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a) \(x^2+4x+3\)
\(=x^2+3x+x+3\)
\(=x\left(x+3\right)+\left(x+3\right)\)
\(=\left(x+1\right)\left(x+3\right)\)
1) \(\left(3x+7\right)^2-\left(2x-3\right)^2=0\)
\(\Leftrightarrow\left(3x+7-2x+3\right)\left(3x+7+2x-3\right)=0\)
\(\Leftrightarrow\left(x+10\right)\left(5x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+10=0\\5x+4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-10\\x=\frac{-4}{5}\end{cases}}\)
Vạy ...
phần 2 tương tự áp dụng \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
\((4x-1)^2-(5-3x)^2=0\)
\(\Leftrightarrow(4x-1-5-3x)(4x+1+5-3x)=0\)
\(\Leftrightarrow(x-6)(x+6)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-6=0\\x+6=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
Vậy : ...
\(x^2-x-1=0\)
\(\Leftrightarrow\left(x^2-2\cdot\frac{1}{2}\cdot x+\frac{1}{4}\right)-\frac{5}{4}=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2-\frac{5}{4}=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=\frac{5}{4}\)
\(\Leftrightarrow x=\frac{\sqrt{5}}{2}+\frac{1}{2};x=\frac{-\sqrt{5}}{2}+\frac{1}{2}\)
\(x^2-2x-1=0\)
\(\Leftrightarrow\left(x^2-2x+1\right)-2=0\)
\(\Leftrightarrow\left(x-1\right)^2-2=0\)
\(\Leftrightarrow\left(x-1\right)^2=2\)
\(\Leftrightarrow x=\sqrt{2}+1;x=-\sqrt{2}+1\)
1/ \(\left(9x^2-25\right)-\left(6x-10\right)=0\)
\(\Leftrightarrow9x^2-6x-35=0\)
\(\Leftrightarrow\left(2x-1\right)^2-36=0\)
\(\Leftrightarrow\left(2x-7\right)\left(2x+6\right)=0\)
2/ \(\left(3x+5\right)^2-4x^2=0\)
\(\Leftrightarrow\left(x+5\right)\left(5x+5\right)=0\)
3/ \(25x^2-\left(4x-3\right)^2=0\)
\(\Leftrightarrow\left(x+3\right)\left(9x-3\right)=0\)
1) ( 9x2 - 25 ) - ( 6x - 10 ) = 0
\(\Leftrightarrow\) [ ( 3x)2 - 52 ] - 2.( 3x + 5 ) = 0
\(\Leftrightarrow\)( 3x - 5 ).( 3x + 5 ) - 2.( 3x - 5 ) = 0
\(\Leftrightarrow\) ( 3x + 5 ).( 3x + 5 - 2 ) = 0
\(\Leftrightarrow\)( 3x + 5 ).( 3x + 3 ) = 0
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+5=0\\3x+3=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=-5\\3x=-3\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{-5}{3}\\x=-1\end{cases}}\)
Vậy x = \(\frac{-5}{3}\) , x = -1
2) ( 3x + 5 )2 - 4x2 = 0
\(\Leftrightarrow\) ( 3x + 5 - 2x ).( 3x + 5 + 2x ) = 0
\(\Leftrightarrow\)( x + 5 ).( 5x + 5 ) = 0
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+5=0\\5x+5=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-5\\x=-1\end{cases}}\)
Vậy x = -5 , x = -1
3) 25x2 - ( 4x - 3 )2 = 0
\(\Leftrightarrow\)( 5x )2 - ( 4x - 3 )2 = 0
\(\Leftrightarrow\) ( 5x - 4x + 3 ).(5x + 4x - 3 ) = 0
\(\Leftrightarrow\)( x + 3 ).( 9x - 3 ) = 0
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+3=0\\9x-3=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-3\\9x=3\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}\)
Vậy x = 3 , x = \(\frac{1}{3}\)
\(A=x^2-7xy+12y^2\)
\(A=x^2-3xy-4xy+12y^2\)
\(A=x\left(x-3y\right)-4y\left(x-3y\right)\)
\(A=\left(x-4y\right)\left(x-3y\right)\)
\(B=x^2-3xy-4y^2\)
\(B=x^2+xy-4xy-4y^2\)
\(B=x\left(x+y\right)-4y\left(x+y\right)\)
\(B=\left(x-4y\right)\left(x+y\right)\)
\(A=x^2-7xy+12y^2\)
\(=x^2-3xy-4xy+12y^2\)
\(=x\left(x-3y\right)-4y\left(x-3y\right)\)
\(=\left(x-4y\right)\left(x-3y\right)\)
\(x^2-xy-12y^2=0\)
\(\Leftrightarrow\left(x^2+3xy\right)-\left(4xy-12y^2\right)=0\)
\(\Leftrightarrow x\left(x+3y\right)-4y\left(x+3y\right)=0\)
\(\Leftrightarrow\left(x+3y\right)\left(x-4y\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-3y\\x=4y\end{cases}}\)
TH1:\(x=-3y\)
\(A=\frac{3\cdot\left(-3y\right)+2y}{3\left(-3y\right)-2y}=\frac{-9y+2y}{-9y-2y}=\frac{-7y}{-11y}=\frac{7}{11}\)
TH2:\(x=4y\)
\(A=\frac{3\cdot4y+2y}{3\cdot4y-2y}=\frac{12y+2y}{12y-2y}=\frac{14y}{10y}=\frac{7}{5}\)
a. Biểu thức ko thể biểu diễn dưới dạng tích của các thừa số
b. (x-1)(4x+1)
c. -(3z^2-5y^2-6xy-3x^2)
d. x(y^2-2xy+x-9)
e. -(y-x)(y-x+2)
f. y^3+xy^2+3x^2y-y+x^2-x
HỌC TỐT.
Bài 1.
a) x( 8x - 2 ) - 8x2 + 12 = 0
<=> 8x2 - 2x - 8x2 + 12 = 0
<=> 12 - 2x = 0
<=> 2x = 12
<=> x = 6
b) x( 4x - 5 ) - ( 2x + 1 )2 = 0
<=> 4x2 - 5x - ( 4x2 + 4x + 1 ) = 0
<=> 4x2 - 5x - 4x2 - 4x - 1 = 0
<=> -9x - 1 = 0
<=> -9x = 1
<=> x = -1/9
c) ( 5 - 2x )( 2x + 7 ) = ( 2x - 5 )( 2x + 5 )
<=> -4x2 - 4x + 35 = 4x2 - 25
<=> -4x2 - 4x + 35 - 4x2 + 25 = 0
<=> -8x2 - 4x + 60 = 0
<=> -8x2 + 20x - 24x + 60 = 0
<=> -4x( 2x - 5 ) - 12( 2x - 5 ) = 0
<=> ( 2x - 5 )( -4x - 12 ) = 0
<=> \(\orbr{\begin{cases}2x-5=0\\-4x-12=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-3\end{cases}}\)
d) 64x2 - 49 = 0
<=> ( 8x )2 - 72 = 0
<=> ( 8x - 7 )( 8x + 7 ) = 0
<=> \(\orbr{\begin{cases}8x-7=0\\8x+7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{7}{8}\\x=-\frac{7}{8}\end{cases}}\)
e) ( x2 + 6x + 9 )( x2 + 8x + 7 ) = 0
<=> ( x + 3 )2( x2 + x + 7x + 7 ) = 0
<=> ( x + 3 )2 [ x( x + 1 ) + 7( x + 1 ) ] = 0
<=> ( x + 3 )2( x + 1 )( x + 7 ) = 0
<=> x = -3 hoặc x = -1 hoặc x = -7
g) ( x2 + 1 )( x2 - 8x + 7 ) = 0
Vì x2 + 1 ≥ 1 > 0 với mọi x
=> x2 - 8x + 7 = 0
=> x2 - x - 7x + 7 = 0
=> x( x - 1 ) - 7( x - 1 ) = 0
=> ( x - 1 )( x - 7 ) = 0
=> \(\orbr{\begin{cases}x-1=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=7\end{cases}}\)
Bài 2.
a) ( x - 1 )2 - ( x - 2 )( x + 2 )
= x2 - 2x + 1 - ( x2 - 4 )
= x2 - 2x + 1 - x2 + 4
= -2x + 5
b) ( 3x + 5 )2 + ( 26x + 10 )( 2 - 3x ) + ( 2 - 3x )2
= 9x2 + 30x + 25 - 78x2 + 22x + 20 + 9x2 - 12x + 4
= ( 9x2 - 78x2 + 9x2 ) + ( 30x + 22x - 12x ) + ( 25 + 20 + 4 )
= -60x2 + 40x2 + 49
d) ( x + y )2 - ( x + y - 2 )2
= [ x + y - ( x + y - 2 ) ][ x + y + ( x + y - 2 ) ]
= ( x + y - x - y + 2 )( x + y + x + y - 2 )
= 2( 2x + 2y - 2 )
= 4x + 4y - 4
Bài 3.
A = 3x2 + 18x + 33
= 3( x2 + 6x + 9 ) + 6
= 3( x + 3 )2 + 6 ≥ 6 ∀ x
Đẳng thức xảy ra <=> x + 3 = 0 => x = -3
=> MinA = 6 <=> x = -3
B = x2 - 6x + 10 + y2
= ( x2 - 6x + 9 ) + y2 + 1
= ( x - 3 )2 + y2 + 1 ≥ 1 ∀ x,y
Đẳng thức xảy ra <=> \(\hept{\begin{cases}x-3=0\\y^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=3\\y=0\end{cases}}\)
=> MinB = 1 <=> x = 3 ; y = 0
C = ( 2x - 1 )2 + ( x + 2 )2
= 4x2 - 4x + 1 + x2 + 4x + 4
= 5x2 + 5 ≥ 5 ∀ x
Đẳng thức xảy ra <=> 5x2 = 0 => x = 0
=> MinC = 5 <=> x = 0
D = -2/7x2 - 8x + 7 ( sửa thành tìm Max )
Để D đạt GTLN => 7x2 - 8x + 7 đạt GTNN
7x2 - 8x + 7
= 7( x2 - 8/7x + 16/49 ) + 33/7
= 7( x - 4/7 )2 + 33/7 ≥ 33/7 ∀ x
Đẳng thức xảy ra <=> x - 4/7 = 0 => x = 4/7
=> MaxC = \(\frac{-2}{\frac{33}{7}}=-\frac{14}{33}\)<=> x = 4/7
1,\(\left(x-3\right)^3-5\left(x-2\right)+5=0\)
\(\Rightarrow\left(x-3\right)^3-5\left(x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left[\left(x-3\right)^2-5\right]=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\\left(x-3\right)^2-5=0\end{cases}}\)
\(\Rightarrow x=3\) hoặc \(x=\sqrt{5}+3\) hoặc \(x=-\sqrt{5}+3\)
Vậy........
1) \(2x^2-3x-2\)
\(=2x^2-4x+x-2\)
\(=2x\left(x-2\right)+x-2\)
\(=\left(2x+1\right)\left(x-2\right)\)
2) \(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(=3x\left(x+1\right)-10\left(x+1\right)\)
\(=\left(3x-10\right)\left(x+1\right)\)