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\(x^4+y^4+\left(x+y\right)^4=2\left(x^4+y^4+2x^3y+3x^2y^2+2xy^3\right)\)
\(=2\left(\left(x^4+y^4+2x^2y^2\right)+\left(2x^3y+2xy^3\right)+x^2y^2\right)\)
\(=2\left(\left(x^2+y^2\right)^2+2xy\left(x^2+y^2\right)+x^2y^2\right)\)
\(=2\left(x^2+y^2+xy\right)^2\)
Đặt x2 + xy + y2 = a2 ; x + y = b.Ta có :
a4 = (a2)2 = (x2 + xy + y2)2 = x4 + y4 + x2y2 + 2x3y + 2xy2 + 2x2y2 = x4 + y4 + x2y2 + 2xy(x2 + y2 + xy) = x4 + y4 + x2y2 + 2xya2 (1)
mà b = x + y
=> b2 = x2 + y2 + 2xy = a2 + xy => b4 = a4 + x2y2 + 2a2xy .Từ (1) và (2) ,ta có :
2a4 = x4 + y4 + a4 + x2y2 + 2xya2 = x4 + y4 + b4.Thay a2 = x2 + xy + y2 ; b = x + y,ta có đpcm
<=>
\(\left(a^2+b^2\right)\left(x^2+y^2\right)=\left(ax+by\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+b^2x^2+b^2y^2=a^2x^2+2axby+b^2y^2\)
\(\Leftrightarrow a^2y^2-2axby+b^2x^2=0\)
\(\Leftrightarrow\left(ay-bx\right)^2=0\Leftrightarrow ay-bx=0\Leftrightarrow ay=bx\Leftrightarrow\frac{a}{x}=\frac{b}{y}\)
\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)=\left(ax+by+cz\right)^2\)
\(\Leftrightarrow a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2=a^2x^2+b^2y^2+c^2z^2+2\left(axby+bycz+axcz\right)\)
\(\Leftrightarrow\left(a^2y^2-2axby+b^2x^2\right)+\left(a^2z^2-2axcz+x^2c^2\right)+\left(c^2y^2-2bycz+b^2z^2\right)=0\)
\(\Leftrightarrow\left(ay-bx\right)^2+\left(az-cx\right)^2+\left(cy-bz\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}ay=bx\\az=cx\\cy=bz\end{cases}\Leftrightarrow}\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\)
\(VT=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)\)
\(=x\left(x^3+x^2y+xy^2+y^3\right)-y\left(x^3+x^2y+xy^2+y^3\right)\)
\(=x^4+x^3y+x^2y^2+xy^3-x^3y-x^2y^2-xy^3-y^4\)
\(=\left(x^4-y^4\right)+\left(x^3y-x^3y\right)+\left(x^2y^2-x^2y^2\right)+\left(xy^3-xy^3\right)\)
\(=x^4-y^4=VP\)
\(VT=\left(a+b\right)^2-\left(a-b\right)^2=4ab\)
\(=\left(a^2+2ab+b^2\right)-\left(a^2-2ab+b^2\right)\)
\(=a^2+2ab+b^2-a^2+2ab-b^2\)
\(=\left(a^2-a^2\right)-\left(b^2+b^2\right)+\left(2ab+2ab\right)\)
\(=4ab=VP\)
Câu a :
\(VT=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)\)
Nhân 2 vế lại ta được \(x^4-y^4=VP\)
\(\Rightarrowđpcm\)
Câu b :
\(VT=\left(a+b\right)^2-\left(a-b\right)^2=\left(a+b-a+b\right)\left(a+b+a-b\right)=2b.2a=4ab=VP\)
\(\Rightarrowđpcm\)
a)\(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)\)
\(=x^5+x^4y+x^3y^2+x^2y^3+xy^4-x^4y-x^3y^2-x^2y^3-xy^4-y^5\)
\(=x^5-y^5+\left(x^4y\right)+\left(x^3y^2-x^3y^2\right)+\left(x^2y^3-x^2y^3\right)+\left(xy^4-xy^4\right)\)
\(\Rightarrow\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)=x^5-y^5\)
b)\(\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(=a^3-a^2b+ab^2+a^2b-ab^2+b^3\)
\(=a^3+b^3+\left(-a^2b+a^2b\right)+\left(ab^2-ab^2\right)\)
\(\Rightarrow\)\(\left(a+b\right)\left(a^2-ab+b^2\right)=a^3+b^3\)
a) (x - y)(x4 + x3y + x2y2 + xy3 + y4)
= x(x4 + x3y + x2y2 + xy3 + y4) - y(x4 + x3y + x2y2 + xy3 + y4)
= x5 + x4y + x3y2 + x2y3 + xy4 - x4y - x3y2 - x2y3 - xy4 - y5
= x5 - y5
b) (a + b)(a2 - ab + b2)
= a(a2 - ab + b2) + b(a2 - ab + b2)
= a3 - a2b + ab2 + a2b - ab2 + b3
= a3 + b3
a. \(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)\)
\(\Rightarrow x^5+x^4y+x^3y^2+x^2y^3+y^5-yx^4-x^3y^2-x^2y^3-xy^4-y^5=VP\)
\(\Rightarrow dpcm\)
b. \(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)\)
\(\Rightarrow x^5-x^4y+x^3y^2-x^2y^3+xy^4+yx^4-x^3y^2-xy^4+y^5=VP\)
\(\Rightarrow dpcm\)
c.d làm tương tự
Bài làm
a) Biến đổi vế trái, ta được:
\(VT=\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)\)
\(=x^5+x^4y+x^3y^2+x^2y^3+xy^4-x^4y-x^3y^2-x^2y^3-xy^4-y^5\)
\(=\left(x^5-y^5\right)+\left(x^4y-x^4y\right)+\left(x^3y^2-x^3y^2\right)+\left(x^2y^3-x^2y^3\right)+\left(xy^4-xy^4\right)\)
\(=x^5-y^5=VP\left(đpcm\right)\)
b) Biến đổi vế trái, ta có:
\(VT=\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)\)
\(=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5\)
\(=\left(x^5+y^5\right)+\left(-x^4y+x^4y\right)+\left(x^3y^2-x^3y^2\right)+\left(-x^2y^3+x^2y^3\right)+\left(xy^4-xy^4\right)\)
\(=x^5+y^5=VP\left(đpcm\right)\)
c) Biến đổi vế trái, ta có:
\(VT=\left(a+b\right)\left(a^3-a^2b+ab^2-b^3\right)\)
\(=a^4-a^3b+a^2b^2-ab^3+a^3b-a^2b^2+ab^3-b^4\)
\(=\left(a^4-b^4\right)+\left(-a^3b+a^3b\right)+\left(a^2b^2-a^2b^2\right)+\left(-ab^3+ab^3\right)\)
\(=a^4-b^4=VP\left(đpcm\right)\)
d) Đây là hằng đẳng thức, như vế phải hình như bạn viết bị sai, mik sửa là vế phải nha.
\(\left(a+b\right)\left(a^2-ab+b^2\right)=a^3+b^3\)
Biến đổi vế trái, ta có:
\(VT=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(=a^3-a^2b+ab^2+a^2b-ab^2+b^3\)
\(=\left(a^3+b^3\right)+\left(-a^2b+a^2b\right)+\left(ab^2-ab^2\right)\)
\(=a^3+b^3=VP\left(đpcm\right)\)
a) \(x^2+y^2=x^2+y^2+2xy-2xy=\left(x+y\right)^2-2xy\)
b) \(\left(a+b\right)^2-\left(a-b\right)\left(a+b\right)=\left(a+b\right)^2-\left(a^2-b^2\right)=a^2+2ab+b^2-a^2+b^2\)
\(=2ab+2b^2=2b\left(a+b\right)\)
c)\(\left(a+b\right)^2-\left(a-b\right)^2=\left(a+b-a+b\right)\left(a+b+a-b\right)\)
\(=2b.2a=4ab\)
a: \(\left(x+y\right)^2-2xy\)
\(=x^2+2xy+y^2-2xy\)
\(=x^2+y^2\)
b: \(\left(a+b\right)^2-\left(a-b\right)\left(a+b\right)\)
\(=\left(a+b\right)\left(a+b-a+b\right)\)
\(=2b\left(a+b\right)\)
c: \(\left(a+b\right)^2-\left(a-b\right)^2\)
\(=\left(a+b-a+b\right)\left(a+b+a-b\right)\)
\(=4ab\)
Ta có : VP = \(x^4-y^4\)
\(=\left(x^2\right)^2-\left(y^2\right)^2\)
\(=\left(x^2-y^2\right)\left(x^2+y^2\right)\)
\(=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\)
Vp\(=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)\) = VT
Vậy \(x^4-y^4\) \(=\left(x-y\right)\left(x^3+x^2y+xy^2+y^3\right)\) (đpcm)