\(\left(x+1\right)+\left(x+2\right)+....+\left(x+100\right)=50\)

\(100x+\left(1+2+3+....+100\right)=50\)

\(100x+5050=50\)

\(100x=50-5050\)

\(x=\left(-5000\right):100\)

\(x=-50\)

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 .

=> \(\left\{{}\begin{matrix}7xy=12\left(x+y\right)\\45y=20\left(y+5\right)\\40x=15\left(5+x\right)\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=4\\x=3\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x=3\\y=4\\z=5\end{matrix}\right.\)

*Lâu lâu mới làm 1 câu...nếu sai xin thông cảm...> . < ...*

Do \(\overline{X}=\dfrac{140}{2}=7\) nên :

\(\dfrac{5\cdot2+6x+7y+9\cdot3}{2+x+y+3}=\dfrac{140}{2}=7\)

\(\dfrac{37+6x+7y}{5+x+y}=\dfrac{140}{2}\)

\(\Leftrightarrow6x+7y+37=7\cdot\left(5+x+y\right)\)

\(\Leftrightarrow6x+7y+37=35+7x+7y\)

\(\Leftrightarrow6x-7x+7y-7y=35-37\)

\(\Leftrightarrow\left(-x\right)=\left(-2\right)\)

\(\Rightarrow x=2\)

\(\Rightarrow y=13\)

\(H-\left(3x^2y^2-7xy+3\right)=-5x^2y^2+7xy-y^4-5\)

=> \(H=\left(-5x^2y^2+7xy-y^4-5\right)+\left(3x^2y^2-7xy+3\right)\)

=> \(H=-2x^2y^2-y^4-2\)

Ta có \(-2x^2y^2\le0\)với mọi giá trị của x

\(-y^4\le0\)với mọi giá trị của x

=> \(-2x^2y^2-y^4-2< 0\)với mọi giá trị của x

Vậy tại mọi giá trị của x, y thì H luôn âm (đpcm)