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\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\begin{array}{l} a){\left( {ab - 1} \right)^2} + {\left( {a + b} \right)^2}\\ = {a^2}{b^2} - 2ab + 1 + {a^2} + 2ab + {b^2}\\ = {a^2}{b^2} + 1 + {a^2} + {b^2}\\ = {a^2}\left( {{b^2} + 1} \right) + \left( {{b^2} + 1} \right)\\ = \left( {{a^2} + 1} \right)\left( {{b^2} + 1} \right)\\ c){x^3} - 4{x^2} + 12x - 27\\ = {x^3} - 27 + \left( { - 4{x^2} + 12x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) - 4x\left( {x - 3} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9 - 4x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} - x + 9} \right)\\ b){x^3} + 2{x^2} + 2x + 1\\ = {x^3} + 2{x^2} + x + x + 1\\ = x\left( {{x^2} + 2x + 1} \right) + \left( {x + 1} \right)\\ = x{\left( {x + 1} \right)^2} + \left( {x + 1} \right)\\ = \left( {x + 1} \right)\left( {x\left( {x + 1} \right) + 1} \right)\\ = \left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\\ d){x^4} - 2{x^3} + 2x - 1\\ = {x^4} - 2{x^3} + {x^2} - {x^2} + 2x - 1\\ = {x^2}\left( {{x^2} - 2x + 1} \right) - \left( {{x^2} - 2x + 1} \right)\\ = \left( {{x^2} - 2x + 1} \right)\left( {{x^2} - 1} \right)\\ = {\left( {x - 1} \right)^2}\left( {x - 1} \right)\left( {x + 1} \right)\\ = {\left( {x - 1} \right)^3}\left( {x + 1} \right)\\ e){x^4} + 2{x^3} + 2{x^2} + 2x + 1\\ = {x^4} + 2{x^3} + {x^2} + {x^2} + 2x + 1\\ = {x^2}\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right)\\ = \left( {{x^2} + 2x + 1} \right)\left( {{x^2} + 1} \right)\\ = {\left( {x + 1} \right)^2}\left( {{x^2} + 1} \right) \end{array} |
Bài 1 :
a, \(\left(a-2\right)^2-b^2=\left(a-2-b\right)\left(a-2+b\right)\)
b, \(2a^3-54b^3=2\left(a^3-27b^3\right)=2\left(a-3b\right)\left(a^2+3ab+9b\right)\)
Bài 2 : tự kết luận nhé, ngại mà lười :(
a, \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)
\(\Leftrightarrow\frac{4x-3}{5}-\frac{5x-4}{3}=\frac{6x-2}{7}+3\)
\(\Leftrightarrow\frac{12x-9-25x+20}{15}=\frac{6x-2+21}{7}\)
\(\Leftrightarrow\frac{-13x-29}{15}=\frac{6x+19}{7}\Rightarrow-91x-203=90x+285\)
\(\Leftrightarrow181x=-488\Leftrightarrow x=-\frac{488}{181}\)
b, \(\frac{x+2}{3}+\frac{3\left(2x-1\right)}{4}-\frac{5x-3}{6}=x+\frac{5}{12}\)
\(\Leftrightarrow\frac{4x+8+9\left(2x-1\right)}{12}-\frac{10x-6}{12}=\frac{12x+5}{12}\)
\(\Rightarrow4x+8+18x-9-10x+6=12x+5\)
\(\Leftrightarrow12x+5=12x+5\Leftrightarrow0x=0\)
Vậy phương trình có vô số nghiệm
c, \(\left|2x-3\right|=4\)
Với \(x\ge\frac{3}{2}\)pt có dạng : \(2x-3=4\Leftrightarrow x=\frac{7}{2}\)
Với \(x< \frac{3}{2}\)pt có dạng : \(2x-3=-4\Leftrightarrow x=-\frac{1}{2}\)
d, \(\left|3x-1\right|-x=2\Leftrightarrow\left|3x-1\right|=x+2\)
Với \(x\ge\frac{1}{3}\)pt có dạng : \(3x-1=x+2\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Với \(x< \frac{1}{3}\)pt có dạng : \(3x-1=-x-2\Leftrightarrow4x=-1\Leftrightarrow x=-\frac{1}{4}\)
\(a,x^2-5x\)
\(=x\left(x-5\right)\)
\(b,5x\left(x+5\right)+4x+20\)
\(=5x\left(x+5\right)+4\left(x+5\right)\)
\(=\left(5x+4\right)\left(x+5\right)\)
\(c,7x\left(2x-1\right)-4x+2\)
\(=7x\left(2x-1\right)-2\left(2x-1\right)\)
\(=\left(7x-2\right)-\left(2x-1\right)\)
\(d,x^2-16+2\left(x+4\right)\)
\(=x^2-16+2x+8\)
\(=x\left(x-2\right)-8\) ( Ý này thì k chắc lắm, sai thông cảm :)) )
\(e,x^2-10x+9\)
\(=x^2-x-9x+9\)
\(=x\left(x-1\right)-9\left(x-1\right)\)
\(=\left(x-9\right)\left(x-1\right)\)
\(f,\left(2x-1\right)^2-\left(x-3\right)^2=0\) ( mk đoán bài này là tìm x, sai thì bảo mk để mk sửa nhé )
\(\Rightarrow\left(2x-1\right)^2=\left(x-3\right)^2\)
\(\Leftrightarrow\pm\left(2x-1\right)=\pm\left(x-3\right)\)
\(\Rightarrow\hept{\begin{cases}2x-1=x-3\\-\left(2x-1\right)=-\left(x-3\right)\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x-1-x+3=0\\-2x+1-x+3=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+2=0\\-3x+4=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\left(-2\right)\\x=\frac{4}{3}\end{cases}}\)
Vậy ...
Bài 1:
1 (x+3)2=x2+6x+9
2
a, 2x2(3x-5x3)+10x5-5x3=6x3-10x5+10x5-5x3=x3
b, (x+3)(x2-3x+9)+(x-9)(x+3)=(x3+27)+(x2-6x-27)=x3+x2-6x
Bài 2:
a, x2-25x=0
\(\Leftrightarrow x\left(x-25\right)=0\)
\(\Leftrightarrow\begin{cases}x=0\\x-25=0\end{cases}\)
\(\Leftrightarrow\begin{cases}x=0\\x=25\end{cases}\)
b, (4x-1)2-9=0
\(\Leftrightarrow\left(4x-1-3\right)\left(4x-1+3\right)=0\)
\(\Leftrightarrow\left(4x-4\right)\left(4x+2\right)=0\)
\(\Leftrightarrow4\left(x-1\right)2\left(2x+1\right)=0\)
\(\Leftrightarrow8\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\begin{cases}x-1=0\\2x+1=0\end{cases}\)
\(\Leftrightarrow\begin{cases}x=1\\x=\frac{-1}{2}\end{cases}\)
Bài 3:
a, 3x2-18x+27=3(x2-6x+9)=3(x-3)2
b, xy-y2-x+y=y(x-y)-(x-y)=(y-1)(x-y)
c, x2-5x-6=x2-6x+x-6=x(x-6)+(x-6)=(x+1)(x-6)
Bài 4:
a, ( 12x3y3-3x2y3+4x2y4):6x2y3=(12x3y3:6x2y3)-(3x2y3:6x2y3)+(4x2y4:6x2y3)
=2x-1/2 + 2/3y
b, bạn ơi mình không biết cách vẽ đường kẻ để chia ý , nếu bạn biết thì chỉ cho mình rồi mình làm cho
Bài 5 :
b, A = x(2x-3)
A= 2x2-3x
A= 2(x2-3/2x)
A= 2(x2-2x3/4+9/16-9/16)
A=2[(x-3/4)2-9/16]
A=2(x-3/4)2-9/8
A=2(x-3/4)2+(-9/8)
Vì (x-3/4)2 \(\ge\)0 \(\forall x\)
-> 2(x-3/4)2 \(\ge0\forall x\)
-> 2(x-3/4)2+(-9/8)\(\ge-\frac{9}{8}\forall x\)
Vậy MinA= -9/8
Bài 1:
1. Khai triển hằng đẳng thức
(x+3)2 = x2+6x+9
2. Thực hiện phép tính
a) 2x2(3x-5x3)+10x5-5x3
=6x3-10x5+10x5-5x3
=x3
b)(x+3)(x2-3x+9)+(x-9)(x+3)
=(x3+27)+(x2+3x-9x-27)
=x3+27+x2+3x-9x-27
=x3+x2-6x
Bài 2:
a) x2-25x=0
\(\Leftrightarrow\)x(x-25)=0
\(\Leftrightarrow\) \(\left[\begin{matrix}x=0\\x-25=0\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}x=0\\x=25\end{matrix}\right.\)
Vậy x=0 hoặc x=25
b)(4x-1)2 - 9=0
\(\Leftrightarrow\)(4x-1+3)(4x-1-3)=0
\(\Leftrightarrow\)(4x+2)(4x-4)=0
\(\Leftrightarrow\)2(2x+1)(2x-2)=0
\(\Leftrightarrow\left[\begin{matrix}2x+1=0\\2x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}x=\frac{-1}{2}\\x=1\end{matrix}\right.\)
Vậy x=1 hoặc x=\(\frac{-1}{2}\)
Bài 3:
a) 3x2-18x+27
=3(x2-6x+9)
=3(x-3)2
b) xy-y2-x+y
=(xy-y2)-(x-y)
=y(x-y)-(x-y)
=(x-y)(y-1)
c) x2-5x-6
=x2-6x+x-6
=(x2-6x)+(x-6)
=x(x-6)+(x-6
=(x-6)(x+1)
Bài 4:
a) (12x3y3-3x2y3+4x2y4) : 6x2y3
=x2y3(12x-3+4y): 6x2y3
=(12x-3+4y) : 6
= (12x : 6)-(3 : 6)+(4y : 6)
=2x-\(\frac{1}{2}\)+\(\frac{2y}{3}\)
b) (6x3-19x2+23x-12) : (2x-3)
=(3x2-5x+4)(2x-3) : (2x-3)
=3x2-5x+4
\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)-4x^2=\left(x-y-x-y\right)^2-\left(2x\right)^2=\left(-2y\right)^2-\left(2x\right)^2\)
\(=\left(2y-2x\right)\left(2y+2x\right)=2\left(y-x\right)2\left(y+x\right)=4\left(x+y\right)\left(y-x\right)\)
\(x^3-x^2y+3x-3y=x^2\left(x-y\right)+3\left(x-y\right)=\left(x-y\right)\left(x^2+3\right)\)
\(x^3-2x^2-4xy^2+x=x\left(x^2-2x+1-4y^2\right)=x\left[\left(x-1\right)^2-\left(2y\right)^2\right]=x\left(x+2y-1\right)\left(x-2y-1\right)\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)
Đặt \(x^2+7x+10=t\), ta có:
\(t\left(t+2\right)-8=t^2+2t-8=t^2-2t+4t-8=t\left(t-2\right)+4\left(t-2\right)=\left(t-2\right)\left(t+4\right)\)
\(=\left(x^2+7x+10+4\right)\left(x^2+7x+10-2\right)=\left(x^2+7x+14\right)\left(x^2+7x-8\right)\)
\(a.\left(3x+2\right)^2-\left(3x-2\right)^2=5x+38\\\Leftrightarrow 9x^2+12x+4-9x^2+12x-4=5x+38\\ \Leftrightarrow24x-5x=38\\ \Leftrightarrow19x=38\\\Leftrightarrow x=2\)
Vậy nghiệm của phương trình trên là \(2\)
\(b.3\left(x-2\right)^2+9\left(x-1\right)=3\left(x^2+x-3\right)\\\Leftrightarrow 3\left(x^2-4x+4\right)+9x-9=3x^2+3x-9\\ \Leftrightarrow3x^2-3x^2-12x+9x-3x=-12+9-9\\ \Leftrightarrow-6x=-12\\\Leftrightarrow x=2\)
Vậy nghiệm của phương trình trên là \(2\)
\(c.\left(x-1\right)^3-x\left(x+1\right)^2=5x\left(2-x\right)-11\left(x-2\right)\\ \Leftrightarrow x^3-3x^2+3x-1-x\left(x^2+2x+1\right)=10x-5x^2-11x+22\\ \Leftrightarrow x^3-3x^2+3x-1-x^3-2x^2-x=10x-5x^2-11x+22\\\Leftrightarrow x^3-x^3-3x^2-2x^2+5x^2+3x-x-10x+11x=1+22\\ \Leftrightarrow3x=23\\\Leftrightarrow x=\frac{23}{3}\)
Vậy nghiệm của phương trình trên là \(\frac{23}{3}\)
\(d.\left(x+3\right)^2-\left(x-3\right)^2=6x+18\\ \Leftrightarrow x^2+6x+9-x^2+6x-9=6x+18\\ \Leftrightarrow12x-6x=18\\ \Leftrightarrow6x=18\\ \Leftrightarrow x=3\)
Vậy nghiệm của phương trình trên là \(3\)
\(e.\left(x+1\right)\left(x^2-x+1\right)-2x=x\left(x-1\right)\left(x+1\right)\\\Leftrightarrow x^3+1-2x=x\left(x^2-1\right)\\\Leftrightarrow x^3+1-2x=x^3-x\\ \Leftrightarrow x^3-x^3-2x+x=-1\\ \Leftrightarrow-x=-1\\ \Leftrightarrow x=1\)
Vậy nghiệm của phương trình trên là \(1\)
\(f.\left(x-2\right)^3+\left(3x-1\right)\left(3x+1\right)=\left(x+1\right)^3\\\Leftrightarrow x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1\\ \Leftrightarrow x^3-x^3-6x^2+9x^2-3x^2+12x-3x=8+1+1\\ \Leftrightarrow9x=10\\ \Leftrightarrow x=\frac{10}{9}\)
Vậy nghiệm của phương trình trên là \(\frac{10}{9}\)
1.
a, \(\left(x+3\right)\left(x-3\right)-\left(x-3\right)^2\)
\(=\left(x-3\right)\left(x+3-x+3\right)\)
\(=9\left(x-3\right)=9x-27\)
b, \(\left(2x+1\right)^2+2\left(2x+1\right)\left(x-1\right)+\left(x-1\right)^2\)
\(=\left(2x+1+x-1\right)^2=9x^2\)
c, \(x\left(x-3\right)\left(x+3\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x\left(x^2-9\right)-\left(x^4-1\right)\)
\(=x^3-9x-x^4+1=-x^4+x^3-9x+1\)
a: \(=-\left(x^2+10x-11\right)\)
\(=-\left(x^2+10x+25-36\right)\)
\(=-\left(x+5\right)^2+36< =36\)
Dấu '=' xảy ra khi x=-5
b: \(=-\left(x^2-6x+5\right)\)
\(=-\left(x^2-6x+9-4\right)\)
\(=-\left(x-3\right)^2+4< =4\)
Dấu '=' xảy ra khi x=3
c: \(=-2\left(x^2-x+\dfrac{5}{2}\right)\)
\(=-2\left(x^2-x+\dfrac{1}{4}+\dfrac{9}{4}\right)\)
\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}< =-\dfrac{9}{2}\)
Dấu '=' xảy ra khi x=1/2
d: \(=2x+8-x^2-4x\)
\(=-x^2-2x+8\)
\(=-\left(x^2+2x-8\right)\)
\(=-\left(x^2+2x+1-9\right)\)
\(=-\left(x+1\right)^2+9< =9\)
Dấu '=' xảy ra khi x=-1
1/ Ta có : \(P\left(x\right)=-x^2+13x+2012=-\left(x-\frac{13}{2}\right)^2+\frac{8217}{4}\le\frac{8217}{4}\)
Dấu "=" xảy ra khi x = 13/2
Vậy Max P(x) = 8217/4 tại x = 13/2
2/ Ta có : \(x^3+3xy+y^3=x^3+3xy.1+y^3=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1\)
3/ \(a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)
\(\Leftrightarrow ab+bc+ac=-\frac{1}{2}\) \(\Leftrightarrow\left(ab+bc+ac\right)^2=\frac{1}{4}\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\frac{1}{4}\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}\)(vì a+b+c=0)
Ta có : \(a^2+b^2+c^2=1\Leftrightarrow\left(a^2+b^2+c^2\right)^2=1\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1\)
\(\Leftrightarrow a^4+b^4+c^4=1-2\left(a^2b^2+b^2c^2+c^2a^2\right)=1-\frac{2.1}{4}=\frac{1}{2}\)