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a/ \(\left|2x-1,6\right|-2,3=1,4\)
\(\Leftrightarrow\left|2x-1,6\right|=3,7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1,6=3,7\\2x-1,6=-3,7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=5,3\\2x=-2,1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2,65\\x=-1,05\end{matrix}\right.\)
Vậy ....
b/ \(5,4-\left|3x-1,2\right|=5,5\)
\(\Leftrightarrow\left|3x-1,2\right|=-0,1\)
Mà \(\left|3x-1,2\right|\ge0\)
\(\Leftrightarrow x\in\varnothing\)
c/ \(\left|x+1,3\right|+\left|x+2,4\right|=4x\)
Mà \(\left\{{}\begin{matrix}\left|x+1,3\right|\ge0\\\left|x+2,4\right|\ge0\end{matrix}\right.\) \(\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x+1,3+x+2,4=4x\)
\(\Leftrightarrow2x+3,7=4x\)
\(\Leftrightarrow3,7=4x-2x\)
\(\Leftrightarrow2x=3,7\)
\(\Leftrightarrow x=1,85\)
Vậy ....
d/ \(\left|x-1,2\right|+\left|2,5-x\right|=0\)
Mà \(\left\{{}\begin{matrix}\left|x-1,2\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x-1,2\right|=0\\\left|2,5-x\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1,2=0\\2,5-x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1,2\\x=2,5\end{matrix}\right.\) (loại)
Vậy ..
a, \(\left|2x-1,6\right|-2,3=1,4\)
\(\Rightarrow\left|2x-1,6\right|=3,7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1,6=3,7\\2x-1,6=-3,7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2,65\\x=-1,05\end{matrix}\right.\)
b,\(5,4-\left|3x-1,2\right|=5,5\)
\(\Rightarrow\left|3x-1,2\right|=-0,1\) (vô lí)
Vì \(\left|x\right|\ge0\) mà \(\left|3x-1,2\right|< 0\)
Vậy, không có giá trị của x thỏa mãn.
c, \(\left|x+1,3\right|+\left|x+2,4\right|=4x\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x+1,3\right|\ge0\\\left|x+2,4\right|\ge0\end{matrix}\right.\Leftrightarrow4x\ge0\)
\(\Leftrightarrow x+1,3+x+2,4=4x\)
\(\Leftrightarrow x+x+1,3+2,4=4x\)
\(\Leftrightarrow2x+3,7=4x\)
\(\Leftrightarrow2x-4x=-3,7\)
\(\Leftrightarrow-2x=-3,7\)
\(\Leftrightarrow x=\dfrac{3,7}{2}\)
d, \(\left|x-1,2\right|+\left|2,5-x\right|=0\)
\(\Rightarrow\left\{{}\begin{matrix}\left|x-1,2\right|\ge0\\\left|2,5-x\right|\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x-1,2\right|=0\\\left|2,5-x\right|=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1,2=0\\2,5-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1,2\\x=2,5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{3}x-\dfrac{8}{15}=0\\2,5+\left(-\dfrac{7}{5}\right)\div x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{3}x=0+\dfrac{8}{15}\\-\dfrac{7}{5}\div x=0-2,5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{3}x=\dfrac{8}{15}\\-\dfrac{7}{5}\div x=-2,5\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{15}\div\dfrac{1}{3}\\x=-\dfrac{7}{5}\div\left(-2,5\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{8}{5}\\x=\dfrac{14}{25}\end{matrix}\right.\)
Bài 1:
- \(E=5,5.\left(2-3,6\right)\)
C1: \(E=5,5.\left(2-3,6\right)=-1,6.5,5=-8,8\)
C2: \(E=5,5.\left(2-3,6\right)=5,5.2-5,5.3,6=11-19,8=-8,8\)
- \(F=-3,1.\left(3-5,7\right)\)
C1: \(F=-3,1.\left(3-5,7\right)=-3,1.\left(-2,7\right)=8,37\)
C2: \(F=-3,1.\left(3-5,7\right)=-3,1.3+\left[\left(-3,1\right).\left(-5,7\right)\right]=-9,3+17,67=8,37\)
Bài 2:
a, \(\left|2,5-x\right|=1,3\Rightarrow\left[{}\begin{matrix}2,5-x=1,3\\2,5-x=-1,3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1,2\\x=3,8\end{matrix}\right.\)
\(\Rightarrow\left\{x_1=1,2;x_2=3,8\right\}\)
b, \(1,6-\left|x-0,2\right|=0\Rightarrow\left|x-0,2\right|=1,6\)
\(\Rightarrow\left[{}\begin{matrix}x-0,2=1,6\\x-0,2=-1,6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1,8\\x=-1,4\end{matrix}\right.\)
\(\Rightarrow\left\{x_1=1,8;x_2=-1,4\right\}\)
c, \(\left|x-1,5\right|+\left|2,5-x\right|=0\)
Không có giá trị của x thỏa mãn.
a) \(-4,3+\left[\left(-7,5\right)+4,3\right]=-4,3+-3,2=-7,5\)
b) \(45,3+\left[7,3+\left(-22\right)\right]=45,3+\left(-14,7\right)=30,6\)
c) \(\left[\left(-11,7\right)+5,5\right]+\left[11,7+\left(-2,5\right)\right]\)
\(=-6,2+9,2=3\)
\(\left|x-2,5\right|-\frac{3}{4}=0\)
\(\Leftrightarrow\left|x-\frac{1}{4}\right|=\frac{3}{4}\)
\(\Leftrightarrow x-\frac{1}{4}=\pm\frac{3}{4}\)
TH1:\(x-\frac{1}{4}=\frac{3}{4}\)
\(\Leftrightarrow x=\frac{3}{4}+\frac{1}{4}=1\)
TH2:\(x-\frac{1}{4}=-\frac{3}{4}\)
\(\Leftrightarrow x=-\frac{3}{4}+\frac{1}{4}=-\frac{1}{2}\)
Vậy \(x=1;-\frac{1}{2}\)
Lời giải:
Ta thấy:
$(7x-5y)^{2018}\geq 0, \forall x,y$
$(3x-2z)^{2020}\geq 0, \forall x,z$
$(xy+yz+xz-4500)^{2022}\geq 0, \forall x,y,z$
Do đó để tổng $(7x-5y)^{2018}+(3x-2z)^{2020}+(xy+yz+xz-4500)^{2022}=0$ thì:
$(7x-5y)^{2018}=(3x-2z)^{2020}=(xy+yz+xz-4500)^{2022}=0$
$\Leftrightarrow$ \(\left\{\begin{matrix} 7x=5y(1)\\ 3x=2z(2)\\ xy+yz+xz=4500(3)\end{matrix}\right.\)
Từ $(1);(2)\Rightarrow y=\frac{7}{5}x; z=\frac{3}{2}x$
Thay vào $(3)$:
$x.\frac{7}{5}x+\frac{7}{5}x.\frac{3}{2}x+x.\frac{3}{2}x=4500$
$\Leftrightarrow x^2=900\Rightarrow x=\pm 30$
Nếu $x=30\Rightarrow y=42; z=45$
Nếu $x=-30\Rightarrow y=-42; z=-45$
Vì |3,4-x| lớn hơn hoặc bằng 0 với mọi x
=> 1,7+|3,4-x| lớn hơn hoặc bằng 1,7+0
=> A lớn hơn hoặc bằng 1,7
Dấu "=" xảy ra <=> |3,4-x|=0
=>3,4-x=0
=> x= 3,4
Vậy min A= 1,7 khi x= 3,4
\(A=1,7+\left|3,4-x\right|\)
mà \(\left|3,4-x\right|\ge0\forall x\Rightarrow A\ge1,7\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow3,4-x=0\Leftrightarrow x=3,4\)
\(N=\left|x+3,2\right|-2,5\)
mà \(\left|x+3,2\right|\ge0\forall x\Rightarrow N\ge-2,5\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x+3,2=0\Leftrightarrow x=-3,2\)
\(P=5,5+\left|2x-0,5\right|\)
mà \(\left|2x-0,5\right|\ge0\forall x\Rightarrow P\ge5,5\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow2x-0,5=0\Leftrightarrow x=0,25\)
\(2,5-3x=5,5.2022^0\)
\(=>2,5-3x=5,5.1\)
\(=>2,5-3x=5,5\)
\(=>3x=2,5-5,5\)
\(=>3x=-3\)
\(=>x=\left(-3\right):3\)
\(=>x=\dfrac{-3}{3}=-1\)
Vậy...
\(#NqHahh\)
\(2,5-3x=5,5\cdot2022^0\)
\(2,5-3x=5,5\cdot1\)
\(2,5-3x=5,5\)
\(3x=2,5-5,5\)
\(3x=-3\)
\(x=-3:3\)
\(x=-1\)
Vậy \(x=-1\)