Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
|3x-4|+|6y-8|+|z+7|=0
=>3x-4=0 hoặc 6y-8=0 hoặc z+7=0
=>x=4:3;y=8:6;z=-7
\(x^7+x^5+1\)
\(=x^7+x^6+x^5-x^6-x^5-x^4+x^5+x^4+x^3-x^3-x^2-x\) \(+x^2+x+1\)
\(=\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+\left(x^5+x^4+x^3\right)-\) \(\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\)
\(=x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)\) \(+\left(x^2+x+1\right)\)
\(=\)\(\left(x^5-x^4+x^3-x+1\right)\left(x^2+x+1\right)\)
= (x-y)(x+y)-(x+y)
=(x+y)(x-y-1)
học tốt
Để \(\left|x+3\right|+\left(y-4\right)^2+\left|z-9\right|=0\)
\(\Leftrightarrow\hept{\begin{cases}\left|x+3\right|=0\\\left(y-4\right)^2=0\\\left|z-9\right|=0\end{cases}\Leftrightarrow\hept{\begin{cases}x+3=0\\y-4=0\\z-9=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-3\\y=4\\z=9\end{cases}}}\)
| x +3 | + (y-4)2 + | z - 9| = 0
Do | x + 3 | \(\ge\)0 \(\forall\)x
( y - 4)2 \(\ge\)0 \(\forall\)y
| z - 9|\(\ge\)0 \(\forall\)z
\(\Rightarrow\) | x+3 | + ( y-4 )2 + | z-9 | \(\ge\)0 \(\forall\)x,y,z
Dấu '' = '' xảy ra khi :
\(\hept{\begin{cases}\\\\\end{cases}}\)| x+3| = 0 ( y-4 )2 = 0 | z-9 | =0
\(\hept{\begin{cases}\\\\\end{cases}}\)x + 3 = 0 ; y -4 = 0 ; z - 9 = 0
\(\hept{\begin{cases}\\\\\end{cases}}\)x = -3 ; y = 4 ; z = 9
Vậy x = -3, y = 4, z = 9
\(x^2+4x+3=0\)
\(\Leftrightarrow x^2+x+3x+3=0\)
\(\Leftrightarrow x.\left(x+1\right)+3.\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right).\left(x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=-3\end{cases}}}\)
Vậy x = -1 hoặc x = -3
thêm x2 + y2 + z2 = 1 nha
HT nha vinh
1) \(x^2-x-y^2-y=\left(x^2-y^2\right)-\left(x+y\right)=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\)
\(x^2-2xy+y^2-z^2=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\)
2)\(5x-5y+ax-ay=5\left(x-y\right)+a\left(x-y\right)=\left(x-y\right)\left(a+5\right)\)
\(a^3-a^2x-ay+xy=a^2\left(a-x\right)-y\left(a-x\right)=\left(a-x\right)\left(a^2-y\right)\)