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\(\left(3x+1\right)\left(x-3\right)=\left(3x+1\right)\left(2x-5\right)\)
\(\Leftrightarrow\left(3x+1\right)\left(x-3\right)-\left(3x+1\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x-3-2x+5\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(2-x\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}3x+1=0\\2-x=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[\begin{matrix}3x=-1\\x=2\end{matrix}\right.\)
\(\Leftrightarrow\left[\begin{matrix}x=-\frac{1}{3}\\x=2\end{matrix}\right.\)
Vậy tập nghiệm của pt là \(S=\left\{-\frac{1}{3};2\right\}\)
Có : \(\left(3x+1\right)\left(x-3\right)=\left(3x+1\right)\left(2x-5\right)\)
\(\Leftrightarrow\) \(\left(3x+1\right)\left(x-3\right)-\left(3x+1\right)\left(2x-5\right)=0\)
\(\Leftrightarrow\) \(\left(3x+1\right)\left(x-3-2x+5\right)=0\)
\(\Leftrightarrow\) \(\left(3x+1\right)\left(-x+2\right)=0\)
\(\Leftrightarrow\) \(\left[\begin{matrix}3x+1=0\\-x+2=0\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[\begin{matrix}3x=-1\\-x=-2\end{matrix}\right.\) \(\Leftrightarrow\) \(\left[\begin{matrix}x=\frac{-1}{3}\\x=2\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{-1}{3};2\right\}\)
a, \(5\left|2x-1\right|-3=7\Leftrightarrow5\left|2x-1\right|=10\Leftrightarrow\left|2x-1\right|=2\)
TH1 : \(2x-1=2\Leftrightarrow x=\frac{3}{2}\)
TH2 : \(2x-1=-2\Leftrightarrow x=-\frac{1}{2}\)
b, \(\left(2x+3\right)\left(x-2\right)-x^2+4=0\Leftrightarrow\left(2x+3\right)\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x+3-x-2\right)=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow x=-1;x=2\)
c, \(\frac{2x-3}{2}< \frac{1-3x}{-5}\Leftrightarrow\frac{2x-3}{2}+\frac{1-3x}{5}< 0\)
\(\Leftrightarrow\frac{10x-15+2-6x}{10}< 0\Rightarrow4x-13< 0\Leftrightarrow x< \frac{13}{4}\)
\(\left(x-1\right)^2-\left(x+1\right)^2=2\left(x+3\right)\)
\(\Leftrightarrow\left(x-1+x+1\right)\left(x-1-x-1\right)=2\left(x+3\right)\)
\(\Leftrightarrow2x\left(-2\right)=2\left(x+3\right)\)
\(\Leftrightarrow-4x=2x+6\)
\(\Leftrightarrow-6x=6\)
\(\Leftrightarrow x=-1\)
2) \(\left(2x-1\right)^2-\left(2x+1\right)^2=4\left(x-3\right)\)
\(\Leftrightarrow\left(2x-1+2x+1\right)\left(2x-1-2x-1\right)-4\left(x-3\right)=0\)
\(\Leftrightarrow4x\left(-2\right)-4x+12=0\)
\(\Leftrightarrow-12x=-12\)
\(\Leftrightarrow x=1\)
3)\(\left(2x+3\right)^2-\left(2x+3\right)\left(2x-4\right)+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(2x+3\right)\left(2x+3-2x+4\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow7\left(2x+3\right)+x^2-4x+4=0\)
\(\Leftrightarrow x^2+10x+25=0\)
\(\Leftrightarrow\left(x+5\right)^2=0\)
\(\Leftrightarrow x=-5\)
4) \(8x^3-\left(x+1\right)^3=3x-3\)
\(\Leftrightarrow8x^3-\left(x^3+3x+3x^2+1\right)-3x+3=0\)
\(\Leftrightarrow7x^3-3x^2-6x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(7x^2+4x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\frac{-2+3\sqrt{2}}{7}\\x=\frac{-2-3\sqrt{2}}{7}\end{matrix}\right.\)
5)\(\left(3x-2\right)\left(9x^2+6x+4\right)-\left(3x-1\right)\left(9x^2-3x+1\right)=x-4\)
\(\Leftrightarrow\left(3x\right)^3-2^3-\left(\left(3x\right)^3-1^3\right)=x-4\)
\(\Leftrightarrow27x^3-8-\left(27x^3-1\right)=x-4\)
\(\Leftrightarrow-7=x-4\)
\(\Leftrightarrow x=-3\)
a) 5 - (x - 6) = 4(3 - 2x)
<=> 5 - x + 6 = 12 - 8x
<=> -x + 8x = 12 - 11
<=> 7x = 1
<=> x = 1/7
Vậy S = {1/7}
b) 2x(x - 3) + 5(x - 3) = 0
<=> (2x + 5)(x - 3) = 0
<=> \(\orbr{\begin{cases}2x+5=0\\x-3=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-\frac{5}{2}\\x=3\end{cases}}\)
Vậy S = {-5/2; 3}
c)ĐK: x \(\ne\)1; x \(\ne\)2
\(\frac{3x-5}{x-2}-\frac{2x-5}{x-1}=1\)
<=> \(\frac{\left(3x-5\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}-\frac{\left(2x-5\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}=\frac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)}\)
<=> 3x2 - 8x + 5 - 2x2 + 9x - 10 = x2 - 3x + 2
<=> x2 + x - 5 = x2 - 3x + 2
<=> x2 + x - x2 + 3x = 2 + 5
<=> 4x = 7
<=> x = 7/4
Vậy S = {7/4}
b) \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)
\(\Leftrightarrow\)\(\left(x^2+3x\right)\left(x^2+3x+2\right)-24=0\)
Đặt \(x^2+3x=t\) ta có:
\(t\left(t+2\right)-24=0\)
\(\Leftrightarrow\)\(t^2+2t-24=0\)
\(\Leftrightarrow\)\(\left(1-4\right)\left(1+6\right)=0\)
đến đây bn giải tiếp
câu 1:
a)x-1=5-x\(\Leftrightarrow\)x+x=5+1\(\Leftrightarrow\)2x=6\(\Leftrightarrow\)x=3
Vậy tập nghiệm của PT (a) là S={3}
b)3+x=2-x\(\Leftrightarrow\)x+x=2-3\(\Leftrightarrow\)2x=-1\(\Leftrightarrow\)x=-0,5
Vậy tập nghiệm của PT (b) là:S={-0,5}
câu 2:
a) 3x+7=2x-3\(\Leftrightarrow\)3x-2x=-3-7\(\Leftrightarrow\)x=-10
Vậy tập nghiệm của PT (a) là:S={-10}
b)4-(x-2)=(3-2x)\(\Leftrightarrow\)4-x+2=3-2x\(\Leftrightarrow\)-x+2x=-4+3-2\(\Leftrightarrow\)x=-3
Vậy tập nghiệm của PT (b) là:S={-3}
Câu 3:
a)\(\dfrac{5x-4}{2}=\dfrac{16x+1}{7}\Leftrightarrow\dfrac{7\left(5x-4\right)}{14}=\dfrac{2\left(16x+1\right)}{14}\)
\(\Leftrightarrow\)35x-28=32x+2\(\Leftrightarrow\)35x-32x=2+28\(\Leftrightarrow\)3x=30\(\Leftrightarrow\)x=10
Vậy tập nghiệm của PT (a) là :S={10}
b)\(\dfrac{12x+5}{3}=\dfrac{2x-7}{4}\Leftrightarrow\dfrac{4\left(12x+5\right)}{12}=\dfrac{3\left(2x-7\right)}{12}\)
\(\Leftrightarrow\)48x+20=6x-21\(\Leftrightarrow\)48x-6x=-20-21\(\Leftrightarrow\)42x=-41\(\Leftrightarrow\)x=\(-\dfrac{41}{42}\)
Vậy tập nghiệm của PT (b) là:S={\(-\dfrac{41}{42}\)}
a, (x-3)(2x+2) - (2x+1)(x-3)+12 =0
(x-3)(2x +2-2x-1) +12 = 0
(x-3) . 1 +12=0
x - 3 +12 =0
x = 9
a) |2x-1|=x+3
Nếu x\(\ge\)\(\dfrac{1}{2}\) thì: 2x-1=x+3
\(\Leftrightarrow\)x=4 (t/m)
Nếu x<\(\dfrac{1}{2}\) thì: 2x-1=-x-3
\(\Leftrightarrow\)x=\(\dfrac{-2}{3}\) (t/m)
b) |x+2|=|3x-1|
\(\Leftrightarrow\) (x+2)2=(3x-1)2\(\Leftrightarrow\)x2+4x+4=9x2-6x+1
\(\Leftrightarrow\)-8x2+10x+3=0\(\Leftrightarrow\)-8x2-2x+12x+3=0
\(\Leftrightarrow\)(4x+1)(-2x+3)=0\(\Leftrightarrow\)x\(\in\){\(\dfrac{-1}{4}\);\(\dfrac{3}{2}\)}
c)|x+1|+|x-2|=4
Lập bảng:
x | -1 2
x+1| -x-1 0 x+1 | x+1
x-2 | -x+2 | -x+2 0 x-2
VT | -2x+1 | 3 | 2x-1
Nếu x<-1 thì -2x+1=4\(\Leftrightarrow\)x=\(\dfrac{-3}{2}\) (t/m)
Nếu -1\(\le\)x<2 thì không có giá trị nào của x
Nếu 2\(\le\)x thì 2x-1=4\(\Leftrightarrow\)x=\(\dfrac{5}{2}\) (t/m)
Vậy x\(\in\){\(\dfrac{-3}{2};\dfrac{5}{2}\)}