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a) \(x^2+4x+3=\left(x^2+4x+4\right)-1=\left(x+2\right)^2-1^2=\left(x+1\right)\left(x+3\right)\) (mình sửa lại)
b) \(x^2+8x-9=\left(x^2+8x+16\right)-25=\left(x+4\right)^2-5^2=\left(x-1\right)\left(x+9\right)\)
c) \(3x^2+6x-9=3\left[\left(x^2+2x+1\right)-4\right]=3\left[\left(x+1\right)^2-2^2\right]=3\left(x-1\right)\left(x+3\right)\)
d) \(2x^2+x-3=2x^2-4x+2+5x-5=2\left(x^2-2x+1\right)+5\left(x-1\right)=2\left(x-1\right)^2+5\left(x-1\right)=\left(x-1\right)\left(2x+3\right)\)
\(a,\frac{1}{64}x^6-125y^3\)
\(=\left(\frac{1}{2}x\right)^6-\left(5y\right)^3\)
\(=\left(\frac{1}{4}x^2\right)^3-\left(5y\right)^3\)
\(\left(\frac{1}{4}x^2-5y\right)\left[\left(\frac{1}{4}x^2\right)^2+\left(\frac{1}{4}x^2\right).5y+25y^2\right]\)
\(b,27a^3-54a^2b+36ab^2-8b^3\)
\(=\left(3a\right)^3-3.2.\left(3a\right)^2b+3.3a.\left(2b\right)^2-\left(2b\right)^3\)
\(=\left(3a-2b\right)^3\)
\(c,x^6-x^6\)
\(=0\)
\(d,10x-25-x^2\)
\(=-x^2+10x-25\)
\(=-\left(x^2-10x+25\right)\)
\(=-\left(x-5\right)^2\)
Bài làm:
a) \(x^6-6x^4+12x^2-8\)
\(=\left(x^2-2\right)^3\)
b) \(x^2+16-8x=\left(x-4\right)^2\)
c) \(10x-x^2-25=-\left(x-5\right)^2\)
d) \(9\left(a-b\right)^2-4\left(x-y\right)^2\)
\(=\left[3\left(a-b\right)\right]^2-\left[2\left(x-y\right)\right]^2\)
\(=\left(3a-3b-2x+2y\right)\left(3a-3b+2x-2y\right)\)
e) \(\left(x+y\right)^2-2xy+1\)
\(=x^2+2xy+y^2-2xy+1\)
\(=x^2+y^2+1\)
sai sai
a. \(x^6-6x^4+12x^2-8=\left(x^2\right)^3-3\left(x^2\right)^2.2+3x^22-2^3=\left(x^2-2\right)^3\)
b. \(x^2+16-8x=x^2-8x+4^2=\left(x-4\right)^2\)
c. \(10x-x^2-25=10x-x^2-5^2=-\left(x-5\right)^2\)
d. \(9\left(a-b\right)^2-4\left(x-y\right)^2=\left[3\left(x-y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x+y\right)\right]\)
\(=\left(3x-3y-2x-2y\right)\left(3x-3y+2x+2y\right)=\left(x-5y\right)\left(5x-y\right)\)
e. \(\left(x+y\right)^2-2xy+1=x^2+2xy+y^2-2xy+1=x\left(x+2y\right)-y\left(y+2x\right)+2y^2+1\)
\(=x\left(x+y\right)-y\left(y+x\right)+xy-yx+2y^2+x=\left(x-y\right)\left(x+y\right)+2y^2+x\)
Bài 1:
\(6x^2-2\left(x-y\right)^2-6y^2\)
\(=6\left(x-y\right)\left(x+1\right)-2\left(x-y\right)^2\)
\(=2\left(x-y\right)\left(3x+3-x+y\right)\)
\(=2\left(x-y\right)\left(2x+3+y\right)\)
Bài 2:
\(P=\left(3x-1\right)^2+2\left(3x-1\right)\left(x+1\right)+\left(x+1\right)^2\)
\(=\left(3x-1-x-1\right)^2\)
\(=\left(2x-2\right)^2\)(1)
b) Thay \(x=\frac{9}{4}\)vào (1) ta được:
\(\left(2.\frac{9}{4}-2\right)^2\)
\(=\frac{25}{4}\)
Vậy giá trị của P \(=\frac{25}{4}\)khi \(x=\frac{9}{4}\)
Bài 3:
Ta có: \(M=x^2+4x+5\)
\(=\left(x+2\right)^2+1\)
Vì \(\left(x+2\right)^2\ge0;\forall x\)
\(\Rightarrow\left(x+2\right)^2+1\ge0+1;\forall x\)
Hay \(M\ge1;\forall x\)
Dấu"="xảy ra \(\Leftrightarrow\left(x+2\right)^2=0\)
\(\Leftrightarrow x=-2\)
Vậy \(M_{min}=1\Leftrightarrow x=-2\)
Bài 1 : trên là sai nha mình làm lại
\(6x^2-2\left(x-y\right)^2-6y^2\)
\(=6\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)
\(=2\left(x-y\right)\left(3x+3y-x+y\right)\)
\(=2\left(x-y\right)\left(2x+4y\right)\)
\(=4\left(x-y\right)\left(x+2y\right)\)
a) ĐKXĐ: x khác +2
\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)
<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)
<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22
<=> x^2 - 7x - 2 = 2x - 22
<=> x^2 - 7x - 2 - 2x + 22 = 0
<=> x^2 - 9x + 20 = 0
<=> (x - 4)(x - 5) = 0
<=> x - 4 = 0 hoặc x - 5 = 0
<=> x = 4 hoặc x = 5
làm nốt đi
a: \(\Leftrightarrow x^2\left(x^2+x-12\right)=0\)
\(\Leftrightarrow x^2\left(x+4\right)\left(x-3\right)=0\)
hay \(x\in\left\{0;-4;3\right\}\)
d: \(\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24=0\)
\(\Leftrightarrow\left(x^2+5x-6\right)\left(x^2+5x+4\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-1\right)\left(x+1\right)\left(x+4\right)=0\)
hay \(x\in\left\{-6;1;-1;-4\right\}\)
f: \(x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)^2-2\left(x^2+x\right)-24=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
hay \(x\in\left\{-3;2\right\}\)
Câu 1:
a) \(2x^2+5x-3=\left(2x^2+6x\right)-\left(x+3\right)\)
\(=2x\left(x+3\right)-\left(x+3\right)=\left(x+3\right)\left(2x-1\right)\)
b) \(x^4+2009x^2+2008x+2009\)
\(=\left(x^4-x\right)+\left(2009x^2+2009x+2009\right)\)
\(=x\left(x-1\right)\left(x^2+x+1\right)+2009\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+2009\right)\)
c) \(\left[\left(x+2\right)\left(x+8\right)\right]\left[\left(x+4\right)\left(x+6\right)\right]=-16\) (đã sửa đề)
\(\Leftrightarrow\left(x^2+10x+16\right)\left(x^2+10x+24\right)+16=0\)
\(\Leftrightarrow\left(x^2+10x+20\right)^2-16+16=0\)
\(\Leftrightarrow\left(x^2+10x+20\right)^2=0\)
\(\Leftrightarrow\left(x+5\right)^2-5=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-5-\sqrt{5}\\x=-5+\sqrt{5}\end{cases}}\)
Câu 1.
a) 2x2 + 5x - 3 = 2x2 + 6x - x - 3 = 2x( x + 3 ) - ( x + 3 ) = ( x + 3 )( 2x - 1 )
b) x4 + 2009x2 + 2008x + 2009
= x4 + 2009x2 + 2009x - x + 2009
= ( x4 - x ) + ( 2009x2 + 2009x + 2009 )
= x( x3 - 1 ) + 2009( x2 + x + 1 )
= x( x - 1 )( x2 + x + 1 ) + 2009( x2 + x + 1 )
= ( x2 + x + 1 )[ x( x - 1 ) + 2009 ]
= ( x2 + x + 1 )( x2 - x + 2009 )
c) ( x + 2 )( x + 4 )( x + 6 )( x + 8 ) = 16 ( xem lại đi chứ không phân tích được :v )
Câu 2.
3x2 + x - 6 - √2 = 0
<=> ( 3x2 - 6 ) + ( x - √2 ) = 0
<=> 3( x2 - 2 ) + ( x - √2 ) = 0
<=> 3( x - √2 )( x + √2 ) + ( x - √2 ) = 0
<=> ( x - √2 )[ 3( x + √2 ) + 1 ] = 0
<=> \(\orbr{\begin{cases}x-\sqrt{2}=0\\3\left(x+\sqrt{2}\right)+1=0\end{cases}}\)
+) x - √2 = 0 => x = √2
+) 3( x + √2 ) + 1 = 0
<=> 3( x + √2 ) = -1
<=> x + √2 = -1/3
<=> x = -1/3 - √2
Vậy S = { √2 ; -1/3 - √2 }
Câu 3.
A = x( x + 1 )( x2 + x - 4 )
= ( x2 + x )( x2 + x - 4 )
Đặt t = x2 + x
A = t( t - 4 ) = t2 - 4t = ( t2 - 4t + 4 ) - 4 = ( t - 2 )2 - 4 ≥ -4 ∀ t
Dấu "=" xảy ra khi t = 2
=> x2 + x = 2
=> x2 + x - 2 = 0
=> x2 - x + 2x - 2 = 0
=> x( x - 1 ) + 2( x - 1 ) = 0
=> ( x - 1 )( x + 2 ) = 0
=> x = 1 hoặc x = -2
=> MinA = -4 <=> x = 1 hoặc x = -2
1)\(8x^6-\frac{1}{125}y^3=\left(2x^2\right)^3-\left(\frac{1}{5}y\right)^3\)
Bạn tự lm tiếp.AD HĐT số (7)
2)\(\left(x+4\right)^3-64=\left(x+4\right)^3-4^3\)
AD HĐT số (7).Tự lm tiếp
3)\(x^6+1=\left(x^2\right)^3+1\)
AD HĐT số (7).Tự lm tiếp
4)\(x^9+1=\left(x^3\right)^3+1\)
AD HĐT số (7).Tự lm tiếp
5,\(x^{12}-y^4=\left(x^6\right)^2-\left(y^2\right)^2\)
AD HĐT số (3).Tự lm tiếp
6)\(x^3+6x^2+12x+8=\left(x+2\right)^3\)
AD HĐT số (4)
7)\(x^3-15x^2+75x-125=\left(x-5\right)^3\)
AD HĐT số (5)
8)\(27a^3-54a^2b+36ab^2-8b^3\)
\(=\left(3a\right)^3-3.\left(3a\right)^2.2b+3.3a.\left(2b\right)^2-\left(2b\right)^3\)
\(=\left(3a-2b\right)^3\)
AD HĐT số (5)