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a) Có \(\left|x-3y\right|^5\ge0\);\(\left|y+4\right|\ge0\)
\(\rightarrow\left|x-3y\right|^5+\left|y+4\right|\ge0\)
mà \(\left|x-3y\right|^5+\left|y+4\right|=0\)
\(\rightarrow\left\{{}\begin{matrix}\left|x-3y\right|^5=0\\\left|y+4\right|=0\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}x=3y\\y=-4\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\)
b) Tương tự câu a, ta có:
\(\left\{{}\begin{matrix}\left|x-y-5\right|=0\\\left(y-3\right)^4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=3\end{matrix}\right.\)
c. Tương tự, ta có:
\(\left\{{}\begin{matrix}\left|x+3y-1\right|=0\\\left|y+2\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1-3y\\y=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=-2\end{matrix}\right.\)
a. \(\left|x-3y\right|^5\ge0,\left|y+4\right|\ge0\Rightarrow\left|x-3y\right|^5+\left|y+4\right|\ge0\) \(\Rightarrow VT\ge VP\)
Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x-3y\right|^5=0\\\left|y+4\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3y\\y=-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\) Vậy...
b. \(\left|x-y-5\right|\ge0,\left(y-3\right)^4\ge0\Rightarrow\left|x-y-5\right|+\left(y-3\right)^4\ge0\) \(\Rightarrow VT\ge VP\)
Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x-y-5\right|=0\\\left(y-3\right)^4=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=3\end{matrix}\right.\) Vậy ...
c. \(\left|x+3y-1\right|\ge0,3\cdot\left|y+2\right|\ge0\Rightarrow\left|x+3y-1\right|+3\left|y+2\right|\ge0\) \(\Rightarrow VT\ge VP\) Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x+3y-1\right|=0\\3\left|y+2\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1-3y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-\left(-2\right)\cdot3=7\\y=-2\end{matrix}\right.\) Vậy...
\(\frac{x}{4}:2=4:\frac{x}{2}\)
\(\Leftrightarrow\frac{x}{8}=\frac{8}{x}\)
\(\Leftrightarrow x^2=64\)
\(\Rightarrow\left[\begin{array}{nghiempt}x=8\\x=-8\end{array}\right.\)
Vậy \(x\in\left\{8,-8\right\}\)
x/4 : 2 =4 : x/2
=>x/4 . 1/2 = 4.2/x
=>x/8=8/x
=> x.x=8.8
=>x^2=64
=>x=8 , x =-8
\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{x}\left(1+2+...+x\right)\)
\(=1+\frac{1}{2}\cdot\frac{2\cdot3}{2}+\frac{1}{3}\cdot\frac{3\cdot4}{2}+\frac{1}{4}+\frac{4\cdot5}{2}+...+\frac{1}{x}\cdot\frac{x\left(x+1\right)}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{x+1}{2}\)
\(=\frac{1}{2}\left(2+3+4+...+x+1\right)\)
\(=\frac{1}{2}\cdot\frac{\left(x+1+2\right)\left(x+1-2+1\right)}{2}\)
\(=\frac{1}{2}\cdot\frac{x\left(x+3\right)}{2}=\frac{x\left(x+3\right)}{4}\).
a) \(\left|x\right|=\dfrac{3}{7}\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-3}{7}\\x=\dfrac{3}{7}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{3}{7};\dfrac{-3}{7}\right\}\)
b) \(\left|x\right|=0\)
\(\Rightarrow x=0\)
Vậy x=0
c) \(\left|x\right|=-8,7\)
Vì \(\left|x\right|\ge0\)
\(\Rightarrow x=\varnothing\)
Áp dụng tính chất `|P|>=P,|P|>=-P`
`=>{(|x+5|>=x+5),(|x+1|>=-x-1):}`
`=>|x+5|+|x+1|>=x+5-x-1=4`
Mặt khác:`|x+3|>=0`
`=>|x+1|+|x+3|+|x+5|>=4(đpcm)`
Dấu "=" xảy ra khi `x=-3`
x^2+1>=1
=>(x^2+1)^2>=1
y^2+2>=2
=>(y^2+2)^4>=16
=>(x^2+1)^2+(y^2+2)^4>=17
=>(x^2+1)^2+(y^2+2)^4-2>=15
Dấu = xảy ra khi x=y=0
\(\left(x-\frac{3}{4}\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{3}{4}=0\\x+2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=-2\end{cases}}\)
Vậy x = 3/4 hoặc x = -2
\(\left(x-\frac{3}{4}\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{3}{4}=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{4}\\x=-2\end{cases}}}\)
Vậy x = \(\frac{3}{4}\) hoặc x = - 2