Ta có:
\(\left\{\begin{matrix}x\left(x+y\right)=90\left(1\right)\\y\left(y+x\right)=54\left(2\right)\end{matrix}\right.\)
Cộng từng vế hai \(BĐT\left(1\right);\left(2\right)\) ta có:
\(x\left(x+y\right)+y\left(y+x\right)=90+54\)
\(\Rightarrow\left(x+y\right)^2=144\)
\(\Rightarrow x+y=\sqrt{144}\)
\(\Rightarrow x+y=\pm12\)
Nếu \(x+y=12\)
\(\Rightarrow\left\{\begin{matrix}x\left(x+y\right)=90\Leftrightarrow x=\frac{15}{2}\\y\left(y+x\right)=54\Leftrightarrow y=\frac{9}{2}\end{matrix}\right.\)
\(\Rightarrow A=\left|x+y\right|=\left|\frac{15}{2}+\frac{9}{2}\right|=\left|\frac{24}{12}\right|=\left|12\right|=12\)
Nếu \(x+y=-12\)
\(\Rightarrow\left\{\begin{matrix}x\left(x+y\right)=90\Leftrightarrow x=\frac{-15}{2}\\y\left(y+x\right)=54\Leftrightarrow y=\frac{-9}{2}\end{matrix}\right.\)
\(\Rightarrow A=\left|x+y\right|=\left|\frac{-15}{2}+\frac{-9}{2}\right|=\left|\frac{-24}{2}\right|=\left|-12\right|=12\)
Vậy \(A=\left|x+y\right|=12\)
Ta có: x(x+y)=9 (1)
y(x+y)=54 (2)
cộng từng vế 2 BĐT (1);(2) ta có:
x(x+y)+y(x+y)=90+54
=>(x+y)2=144
=>
TH1:x+y=12
Tacó:x(x+y)=90=>x=90:12=15/2
y(x+y)=54=>y=54:12=9/2
khi đó A=|x+y|=|15
2 +
9
2 |=|
24
12 |=12 (thỏa mãn)
TH2:x+y=-12
Ta có x(x+y)=90=>x=90:(-12)=-15/2
y(x+y)=54=>y=54:(-12)=-9/2
khi đó A=|x+y|=|
−15
2 +
−9
2 |=|−
24
2 |=|−12|=12 (thỏa mãn)
Vậy |x+y|=12
Có : x(x+y)=90 ; y(y+x)=54
=> x(x+y)+y(y+x)=90+54=144
=>(x+y)^2=144
=>x+y=\(\pm\)12
=>\(\left|x+y\right|\)=12
SÁNG NAY MIK VỪA THI XONG !
