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a) \(\left(x+y+z\right)\left(xy+yz+xz\right)-xyz\)
\(=\left(y+z\right)\left(xy+yz+zx\right)+x^2y+x^2z+xyz-xyz\)
\(=\left(y+z\right)\left(xy+yz+zx\right)+x^2\left(y+z\right)\)
\(=\left(y+z\right)\left(xy+yz+zx+x^2\right)\)
\(=\left(y+z\right)\left[y\left(x+z\right)+x\left(z+x\right)\right]\)
\(=\left(y+z\right)\left(x+z\right)\left(x+y\right)\)
b) \(\left(x^2+y^2+5\right)^2-4x^2y^2-16xy-16\)
\(=\left(x^2+y^2+5\right)^2-\left(4x^2y^2+16xy+16\right)\)
\(=\left(x^2+y^2+5\right)^2-\left(2xy+4\right)^2\)
\(=\left(x^2+y^2+5-2xy-4\right)\left(x^2+y^2+5+2yx+4\right)\)
\(=\left(x^2+y^2+5-2xy-4\right)\left(x^2+y^2+5+2yx+4\right)\)
c)sai đề.
đặt \(x^2+x+1=t\)
\(\Rightarrow\left(x^2+x+1\right)^2+\left(x^2+x+2\right)-12\)
\(=t^2+t+1-12\)
.........................................
mình sửa đề không biết có đúng hay không nên mình chỉ nêu hướng làm thôi. bạn thông cảm.
d) \(x^2-x-2001.2002\)
\(=x\left(x+2001\right)-2002\left(x+2001\right)\)
\(=\left(x-2002\right)\left(x+2001\right)\)
Bài 1:
a)\(A=x\left(x^2-y\right)-x^2\left(x+y\right)+y\left(x^2-x\right)\)\(=x^3-xy-x^3-x^2y+yx^2-yx=-2xy\)
Thay x=1/2 và y=-100 vào biểu thức A ta được \(A=-2.\frac{1}{2}.\left(-100\right)=100\)
b)\(B=\left(x^2-5\right)\left(x+3\right)+\left(x+4\right)\left(x-x^2\right)=x^3+3x^2-5x-15-x^3-3x^2+4x\)=-x-15
Thay x=-1 vào biểu thức B ta được B=-(-1)-15=1-15=-14
a, x^4+x^3+x+1
=(x^4+x^3)+(x+1)
=x^3(x+1)+(x+1)
=(x+1)(x^3+1)
=(x+1)^2(x^2+x+1)
b,x^4-x^3-x^2+1
=(x^4-x^3)-(x^2-1)
=x^3(x-1)-(x-1)(x+1)
=(x-1)(x^3-x+1)
a) \(=x^2+2xy+y^2-x^2+y^2=2xy+2y^2=2y\left(x+y\right)\)
b) \(=\left(x^2-4y^2\right)-\left(2x+4y\right)=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)=\left(x+2y\right)\left(x-2y-2\right)\)
c) \(=3\left[\left(x^2+2xy+y^2\right)-z^2\right]=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y+z\right)\left(x+y-z\right)\)
d) \(=\left(2xy+1+2x+y\right)\left(2xy+1-2x-y\right)\)
e) \(=\left(x-3\right)\left(x^2+3x+9\right)-2x\left(x-3\right)=\left(x-3\right)\left(x^2+x+9\right)\)
f) \(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)=\left(x+5\right)\left(x^2-6x+25\right)\)
a) \(\left(x+y\right)^2-\left(x^2-y^2\right)\)
\(=x^2+2xy+y^2-x^2+y^2\)
\(=2y^2+2xy\)
\(=2y\left(x+y\right)\)
c) \(3x^2+6xy+3y^2-3z^2\)
\(=3\left(x^2+2xy+y^2-x^2\right)\)
\(=3\left[\left(x+y\right)^2-z^2\right]\)
\(=3\left(x+y+z\right)\left(x+y-z\right)\)
d) \(\left(2xy+1\right)^2-\left(2x+y\right)^2\)
\(=\left(2xy+1+2x+y\right)\left(2xy+1-2x-y\right)\)
\(=\left[\left(2xy+2x\right)+\left(y+1\right)\right]\left[\left(2xy-2x\right)-\left(y-1\right)\right]\)
\(=\left[2x\left(y+1\right)+\left(y+1\right)\right]\left[2x\left(y-1\right)-\left(y-1\right)\right]\)
\(=\left(2x+1\right)\left(y+1\right)\left(2x-1\right)\left(y-1\right)\)
\(=\left(4x^2-1\right)\left(y^2-1\right)\)
Bài 1:
a: \(6x^2-11x+3\)
\(=6x^2-9x-2x+3\)
\(=3x\left(2x-3\right)-\left(2x-3\right)\)
\(=\left(2x-3\right)\left(3x-1\right)\)
b: \(2x^2+3x-27\)
\(=2x^2+9x-6x-27\)
\(=x\left(2x+9\right)-3\left(2x+9\right)\)
\(=\left(2x+9\right)\left(x-3\right)\)
c: \(x^2-10x+24\)
\(=x^2-4x-6x+24\)
\(=x\left(x-4\right)-6\left(x-4\right)\)
\(=\left(x-4\right)\left(x-6\right)\)
d: \(49x^2+28x-5\)
\(=49x^2+28x+4-9\)
\(=\left(7x+2\right)^2-9\)
\(=\left(7x-1\right)\left(7x+5\right)\)
e: \(2x^2-5xy-3y^2\)
\(=2x^2-6xy+xy-3y^2\)
\(=2x\left(x-3y\right)+y\left(x-3y\right)\)
\(=\left(x-3y\right)\left(2x+y\right)\)
Câu 1:
a/ (-5x3)(2x2+3x-5)
=-10x5-15x4+25x3
b/(2x-1)x
=2x2-x
c/(x-y)(3x2+4xy)
=3x3+4x2y-3x2y-4xy2
=3x3 +x2y-4xy2
Câu 2:
a/ x3-2x2+x
=x(x2-2x+1)
=x(x-1)2
b/x2-x-12
=x2 +3x-4x-12
=(x2 +3x)+(-4x-12)
=x(x+3)-4(x+3)
=(x+3)(x-4)
c/ 2x-6
=2(x-3)
e/ x2+4x+4-y2
=(x2+4x+4)-y2
=(x+2)2-y2
=(x+2-y)(x+2+y)
d/ x2-2xy+y2-16
=(x2-2xy+y2)-16
=(x-y)2-16
=(x-y-4)(x-y+4)
Câu 3:
a: \(=\dfrac{5xy-4+3xy+4}{2x^2y^3}=\dfrac{8xy}{2x^2y^3}=\dfrac{4}{xy^2}\)
b: \(=\dfrac{y-12}{6\left(y-6\right)}+\dfrac{6}{y\left(y-6\right)}\)
\(=\dfrac{y^2-12y+36}{6y\left(y-6\right)}=\dfrac{y-6}{6y}\)
c: \(=\dfrac{3x+1-2x+3}{x+y}=\dfrac{x+4}{x+y}\)
d: \(=\dfrac{4x+7+5x+7}{9}=\dfrac{9x+14}{9}\)
e: \(=\dfrac{5\left(x+2\right)}{2\left(2x-1\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-5\left(x-2\right)}{2x-1}\)
GIÚP MÌNH VỚI ĐỀ BÀI LÀ RÚT GỌN THÔI NHA THUỘC KIỂU HẰNG ĐẲNG THỨC 6 VÀ 7 GIÚP MÌNH VỚI MÌNH CẦN GẤP TRONG TỐI NAY GIÚP VỚI
a) \(x^3-2x^2+x-xy^2\)
= \(x\left(x^2-2x+1-y^2\right)\)
\(=x\left(x-1-y\right)\left(x-1+y\right)\)
b) \(x^3-4x^2-12x+27\)
\(=x^3+27-4x^2-12x\)
\(=\left(x+3\right)\left(x^2-3x+9\right)-4x\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
c) \(x^2-2xy+y^2+1\)
\(=\left(x-1\right)^2+1\)
d) \(2x^2+4x+2-2y^2\)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left(x+1+y\right)\left(x+1-y\right)\)
e) \(2xy-x^2-y^2+16\)
\(=-\left(x^2-2xy+y^2-4^2\right)\)
\(=-\left(x-y-4\right)\left(x-y+4\right)\)
đăng 3 câu /1 lần hỏi thôi