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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:
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) \(=2xy^2\left(x^2+8x+15\right)\)
\(=2xy^2\left[\left(x^2+8x+16\right)-1\right]\)
\(=2xy^2\left[\left(x+4\right)^2-1\right]\)
\(=2xy^2\left(x+4+1\right)\left(x+4-1\right)\)
\(=2xy^2\left(x+5\right)\left(x-3\right)\)
mấy câu sau tự làm nha :*
b,=(x^2-10x+25)-4
=(x-5)^2-2^2
=(x-5-2)(x-5+2)
=(x-7)(x-3)
\(a.\left(x^2+\frac{2}{5}y\right)\left(x^2-\frac{2}{5}y\right)\\ =x^4-\frac{4}{25}y^2\)
\(b.\left(2x+y^2\right)^3\\ =8x^3+12x^2y^2+6xy^4+y^6\)
\(c.\left(3x^2-2y\right)^3\\ =27x^6-54x^4y+36x^2y^2-8y^3\)
\(\left(x+4\right)\left(x^2-4x+16\right)\\ =x^3+64\)
\(e.\left(x^2-\frac{1}{3}\right)\left(x^4+\frac{1}{3}x^2+\frac{1}{9}\right)\\ =x^6-\frac{1}{27}\)
a: \(x^2-9-x^2\left(x^2-9\right)\)
\(=\left(x^2-9\right)-x^2\left(x^2-9\right)\)
\(=\left(x^2-9\right)\left(1-x^2\right)\)
\(=\left(1-x\right)\left(1+x\right)\left(x-3\right)\left(x+3\right)\)
b: \(x^2\left(x-y\right)+y^2\left(y-x\right)\)
\(=x^2\left(x-y\right)-y^2\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-y^2\right)\)
\(=\left(x-y\right)\left(x-y\right)\left(x+y\right)=\left(x-y\right)^2\cdot\left(x+y\right)\)
c: \(x^3+27+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)\)
\(=\left(x+3\right)\left(x^2-3x+9+x-9\right)\)
\(=\left(x+3\right)\left(x^2-2x\right)=x\left(x-2\right)\left(x+3\right)\)
d: \(x^2+5x+6\)
\(=x^2+2x+3x+6\)
\(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
e: \(3x^2-4x-4\)
\(=3x^2-6x+2x-4\)
\(=3x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(3x+2\right)\)
g: \(x^4+64y^4\)
\(=x^4+16x^2y^2+64y^4-16x^2y^2\)
\(=\left(x^2+8y^2\right)^2-\left(4xy\right)^2\)
\(=\left(x^2+8y^2-4xy\right)\left(x^2+8y^2+4xy\right)\)
h: \(a^2+b^2+2a-2b-2ab\)
\(=a^2-2ab+b^2+2a-2b\)
\(=\left(a-b\right)^2+2\left(a-b\right)=\left(a-b\right)\left(a-b+2\right)\)
i: \(\left(x+1\right)^2-2\left(x+1\right)\left(y-3\right)+\left(y-3\right)^2\)
\(=\left(x+1-y+3\right)^2\)
\(=\left(x-y+4\right)^2\)
k: \(x^2\left(x+1\right)-2x\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-2x+1\right)\)
\(=\left(x+1\right)\left(x-1\right)^2\)