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a) ta có
|9+x| = 9+x thì 9+x ≥ 0 ⇔ x ≥ -9
|9+x|=-(9-x)thì 9+x <0 ⇔ x<-9
th1 với x ≥ -9
9+x=2x
⇔ 9=2x-x
⇔ 9=x (tmđk)
th2 với x < -9
-(9+x)=2x
⇔ -9-x=2x
⇔ -x-2x=9
⇔ -3x=9
⇔ x=-2 (ktm)
vậy phương trình có tập nghiệm là S+{ 9}
b) Với : x < -6 , phương trình có dạng :
- x - 6 = 2x + 9
<=> -3x = 15
<=> x = - 5 ( không thỏa mãn )
Với : x ≥ - 6 , phương trình có dạng :
x + 6 = 2x + 9
<=> x = - 3 ( thỏa mãn)
Vậy , phương trình nhận : x = - 3 làm nghiệm duy nhất
c) Với : x < 0 , phương trình có dạng :
- 5x = 3x - 2
<=> -8x = -2
<=> x = \(\dfrac{1}{4}\) ( không thỏa mãn )
Với : x ≥ 0 , phương trình có dạng :
5x = 3x - 2
<=> 2x = -2
<=> x = -1 ( không thỏa mãn )
Vậy, phương trình đã cho vô nghiệm
n) \(\left|3-x\right|+x^2-x\left(x+4\right)=0\)
\(\Rightarrow\left|3-x\right|+x^2-x^2-4x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3-x-4x=0\left(đk:3-x\ge0\right)\\-\left(3-x\right)-4x=0\left(đk:3-x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(đk:x\le3\right)\\x=-1\left(đk:x>3\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x=\dfrac{3}{5}\)
m) \(\left(x-1\right)^2+\left|x+21\right|-x^2-13=0\)
\(\Rightarrow x^2-2x+1+\left|x+21\right|-x^2-13=0\)
\(\Leftrightarrow-2x-12+\left|x+21\right|=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-2x-12+x+21=0\left(đk:x+21\ge0\right)\\-2x-12-\left(x+21\right)=0\left(đk:x+21< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=9\left(đk:x\ge-21\right)\\x=-11\left(đk:x< -21\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=9\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x=9\)
e) \(\left|5x\right|=3x-2\)
\(\Rightarrow5\cdot\left|x\right|=3x-2\)
\(\Leftrightarrow5\cdot\left|x\right|-3x=-2\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-3x=-2\left(đk:x\ge0\right)\\5\cdot\left(-x\right)-3x=-2\left(đk:x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(đk:x\ge0\right)\\x=\dfrac{1}{4}\left(đk:x< 0\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x\in\varnothing\)
g) \(\left|-2,5x\right|=x-12\)
\(\Rightarrow2,5\cdot\left|x\right|=x-12\)
\(\Leftrightarrow2x5\cdot\left|x\right|-x=-12\)
\(\Leftrightarrow\left[{}\begin{matrix}2,5x-x=-12\left(đk:x\ge0\right)\\2,5\cdot\left(-x\right)-x=-12\left(đk:x< 0\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-8\left(đk:x\ge0\right)\\x=\dfrac{24}{7}\left(đk:x< 0\right)\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\in\varnothing\\x\in\varnothing\end{matrix}\right.\)
Vậy \(x\in\varnothing\)
a, \(5\left(m+3x\right)\left(x+1\right)-4\left(1+2x\right)=80\)
Phương trình nhận \(x=2\)làm nghiệm nên :
\(5\left(m+3.2\right)\left(2+1\right)-4\left(1+2.2\right)=80\)
\(\Leftrightarrow15m+90-20=80\)
\(\Leftrightarrow15m=80+20-90\)
\(\Leftrightarrow15m=10\Leftrightarrow m=1,5\)
....
b, \(3\left(2x+m\right)\left(3x+2\right)-2\left(3x+1\right)^2=43\)
Phương trình nhận \(x=1\)làm nghiệm nên :
\(3\left(2.1+m\right)\left(3.1+2\right)-2\left(3.1+1\right)^2=43\)
\(\Leftrightarrow30+15m-32=43\)
\(\Leftrightarrow15m=43+32-30\)
\(\Leftrightarrow15m=45\Leftrightarrow m=3\)
....
\(\frac{315-x}{101}+\frac{313-x}{103}+\frac{311-x}{105}+\frac{309-x}{107}+4=0\)
\(\Leftrightarrow\frac{315-x}{101}+1+\frac{313-x}{103}+1+\frac{311-x}{105}+1+\frac{309-x}{107}+1=0\)
\(\Leftrightarrow\frac{416-x}{101}+\frac{416-x}{103}+\frac{416-x}{105}+\frac{416-x}{107}=0\)
\(\Leftrightarrow\left(416-x\right)\left(\frac{1}{101}+\frac{1}{103}+\frac{1}{105}+\frac{1}{107}\right)=0\)
\(\Leftrightarrow416-x=0\)
\(\Leftrightarrow x=416\)
a) 5(m + 3x)(x + 1) - 4(1 + 2x) = 80
Phương trình có nghiệm x = 2:
5(m + 3.2)(2 + 1) - 4(1 + 2.2) = 80
<=> 5(m + 6).3 - 4.5 = 80
<=> 15(m + 6) - 4.5 = 80
<=> 15(m + 6) - 20 = 80
<=> 15(m + 6) = 80 + 20
<=> 15(m + 6) = 100
<=> m + 6 = 100 : 15
<=> m + 6 = 20/3
<=> m = 20/3 - 6
<=> m = 2/3
b) 3(2x + m)(3x + 2) - 2(3x + 1)2 = 43
Phương trình có nghiệm x = 1:
3(2.1 + m)(3.1 + 2) - 2(3.1 + 1)2 = 43
<=> 3(2 + m).5 - 2.16 = 43
<=> 15(2 + m) - 32 = 43
<=> 15(2 + m) = 43 + 32
<=> 15(2 + m) = 75
<=> 2 + m = 75 : 15
<=> 2 + m = 5
<=> m = 5 - 2
<=> m = 3
1.
a, \(\left(x+3\right)\left(x-3\right)-\left(x-3\right)^2\)
\(=\left(x-3\right)\left(x+3-x+3\right)\)
\(=9\left(x-3\right)=9x-27\)
b, \(\left(2x+1\right)^2+2\left(2x+1\right)\left(x-1\right)+\left(x-1\right)^2\)
\(=\left(2x+1+x-1\right)^2=9x^2\)
c, \(x\left(x-3\right)\left(x+3\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x\left(x^2-9\right)-\left(x^4-1\right)\)
\(=x^3-9x-x^4+1=-x^4+x^3-9x+1\)
a ) \(\left(5x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{5}\\x=7\end{matrix}\right.\)
b ) \(\left(x^2-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=-3\end{matrix}\right.\)
c )\(x^2-x-6=0\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=3\end{matrix}\right.\)
d ) \(x^2+x-12=0\)
\(\Leftrightarrow x^2-4x+3x-12\)
\(\Leftrightarrow\left(x+3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
e ) \(15\left(x+9\right)\left(x-3\right)\left(x+21\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-9\\x=3\\x=-21\end{matrix}\right.\)
g ) \(\left(x^2+1\right)\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x+2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(loại\right)\\x=-2\end{matrix}\right.\)
i ) \(x^4+2x^3-2x^2+2x-3=0\)
\(\Leftrightarrow x^4+3x^3-x^3-3x^2+x^2+3x-x-3=0\)
\(\Leftrightarrow x^3\left(x+3\right)-x^2\left(x+3\right)+x\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x^3-x^2+x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x^2+1\right)\left(x-1\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-1\left(loại\right)\\x=1\\x=-3\end{matrix}\right.\)
h) \(x^2+5x+6=0\)
\(\Leftrightarrow x^2+3x+2x+6=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)