\(C=2x-2y+13x^3y^2\left(x-y\right)+15\left(y^2x-x^2y\right)+\left(\dfrac{2015}{2016}\right)^0bitx-y=0\)
mong các bạn giúp minh
Tính: \(C=2x-2y+13x^3y^2.\left(x-y\right)+15\left(y^2x-x^2y\right)+\left(\dfrac{2015}{2016}\right)^0\)
biết x - y = 0
Ta có:
\(C=2\left(x-y\right)+13x^3y^2\left(x-y\right)-15xy\left(x-y\right)+1\)
=\(0+0+0+1=1\)
\(C=2x-2y+13x^3y^2\left(x-y\right)+15\left(y^2x-x^2y\right)+\left(\dfrac{2015}{2016}\right)^0\)
\(=2\left(x-y\right)+13x^3y^2\left(x-y\right)-15xy\left(x-y\right)\)
\(=0+0+1=1\)
~^~
\(C=2x-2y+13x^3y^2\left(x-y\right)+15\left(y^2x-x^2y\right)+\left(\frac{2015}{2016}\right)^0\)
\(C = 2.(x-y)+13x^3y^2.(x-y)+15.xy.\)
\((y-x) +1\)
\(C = 2.( x- y )+13x^3y^2.(x-y)-15.xy.\)
\(( x - y )+1\)
\(C = (x - y)(2 + 13x^3y^2 - 15 ) +1\)
\(C =(x- y)(13x^3y^2 - 13 )+ 1\)
Tính giá trị của biểu thức sau:
C = \(2x-2y+13x^3y^2\left(x-y\right)+15\left(y^2x-x^2y\right)+\left(\dfrac{2015}{2016}\right)^0\), biết \(x-y=0\)
Ta có:\(C=2\left(x-y\right)+13x^3y^2\left(x-y\right)+15xy\left(y-x\right)+1\)Thế \(x-y=0\) vào C ta được:
\(C=0+0+0+1\)
C = 0
Giải các hệ phương trình sau:a) \(\left\{{}\begin{matrix}\left(2x-y\right)^2-6x+3y=0\\x+2y=0\end{matrix}\right.\);b) \(\left\{{}\begin{matrix}\sqrt{\dfrac{2x-y}{x+y}}+\sqrt{\dfrac{x+y}{2x-y}}=2\\3x+y=14\end{matrix}\right.\)
a.
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)^2-3\left(2x-y\right)=0\\x+2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(2x-y\right)\left(2x-y-3\right)=0\\x+2y=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}2x-y=0\\x+2y=0\end{matrix}\right.\\\left\{{}\begin{matrix}2x-y-3=0\\x+2y=0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\\\left\{{}\begin{matrix}x=\dfrac{6}{5}\\y=-\dfrac{3}{5}\end{matrix}\right.\end{matrix}\right.\)
b.
ĐKXĐ: \(\dfrac{2x-y}{x+y}>0\)
Đặt \(\sqrt{\dfrac{2x-y}{x+y}}=t>0\) pt đầu trở thành:
\(t+\dfrac{1}{t}=2\Leftrightarrow t^2-2t+1=0\)
\(\Leftrightarrow t=1\Leftrightarrow\sqrt{\dfrac{2x-y}{x+y}}=1\)
\(\Leftrightarrow2x-y=x+y\Leftrightarrow x=2y\)
Thay xuống pt dưới:
\(6y+y=14\Rightarrow y=2\)
\(\Rightarrow x=4\)
a, \(\text{[}\left(x-y\right)^3+3\left(x-y\right)\text{]}:\dfrac{1}{3}\left(x-y\right)\)
b, \(\left(8x^3-27y^3\right):\left(2x-3y\right)\)
c, \(\text{[}5\left(x+2y\right)^6-6\left(x+2y\right)^5\text{]}:2\left(x+2y\right)^4\)
a: \(=\left(x-y\right)^3:\dfrac{1}{3}\left(x-y\right)+3\left(x-y\right):\dfrac{1}{3}\left(x-y\right)\)
=3(x-y)^2+9
b: \(=\dfrac{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}{2x-3y}=4x^2+6xy+9y^2\)
c: \(=\dfrac{5\left(x+2y\right)^6}{2\left(x+2y\right)^4}-\dfrac{6\left(x+2y\right)^5}{2\left(x+2y\right)^4}=\dfrac{5}{2}\left(x+2y\right)^2-3\left(x+2y\right)\)
cho các số x,y thỏa mãn đẳng thức \(3x^2+3y^2+4xy+2x-2y+2=0\\ \)
tính giá trị biểu thức M=\(\left(x+y\right)^{2016}+\left(x+2\right)^{2017}+\left(y-1\right)^{2018}\)
Ta có: \(3x^2+3y^2+4xy+2x-2y+2=0\)
\(\Leftrightarrow x^2+2x+1+y^2-2y+1+2x^2+4xy+2y^2=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+2\left(x^2+2xy+y^2\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(y-1\right)^2+2\left(x+y\right)^2=0\)
Ta có: \(\left(x+1\right)^2\ge0\forall x\)
\(\left(y-1\right)^2\ge0\forall y\)
\(2\left(x+y\right)^2\ge0\forall x,y\)
Do đó: \(\left(x+1\right)^2+\left(y-1\right)^2+2\left(x+y\right)^2\ge0\forall x,y\)
Dấu '=' xảy ra khi
\(\left\{{}\begin{matrix}x+1=0\\y-1=0\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=1\\-1+1=0\left(đúng\right)\end{matrix}\right.\)
Thay x=-1 và y=1 vào biểu thức \(M=\left(x+y\right)^{2016}+\left(x+2\right)^{2017}+\left(y-1\right)^{2018}\), ta được:
\(M=\left(-1+1\right)^{2016}+\left(-1+2\right)^{2017}+\left(1-1\right)^{2018}\)
\(=0^{2016}+1^{2017}+0^{2018}=1\)
Vậy: M=1
giải hệ pt :
a,\(\left\{{}\begin{matrix}x^3y\left(1+y\right)+x^2y^2\left(2+y\right)+xy^3-30=0\\x^2y+x\left(1+y+y^2\right)+y-11=0\end{matrix}\right.\)
b,\(\left\{{}\begin{matrix}xy^2-2y+3x^2=0\\y^2+x^2y+2x=0\end{matrix}\right.\)
c,\(\left\{{}\begin{matrix}3xy+2y=5\\2xy\left(x+y\right)+y^2=5\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^3y^2+x^2y^3+x^3y+2x^2y^2+xy^3-30=0\\x^2y+xy^2+xy+x+y-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2y^2\left(x+y\right)+xy\left(x+y\right)^2-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy\left(x+y\right)\left[xy+x+y\right]-30=0\\xy\left(x+y\right)+xy+x+y-11=0\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}xy\left(x+y\right)=u\\xy+x+y=v\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}uv-30=0\\u+v-11=0\end{matrix}\right.\) \(\Rightarrow\left(u;v\right)=\left(6;5\right);\left(5;6\right)\)
TH1: \(\left\{{}\begin{matrix}xy\left(x+y\right)=6\\xy+x+y=5\end{matrix}\right.\)
Theo Viet đảo \(\Rightarrow\left\{{}\begin{matrix}x+y=3\\xy=2\end{matrix}\right.\) \(\Rightarrow\left(x;y\right)=\left(1;2\right);\left(2;1\right)\)hoặc \(\left\{{}\begin{matrix}x+y=2\\xy=3\end{matrix}\right.\)(vô nghiệm)
TH2: \(\left\{{}\begin{matrix}xy\left(x+y\right)=5\\xy+x+y=6\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y=5\\xy=1\end{matrix}\right.\) \(\Rightarrow...\) hoặc \(\left\{{}\begin{matrix}x+y=1\\xy=5\end{matrix}\right.\) (vô nghiệm)
2 câu dưới hình như em hỏi rồi?
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: \(=x^4-4y^2\)
khai triển các biểu thức sau:
\(a.\left(2x+3y\right)^2\)
\(b.2\left(\dfrac{1}{2}x^2+y\right)\left(x^2-2y\right)\)
\(c.\left(x+y+z\right)^2\)
a. (2x+3y)2= (2x)2+2.2x.3y+(3y)2
=4x2+12xy+9y2
b. 2(\(\dfrac{1}{2}\)x2+y)(x2-2y)
=(x2+2y)(x2-2y)
=x4-4y2
c, (x+y+z)2= [(x+y)+z]2
=(x+y)2+2(x+y)z+z2
=x2+2xy+y2+2xz+2yz+z2
=x2+y2+z2+2xy+2yz+2xz