Giải hệ
a) \(\left\{{}\begin{matrix}x^2+y^2-2y-6+2\sqrt{2y+3}=0\\\left(x-y\right)\left(x^2+xy+y^2+3\right)=3\left(x^2+y^2\right)+2\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}x^2y+2y+x=4xy\\\dfrac{1}{x^2}+\dfrac{1}{xy}+\dfrac{x}{y}=3\end{matrix}\right.\)
Bài 3. Chứng minh các đẳng thức sau:
a. \(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)=x^5-y^5\)
b. \(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)=x^5+y^5\)
c. \(\left(a+b\right)\left(a^3-a^2b+ab^2-b^3\right)=a^4-b^4\)
đ. \(\left(a+b\right)\left(a^2-ab+b^2\right)=a^3-b^3\)
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:
\(VT=\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(=a^3-a^2b+ab^2+a^2b-ab^2+b^3\)
\(=\left(a^3+b^3\right)+\left(-a^2b+a^2b\right)+\left(ab^2-ab^2\right)\)
\(=a^3+b^3=VP\left(đpcm\right)\)
c)
VT=(a+b)(a3-a2b+ab2-b3)=a4-a3b+a2b2-ab3+a3b-a2b2+ab3-b4 =a4-b4=VP
=> Đpcm
d) VT=(a+b)(a2-ab+b2)=a3-a2b+ab2+a2b-ab2+b3=a3+b3 khác VT
=> đẳng thức ko đúng
Thực hiện phép tính:
a) \(\dfrac{2}{5}xy\left(x^2y-5x+10y\right)\)
b) \(\left(x^2-1\right)\left(x^2+2x+y\right)\)
c) \(\left(x+3y\right)^2\)
d) \(\left(4x-y\right)^3\)
e) \(\left(x^2-2y\right)\left(x^2+2y\right)\)
g) \(18x^4y^2z:10x^4y\)
h) \(\left(x^3y^3+\dfrac{1}{2}x^2y^3-x^3y^2\right):\dfrac{1}{3}x^2y^2\)
i) \(\left(6x^3-7x^2-x+2\right):\left(2x+1\right)\)
k) \(\dfrac{5x-1}{3x^2y}+\dfrac{x+1}{3x^2y}\)
l) \(\dfrac{3x+1}{x^2-3x+1}+\dfrac{x^2-6x}{x^2-3x+1}\)
m) \(\dfrac{2x+3}{10x-4}+\dfrac{5-3x}{4-10x}\)
n) \(\dfrac{x}{x^2+2x+1}+\dfrac{3}{5x^2-5}\)
o) \(\dfrac{x^2+2}{x^3-1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\)
p) \(\dfrac{4x+2}{15x^3y}\dfrac{5y-3}{9x^2y}+\dfrac{x+1}{5xy^3}\)
q) \(\dfrac{2x-7}{10x-4}-\dfrac{3x+5}{4-10x}\)
r) \(\dfrac{3}{2x+6}-\dfrac{x-6}{2x^2+6x}\)
x) \(\dfrac{4y^2}{11x^4}.\left(-\dfrac{3x^2}{8y}\right)\)
y) \(\dfrac{x^2-4}{3x+12}.\dfrac{x+4}{2x-4}\)
z) \(\left(x^2-25\right):\dfrac{2x+10}{3x-7}\)
t) \(\left(\dfrac{2x+1}{2x-1}-\dfrac{2x-1}{2x+1}\right):\dfrac{4x}{10x-5}\)
w) \(\left(\dfrac{1}{x^2+x}-\dfrac{2-x}{x+1}\right):\left(\dfrac{1}{x}+x-2\right)\)
c: \(=x^2+6xy+9y^2\)
e: \(=x^4-4y^2\)
Giải hệ phương trình:
1. \(\left\{{}\begin{matrix}x+3=2\sqrt{\left(3y-x\right)\left(y+1\right)}\\\sqrt{3y-2}-\sqrt{\dfrac{x+5}{2}}=xy-2y-2\end{matrix}\right.\)
2. \(\left\{{}\begin{matrix}\sqrt{2y^2-7y+10-x\left(y+3\right)}+\sqrt{y+1}=x+1\\\sqrt{y+1}+\dfrac{3}{x+1}=x+2y\end{matrix}\right.\)
3. \(\left\{{}\begin{matrix}\sqrt{4x-y}-\sqrt{3y-4x}=1\\2\sqrt{3y-4x}+y\left(5x-y\right)=x\left(4x+y\right)-1\end{matrix}\right.\)
4. \(\left\{{}\begin{matrix}9\sqrt{\dfrac{41}{2}\left(x^2+\dfrac{1}{2x+y}\right)}=3+40x\\x^2+5xy+6y=4y^2+9x+9\end{matrix}\right.\)
5. \(\left\{{}\begin{matrix}\sqrt{xy+\left(x-y\right)\left(\sqrt{xy}-2\right)}+\sqrt{x}=y+\sqrt{y}\\\left(x+1\right)\left[y+\sqrt{xy}+x\left(1-x\right)\right]=4\end{matrix}\right.\)
6. \(\left\{{}\begin{matrix}x^4-x^3+3x^2-4y-1=0\\\sqrt{\dfrac{x^2+4y^2}{2}}+\sqrt{\dfrac{x^2+2xy+4y^2}{3}}=x+2y\end{matrix}\right.\)
7. \(\left\{{}\begin{matrix}x^3-12z^2+48z-64=0\\y^3-12x^2+48x-64=0\\z^3-12y^2+48y-64=0\end{matrix}\right.\)
Rút gọn các phân thức sau:
a) \(\dfrac{6x^2y^2}{8xy^{ }5}\)
b) \(\dfrac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}\)
c) \(\dfrac{2x^2+2x
}{x+1}\)
d) \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)
e) \(\dfrac{36\left(x-2\right)^3}{32-16x}\)
a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)
b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)
c) \(=\dfrac{2x\left(x+1\right)}{x+1}=2x\)
d) \(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)
Rút gọn rồi tính giá trị của biểu thức khi x=1;y=\(-3\frac{1}{4}\)
\(\frac{\left(x-y\right)^2+xy}{\left(x+y\right)^2-xy}\)\(\left[1:\frac{x^5+y^5+x^3y^2+x^2y^3}{\left(x^3y^3\right)\left(x^3+y^3+x^2y+xy^2\right)}\right]\)
Chứng minh đẳng thức:
a)\(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)=x^5-y^5\)
b)\(\left(a+b\right)\left(a^2-ab+b^2\right)=a^3+b^3\)
Cảm ơn mn trước nha!
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
R/gọn: \(\left(\dfrac{x^2}{x+y}+y\right).\left(\dfrac{1}{x^2-xy}-\dfrac{3y^2}{x^4-xy^3}-\dfrac{y}{x^3+x^2y+xy^2}\right)\)
\(\left(\dfrac{x^2}{x+y}+y\right).\left(\dfrac{1}{x^2-xy}-\dfrac{3y^3}{x^4-xy^3}-\dfrac{y}{x^3+x^2y+xy^2}\right)\)
\(=\left(\dfrac{x^2+xy+y^2}{x+y}\right).\left(\dfrac{1}{x\left(x-y\right)}-\dfrac{3y^2}{x\left(x^3-y^3\right)}-\dfrac{y}{x\left(x^2+xy+y^2\right)}\right)\)\(=\left(\dfrac{x^2+xy+y^2}{x+y}\right).\left(\dfrac{x^2+xy+y^2}{x\left(x^3-y^3\right)}-\dfrac{3y^2}{x\left(x^3-y^3\right)}-\dfrac{xy-y^2}{x\left(x^3-y^3\right)}\right)\)
\(=\dfrac{x\left(x^3-y^3\right)}{x^3-xy^2}.\dfrac{x^2+xy+y^2-3y^2-xy+y^2}{x\left(x^3-y^3\right)}\\ =\dfrac{x^2-y^2}{x\left(x^2-y^2\right)}=\dfrac{1}{x}\)
mình viết trên máy tinh hơi xấu bạn thông cảm nhé!!!Nếu ko chê có thể xem cách giải này!
Chứng minh đẳng thức:
a) \(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)=x^5-y^5\)
(x-y)(x^4+x^3y+x^2y^2+xy^3+y^4)
= 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^4y-x^4y)+(x^3y^2-x^3y^2)+(x^2y^3)+(xy^4-xy^4)+x^5-y^5
= 0+0+0+0+x^5-y^5
= x^5-y^5
Vay (x-y)(x^4+x^3y+x^2y^2+xy^3+y^4) = x^5-y^5
