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a, |x + 2| + 6 = 2x
\(\Rightarrow\left|x+2\right|=2x-6\) (1)
Vì |x + 2| \(\ge\) 0 \(\Rightarrow2x-6\ge0\Rightarrow2x\ge6\Rightarrow x\ge3\)
Từ (1) \(\Rightarrow\left[{}\begin{matrix}x+2=2x-6\\x+2=-\left(2x-6\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2x=-6-2\\x+\left(-2x\right)=6-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=-8\\-x=4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{8}{3}\\x=-4\end{matrix}\right.\)(Loại)
Vậy không có giá trị x thỏa mãn
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X : 6 x 7,2 + X x 1,3 + X : 2 + 15 = 19,95
X x \(\frac{1}{6}\) x \(\frac{36}{5}\)+ X x \(\frac{13}{10}\)+ X x \(\frac{1}{2}\) = 19,95 - 15 = 4,95 = \(\frac{99}{20}\)
X x \(\frac{6}{5}\)+ X x \(\frac{13}{10}\)+ X x \(\frac{1}{2}\)= \(\frac{99}{20}\)
X x ( \(\frac{6}{5}+\frac{13}{10}+\frac{1}{2}\)) = \(\frac{99}{20}\)
X x 3 = \(\frac{99}{20}\)
X = \(\frac{99}{20}:3\)
X = \(\frac{33}{20}\)
X : 6 x 7,2 + 1,3 x X + X : 2 + 15 = 19,95
(X x 7,2) / 6 + (7,8 x X) / 6 + (X x 3) / 6 + 90/6 = 119,7/6
X x 7,2 + 7,8 x X + X x 3 + 90 = 119,7
X x (7,2 + 7,8 + 3) = 119,7 - 90
X x 18 = 29,7
X = 29,7 : 18
X = 1,65
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a: \(\Leftrightarrow x\in\left\{0;-3\right\}\)
b: =>(x-3)(x-5)=0
hay \(x\in\left\{3;5\right\}\)
c: =>x(x-4)=0
hay \(x\in\left\{0;4\right\}\)
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Bài 1L
a) \(\left(x-7\right)\left(x+3\right)< 0\)
TH1:
\(\hept{\begin{cases}x-7>0\\x+3< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>7\\x< -3\end{cases}}}\)( loại )
TH2:
\(\hept{\begin{cases}x-7< 0\\x+3>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< 7\\x>-3\end{cases}\Leftrightarrow}-3< x< 7}\)( chọn )
Vậy \(-3< x< 7\)
Bài 2:
a) \(\left(5x+8\right)-\left(2x-15\right)+21=2x-5\)
\(\Leftrightarrow5x+8-2x+15+21=2x-5\)
\(\Leftrightarrow5x-2x-2x=-5-21-8-15\)
\(\Leftrightarrow x=-49\)
Vậy ...
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a, 3x-(2x+1)=6
3x-2x-1=6
x-1=6
x=7
b,2x-[(-15)+x]-6=16
2x-[(-15)+x]=22
2x-(-15)-x=22
x+15=22
x=7
c,(15-18)2+3x=2.(x-6)
(-3)2+3x=2x-12
9+3x=2x-12
3x-2x=-12-9
x=-21
a: =>|x+2|=2x-6
\(\Leftrightarrow\left\{{}\begin{matrix}x>=3\\\left(2x-6-x-2\right)\left(2x-6+x+2\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=3\\\left(x-8\right)\left(3x-4\right)=0\end{matrix}\right.\Leftrightarrow x=8\)
b: =>1,2x+1,3x=2,95
=>2,5x=2,95
hay x=1,18