(2x-3).(x+1)+x(x-2)=3(x+2)^2
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a)
$|3x-2|=2x\Rightarrow x\geq 0$.
Xét 2 TH:
TH1: $x\geq \frac{2}{3}$ thì pt trở thành:
$3x-2=2x\Leftrightarrow x=2$ (thỏa mãn)
TH2: $0\leq x< \frac{2}{3}$ thì pt trở thành:
$2-3x=2x\Leftrightarrow x=\frac{2}{5}$ (thỏa mãn)
b)
PT $\Rightarrow x\geq 0$
$\Rightarrow |4+2x|=4+2x$. PT trở thành:
$4+2x=4x\Leftrightarrow x=2$ (thỏa mãn)
c)
Xét các TH sau:
TH1: $x\geq \frac{3}{2}$. Khi đó, pt trở thành:
$2x-3=-x+21$
$\Leftrightarrow x=8$ (thỏa mãn)
TH2: $x< \frac{3}{2}$. Khi đó, pt trở thành:
$3-2x=-x+21$
$\Leftrightarrow x=-18$ (thỏa mãn)
d)
Từ PT suy ra $x-2\geq 0\Leftrightarrow x\geq 2(*)$
Khi đó: $|3x-1|=3x-1$. PT trở thành:
$3x-1=x-2$
$\Leftrightarrow 2x=-1<0\Rightarrow x<0$ (mâu thuẫn với $(*)$)
Vậy PT vô nghiệm.
1) \(\Rightarrow16x^2+24x+9+9x^2-24x+16+4-25x^2=x\)
\(\Rightarrow x=29\)
2)
a) \(=x^2-9-x^2+6x-9=6x-18\)
b) \(=\left(3x-1+2x+1\right)^2=\left(5x\right)^2=25x^2\)
( 1/2 + 1/3 ) x 2/5
c1 = 5/6 x 2/5 = 1/3
c2 = 1/2 x 2/5 + 1/3 x 2/5
= 1/5 + 2/15
= 1/3
3/5 x 17/21 x 2/5
c1 := 17/35 x 2/5 = 34/175
c2 : = (3/5 x 2/5) x 17/21
= 6/25 x 17/21
= 34/175?
( 1/3 - 1/5 ) x 1/2
c1 : = 2/15 x 1/2
= 1/15
c2 : = 1/3 x 1/2 - 1/5 x 1/2
= 1/6 - 1/10
= 1/15
\(|-2x+1,5|=\dfrac{1}{4}\Rightarrow-2x+1,5=\pm\dfrac{1}{4}\)
\(-2x+1,5=\dfrac{1}{4}\Rightarrow-2x=1,5-0,25\Rightarrow-2x=1,25\Rightarrow x=1,25:\left(-2\right)\Rightarrow x=...\)
\(-2x+1,5=-\dfrac{1}{4}\Rightarrow-2x=-0,25-1,5\Rightarrow-2x=1,75\Rightarrow x=1,75:\left(-2\right)\Rightarrow x=...\)
\(\dfrac{3}{2}-|1.\dfrac{1}{4}+3x|=\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{3}{2}-\dfrac{1}{4}\Rightarrow|1.\dfrac{1}{4}+3x|=\dfrac{5}{4}\)
\(\Rightarrow1.\dfrac{1}{4}+3x=\pm\dfrac{5}{4}\)
\(1.\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=\dfrac{5}{4}\Rightarrow3x=\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=1\Rightarrow x=3\)
\(1.\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow\dfrac{1}{4}+3x=-\dfrac{5}{4}\Rightarrow3x=-\dfrac{5}{4}-\dfrac{1}{4}\Rightarrow3x=-\dfrac{3}{2}x=...\)
=>2x^2+2x-3x-3+x^2+2x=3x^2+12x+12
=>12x+12=x-3
=>11x=-15
=>x=-15/11
\(\left|x+1\right|-\left|-2x-2\right|=2\)
\(\Leftrightarrow\left|x+1\right|-\left|-2\left(x+1\right)\right|=2\)
\(\Leftrightarrow\left|x+1\right|-2\left|x+1\right|=2\)
\(\Leftrightarrow-\left|x+1\right|=2\)
\(\Leftrightarrow\left|x+1\right|=-2\)
\(\Leftrightarrow\left|x+1\right|+2=0\)
Mà: \(\left|x+1\right|\ge0\forall x\Rightarrow\left|x+1\right|+2\ge2>0\)
\(\Leftrightarrow\left|x+1\right|+2=0\) (vô lí)
Vậy phương trình vô nghiệm:
\(x\in\varnothing\)
=>|x+1|-2|x+1|=2
=>-|x+1|=2
=>|x+1|=-2(vô lý)
Vậy: \(x\in\varnothing\)
`a)|2x+1|=5`
`<=>` \(\left[ \begin{array}{l}2x+1=5\\2x+1=-5\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x=4\\2x=-6\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\)
`b)|2x+1|=0`
`<=>2x+1=0`
`<=>2x=-1`
`<=>x=-1/2`
`c)|2x+1|=7`
`<=>` \(\left[ \begin{array}{l}2x+1=7\\2x+1=-7\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x=6\\2x=-8\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=4\\x=-4\end{array} \right.\)
`d)|2x+5|=|3x-7|`
`<=>` \(\left[ \begin{array}{l}2x+5=3x-7\\2x+5=7-3x\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=12\\5x=2\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=12\\x=\dfrac25\end{array} \right.\)
`e)|2x+7|=1`
`<=>` \(\left[ \begin{array}{l}2x+7=1\\2x+7=-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}2x=-6\\2x=-8\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=3\\x=-4\end{array} \right.\)
`g)|x-2|+|2x-3|=2`
Nếu `x>=2=>|x-2|=x-2,|2x-3|=2x-3`
`pt<=>x-2+2x-3=2`
`<=>3x-5=2`
`<=>3x=7`
`<=>x=7/3(tm)`
Nếu `x<=3/2=>|x-2|=2-x,|2x-3|=3-2x`
`pt<=>2-x+3-2x=2`
`<=>5-3x=2`
`<=>3x=3`
`<=>x=1(tm)`
Nếu `3/2<=x<=2=>|x-2|=2-x,|2x-3|=2x-3`
`pt<=>2-x+2x-3=2`
`<=>x-1=2`
`<=>x=3(l)`
`h)|x+2|+|1-x|=3x+2`
Vì `VT>=0=>3x+2>=0=>x>=-2/3`
`=>|x+2|=x+2`
`pt<=>x+2+|1-x|=3x+2`
`<=>|1-x|=2x(x>=0)`
`<=>` \(\left[ \begin{array}{l}2x=1-x\\2x=x-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}3x=1\\x=-1\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=\dfrac13(TM)\\x=-1(KTM)\end{array} \right.\)
a.
$|2x+1|=5$
\(\Leftrightarrow \left[\begin{matrix}
2x+1=5\\
2x+1=-5\end{matrix}\right.\Leftrightarrow \left[\begin{matrix}
x=2\\
x=-3\end{matrix}\right.\)
b.
$|2x+1|=0$
$\Leftrightarrow 2x+1=0$
$\Leftrightarrow x=-\frac{1}{2}$
c.
$|2x+1|=7$
\(\Leftrightarrow \left[\begin{matrix} 2x+1=7\\ 2x+1=-7\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=3\\ x=-4\end{matrix}\right.\)
\(\left(x+3\right)^2+\left(x-2\right)^2=2x^2\\ \Leftrightarrow x^2+6x+9+x^2-4x+4=2x^2\\ \Leftrightarrow2x=-13\Leftrightarrow x=-\dfrac{13}{2}\)
<=> x2+6x+9+x2+4x+4-2x2 =0
<=>10x+13=0
<=> x= -13/10
Chúc bạn học tốt nha!
\(=2.\left(-1\right)^2.2+4.\left(-1\right)^3.2^3+2.\left(-1\right).2^2\\ =4+\left(-32\right)+\left(-8\right)=\left(-36\right)\)
\(a,50\%x-0,2+x=\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{1}{2}x-0,2+x=\dfrac{4}{5}\)
\(\Leftrightarrow\dfrac{1}{2}x+x=\dfrac{4}{5}+0,2\)
\(\Leftrightarrow\dfrac{3}{2}x=\dfrac{4}{5}+\dfrac{1}{5}\)
\(\Leftrightarrow\dfrac{3}{2}x=1\)
\(\Leftrightarrow x=\dfrac{2}{3}\)
\(b,\left(x-\dfrac{3}{4}\right):\dfrac{1}{2}+\dfrac{3}{2}=\dfrac{25}{2}\)
\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{25}{2}-\dfrac{3}{2}\)
\(\Leftrightarrow\left(x-\dfrac{3}{4}\right).2=\dfrac{22}{2}\)
\(\Leftrightarrow x-\dfrac{3}{4}=11:2\)
\(\Leftrightarrow x=\dfrac{11}{2}+\dfrac{3}{4}\)
\(\Leftrightarrow x=\dfrac{25}{4}\)
(2x - 3)(x + 1) + x(x - 2) = 3(x + 2)2
<=> 2x2 - x - 3 + x2 - 2x = 3(x2 + 4x + 4)
<=> 3x2 - 3x - 3 = 3x2 + 12x + 12
<=> 15x = 15
<=> x = 1
Vậy x = 1 là nghiệm của phương trình
( 2x - 3 )( x + 1 ) + x( x - 2 ) = 3( x + 2 )2
<=> 2x2 - x - 3 + x2 - 2x = 3( x2 + 4x + 4 )
<=> 3x2 - 3x - 3 - 3x2 - 12x - 12 = 0
<=> -15x - 15 = 0
<=> x = -1
Vậy phương trình có một nghiệm x = -1