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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)
\(x^3-3x^2y-9xy^2+5y^3\)
\(=\left(x^3-x^2y\right)+\left(4x^2y-4xy^2\right)-\left(5xy^2-5y^3\right)\)
\(=x^2\left(x-y\right)+4xy\left(x-y\right)-5y^2\left(x-y\right)\)
\(=\left(x-y\right)\left(x^2-4xy-5y^2\right)\)
\(=\left(x-y\right)^2\left(x-5y\right)\)
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) \(x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\)
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, \(3x^3y^2-6x^2y^3+9x^2y^2=3x^2y^2\left(x-y+3\right)\)
Bài 1
d, \(x^2+2xy+y^2-2x-2y+1\)
\(\Rightarrow x^2+y^2=1+2xy-2y-2x\)
\(\Rightarrow\left(x+y-1\right)^2\)
Bài 2:
a, \(\left(x+1\right)\left(x+1\right)=\left(x+2\right)\left(x+5\right)\)
\(\Leftrightarrow\left(x+1\right)^2=x^2+5x+2x+10\)
\(\Leftrightarrow x^2+2x+1=x^2=5x+2x+10\)
\(\Leftrightarrow-5x=9\)
\(\Leftrightarrow x=-\frac{9}{5}\)
b,\(\left(x+3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)
c, \(4x^2-9=0\)
\(\Leftrightarrow4x^2=9\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\\frac{3}{2}\end{matrix}\right.\)
d,\(\left(4x-5\right)^2-\left(3x-4\right)^2=0\)
\(\Leftrightarrow16x^2-40x+25-\left(9x^2-24x+16\right)=0\)
\(\Leftrightarrow16x^2-40x+25-9x^2+24x-16=0\)
\(\Leftrightarrow7x^2-16x+9=0\)
\(\Leftrightarrow x=\frac{-\left(-16\right)\pm\sqrt{\left(-16\right)^2-4.7.9}}{14}\)
\(\Leftrightarrow x=\frac{16\pm\sqrt{256-252}}{14}\)
\(\Leftrightarrow x=\frac{16\pm\sqrt{4}}{14}\)
\(\Leftrightarrow x=\frac{16\pm2}{14}\)
\(\Leftrightarrow x=\left[{}\begin{matrix}\frac{16+2}{14}\\\frac{16-2}{14}\end{matrix}\right.\)
\(\Leftrightarrow x=\left[{}\begin{matrix}\frac{9}{7}\\1\end{matrix}\right.\)
1.a)\(3x-3y+x^2-2xy+y^2\)
\(=3\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3+x-y\right)\)
d)\(x^2+2xy+y^2-2x-2y+1\)
\(=\left(x+y\right)^2-2\left(x+y\right)+1\)
\(=\left(x+y+1\right)^2\)
2.a)\(\left(x+1\right)\left(x+1\right)=\left(x+2\right)\left(x+5\right)\)
\(\Leftrightarrow\left(x+1\right)^2=x^2+5x+2x+10\)
\(\Leftrightarrow x^2+2x+1-x^2-7x-10=0\)
\(\Leftrightarrow-5x-9=0\)
\(\Leftrightarrow-5x=9\)
\(\Leftrightarrow x=-\frac{9}{5}\). Vậy \(S=\left\{-\frac{9}{5}\right\}\)
b)\(\left(x+3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\).Vậy \(S=\left\{-3;-5\right\}\)
c)\(4x^2-9=0\)
\(\Leftrightarrow\left(2x+3\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\2x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\x=\frac{3}{2}\end{matrix}\right.\). Vậy \(S=\left\{\pm\frac{3}{2}\right\}\)
d)\(\left(4x-5\right)^2-\left(3x-4\right)^2=0\)
\(\Leftrightarrow\left(4x-5+3x-4\right)\left(4x-5-3x+4\right)=0\)
\(\Leftrightarrow\left(7x-9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7x-9=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{9}{7}\\x=1\end{matrix}\right.\). Vậy \(S=\left\{1;\frac{9}{7}\right\}\)
3.Ta có:
8x^2-26x+m 2x-3 4x-7 -14x+m m+21
Để \(A\left(x\right)⋮B\left(x\right)\) thì: \(m+21⋮2x-3\)
\(\Rightarrow m+21=0\)
\(\Rightarrow m=-21\)
Vậy...!
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/ - Với \(x>\frac{1}{4}\) PT vô nghiêm
- Với \(x\le\frac{1}{4}\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(1-4x\right)^2\)
\(\Leftrightarrow\left(x^2+4x-2\right)\left(x^2-4x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2+4x-2=0\\x^2-4x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-2+\sqrt{6}\left(l\right)\\x=-2-\sqrt{6}\\x=4\left(l\right)\\x=0\end{matrix}\right.\)
2.
- Với \(x\ge-\frac{1}{4}\Leftrightarrow4x+1=x^2+2x-4\)
\(\Leftrightarrow x^2-2x-5=0\Rightarrow\left[{}\begin{matrix}x=1+\sqrt{6}\\x=1-\sqrt{6}\left(l\right)\end{matrix}\right.\)
- Với \(x< -\frac{1}{4}\)
\(\Leftrightarrow-4x-1=x^2+2x-4\)
\(\Leftrightarrow x^2+6x-3=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-3+2\sqrt{3}\left(l\right)\\x=-3-2\sqrt{3}\end{matrix}\right.\)
3.
- Với \(x\ge\frac{5}{3}\)
\(\Leftrightarrow3x-5=2x^2+x-3\)
\(\Leftrightarrow2x^2-2x+2=0\left(vn\right)\)
- Với \(x< \frac{5}{3}\)
\(\Leftrightarrow5-3x=2x^2+x-3\)
\(\Leftrightarrow2x^2+4x-8=0\Rightarrow\left[{}\begin{matrix}x=-1+\sqrt{5}\\x=-1-\sqrt{5}\end{matrix}\right.\)
4. Do hai vế của pt đều không âm, bình phương 2 vế:
\(\Leftrightarrow\left(x^2-2x+8\right)^2=\left(x^2-1\right)^2\)
\(\Leftrightarrow\left(x^2-2x+8\right)^2-\left(x^2-1\right)^2=0\)
\(\Leftrightarrow\left(2x^2-2x+7\right)\left(-2x+9\right)=0\)
\(\Leftrightarrow-2x+9=0\Rightarrow x=\frac{9}{2}\)
câu 1.
a. \(=\left(x+y\right)\left(x-5\right)\)
b. \(=\left(x+2y\right)^2\)
c. \(=\left(x-1\right)\left(x-6\right)\)
câu 3.
a. \(A=5\left(x+1\right)^2+2010\ge2010\forall x\)
Vậy \(minA=2010\Leftrightarrow x=-1\)
b. \(\Leftrightarrow\left(y+1\right)\left(x-1\right)=11\)
Vì x, y nguyên nên có các TH :
\(\left[{}\begin{matrix}\left\{{}\begin{matrix}y+1=1\\x-1=11\end{matrix}\right.\\\left\{{}\begin{matrix}y+1=11\\x-1=1\end{matrix}\right.\\\left\{{}\begin{matrix}y+1=-1\\x-1=-11\end{matrix}\right.\\\left\{{}\begin{matrix}y+1=-11\\x-1=-1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}y=0\\x=12\end{matrix}\right.\\\left\{{}\begin{matrix}y=10\\x=2\end{matrix}\right.\\\left\{{}\begin{matrix}y=-2\\x=-10\end{matrix}\right.\\\left\{{}\begin{matrix}y=-12\\x=0\end{matrix}\right.\end{matrix}\right.\)
câu 6.
a. giống câu 3
b. \(B=-2\left(x-1\right)^2+7\le7\forall x\in R\)