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Câu hỏi của Đẹp Trai Không Bao Giờ Sai - Toán lớp 7 | Học trực tuyến tương tự
Với mọi x,y ta có :
\(+,\left|x-2y-1\right|\ge0\)
+, \(\left|y-4\right|+2\ge2\Leftrightarrow\dfrac{10}{\left|x-4\right|+2}\le5\)
Dấu "=" xảy ra khi :
\(\left\{{}\begin{matrix}\left|x-2y-1\right|=5\\\dfrac{10}{\left|x-4\right|+2}=5\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}y=4\\x=9\end{matrix}\right.\)
Vậy ..
a) \(\left|3x-\dfrac{1}{2}\right|+\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|=0\)
Do \(\left|3x-\dfrac{1}{2}\right|,\left|\dfrac{1}{4}y+\dfrac{3}{5}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}3x-\dfrac{1}{2}=0\\\dfrac{1}{4}y+\dfrac{3}{5}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=-\dfrac{12}{5}\end{matrix}\right.\)
b) \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|+\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\le0\)
Do \(\left|\dfrac{3}{2}x+\dfrac{1}{9}\right|,\left|\dfrac{5}{7}y-\dfrac{1}{2}\right|\ge0\forall x,y\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{3}{2}x+\dfrac{1}{9}=0\\\dfrac{5}{7}y-\dfrac{1}{2}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{2}{27}\\y=\dfrac{7}{10}\end{matrix}\right.\)
3: \(\left|x-\dfrac{3}{4}\right|-\dfrac{1}{2}=0\)
\(\Leftrightarrow\left|x-\dfrac{3}{4}\right|=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{3}{4}=\dfrac{1}{2}\\x-\dfrac{3}{4}=-\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{4}\\x=\dfrac{1}{4}\end{matrix}\right.\)
`|x+1/3|+|x+2/3|+|x+2/5|+|x+3/2|=33x`
`@TH1: x >= -1/3`
`=>x+1/3+x+2/3+x+2/5+x+3/2=33x`
`=>29x=29/10`
`=>x=1/10` (t/m)
`@TH2: -2/3 <= x < -1/3`
`=>-x-1/3+x+2/3+x+2/5+x+3/2=33x`
`=>31x=67/30`
`=>x=67/930` (ko t/m)
`@TH3:-2/5 <= x < -2/3`
`=>-x-1/3-x-2/3+x+2/5+x+3/2=33x`
`=>33x=9/10`
`=>x=3/110` (ko t/m)
`@TH4:-3/2 <= x < -2/5`
`=>-x-1/3-x-2/3-x-2/5+x+3/2=33x`
`=>35x=1/10`
`=>x=1/350` (ko t/m)
`@TH5: x < -3/2`
`=>-x-1/3-x-2/3-x-2/5-x-3/2=33x`
`=>37x=-29/10`
`=>x=-29/370` (ko t/m)
có VT \(\ge\) 0 với mọi x
=>VP:33x\(\ge\) 0 \(\Rightarrow\) x\(\ge\)0
\(\Rightarrow\) |x+1/3|\(\ge\)0;|x+2/3|\(\ge\) 0;|x+2/5|\(\ge\) 0;|x+3/2|\(\ge\) 0
=> (x+1/3)+(x+2/3)+(x+2/5)+(x+3/2)=33x
=>(x+x+x+x)+(1/3+2/3+2/5+3/2)=33x
=>4x+29/10=33x
=> 29/10=33x-4x
=>29/10=29x
=>x=29/10:29
=>x=1/10
T đã hứa thì t sẽ làm:v
\(3\left|2x+1\right|+4\left|2y-1\right|\le7\)
\(\Rightarrow3\left|2x+1\right|\le7-4\left|2y-1\right|\le7\)
mà: \(\left\{{}\begin{matrix}3 \left|2x+1\right|\ge0\\3\left|2x+1\right|⋮3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}0\le3\left|2x+1\right|\le7\\3\left|2x+1\right|⋮3\end{matrix}\right.\)
Vì x nguyên nên: \(3\left|2x+1\right|\in\left\{0;3;6\right\}\)
\(\Rightarrow\left[{}\begin{matrix}\left|2x+1\right|=0\\\left|2x+1\right|=1\\\left|2x+1\right|=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\left(loại\right)\\\left[{}\begin{matrix}2x+1=1\Leftrightarrow x=0\left(chọn\right)\\2x+1=-1\Leftrightarrow x=-1\left(chọn\right)\end{matrix}\right.\\\left[{}\begin{matrix}2x+1=2\Leftrightarrow x=\dfrac{1}{2}\left(loại\right)\\2x+1=-2\Leftrightarrow x=-\dfrac{3}{2}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
Với \(\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\) thì: \(3\left|2x+1\right|=3\Leftrightarrow4\left|2y-1\right|\le7-3=4\)
Vì \(y\in Z\) nên: \(\left[{}\begin{matrix}4\left|2y-1\right|=4\\4\left|2y-1\right|=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}2y-1=1\Leftrightarrow y=1\left(chọn\right)\\2y-1=-1\Leftrightarrow y=0\left(chọn\right)\end{matrix}\right.\\2y=1\Leftrightarrow y=\dfrac{1}{2}\left(loại\right)\end{matrix}\right.\)
Vậy: \(\left(x;y\right)=\left(0;0\right);\left(0;1\right);\left(-1;1\right);\left(-1;0\right)\)
\(pt\Leftrightarrow\left|x+2\right|+\left|x+\dfrac{3}{5}\right|+\left|x+\dfrac{1}{2}\right|=10x\)
Ta có: \(\left|x+2\right|+ \left|x+\dfrac{3}{5}\right|+\left|x+\dfrac{1}{2}\right|\ge0\Leftrightarrow10x\ge0\Leftrightarrow x\ge0\)
Khi \(x\ge0\) thì: \(x+2+x+\dfrac{3}{5}+x+\dfrac{1}{2}=10x\)
\(\Rightarrow7x+2+\dfrac{3}{5}+\dfrac{1}{2}=\dfrac{31}{10}\Leftrightarrow x=\dfrac{31}{70}\)
Em cảm ơn cô nhiều ạ!