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\(Taco:\)
\(\left(x+y\right)\left(y+z\right)=187\Leftrightarrow xy+xz+yy+yz=187\)
\(\left(y+z\right)\left(z+x\right)=154\Leftrightarrow yz+xy+zz+xz=154\)
\(\left(z+x\right)\left(x+y\right)=238\Leftrightarrow xz+zy+xx+xy=238\)
\(\Rightarrow\left(x+y\right)\left(y+z\right)+\left(x+z\right)\left(x+y\right)+\left(y+z\right)\left(z+x\right)=579\)
\(\Leftrightarrow xy+zx+yy+yz+yz+xy+zz+xz+xz+zy+xx+xy=579\)
\(\Leftrightarrow3\left(xz+xy+yz\right)+x^2+y^2+z^2=579\)
\(\left(z+x\right)\left(x+y\right)-\left(x+y\right)\left(y+z\right)=51\)
\(\Leftrightarrow\left(x+y\right)\left(x-y\right)=x^2-y^2=51\)
\(\left(z+x\right)\left(x+y\right)-\left(y+z\right)\left(x+z\right)=84\)
\(\Leftrightarrow\left(x+z\right)\left(x-z\right)=84\Leftrightarrow x^2-z^2=84\)
\(\Leftrightarrow y^2-z^2=33\)
đến đây tịt
Giải:
Đặt: (x + y) = a ; (y + z) = b ; (z + x) = c
HPT <=> \(\left\{{}\begin{matrix}ab=187\\bc=154\\ca=238\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{187}{a}\\\dfrac{187}{a}\cdot c=154\\c\cdot a=238\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}b=\dfrac{187}{a}\\c=\dfrac{154a}{187}\\\dfrac{154a}{187}\cdot a=238\end{matrix}\right.\) => \(154a^2=238\cdot187=44506\)
=> \(a^2=\dfrac{44506}{154}=289\Rightarrow a=\sqrt{289}=17\)
=> b = \(\dfrac{187}{17}=11\) ; c = \(\dfrac{238}{17}=14\)
Hay \(\left\{{}\begin{matrix}x+y=17\\y+z=11\\z+x=14\end{matrix}\right.\)
\(\Rightarrow x+y+y+z+z+x-17+11+14=42\)
\(\Leftrightarrow2\left(x+y+z\right)=42\Rightarrow x+y+z=21\)
=> \(\left\{{}\begin{matrix}x=21-\left(y+z\right)=21-11=10\\y=21-\left(z+x\right)=21-14=7\\z=21-\left(x+y\right)=21-17=4\end{matrix}\right.\)
Vậy ..........................
Đặt x + y = a ( a > 0 )
y + z = b ( b > 0 )
x + z = c (c > )
Khi đó hệ pt thành :
\(\left\{{}\begin{matrix}ab=187\left(1\right)\\bc=154\left(2\right)\\ac=238\left(3\right)\end{matrix}\right.\)
Nhân (1) (2) (3) vế theo vế được: abc = 2618 (4)
Lần lượt chia (4) cho (1) (2) (3) ta được:
\(\left\{{}\begin{matrix}a=17\\b=11\\c=14\end{matrix}\right.\) hay \(\left\{{}\begin{matrix}x+y=17\\y+z=11\\x+z=14\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-z=6\\x+z=14\end{matrix}\right.\Leftrightarrow x=10\Rightarrow y=7\) và \(z=4\)
Vậy nghiệm của hệ pt là (10;7;4)
Ta có \(\hept{\begin{cases}\text{(x+y)(y+z)=187}\\\text{(y+z)(z+x)=154}\\\text{(z+x)(x+y)=238}\end{cases}}\)\(\Rightarrow\)(x+y)2(y+z)2(z+x)2=187.154.238 \(\Rightarrow\) (x+y)(y+z)(z+x)=2618
\(\Rightarrow\)\(\hept{\begin{cases}z+x=14\\x+y=17\\y+z=11\end{cases}}\) \(\Rightarrow\) 2(x+y+z)=14+17+11=42 \(\Rightarrow\) x+y+z=21 \(\Rightarrow\) \(\hept{\begin{cases}y=7\\z=4\\x=10\end{cases}}\)
đặt x+y=a,y+z=b,z+y=c
hPt trở thành :ab=187,bc=154,ca=238
nhân hết 3 vế với nhau:\(a^2b^2c^2=6853924\)
Suy ra \(abc=2613\)nên c=abc:ab=2613:187=14.b và c tính tương tự
trở về ẩn cũ r giải nốt đi
(1)+(3)-(2) \(\Rightarrow x\left(x+y+z\right)=24\) (4)
\(\left(1\right)+\left(2\right)-\left(3\right)\Rightarrow y\left(x+y+z\right)=48\) (5)
\(\left(2\right)+\left(3\right)-\left(1\right)\Rightarrow z\left(x+y+z\right)=72\) (6)
Cộng vế với vế: \(\Rightarrow\left(x+y+z\right)^2=144\Rightarrow\left[{}\begin{matrix}x+y+z=12\\x+y+z=-12\end{matrix}\right.\)
- Với \(x+y+z=12\) (7) lần lượt chia vế cho vế cho (4); (5); (6) cho (7)
- Với \(x+y+z=-12\) (8) lần lượt chia vế cho vế của (4); (5); (6) cho (8)
b, Ta có : \(\left\{{}\begin{matrix}x+y+z=3\\y+z+t=4\\z+t+x=5\\t+x+y=6\end{matrix}\right.\)
=> \(x+y+z+y+z+t+z+t+x+t+x+y=18\)
=> \(3\left(x+y+z+t\right)=18\)
=> \(x+y+z+t=6\)
=> \(x+y+z+t=x+y+t\)
=> \(z=0\)
=> \(\left\{{}\begin{matrix}x+y=3\\y+t=4\\x+t=5\\x+y+t=6\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x+y=3\\y+t=4\\x+t=5\\y+5=6\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x+1=3\\t+1=4\\x+t=5\\y=1\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=2\\t=3\\x+t=5\\y=1\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=2\\y=1\\z=0\\t=3\end{matrix}\right.\)
a, Ta có : \(\left\{{}\begin{matrix}7xy=12\left(x+y\right)\\9yz=20\left(y+z\right)\\8zx=15\left(z+x\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}7xy-12x-12y=0\\9yz-20y-20z=0\\8zx-15z-15x=0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=\frac{12y}{7y-12}\\y=\frac{20z}{9z-20}\\x=\frac{15z}{8z-15}\end{matrix}\right.\)
=> \(12y\left(8z-15\right)=15z\left(7y-12\right)\)
=> \(96yz-180y=105yz-180z\)
=> \(105yz-96yz=-180y+180z\)
=> \(9yz=-180y+180z\)
=> \(180z-180y=20y+20z\)
=> \(180z-20z=180y+20y=160z=200y\)
=> \(y=\frac{4}{5}z\)
=> \(\frac{20z}{9z-20}=\frac{4z}{5}\)
=> \(4z\left(9z-20\right)=100z\)
=> \(36z^2-180z=0\)
=> \(\left[{}\begin{matrix}z=5\\z=0\end{matrix}\right.\)
TH1 : z = 0 .
=> \(x=y=z=0\)
TH2 : z = 5 .
=> \(\left\{{}\begin{matrix}7xy=12\left(x+y\right)\\45y=20\left(y+5\right)\\40x=15\left(5+x\right)\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}y=4\\x=3\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=3\\y=4\\z=5\end{matrix}\right.\)
