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Bài 2:
a: =>x=0 hoặc x=-3
b: =>x-2=0 hoặc 5-x=0
=>x=2 hoặc x=5
c: =>x-1=0
hay x=1
a, |x - 3| - 5 = 7x
=> |x - 3| = 7x + 5
Đk: 7x + 5 ≥ 0 => x ≥ -5/7
Ta có: |x - 3| = 7x + 5
\(\Rightarrow\orbr{\begin{cases}x-3=7x+5\\x-3=-7x-5\end{cases}\Rightarrow}\orbr{\begin{cases}-6x=8\\8x=-2\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{-4}{3}\left(ktm\right)\\x=\frac{-1}{4}\left(tm\right)\end{cases}}\Rightarrow x=\frac{-1}{4}\)
b, 209 - |x - 209| = x
=> |x - 209| = 209 - x
Đk: 209 - x ≥ 0 => x ≤ 209
Ta có: |x - 209| = 209 - x
\(\Rightarrow\orbr{\begin{cases}x-209=209-x\\x-209=x-209\end{cases}\Rightarrow}\orbr{\begin{cases}2x=418\\0x=0\forall x\le209\end{cases}\Rightarrow\orbr{\begin{cases}x=209\\x\le209\end{cases}}}\)
=> x ≤ 209
c, (x - 1)2008 + (y - 1)2008 + |x + y + z| = 0
Vì (x - 1)2008 ≥ 0 ; (y - 1)2008 ≥ 0 ; |x + y + z| ≥ 0
=> (x - 1)2008 + (y - 1)2008 + |x + y + z| ≥ 0
Dấu " = " xảy ra <=> \(\hept{\begin{cases}\left(x-1\right)^{2008}=0\\\left(y-1\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x-1=0\\y-1=0\\x+y+z=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=1\\1+1+z=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=y=1\\z=-2\end{cases}}}\)
a, 3.2x+1-2=94
B, (3x-1)3=125
C, 2x+2x+1+...........+2x+99=2100-1
. là dấu nhân
MIK CẦN GẤP
HELP!!!!!!!!!!!!!!!!!!!!!!!!!!
\(\hept{\begin{cases}x-\left(y+z\right)=\frac{-1}{12}\\y-\left(x+z\right)=\frac{-1}{2}\\z-\left(x+y\right)=\frac{-7}{12}\end{cases}}\)
\(\Leftrightarrow-\left(x+y+z\right)=\frac{-7}{6}\)
\(\Leftrightarrow\hept{\begin{cases}-\left(x+y\right)=z-\frac{7}{6}\\-\left(x+z\right)=y-\frac{7}{6}\\-\left(y+z\right)=x-\frac{7}{6}\end{cases}}\)
Thay vô tinh tiếp, đc chứ??
Ta có : \(\hept{\begin{cases}\left|5-\frac{2}{3}x\right|\ge0\forall x\\\left|\frac{1}{7}y-3\right|\ge0\forall y\end{cases}}\Leftrightarrow\left|5-\frac{2}{3}x\right|+\left|\frac{1}{7}y-3\right|\ge0\forall x;y\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}5-\frac{2}{3}x=0\\\frac{1}{7}y-3=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{15}{2}\\y=21\end{cases}}\)
b) Ta có \(\hept{\begin{cases}\left|5x+10\right|\ge0\forall x\\\left|6y-9\right|\ge0\forall y\end{cases}}\Leftrightarrow\left|5x+10\right|+\left|6y-9\right|\ge0\forall x;y\)
Dấu "=" xảy ra <=> \(\hept{\begin{cases}5x+10=0\\6y-9=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-2\\y=1,5\end{cases}}\)
Answer:
\(P=\frac{1}{1+x+xy+xyz}+\frac{1}{1+y+yz+yzt}+\frac{1}{1+z+zt+ztx}+\frac{1}{1+t+tx+txy}\)
\(=\frac{1}{1+x+xy+xyz}+\frac{x}{x+xy+xyz+xyzt}+\frac{xy}{xy+xyz+xyzt+xyzt.x}+\frac{xyz}{xyz+xyzt+xyzt.x+xyzt.xy}\)
\(=\frac{1}{1+x+xy+xyz}+\frac{x}{x+xy+xyz+1}+\frac{xy}{xy+xyz+1+x}+\frac{xyz}{xyz+1+x+xy}\)
\(=\frac{1+x+xy+xyz}{1+x+xy+xyz}\)
\(=1\)
Ta có: \(\left(x-\dfrac{1}{5}\right)^{2004}\ge0\forall x\)
\(\left(y+0.4\right)^{100}\ge0\forall y\)
\(\left(z-3\right)^{678}\ge0\forall z\)
Do đó: \(\left(x-\dfrac{1}{5}\right)^{2004}+\left(y+0.4\right)^{100}+\left(z-3\right)^{678}\ge0\forall x,y,z\)
Dấu '=' xảy ra khi
\(\left\{{}\begin{matrix}x-\dfrac{1}{5}=0\\y+0.4=0\\z-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{5}\\y=-\dfrac{2}{5}\\z=3\end{matrix}\right.\)
Vậy: (x,y,z)=\(\left(\dfrac{1}{5};-\dfrac{2}{5};3\right)\)
