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a) \(x^2-81=\left(x-9\right)\left(x+9\right)\)
b) \(4x^2-25=\left(2x-5\right)\left(2x+5\right)\)
c) \(x^4-y^4=\left(x^2-y^2\right)\left(x^2+y^2\right)=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)
d) \(x^2+6xy+9y^2=\left(x+3y\right)^2\)
e) \(6x-9-x^2=-\left(x^2-6x+9\right)=-\left(x-3\right)^2\)
f) \(x^2-4x^2+4y^2+4xy=\left(x^2+4xy+4y^2\right)-4x^2=\left(x+2y\right)^2-4x^2\\ =\left(x+2y+2x\right)\left(x+2y-2x\right)=\left(3x+2y\right)\left(2y-x\right)\)
g) \(\left(a+b\right)^3+\left(a-b\right)^3=\left(a+b+a-b\right)\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)
\(=2a\left(a^2+2ab+b^2-a^2+b^2+a^2-2ab+b^2\right)=2a\left(a^2+3b^2\right)\)
h) \(\left(3x+1\right)^2-\left(x+1\right)^2=\left(3x+1+x+1\right)\left(3x+1-x-1\right)\\ =\left(4x+2\right)\cdot2x=4x\left(2x+1\right)\)
a) 9x^2+6xy+y^2
=(3x+y)2
b)6x - 9 - x^2
=-(x2-6x+9)
=-(x-3)2
c) x^2 + 4y^2 + 4xy
=(x+2y)2
Bạn tải ứng dụng PhotoMath về nha. Ứng dụng này sẽ giải toán số chi tiết
a) \(x^3-4x^2-12x+27\)
\(=\left(x^3+27\right)-\left(4x^2+12x\right)\)
\(=\left(x+3\right)\left(x^2-3x+9\right)-4x\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
b) \(x^3-3x^2-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x^2-4\right)\left(x-3\right)\)
\(=\left(x+2\right)\left(x-2\right)\left(x-3\right)\)
a) \(9x^2+6xy+y^2=\left(3x+y\right)^2\)
b) \(6x-9-x^2=-\left(x-3\right)^2\)
9x2 + 6xy + y2
= (3x)2 + 2.3x.y + y2
= (3x + y)2
b) 6x - 9 - x2
= -(x2 - 6x + 9)
= -(x - 3)2
Bài 1:
a) \(9x^2+6xy+y^2=\left(3x+y\right)^2\)
b) \(6x-9-x^2=-9+6x-x^2=-\left(3-x\right)^2\)
c) \(x^2+4y^2+4xy=x^2+4xy+4y^2=\left(x+2y\right)^2\)
Chúc bạn học tốt!
1)
a) \(9x^2+6xy+y^2=\left(3x+y\right)^2\)
b) \(6x-9-x^2=-\left(x^2-6x+9\right)=-\left(x-3\right)^2\)
c) \(x^2+4y^2+4xy=\left(x+2y\right)^2\)
2)
a) \(x^3-0,25x=0\)
Bài này có nghiệm x khủng bố lắm, có lẽ đề sai rồi. Nếu đề là 0,125 thì còn làm được...
b) \(x^2-10x=-25\)
\(x^2-10x+25=0\)
\(\left(x-5\right)^2=0\)
\(x-5=0\Rightarrow x=5\)
a)\(6x-9-x^2\)
\(=-\left(x^2+6x+9\right)\)
\(=-\left(x+3\right)^2\)
b)\(x^2+4y^2+4xy\)
\(=\left(x+2y\right)^2\)
c)\(x^2+8x+16\)
\(=\left(x+4\right)^2\)
d)\(9x^2-12xy+4y^2\)
\(=\left(3x-2y\right)^2\)
e)\(-25x^2y^2+10xy-1\)
\(=-\left(25x^2y^2-10xy+1\right)\)
\(=-\left(5xy-1\right)^2\)
f)\(4x^2-4x+1\)
\(=\left(2x-1\right)^2\)
j)\(x^2+6x+9\)
\(=\left(x+3\right)^2\)
h)\(9x^2-6x+1\)
\(=\left(3x-1\right)^2\)
#H
a, 6x - 9 - x2 = - x2 + 6x - 9 = - (x2 - 6x + 9) = - (x - 3)2
b, x2 + 4y2 + 4xy = x2 + 2. x . 2y + (2y)2 = (x + 2y)2
c, x2 + 8x + 16 = x2 + 2 . x . 4 + 42 = (x + 4)2
d, 9x2 - 12xy + 4y2 = (3x)2 - 2 . 3x . 2y + (2y)2 = (3x - 2y)2
e, - 25x2y2 + 10xy - 1 = - (25x2y2 - 10xy + 1) = - [(5xy)2 - 2 . 5xy + 1] = - (5xy - 1)2
f, 4x2 - 4x + 1 = (2x)2 - 2 . 2x + 1 = (2x - 1)2
j, x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
h, 9x2 - 6x + 1 = (3x)2 - 2 . 3x + 1 = (3x - 1)2
a) \(4x^2-4xy+y^2-9\)
\(=\left(2x-y\right)^2-3^2\)
\(=\left(2x-y+3\right)\left(2x-y-3\right)\)
b) \(x^2-36+4xy+4y^2\)
\(=\left(4y^2+4xy+x^2\right)-36\)
\(=\left(2y+x\right)^2-6^2\)
\(=\left(2y+x+6\right)\left(2y+x-6\right)\)
c) \(9x^2-12xy-25+4y^2\)
\(=\left(9x^2-12xy+4y^2\right)-25\)
\(=\left(3x-2y\right)^2-5^2\)
\(=\left(3x-2y+5\right)\left(3x-2y-5\right)\)
d) \(25x^2+10x-4y^2+1\)
\(=\left(25x^2+10x+1\right)-4y^2\)
\(=\left(5x+1\right)^2-\left(2y\right)^2\)
\(=\left(5x+2y+1\right)\left(5x-2y+1\right)\)
bài 1
a(x+y)2-(x-y)2
=[(x+y)-(x-y)][(x+y)+(x-y)]
=(x+y-x+y)(x+y+x-y)
=2y.2x
b,(3x+1)2-(x+1)2
=[(3x+1)-(x+1)][(3x+1)+(x+1)]
=(3x+1-x-1)(3x+1+x+1)
=2x.(4x+2)
4x.(x+10
bài 2
x3-0,25x=0
=>x(x2-0,25)=0
=>x=0 hoặc x2-0,25=0
=> x=0 hoặc x=\(\pm0,5\)
phân tích thành nhân tử:
\(x^2-9=x^2-3^2=\left(x+3\right)\left(x-3\right)\)
\(4x^2-25=\left(2x\right)^2-5^2=\left(2x+5\right)\left(2x-5\right)\)
\(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2\)\(=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
\(=\left(x+y\right)\left(x-y\right)\left(x^2+xy+y^2\right)\left(x^2-xy+y^2\right)\)
\(9x^2+6xy+y^2=\left(3x\right)^2+2\cdot3x\cdot1+y^2=\left(3x+y\right)^2\)
\(x^2+4y^2+4xy=x^2+2\cdot x\cdot2y+\left(2y\right)^2=\left(x+2y\right)^2\)
a. \(x^3-0.25x=0\Rightarrow x\left(x^2-\frac{1}{4}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x^2-\frac{1}{4}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\x^2=\frac{1}{4}\end{cases}}}\) \(\Rightarrow\orbr{\begin{cases}x=0\\\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{-1}{2}\end{cases}}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=0\\\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{-1}{2}\end{cases}}\end{cases}}\)=> \(x\in\left\{0;\frac{1}{2};\frac{-1}{2}\right\}\)
b, \(x^2-10x=-25\)\(\Rightarrow x^2-10x+25=0\)
\(\Rightarrow\left(x-5\right)^2=0\Rightarrow x-5=0\Rightarrow x=5\)
a, \(x^2-9=x^2-3x+3x-9\)
\(=x\left(x-3\right)+3\left(x-3\right)=\left(x-3\right)\left(x+3\right)\)
b, \(4x^2-25=\left(2x\right)^2-5^2=\left(2x-5\right)\left(2x+5\right)\)
c, \(x^6-y^6=\left(x^3\right)^2-\left(y^3\right)^2=\left(x^3+y^3\right)\left(x^3-y^3\right)\)
d, \(9x^2+6xy+y^2=\left(3x\right)^2+2\left(3xy\right)+y^2\) \(=\left(3x+y\right)^2\)
e, \(6x-9-x^2=6x-18+9-x^2\) \(=6\left(x-3\right)-\left(x-3\right)\left(x+3\right)\)
\(=\left(x-3\right)\left(6-x-3\right)=\left(x-3\right)\left(3-x\right)\)
f, \(x^2+4y^2+4xy=x^2+2\left(2xy\right)+\left(2y\right)^2\)
\(\left(x+2y\right)^2\)
\(\)