}{2}\right)^2-\frac{49}{4}\ge\frac{-49}{4}\forall x\)
Dấu '=' xảy ra khi
\(\left(x-\frac{7}{2}\right)^2=0\Leftrightarrow x-\frac{7}{2}=0\Leftrightarrow x=\frac{7}{2}\)
Vậy: GTNN của đa thức \(G=x^2-7x\) là \(\frac{-49}{4}\) khi \(x=\frac{7}{2}\)
\(A=x^2-6x+11\)
\(A=x^2-2.x.3+3^2-3^2+11\)
\(A=\left(x^2-6x+3^2\right)-3^2+11\)
\(A=\left(x-3\right)^2+2\)
Vì \(\left(x-3\right)^2\ge0\forall x\)
=>\(\left(x-3\right)^2\ge0\ge2\forall x\)
Min A = 2 khi \(\left(x-3\right)^2=0\)
=> \(x-3=0hayx=3\)
Vậy Min A = 2 khi x = 3
\(B=x^2-4x+3\)
\(B=x^2-2.x.2+2^2-2^2+3\)
\(B=\left(x^2-4x+2^2\right)-4+3\)
\(B=\left(x-2\right)^2-1\)
=> \(\left(x-2\right)^2-1\ge0\forall x\)
MIn B = -1 khi \(\left(x-2\right)^2=0\)
=>\(\left(x-2\right)=0hayx=2\)
Vậy Min B = -1 khi x= 2