Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: \(x^2-y+\frac{1}{4}=y^2-x+\frac{1}{4}=0\)
\(\Rightarrow\left(x^2-x+\frac{1}{4}\right)+\left(y^2-y+\frac{1}{4}\right)=0\)
\(\Rightarrow\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}x-\frac{1}{2}=0\\y-\frac{1}{2}=0\end{cases}\Rightarrow}x=y=\frac{1}{2}\)
Vậy \(x=y=\frac{1}{2}\)
\(2x^2+y^2+9=6x+2xy\Leftrightarrow\left(x^2-2xy+y^2\right)+\left(x^2-6x+9\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(x-3\right)^2=0\Leftrightarrow\hept{\begin{cases}x-3=0\\x-y=0\end{cases}}\Leftrightarrow x=y=3\)
\(\Rightarrow A=x^{2019}.y^{2020}-x^{2020}.y^{2019}+\frac{1}{9xy}=\frac{1}{27}\)
\(x^2+y^2=1\Leftrightarrow\frac{^4}{a}+\frac{y^4}{b}=\frac{x^2+y^2}{a+b}\)
Theo tính chất tỉ lệ thức
\(\frac{x^2+y^2}{a+b}=\frac{x^2}{a}=\frac{y^2}{b}\left(a;b\ne0\right)\)
\(\frac{x^{2012}}{a^{1006}}+\frac{y^{2012}}{b^{1006}}=\left(\frac{x^2}{a}\right)^{1006}+\left(\frac{y^2}{b}\right)^{1006}=2.\left(\frac{x^2+y^2}{a+b}\right)^{2006}=\frac{2}{\left(a+b\right)^{2006}}\left(đpcm\right)\)
Bài 4 :
Thay x=y+5 , ta có :
a ) ( y+5)*(y5+2)+y*(y-2)-2y*(y+5)+65
=(y+5)*(y+7)+y^2-2y-2y^2-10y+65
=y^2+7y+5y+35-y^2-2y-2y^2-10y+65
= 100
Bài 5 :
A = 15x-23y
B = 2x-3y
Ta có : A-B
= ( 15x -23y)-(2x-3y)
=15x-23y-2x-3y
=13x-26y
=13x*(x-2y) chia hết cho 13
=> Nếu A chia hết cho 13 thì B chia hết cho 13 và ngược lại
1/ Ta có : P\left(x\right)=-x^2+13x+2012=-\left(x-\frac{13}{2}\right)^2+\frac{8217}{4}\le\frac{8217}{4}P(x)=−x2+13x+2012=−(x−213)2+48217≤48217
Dấu "=" xảy ra khi x = 13/2
Vậy Max P(x) = 8217/4 tại x = 13/2
1/ Ta có : P\left(x\right)=-x^2+13x+2012=-\left(x-\frac{13}{2}\right)^2+\frac{8217}{4}\le\frac{8217}{4}P(x)=−x2+13x+2012=−(x−213)2+48217≤48217
Dấu "=" xảy ra khi x = 13/2
Vậy Max P(x) = 8217/4 tại x = 13/2
2/ Ta có : x^3+3xy+y^3=x^3+3xy.1+y^3=x^3+y^3+3xy\left(x+y\right)=\left(x+y\right)^3=1x3+3xy+y3=x3+3xy.1+y3=x3+y3+3xy(x+y)=(x+y)3=1
3/ a+b+c=0\Leftrightarrow\left(a+b+c\right)^2=0\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ac\right)=0a+b+c=0⇔(a+b+c)2=0⇔a2+b2+c2+2(ab+bc+ac)=0
\Leftrightarrow ab+bc+ac=-\frac{1}{2}⇔ab+bc+ac=−21 \Leftrightarrow\left(ab+bc+ac\right)^2=\frac{1}{4}\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=\frac{1}{4}⇔(ab+bc+ac)2=41⇔a2b2+b2c2+c2a2+2abc(a+b+c)=41
\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}⇔a2b2+b2c2+c2a2=41(vì a+b+c=0)
Ta có : a^2+b^2+c^2=1\Leftrightarrow\left(a^2+b^2+c^2\right)^2=1\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1a2+b2+c2=1⇔(a2+b2+c2)2=1⇔a4+b4+c4+2(a2b2+b2c2+c2a2)=1
\Leftrightarrow a^4+b^4+c^4=1-2\left(a^2b^2+b^2c^2+c^2a^2\right)=1-\frac{2.1}{4}=\frac{1}{2}⇔a4+b4+c4=1−2(a2b2+b2c2+c2a2)=1−42.1=21
\(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\Leftrightarrow\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(a^2+b^2\right)^2}{a+b}\)
\(\Leftrightarrow\frac{x^4b+y^4a}{ab}=\frac{\left(x^4+y^4+2x^2y^2\right)}{a+b}\Rightarrow x^4ab+x^4b^2+y^4ab+y^4a^2=x^4ab+y^4ab+2x^2y^2ab\)
\(\Leftrightarrow x^4b^2+y^4a^2-2x^2y^2ab=0\Leftrightarrow\left(x^2b-y^2a\right)^2=0\Leftrightarrow x^2b=y^2a\Leftrightarrow\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}\)
\(\Rightarrow\frac{x^{2010}}{a^{1006}}+\frac{y^{2012}}{b^{1006}}=\frac{2\left(x^2+y^2\right)^{1006}}{\left(a+b\right)^{1006}}=\frac{2}{\left(a+b\right)^{1006}}\)
Ta có:
\(\frac{x^4}{a}+\frac{y^4}{b}\ge\frac{\left(x^2+y^2\right)^2}{a+b}=\frac{1}{a+b}\)
Dấu = xảy ra khi .... Làm tiếp nhé
ta có: \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\)=> \(\frac{bx^4+ay^4}{ab}=\frac{\left(x^2+y^2\right)^2}{a+b}\) (vì x^2 +y^2 =1)
=>\(abx^4+b^2x^4+aby^4+a^2y^4\) = \(ab\left(x^4+2x^2y^2+y^4\right)\)
=>\(abx^4+b^2x^4+aby^4+a^2y^4\) = \(abx^4+2abx^2y^2+aby^4\)
=> \(b^2x^4-2abx^2y^2+a^2y^4=0\)
=>\(\left(bx^2-ay^2\right)^2=0\)=>\(bx^2=ay^2\Rightarrow\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)
=> \(\frac{x^{2012}}{a^{1006}}=\frac{1}{\left(a+b\right)^{1006}}\) và \(\frac{y^{2012}}{b^{1006}}=\frac{1}{\left(a+b\right)^{1006}}\)
=>\(\frac{x^{2012}}{a^{1006}}+\frac{y^{2012}}{b^{1006}}=\frac{2}{\left(a+b\right)^{1006}}\)