Phân tích các đa thức sau thành nhân tử
\(a,\left(x^2+1\right)^2-6\left(x^2+1\right)+9\)
\(b,-25x^6-y^8+10x^3y^4\)
\(c,49\left(y-4\right)^2-9\left(y+2\right)^2\)
Giúp tui gấp !
Bài 1: Phân tích đa thức thành nhân tử:
1) \(3x^3y^2-6xy\)
2) \(\left(x-2y\right).\left(x+3y\right)-2.\left(x-2y\right)\)
3) \(\left(3x-1\right).\left(x-2y\right)-5x.\left(2y-x\right)\)
4) \(x^2-y^2-6y-9\)
5) \(\left(3x-y\right)^2-4y^2\)
6) \(4x^2-9y^2-4x+1\)
8) \(x^2y-xy^2-2x+2y\)
9) \(x^2-y^2-2x+2y\)
Bài 2: Tìm x:
1) \(\left(2x-1\right)^2-4.\left(2x-1\right)=0\)
2) \(9x^3-x=0\)
3) \(\left(3-2x\right)^2-2.\left(2x-3\right)=0\)
4) \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
Bài 2:
1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)
=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)
=>(2x-1)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
2: \(9x^3-x=0\)
=>\(x\left(9x^2-1\right)=0\)
=>x(3x-1)(3x+1)=0
=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)
=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)
=>(2x-3)(2x-3-2)=0
=>(2x-3)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
=>\(2x^2+10x-5x-25-10x+25=0\)
=>\(2x^2-5x=0\)
=>\(x\left(2x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)
Bài 1:
1: \(3x^3y^2-6xy\)
\(=3xy\cdot x^2y-3xy\cdot2\)
\(=3xy\left(x^2y-2\right)\)
2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+3y-2\right)\)
3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)
\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)
\(=(x-2y)(3x-1+5x)\)
\(=\left(x-2y\right)\left(8x-1\right)\)
4: \(x^2-y^2-6y-9\)
\(=x^2-\left(y^2+6y+9\right)\)
\(=x^2-\left(y+3\right)^2\)
\(=\left(x-y-3\right)\left(x+y+3\right)\)
5: \(\left(3x-y\right)^2-4y^2\)
\(=\left(3x-y\right)^2-\left(2y\right)^2\)
\(=\left(3x-y-2y\right)\left(3x-y+2y\right)\)
\(=\left(3x-3y\right)\left(3x+y\right)\)
\(=3\left(x-y\right)\left(3x+y\right)\)
6: \(4x^2-9y^2-4x+1\)
\(=\left(4x^2-4x+1\right)-9y^2\)
\(=\left(2x-1\right)^2-\left(3y\right)^2\)
\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)
8: \(x^2y-xy^2-2x+2y\)
\(=xy\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-2\right)\)
9: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
Phân tích các đa thức sau thành nhân tử :
a/ \(10x\left(x-y\right)-6y\left(y-x\right)\)
b/ \(14x^2y-21xy^2+28x^3y^2\)
c/ \(x^2-4+\left(x-2\right)^2\)
d/ \(\left(x+1\right)^2-25\)
e/ \(x^2-4y^2-2x+4y\)
f/ \(x^2-25-2xy+y^2\)
g/ \(x^3-2x^2+x-xy^2\)
h/ \(x^3-4x^2-12x+27\)
i/ \(x^2+5x-6\)
m/ \(6x^2-7x+2\)
n/ \(4x^4+81\)
\(a.10x\left(x-y\right)-6y\left(y-x\right)\\ =10x\left(x-y\right)+6y\left(x-y\right)\\ =\left(10x-6y\right)\left(x-y\right)\\ =2\left(5x-3y\right)\left(x-y\right)\)
\(b.14x^2y-21xy^2+28x^3y^2\\ =7xy\left(x-y+xy\right)\)
\(c.x^2-4+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2\right)+\left(x-2\right)^2\\ =\left(x-2\right)\left(x+2+x-2\right)\\ =2x\left(x-2\right)\)
\(d.\left(x+1\right)^2-25\\ =\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)
Phân tích đa thức thành nhân tử:
a)\(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)
b)\(49\left(y-4\right)^2-9y^2-3y-36\)
c)\(x\left(x-y\right)+y\left(y-x\right)\)
a/Dùng hằng đẳng thức A2-B2=(A+B)(A-B) phân tích được ngay
\(\left(x-y+4\right)^2-\left(2x+3y-1\right)^2\)
\(=\left(x-y+4+2x+3y-1\right)\left(x-y+4-2x-3y+1\right)\)
=\(\left(3x-2y+3\right)\left(4-x-4y\right)\)
b/Chắc chỉ phân tích hằng đẳng thức (A-B)2=A2-2ab+B2
\(49\left(y-4\right)^2-9y^2-3y-36=49y^2-392y+784-9y^2-3y-36\)
\(=40y^2-395y+748\)
Mình dùng biệt thức cho ra nghiệm vô tỉ, không biết cho phải tại mình tính sai hay đề thiếu nữa
c/Khai triển biểu thức ban đầu ta được
\(x\left(x-y\right)+y\left(y-x\right)=x^2-xy+y^2-xy=x^2-2xy+y^2=\left(x-y\right)^2\)
Phân tích các đa thức sau thành nhân tử
a)\(5x\left(x-2y\right)+2\left(2y-x\right)^2\)
b)\(7x\left(y-4\right)^2-\left(4-y\right)^3\)
c)\(\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9\left(8-4x\right)\)
Bài 1 : Phân tích đa thức thành nhân tử
\(a,5x\left(x-2y\right)+2\left(2y-x\right)^2\)
\(b,7x\left(y-4\right)^2-\left(4-x\right)^3\)
\(c,\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9\left(8-4x\right)\)
phân tích đa thức sau thành nhân tử dựa vào phương pháp nhóm hạng tử :
a) \(x^4+25x^2+20x-4\)
b)\(x^2\left(x^2-6\right)-x^2+9\)
c)\(ab\left(x^2+y^2\right)-xy\left(a^2+b^2\right)\)
a: \(x^4+25x^2+20x-4\)
\(=x^4-5x^3+2x^2+5x^3-25x^2+10x-2x^2+10x-4\)
\(=x^2\left(x^2-5x+2\right)+5x\left(x^2-5x+2\right)-2\left(x^2-5x+2\right)\)
\(=\left(x^2-5x+2\right)\left(x^2+5x-2\right)\)
b: \(=x^4-6x^2-x^2+9\)
\(=\left(x^2-3\right)^2-x^2\)
\(=\left(x^2-x-3\right)\left(x^2+x-3\right)\)
c: \(=abx^2+aby^2-a^2xy-b^2xy\)
\(=\left(abx^2-b^2xy\right)+\left(aby^2-a^2xy\right)\)
\(=xb\left(ax-by\right)+ay\left(by-ax\right)\)
\(=\left(ax-by\right)\cdot\left(xb-ay\right)\)
Phân tích đa thức thành nhân tử
\(a.\left(x^2+4x-3\right)^2-5x\left(x^2+4x-3\right)+6x^2\)
B. \(x^2-2xy+y^2+3x-3y-4\)
\(c.\left(12x^2-12xy+3y^2\right)-10\left(2x-y\right)+8\)
\(d.\left(x^2-2x\right)\left(x^2-2x-1\right)-6\)
Phân tích các đa thức sau thành nhân tử:
\(49\left(y-4\right)^2-9\left(y+2\right)^2\)
[7(y-4)]^2 -[3(y+2)]^2 = [7(y-4)-3(y+2)]*[7(y-4)+3(y+2)]
= (4y-34)(10y-22)=2(2y-17)(5y-11)
mình nghĩ thế sai thâu nha !
Phân tích đa thức thành nhân tử
a) \(\left(x-5\right)^2-16\)
b)\(25-\left(3-x\right)^2\)
c)\(\left(7x-4\right)^2-\left(2x+1\right)^2\)
d)\(49\left(y-4\right)^2-9\left(y+2\right)^2\)
e)\(8x^3+\frac{1}{27}\)
f)\(125-x^6\)
GIÚP MK VS MK ĐG CẦN GẤP AI NHANH MK TICK CHO 3 TICK
a) Ta có : (x - 5)2 - 16
= (x - 5)2 - 42
= (x - 5 - 4)(x - 5 + 4)
= (x - 1)(x - 9)
b) 25 - (3 - x)2
= 52 - (3 - x)2
= (5 - 3 + x)(5 + 3 - x)
= (x + 2)(8 - x)
c) (7x - 4)2 - (2x + 1)2
= (7x - 4 - 2x - 1)(7x - 4 + 2x + 1)
= (5x - 5)(9x - 3)
= 5(x - 1)3(3x - 1)
= 15(x - 1)(3x - 1)
\(49.\left(y-4\right)^2-9\left(y+2\right)^2\)
\(=\left[7.\left(y-4\right)\right]^2-\left[3.\left(y+2\right)\right]^2\)
\(=\left[7.\left(y-4\right)-3.\left(y+2\right)\right].\left[7.\left(y-4\right)+3.\left(y+2\right)\right]\)
\(=\left(7y-28-3y-6\right).\left(7y-28+3y+6\right)\)
\(=\left(4y-34\right).\left(10y-22\right)\)
\(=4.\left(y-17\right).\left(5y-11\right)\)
\(8x^3+\frac{1}{27}=\left(2x\right)^3+\left(\frac{1}{3}\right)^3=\left(2x+\frac{1}{3}\right).\left(4x^2+\frac{2}{3}x+\frac{1}{9}\right)\)
\(125-x^6=5^3-\left(x^2\right)^3=\left(5-x^2\right).\left(25+5x^2+x^4\right)=\left(\sqrt{5}-x\right).\left(\sqrt{5}+x\right).\left(25+5x^2+x^4\right)\)