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Bài 1:
a) \(5x-15y=5\left(x-3y\right)\)
b) \(\dfrac{3}{5}x^2+5x^4-x^2y=x^2\left(\dfrac{3}{5}+5x^2-y\right)\)
c) \(14x^2y^2-21xy^2+28x^2y=7xy\left(2xy-3y+4x\right)\)
d) \(\dfrac{2}{7}x\left(3y-1\right)-\dfrac{2}{7}y\left(3y-1\right)=\dfrac{2}{7}\left(3y-1\right)\left(x-y\right)\)
e) \(x^3-3x^2+3x-1=\left(x-1\right)^3\)
f) \(\left(x+y\right)^2-4x^2=\left(-x+y\right)\left(3x+y\right)\)
g) \(27x^3+\dfrac{1}{8}=\left(3x+\dfrac{1}{2}\right)\left(6x^2+1,5x+\dfrac{1}{4}\right)\)
h) \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3\)
\(=6x^2y+2y^3=2y\left(3x^2+y\right)\)
Bài 2:
a) \(x^2\left(x+1\right)+2x\left(x+1\right)=0\)
\(\Rightarrow x\left(x+1\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\Rightarrow x=-1\\x+2=0\Rightarrow x=-2\end{matrix}\right.\)
b) \(x\left(3x-2\right)-5\left(2-3x\right)=0\)
\(\Rightarrow x\left(3x-2\right)+5\left(3x-2\right)=0\)
\(\Rightarrow\left(3x-2\right)\left(x+5\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x-2=0\Rightarrow x=\dfrac{2}{3}\\x+5=0\Rightarrow x=-5\end{matrix}\right.\)
c) \(\dfrac{4}{9}-25x^2=0\)
\(\Rightarrow\left(\dfrac{2}{3}-5x\right)\left(\dfrac{2}{3}+5x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}-5x=0\Rightarrow x=\dfrac{2}{15}\\\dfrac{2}{3}+5x=0\Rightarrow x=\dfrac{-2}{15}\end{matrix}\right.\)
d) Có tới 2 dấu "=".
bài 1 dễ mk ko lm nữa nhé
bafi2:
a,x(x+1)(x+2)=0
x=0 ; x=-1 ; x=-2
b,x(3x-2)+5(3x-2)=0
(x+5)(3x-2)=0
x=-5 ; x=2/3
c,
(2/3)2- (5x)2=0
(2/3-5x)(2/3+5x)=0
x=+-2/15
d, X2-2*1/2x+(1/2)2=0
(X-1/2)22=0
X=1/2
a) x3 +x+2
=\(\left(x^3+x^2\right)-\left(x^2+x\right)+\left(2x+2\right)\)
=\(\left(x+1\right)\left(x^2-x+2\right)\)
b) x3-2x-1
=\(\left(x^3+x^2\right)-\left(x^2+x\right)-\left(x+1\right)\)
=\(\left(x+1\right)\left(x^2-x-1\right)\)
c) x3+3x2-4
=\(\left(x^3-x^2\right)+\left(4x^2+4x\right)-\left(4x+4\right)\)
=\(\left(x-1\right)\cdot\left(x^2+4x-4\right)\)
d) x3+3x2y-9xy2+5y3
=\(\left(x^3-x^2y\right)+\left(4x^2y-4xy^2\right)-\left(5xy^2-5y^3\right)\)
=\(\left(x-y\right)\left(x^2+4xy-5y^2\right)\)
=\(\left(x-y\right)^2\left(x-5y\right)\)
a)
\(x^3+x+2\)
\(=\left(x^3+x^2\right)-\left(x^2+x\right)+\left(2x+2\right)\)
\(=x^2\left(x+1\right)-x\left(x+1\right)+2\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+2\right)\)
b)
\(x^3-2x-1\)
\(=\left(x^3+x^2\right)-\left(x^2+x\right)-\left(x+1\right)\)
\(=x^2\left(x+1\right)-x\left(x+1\right)-\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x-1\right)\)
c)
\(x^3-3x^2-4\)
\(=\left(x^3-x^2\right)+\left(4x^2-4x\right)+\left(4x-4\right)\)
\(=x^2\left(x-1\right)+4x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+2.2.x+2^2\right)\)
\(=\left(x-1\right)\left(x+2\right)^2\)
d)
\(x^3-3x^2y-9xy^2+5y^3\)
\(=\left(x^3-x^2y\right)+\left(4x^2y-4xy^2\right)-\left(5xy^2-5y^3\right)\)
\(=x^2\left(x-y\right)+4xy\left(x-y\right)-5y^2\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-4xy-5y^2\right)\)
\(=\left(x-y\right)^2\left(x-5y\right)\)
Quy tắc xét tính chẵn lẻ của hàm số:
Chẵn \(\Leftrightarrow\left\{{}\begin{matrix}x\in D\Rightarrow-x\in D\\f\left(x\right)=f\left(-x\right)\end{matrix}\right.\)
Lẻ \(\Leftrightarrow\left\{{}\begin{matrix}x\in D\Rightarrow-x\in D\\f\left(x\right)=-f\left(-x\right)\end{matrix}\right.\)
a/ \(g=2x^4-x^2+5\)
\(x\in D=R\Rightarrow-x\in D\)
\(g\left(-x\right)=2\left(-x\right)^4-\left(-x\right)^2+5=2x^4-x^2+5=g\left(x\right)\)
=> hàm số chẵn
b/ \(y=x^3+3x\)
\(x\in D=R\Rightarrow-x\in D\)
\(y\left(-x\right)=\left(-x\right)^3+3\left(-x\right)=-x^3-3x=-\left(x^3+3x\right)\)
\(\Rightarrow y\left(x\right)=-y\left(-x\right)\)
=> hàm số lẻ
c/ \(y=x^3+3x+1\)
\(x\in D=R\Rightarrow-x\in D\)
\(y\left(-x\right)=\left(-x\right)^3+3\left(-x\right)+1=-x^3-3x+1\)
\(\Rightarrow\left\{{}\begin{matrix}y\left(x\right)\ne y\left(-x\right)\\y\left(x\right)\ne-y\left(-x\right)\end{matrix}\right.\)
=> hàm số ko chẵn ko lẻ
d/ \(y=x^4-3\)
\(x\in D=R\Rightarrow-x\in D\)
\(y\left(-x\right)=\left(-x\right)^4-3=x^4-3=y\left(x\right)\)
=> hàm số chẵn
e/ \(y=3x^4-\left|x\right|+2\)
\(x\in D=R\Rightarrow-x\in D\)
\(y\left(-x\right)=3\left(-x\right)^4-\left|-x\right|+2=3x^4-\left|x\right|+2=y\left(x\right)\)
=> hàm số chẵn
f/ \(x\in D=R\Rightarrow-x\in D\)
\(y\left(-x\right)=\left|-x-1\right|+\left|-x+1\right|=\left|x+1\right|+ \left|x-1\right|=y\left(x\right)\)
=> hàm số chẵn
Các câu sau làm tương tự
a/ \(g\left(-x\right)=2\left(-x\right)^4-\left(-x\right)^2+5=2x^4-x^2+5=g\left(x\right)\)
Hàm chẵn
b/ \(y\left(-x\right)=\left(-x\right)^3+3\left(-x\right)=-x^3-3x=-\left(x^3+3x\right)=-y\left(x\right)\)
Hàm lẻ
c/ \(y\left(-x\right)=-x^3-3x+1\)
Hàm ko chẵn ko lẻ
d/ \(y\left(-x\right)=x^4-3=y\left(x\right)\) hàm chẵn
e/ \(y\left(-x\right)=3x^4-\left|x\right|+2=y\left(x\right)\) hàm chẵn
f/ \(y\left(-x\right)=\left|-x-1\right|+\left|-x+1\right|=\left|x+1\right|+\left|x-1\right|=y\left(x\right)\)
Hàm chẵn
g/ \(y\left(-x\right)=\left|-x-1\right|-\left|-x+1\right|=\left|x+1\right|-\left|x-1\right|=-y\left(x\right)\)
Hàm lẻ
h/ Hàm ko chẵn ko lẻ
1)Thấy: x=0;y=0 không phải là nghiệm của hệ.
\(\begin{cases}x^3-8x=y^3+2y\\x^2-3=3\left(y^2+1\right)\end{cases}\)
\(\Leftrightarrow\begin{cases}x^3-8x=y^3+2y\\x^2=3\left(y^2+2\right)\end{cases}\)
\(\Leftrightarrow\begin{cases}x^3-8x=y\left(y^2+2\right)\\x^2y=3y\left(y^2+2\right)\end{cases}\)
Trừ vế theo vế hai phương trình,đc:
\(x^3-8x-\frac{x^2y}{3}=0\Leftrightarrow y=\frac{3\left(x^3-8x\right)}{x^2}\)
\(\Leftrightarrow y=\frac{3\left(x^2-8\right)}{x}\).Thay \(y=\frac{3\left(x^2-8\right)}{x}\) vào pt 2 đc:
\(26x^4-426x^2-1728=0\)
\(\Leftrightarrow\begin{cases}x^2=9\\x^2=\frac{96}{13}\end{cases}\) dễ nhé 
a) (x + 3)(x2 – 3x + 9) – (54 + x3)
= x3 + 33 – (54 + x3) (Áp dụng HĐT (6) với A = x và B = 3)
= x3 + 27 – 54 – x3
= –27
b) (2x + y)(4x2 – 2xy + y2) – (2x – y)(4x2 + 2xy + y2)
= (2x + y)[(2x)2 – 2x.y + y2] – (2x – y)[(2x)2 + 2x.y + y2]
= [(2x)3 + y3] – [(2x)3 – y3]
= (2x)3 + y3 – (2x)3 + y3
= 2y3
a, I(-4;3), R=\(\sqrt{17}\)
b, I(3;2), R=7
c, 16x2+16y2+16x-8y-11=0 <=> \(x^2+y^2+x-\frac{1}{2}y-\frac{11}{16}=0\)
\(\Rightarrow I\left(\frac{-1}{2};\frac{1}{4}\right),R=1\)
d, I(-4;-7), \(R=\sqrt{15}\)
e, 3x2 + 3y2 + 6x - 12y - 9 = 0<=> x2+y2+2x-4y-3=0
\(\Rightarrow I\left(-1;2\right),R=2\sqrt{2}\)
f, I(-5;-7), R=\(\sqrt{15}\)
a) \(x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\)
b) \(x^2-2x-15=\left(x^2-2x+1\right)-16=\left(x-1\right)^2-4^2=\left(x-1-4\right)\left(x-1+4\right)=\left(x-5\right)\left(x+3\right)\)
c) \(5x^2y^3-25x^3y^4+10x^3y^3=5x^2y^3\left(1-5xy+2x\right)\)
d) \(12x^2y-18xy^2-30y^2=6\left(2x^2y-3xy^2-5y^2\right)\)
e, ntc: x-y
f, đối dấu --> ntc
g, như ý f
h, \(36-12x+x^2=\left(6-x\right)^2=\left(x-6\right)^2\)
i, \(3x^3y^2-6x^2y^3+9x^2y^2=3x^2y^2\left(x-y+3\right)\)
thanks