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\(C=x^2-y^2\)
Tương tự câu \(A=x^2+y^2\)
\(D=x^4+y^4\)
Thay x + y = 17; x.y = 60 vào \(\left(x+y\right)^2=x^2+2xy+y^2\):
172 = x2 + 2.60 + y2
289 = x2 + 120 + y2
\(\Leftrightarrow x^2+y^2=169\)
Lại có:
\(\left(x^2+y^2\right)^2=x^4+y^4+2x^2y^2\)
\(\left(x^2+y^2\right)^2=x^4+y^4+\left(2xy\right)^2\)
Thay \(x^2+y^2=169;x.y=60\)vào biểu thức trên:
1692 = x4 + y4 + 2 . 602
\(\Leftrightarrow x^4+y^4=28561-7200\)
\(\Leftrightarrow x^4+y^4=21361\)
C1: Ta có: \(x-y=7\Leftrightarrow\left(x-y\right)^2=49\Leftrightarrow x^2-2xy+y^2=49\Leftrightarrow x^2+y^2=49+2xy=49+2.60=169\)
=>\(B=x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=7\left(169+60\right)=7.229=1603\)
C2: \(B=x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=\left(x-y\right)\left[\left(x-y\right)^2+3xy\right]=7\left(7^2+3.60\right)=7.229=1603\)
\(\frac{x^2+y^2}{xy}=\frac{10}{3}\Rightarrow3x^2+3y^2-10xy=0\)
\(\Rightarrow\left(3x^2-9xy\right)-\left(xy-3y^2\right)=0\Rightarrow3x\left(x-3y\right)-y\left(x-3y\right)=0\)
\(\Rightarrow\left(x-3y\right)\left(3x-y\right)=0\Rightarrow3x-y=0\left(y>x>0\Rightarrow x-3y< 0\right)\Rightarrow3x=y\)
\(M=\frac{x-y}{x+y}=\frac{x-3x}{x+3x}=\frac{-2x}{4x}=-\frac{1}{2}\)
Ta có:
\(x-y=7\)
\(\Leftrightarrow\left(x-y\right)^2=49\)
\(\Leftrightarrow x^2-2xy+y^2=49\)
\(\Leftrightarrow x^2+y^2-2.60=49\)
\(\Leftrightarrow x^2+y^2-120=49\)
\(\Leftrightarrow x^2+y^2=169\)\
Vậy...
Có:\(x+y=30\Rightarrow\left(x+y\right)^2=900\Rightarrow x^2+y^2+2xy=900\Rightarrow x^2+y^2=900-2.216=468\)(Vì xy=216)
Lại có: \(\left(x-y\right)^2=x^2+y^2-2xy=468-2.216=0\Rightarrow x-y=0\)
\(A=x^2-y^2=\left(x+y\right)\left(x-y\right)=30.0=0\)
Lời giải:
Ta có:
$C=x^2-y^2=(x-y)(x+y)=7(x+y)=7\sqrt{(x+y)^2}$
$=7\sqrt{(x-y)^2+4xy}=7\sqrt{7^2+4.60}=119$
$D=x^4+y^4=(x^2-y^2)^2+2(xy)^2=C^2+2(xy)^2=119^2+2.60^2=21361$
Làm lại : \(x-y=7\Rightarrow x^2+y^2=49+2xy=49+120=169\)
Ta có : \(x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right)=7.\left(169+60\right)=1603\)
Ta có:\(\left(x-y\right)^2+2xy=x^2-2xy+y^2+2xy=x^2+y^2\)
\(\Rightarrow x^2+y^2=\left(x-y\right)^2+2xy\)
\(=7^2+2.60=49+120=169\)
\(A=\left(x-y\right)\left(x+y\right)=7\left(x+y\right)\)
Có \(\left(x-y\right)^2=49\)
\(\Leftrightarrow x^2+y^2-2xy=49\)
\(\Leftrightarrow\left(x^2+y^2+2xy\right)-4xy=49\)
\(\Leftrightarrow\left(x+y\right)^2=289\)
\(\Leftrightarrow x+y=17\)
\(\Rightarrow A=7.17=119\)
Vậy ....