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Bài 1: Giải các phương trình sau:
a) 3(2,2-0,3x)=2,6 + (0,1x-4)
<=> 6.6 - 0.9x = 2,6 + 0,1x - 4
<=> - 0.9x - 0,1x = -6.6 -1,4
<=> -x = -8
<=> x = 8
Vậy x = 8
b) 3,6 -0,5 (2x+1) = x - 0,25(22-4x)
<=> 3,6 - x - 0,5 = x - 5,5 + x
<=> - x - 3,1 = -5,5
<=> - x = -2.4
<=> x = 2.4
Vậy x = 2.4
a) 1,2-(x-0,8)= -2(0,9+x)
VT=-(x-2)
VP=\(-\frac{10x+9}{5}\)
pt trở thành:-(x-2)=\(-\frac{10x+9}{5}\)
<=>2-x=-2x-1,8
<=>5x=-19
<=>x=-3,8
a) 1,2-(x-0,8)= -2(0,9+x)
VT=-(x-2)
VP=$-\frac{10x+9}{5}$−10x+95
pt trở thành:-(x-2)=$-\frac{10x+9}{5}$−10x+95
<=>2-x=-2x-1,8
<=>5x=-19
<=>x=-3,8
a) 1,2 - ( x - 0,8 ) = -2( 0,9+ x )
<=> 1,2 - x + 0,8 = -1.8 - 2x
<=> x = -3,8
Vậy x = -3,8
b) 2,3x - 2(0,7 + 2x ) = 3,6 - 1,7x
<=> 2,3x - 1,4 - 4x = 3,6 - 1,7x
<=> -3,4x = 5
<=> x = \(\dfrac{-25}{17}\)
Vậy x = \(\dfrac{-25}{17}\)
c) 3(2,2 - 0,3x ) = 2,6 + (0,1x - 4 )
<=> 6,6 - 0,9x = 2,6 + 0,1x - 4
<=> -x = -8
<=> x = 8
Vậy x = 8
d) 3,6 - 0,5(2x + 1) = x- 0,25(2-4x)
<=> 3,6 - x - 0.5 = x - 0,5 + x
<=> -3x = -3,6
<=> x = 1.2
Vậy x = 1.2
Áp dụng công thức: \(A\left(x\right).B\left(x\right)=0\Leftrightarrow\left[{}\begin{matrix}A\left(x\right)=0\\B\left(x\right)=0\end{matrix}\right.\)
a) \(PT\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\4x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{4}\end{matrix}\right.\)
Vậy: \(S=\left\{\dfrac{2}{3};-\dfrac{5}{4}\right\}\)
b) \(PT\Leftrightarrow\left[{}\begin{matrix}2,3x-6,9=0\\0,1x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-20\end{matrix}\right.\)
Vậy: \(S=\left\{3;20\right\}\)
c) Vì \(x^2+1\ge1>0\forall x\)
\(\Rightarrow4x+2=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy: \(S=\left\{-\dfrac{1}{2}\right\}\)
d) \(PT\Leftrightarrow\left[{}\begin{matrix}2x+7=0\\x-5=0\\5x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\\x=-\dfrac{1}{5}\end{matrix}\right.\)
Vậy: \(S=\left\{-\dfrac{7}{2};5;-\dfrac{1}{5}\right\}\)
a: =>3x-2=0 hoặc 4x+5=0
=>x=2/3 hoặc x=-5/4
b: =>(x-3)(x+20)=0
=>x=3 hoặc x=-20
c: =>4x+2=0
hay x=-1/2
d: =>2x+7=0 hoặc x-5=0 hoặc 5x+1=0
=>x=-7/2 hoặc x=5 hoặc x=-1/5
\(\Leftrightarrow6,6-0,9x=2,6+0,1x-4\)
\(\Leftrightarrow-0,9x-0,1x=2,6-4-6,6\)
\(\Leftrightarrow-1x=-8\)
\(\Leftrightarrow x=8\)
Vậy \(S=\left\{8\right\}\)
\(PT.\Rightarrow\) \(6,6-0,9x-2,6-0,1x+4=0.\\ \Leftrightarrow-x=8.\Leftrightarrow x=8.\)
\(a,\left(3x-2\right)\left(4x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\4x+5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}3x=2\\4x=-5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{5}{4}\end{cases}}\)
Vậy ............
\(b,\left(2,3x-6,9\right)\left(0,1x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2,3x-6,9=0\\0,1x+2=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}2,3x=6,9\\0,1x=-2\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=-20\end{cases}}\)
Vậy ...........
\(c,\left(4x+2\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4x+2=0\\x^2+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=-2\\x^2=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-0,5\\x\in\varnothing\end{cases}}\)
Vậy .........................
\(d,\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}2x+7=0\\x-5=0\\5x+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x=-7\\x=5\\5x=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{7}{2}\\x=5\\x=-\frac{1}{5}\end{cases}}\)
Vậy ...............
a) \(\left(3x-2\right)\left(4x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-2=0\\4x+5=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{5}{4}\end{cases}}\)
b) \(\left(2,3x-6,9\right)\left(0,1x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2,3x-6,9=0\\0,1x+2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-20\end{cases}}\)
c) \(\left(4x+2\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4x+2=0\\x^2+1=0\end{cases}}\)
\(\Leftrightarrow x=-\frac{1}{2}\) ( do \(x^2+1\ge1>0\forall x\) )
d) \(\left(2x+7\right)\left(x-5\right)\left(5x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2x+7=0\\x-5=0\end{cases}hoặc5x+1=0}\)
\(\Leftrightarrow x\in\left\{-\frac{7}{2},5,-\frac{1}{5}\right\}\)
a: =>|x-7|=3-2x
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(-2x+3\right)^2-\left(x-7\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(2x-3-x+7\right)\left(2x-3+x-7\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =\dfrac{3}{2}\\\left(x+4\right)\left(3x-10\right)=0\end{matrix}\right.\Leftrightarrow x=-4\)
b: =>|2x-3|=4x+9
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(4x+9-2x+3\right)\left(4x+9+2x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=-\dfrac{9}{4}\\\left(2x+12\right)\left(6x+6\right)=0\end{matrix}\right.\Leftrightarrow x=-1\)
c: =>3x+5=2-5x hoặc 3x+5=5x-2
=>8x=-3 hoặc -2x=-7
=>x=-3/8 hoặc x=7/2
\(a,4\left(x-3\right)^2-\left(2x-1\right)^2\ge12\)
\(\Leftrightarrow4x^2-24x+36-4x^2-4x+1\ge12\)
\(\Leftrightarrow-28x+37\ge12\)
\(\Leftrightarrow-28x\ge12-37\)
\(\Leftrightarrow-28x\ge-25\)
\(\Leftrightarrow x\le\dfrac{25}{28}\)
Vậy \(S=\left\{x\left|x\le\dfrac{25}{28}\right|\right\}\)
b, \(\left(x-4\right)\left(x+4\right)\ge\left(x+3\right)^2+5\)
\(\Leftrightarrow x^2-16\ge x^2+6x+9+5\)
\(\Leftrightarrow x^2-x^2-6x\ge9+5+16\)
\(\Leftrightarrow-6x\ge30\)
\(\Leftrightarrow x\le-5\)
Vậy \(S=\left\{x\left|x\le-5\right|\right\}\)
\(c,\left(3x-1\right)^2-9\left(x+2\right)\left(x-2\right)< 5x\)
\(\Leftrightarrow9x^2-6x-1-9x^2+36< 5x\)
\(\Leftrightarrow9x^2-9x^2-6x-5x+36+1< 0\)
\(\Leftrightarrow-11x+37< 0\)
\(\Leftrightarrow-11x< -37\)
\(\Leftrightarrow x>\dfrac{37}{11}\)
vậy \(S=\left\{x\left|x>\dfrac{37}{11}\right|\right\}\)
a)\(1,2-x+0,8=-1,8-2x\)
\(2-x=-1,8-2x\)
\(2x-x=-1,8-2\)
\(x=-3,8\)
Vậy S={-3,8}
b)\(2,3x-1,4-4x=3,6-1,7x\)
\(2,3x-4x+1,7x=3,6+1,4\)
0=5(vô lí)
Vậy S={\(\varnothing\)}
c)\(6,6-0.9=2,6+0,1x-4\)
\(5,7=0,1x-1,4\)
\(-4,3=0,1x\)
\(x=-43\)
Hình như câu c bạn làm sai rồi thì phải.