PHÂN TÍCH:
a) x^2 - 2x - 4y^2 - 4y
b) x^3 - 4x^2 + 12x - 27
c) (x^2 - 3)^2 + 16
d) x^4 - 2x^3 + 2x -1
e) x^4 + 2x^3 -4x -4
Phân tích đa thức thành nhân tử
a.\(16x^3+0,25yz^3\)
b.\(x^4-4x^3+4x^2\)
c.\(x^3+x^2y-xy^2-y^3\)
d.\(x^3+x^2+x+1\)
e.\(x^4-x^2+2x-1\)
f.\(2x^2-18\)
g.\(x^2+8x+7\)
h.\(x^4y^4+4\)
i.\(x^4+4y^4\)
k.\(x^2-2x-15\)
a: \(16x^3+0,25yz^3\)
\(=0,25\cdot x^3\cdot64+0,25\cdot yz^3\)
\(=0,25\left(64x^3+yz^3\right)\)
b: \(x^4-4x^3+4x^2\)
\(=x^2\cdot x^2-x^2\cdot4x+x^2\cdot4\)
\(=x^2\left(x^2-4x+4\right)=x^2\left(x-2\right)^2\)
c: \(x^3+x^2y-xy^2-y^3\)
\(=x^2\left(x+y\right)-y^2\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-y^2\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\cdot\left(x+y\right)^2\)
d: \(x^3+x^2+x+1\)
\(=x^2\left(x+1\right)+\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+1\right)\)
e: \(x^4-x^2+2x-1\)
\(=x^4-\left(x^2-2x+1\right)\)
\(=x^4-\left(x-1\right)^2\)
\(=\left(x^2-x+1\right)\left(x^2+x-1\right)\)
f: \(2x^2-18\)
\(=2\cdot x^2-2\cdot9\)
\(=2\left(x^2-9\right)=2\left(x-3\right)\left(x+3\right)\)
g: \(x^2+8x+7\)
\(=x^2+x+7x+7\)
\(=x\left(x+1\right)+7\cdot\left(x+1\right)=\left(x+1\right)\left(x+7\right)\)
h: \(x^4y^4+4\)
\(=x^4y^4+4x^2y^2+4-4x^2y^2\)
\(=\left(x^2y^2+2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2y^2+2-2xy\right)\left(x^2y^2+2+2xy\right)\)
i: \(x^4+4y^4\)
\(=x^4+4x^2y^2+4y^4-4x^2y^2\)
\(=\left(x^2+2y^2\right)^2-\left(2xy\right)^2\)
\(=\left(x^2-2xy+2y^2\right)\left(x^2+2xy+2y^2\right)\)
k: \(x^2-2x-15\)
\(=x^2-5x+3x-15\)
\(=x\left(x-5\right)+3\left(x-5\right)=\left(x-5\right)\left(x+3\right)\)
Phân tích đa thức thành nhân tử a) x^4 + 2x^3 - 4x - 4 b) x^3 - 4x^2 + 12x - 27 c) xy -4y - 5x + 20
a) `x^4+2x^3-4x-4`
`=(x^4-4)+(2x^3-4x)`
`=(x^2-2)(x^2+2)+2x(x^2-2)`
`=(x^2-2)(x^2+2+2x)`
b) `x^3-4x^2+12x-27`
`=(x^3-27)-(4x^2-12x)`
`=(x-3)(x^2+3x+9)-4x(x-3)`
`=(x-3)(x^2+3x+9-4x)`
`=(x-3)(x^2-x+9)`
c) `xy-4y-5x+20`
`=y(x-4)-5(x-4)`
`=(y-5)(x-4)`
a) Ta có: \(x^4+2x^3-4x-4\)
\(=\left(x^4-4\right)+2x^3-4x\)
\(=\left(x^2-2\right)\left(x^2+2\right)+2x\left(x^2-2\right)\)
\(=\left(x^2-2\right)\left(x^2+2x+2\right)\)
b) Ta có: \(x^3-4x^2+12x-27\)
\(=\left(x-3\right)\left(x^2+3x+9\right)-4x\cdot\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-x+9\right)\)
c) Ta có: \(xy-4y-5x+20\)
\(=y\left(x-4\right)-5\left(x-4\right)\)
\(=\left(x-4\right)\left(y-5\right)\)
1. Phân tích đa thức thành nhân tử:
a. (ab-1)^2+(a+b)^2
b. x^3+2x^2+2x+1
c. x^3-sx^2+12x-27
d. x64-2x^3+2x-1
e. x^4+2x^3+2x^2+2x+1
f. x^2-2x-4y^2-4y
g. x^4+2x^3-4x-4
h. x^2(1-x^2)-4-4x^2
i. (1+2x)(1-2x)-x(x+2)(x-2)
j. x^2+y^2-x^2.y^2+xy-x-y
2.Phân tích đa thức thành nhân tử:
a. a(b^2+c^2+bc)+b(c^2+a^2+ac)+c(a^2+b^2+ab)
b.(a+b+c)(ab+bc+ca)-abc
c. a(a+2b)^3-b(2a+b)^3
GIẢI GIÚP MÌNH GẤP NHA CẢM ƠN CÁC BẠN NHÌU !!!!!!
Phân tích các đa thúc sau thành nhân tử
a. (x^2 + y^2-5)- 4-x^y^2-16xy-16
b.(ab-1)^2+(a+b)^2
c.x^3+2x^2+2x+1
d.x^3-4x^2+12x-27
e.x^4-2x^3+2x-1
f.x^4+2x^3+2x^2+2x+1
g.x^2-2x-4y^2-4y
h.x^4+2x^3-4x-4
i.x^2(1-x^2)-4-4x^2
k.x^2+y^2-x^2y^2+xy-x-y
phân tích thành nhân tử
a)4x^2-12x+9-3(2x-3)(x+1)
b)25-42^2+8xy-4y^2
c)x^3-5x^2+4x-2(x^2-x)
d) 4(x-2)^2-5(x-2)+1
EM CẦN GẤP Ạ
a: \(4x^2-12x+9-3\left(2x-3\right)\left(x+1\right)\)
\(=\left(2x-3\right)^2-\left(2x-3\right)\left(3x+3\right)\)
\(=\left(2x-3\right)\left(2x-3-3x-3\right)\)
\(=-\left(x+6\right)\left(2x-3\right)\)
b: \(25-4x^2+8xy-4y^2\)
\(=25-\left(2x-2y\right)^2\)
\(=\left(5-2x+2y\right)\left(5+2x-2y\right)\)
phân tích đa thức thành nhân tử
a)8x^3+27
b) 4x^2-4x+1-y^2
c) x^4-2x^3+x^2-2x
d) x^2-4y^2+2x+4y
a) \(8x^3+27=\left(2x+3\right)\left(4x^2-6x+9\right)\)
b) \(4x^2-4x+1-y^2=\left(2x-1\right)^2-y^2=\left(2x-1-y\right)\left(2x-1+y\right)\)
c) \(x^4-2x^3+x^2-2x=x^3\left(x-2\right)+x\left(x-2\right)=x\left(x-2\right)\left(x^2-1\right)=x\left(x-2\right)\left(x-1\right)\left(x+1\right)\)
d) \(x^2-4y^2+2x+4y=\left(x-2y\right)\left(x+2y\right)+2\left(x+2y\right)=\left(x+2y\right)\left(x-2y+2\right)\)
1.\(x^2-2x-4y^2-4y=\left(x+2y\right)\left(x-2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)
2.\(x^4+2x^3-4x-4=\left(x^2+2\right)\left(x^2-2\right)+2x\left(x^2-2\right)=\left(x^2-2\right)\left(x^2+2x-2\right)\)
3.\(3x^2-3y^2-2\left(x-y\right)^2=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\left(x-y\right)=\left(x-y\right)\left(3x+3y-2x+2y\right)\)\(=\left(x-y\right)\left(x+5y\right)\)
4.\(x^3-4x^2-9x+36=x^2\left(x-4\right)-9\left(x-4\right)=\left(x-3\right)\left(x+3\right)\left(x-4\right)\)
5.\(\left(x-1\right)\left(2x+1\right)+3\left(x-1\right)\left(x+2\right)\left(2x+1\right)=\left(x-1\right)\left(2x+1\right)\left(1+3x+6\right)\)\(=\left(x-1\right)\left(2x+1\right)\left(3x+7\right)\)
6.\(\left(6x+3\right)-\left(2x-5\right)\left(2x+1\right)=3\left(2x+1\right)-\left(2x-5\right)\left(2x+1\right)\)\(=\left(2x+1\right)\left(3-2x-5\right)=\left(2x+1\right)\left(-2-2x\right)=-2\left(2x+1\right)\left(x+1\right)\)
7.\(\left(x-5\right)^2+\left(x+5\right)\left(x-5\right)+\left(x-5\right)\left(2x+1\right)=\left(x-5\right)\left(x-5+x+5+2x+1\right)\)\(=\left(x-5\right)\left(4x+1\right)\)
8.\(\left(3x-2\right)\left(4x-3\right)+\left(3x-2\right)\left(x-1\right)-2\left(3x-2\right)\left(x+1\right)\)\(=\left(3x-2\right)\left(4x-3+x-1-2x-2\right)=\left(3x-2\right)\left(3x-6\right)=3\left(3x-2\right)\left(x-2\right)\)
Giải phương trình:
1. \(x^4-6x^2-12x-8=0\)
2. \(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)
3. \(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
4. \(2x^2.\sqrt{-4x^4+4x^2+3}=4x^4+1\)
5. \(x^2+4x+3=\sqrt{\dfrac{x}{8}+\dfrac{1}{2}}\)
6. \(\left\{{}\begin{matrix}4x^3+xy^2=3x-y\\4xy+y^2=2\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}\sqrt{x^2-3y}\left(2x+y+1\right)+2x+y-5=0\\5x^2+y^2+4xy-3y-5=0\end{matrix}\right.\)
8. \(\left\{{}\begin{matrix}\sqrt{2x^2+2}+\left(x^2+1\right)^2+2y-10=0\\\left(x^2+1\right)^2+x^2y\left(y-4\right)=0\end{matrix}\right.\)
1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
2.
ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)
\(\dfrac{x}{2x^2+4x+1}+\dfrac{x}{2x^2-4x+1}=\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)
Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)
\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)
Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:
\(\left(1\right)\Leftrightarrow\dfrac{1}{b+8}+\dfrac{1}{b}=\dfrac{3}{5}\)
\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)
\(\Leftrightarrow10b+40=3\left(b+8\right)b\)
\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)
TH1: \(b=2\Leftrightarrow...\)
TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)
giúp mik
tìm x
a 2 (x^3 - 1 ) - 2x^2 ( x +2x^4 ) + ( 4x^5 +4 ) x =6
b (2x)^2 (4x - 2 ) - ( x^3 -8x^3 )=15
chứng tỏ giá trị của biểu thức sau ko phụ thuộc vào giá trị của biến
a P = x ( 2x + 1 ) - x^2 ( x + 2 ) + x^3 - x +3
b Q = x (2x^2 -4x +8 ) +12x^2 (1/3 _1/6x ) -8x +9
\(a,2\left(x^3-1\right)-2x^2\left(x+2x^4\right)+x\left(4x^5+4\right)=6\\ \Leftrightarrow2x^3-2-2x^3-4x^6+4x^6+4x-6=0\\ \Leftrightarrow4x-8=0\\ \Leftrightarrow x=2\\ b,\left(2x\right)^2\left(4x-2\right)-\left(x^3-8x^3\right)=15\\ \Leftrightarrow4x^2\left(4x-2\right)+7x^3-15=0\\ \Leftrightarrow16x^3-8x^2+7x^3-15=0\\ \Leftrightarrow23x^3-8x^2-15=0\\ \Leftrightarrow23x^3-23x^2+15x^2-15x+15x-15=0\\ \Leftrightarrow\left(x-1\right)\left(23x^2+15x-15\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x\in\varnothing\left(23x^2+15x-15>0\right)\end{matrix}\right.\)
Bài 1:
a: Ta có: \(2\left(x^3-1\right)-2x^2\left(2x^4+x\right)+x\left(4x^5+4\right)=6\)
\(\Leftrightarrow2x^3-2-4x^6-2x^3+4x^6+4x=6\)
\(\Leftrightarrow4x=8\)
hay x=2
b: Ta có: \(\left(2x\right)^2\cdot\left(4x-2\right)-\left(x^3-8x^3\right)=15\)
\(\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^3=15\)
\(\Leftrightarrow16x^3-8x^2+7x^3=15\)
\(\Leftrightarrow23x^3-8x^2-15=0\)
\(\Leftrightarrow23x^3-23x^2+15x^2-15=0\)
\(\Leftrightarrow23x^2\left(x-1\right)+15\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(23X^2+15x+15\right)=0\)
\(\Leftrightarrow x-1=0\)
hay x=1
Bài 2:
a: Ta có: \(P=x\left(2x+1\right)-x^2\left(x+2\right)+x^3-x+3\)
\(=2x^2+x-x^3-2x^2+x^3-x+3\)
=3
b: ta có: \(Q=x\left(2x^2-4x+8\right)+12x^2\left(\dfrac{1}{3}-\dfrac{1}{6}x\right)-8x+9\)
\(=2x^3-4x^2+8x+4x^2-2x^3-8x+9\)
=9