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16 tháng 2 2022
\(a,\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x-6=0\Leftrightarrow x=6\\2x-5=0\Leftrightarrow x=\dfrac{5}{2}\\3x+9=0\Leftrightarrow x=-3\end{matrix}\right.\)
\(b,2x\left(x-3\right)+5\left(x-3\right)=0\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\2x+5=0\Leftrightarrow x=-\dfrac{5}{2}\end{matrix}\right.\)
\(c,x^2-4-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
\(x=-7\left(2m-5\right)x-2m^2+8\Leftrightarrow x+7\left(2m-5\right)=8-2m^2\Leftrightarrow x\left(14m-34\right)=8-2m^2\)
\(ycđb\Leftrightarrow14m-34\ne0\Leftrightarrow m\ne\dfrac{34}{14}\)\(\Rightarrow x=\dfrac{8-2m^2}{14m-34}\)
\(3.17\Leftrightarrow4x^2-4x+1-2x-1=0\Leftrightarrow4x^2-6x=0\Leftrightarrow x\left(4x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
DT
16 tháng 2 2022
3.15:
a, \(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\2x-5=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=\dfrac{5}{2}\\x=-\dfrac{9}{3}=-3\end{matrix}\right.\)
b, \(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
c, \(\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
3.16
\(\Leftrightarrow\left(2m-5\right).-7-2m^2+8=0\)
\(\Leftrightarrow-14m+35-2m^2+8=0\)
\(\Leftrightarrow-14m-2m^2+43=0\)
\(\Leftrightarrow-2\left(7m+m^2\right)=-43\)
\(\Leftrightarrow m\left(7-m\right)=\dfrac{43}{2}\)
\(\Leftrightarrow\dfrac{m\left(7-m\right)}{1}-\dfrac{43}{2}=0\)
\(\Leftrightarrow\dfrac{14m-2m^2}{2}-\dfrac{43}{2}=0\)
pt vô nghiệm
10 tháng 5 2022
a: |x+9|=2
=>x+9=2 hoặc x+9=-2
=>x=-7 hoặc x=-11
b: |2x-3|=x-3
\(\Leftrightarrow\left\{{}\begin{matrix}x>=3\\\left(2x-3-x+3\right)\left(2x-3+x-3\right)=0\end{matrix}\right.\Leftrightarrow x=3\)
BL
1
NH
2 tháng 3 2023
\(\dfrac{x+3}{x-3}-\dfrac{x}{x+3}=\dfrac{2x^2+9}{x^2-9}\left(x\ne-3;x\ne3\right)\\ < =>\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}-\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{2x^2+9}{\left(x-3\right)\left(x+3\right)}\)
suy ra
`x^2 +6x+9-x^2 +3x=2x^2 +9`
`<=> 2x^2 - x^2 +x^2 - 6x -3x +9 -9=0`
`<=> 2x^2 -9x=0`
`<=> x(2x-9)=0`
\(< =>\left[{}\begin{matrix}x=0\\2x-9=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{9}{2}\left(tm\right)\end{matrix}\right.\)
LL
2
8 tháng 3 2022
\(\Leftrightarrow\left(2x+5\right)\left(x-3\right)-\left(x-3\right)\left(x+3\right)=0\)
=>(x-3)(2x+5-x-3)=0
=>(x-3)(x+2)=0
=>x=3 hoặc x=-2
8 tháng 3 2022
x2-9=(x-3)(2x-5)
(=) (x-3)(x+3)=(x-3)(2x-5)
(=) (x-3)(x+3)-(x-3)(2x-5)=0
(=) (x-3)(x+3-2x+5)=0
(=) (x-3)(8-x)=0
(=)x-3=0 hoặc 8-x=0
(=)x=0 hoặc x=8
Vậy S=\(\left\{0;8\right\}\)
PT
1
20 tháng 9 2020
1) \(\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{4x+15}{9-x^2}\)
ĐKXĐ : \(x\ne\pm3\)
\(\Leftrightarrow\frac{x-1}{x+3}-\frac{x}{x-3}=\frac{-4x-15}{x^2-9}\)
\(\Leftrightarrow\frac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{x\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\frac{x^2-4x+3}{\left(x-3\right)\left(x+3\right)}-\frac{x^2+3x}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\frac{x^2-4x+3-x^2-3x}{\left(x-3\right)\left(x+3\right)}=\frac{-4x-15}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow-7x+3=-4x-15\)
\(\Leftrightarrow-7x+4x=-15-3\)
\(\Leftrightarrow-3x=-18\)
\(\Leftrightarrow x=6\)( tmđk )
Vậy x = 6 là nghiệm của phương trình
2) 2x + 3 < 6 - ( 3 - 4x )
<=> 2x + 3 < 6 - 3 + 4x
<=> 2x - 4x < 6 - 3 - 3
<=> -2x < 0
<=> x > 0
Vậy nghiệm của bất phương trình là x > 0
N
2
16 tháng 9 2020
Ta có: \(x^2+2x+2x\sqrt{x+3}=9-\sqrt{x+3}\) \(\left(ĐK:x\ge-3\right)\)
\(\Leftrightarrow\left(x^2+2x\sqrt{x+3}+x+3\right)+x+\sqrt{x+3}=12\)
\(\Leftrightarrow\left(x+\sqrt{x+3}\right)^2+\left(x+\sqrt{x+3}\right)-12=0\)
\(\Leftrightarrow\left(x+\sqrt{x+3}\right)\left(x+\sqrt{x+3}+1\right)-12=0\)
Đặt \(a=x+\sqrt{x+3}\)\(\Leftrightarrow\)\(a+1=x+\sqrt{x+3}+1\)
Ta lại có: \(a.\left(a+1\right)-12=0\)
\(\Leftrightarrow a^2+a-12=0\)
\(\Leftrightarrow a^2-3a+4a-12=0\)
\(\Leftrightarrow a\left(a-3\right)+4\left(a-3\right)=0\)
\(\Leftrightarrow\left(a+4\right)\left(a-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a+4=0\\a-3=0\end{cases}}\)
+ \(a+4=0\)\(\Leftrightarrow\)\(x+\sqrt{x+3}+4=0\)
\(\Leftrightarrow\)\(x+4=-\sqrt{x+3}\)
\(\Leftrightarrow\)\(\left(x+4\right)^2=\left(-\sqrt{x+3}\right)^2\)
\(\Leftrightarrow\)\(x^2+8x+16=x+3\)
\(\Leftrightarrow\)\(x^2+7x+13=0\)
\(\Leftrightarrow\)\(\left(x^2+7x+\frac{49}{4}\right)+\frac{3}{4}=0\)
\(\Leftrightarrow\)\(\left(x+\frac{7}{2}\right)^2+\frac{3}{4}=0\)
Vì \(\left(x+\frac{7}{2}\right)^2+\frac{3}{4}>0\forall x\)mà \(\left(x+\frac{7}{2}\right)^2+\frac{3}{4}=0\)
\(\Rightarrow\)Phương trình \(\left(x+\frac{7}{2}\right)^2+\frac{3}{4}=0\)vô nghiệm
+ \(a-3=0\)\(\Leftrightarrow\)\(x+\sqrt{x+3}-4=0\)
\(\Leftrightarrow\)\(x-3=-\sqrt{x+3}\)
\(\Leftrightarrow\)\(\left(x-3\right)^2=\left(-\sqrt{x+3}\right)^2\)
\(\Leftrightarrow\)\(x^2-6x+9=x+3\)
\(\Leftrightarrow\)\(x^2-7x+6=0\)
\(\Leftrightarrow\)\(\left(x^2-x\right)-\left(6x-6\right)=0\)
\(\Leftrightarrow\)\(x.\left(x-1\right)-6.\left(x-1\right)=0\)
\(\Leftrightarrow\)\(\left(x-6\right).\left(x-1\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-6=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=6\left(TM\right)\\x=1\left(TM\right)\end{cases}}\)
Vậy \(S=\left\{1;6\right\}\)
30 tháng 7 2017
/x-3/=9-2x
= (1) x-3 = 9-2x
= 3x=12
x=4
(2) x+3= 9-2x
= 3x= 6
x=2
7 tháng 2 2023
=>(2x-3)(2x+3)(x-4)-(2x-3)(x-4)(x+4)=0
=>(2x-3)(x-4)(2x+3-x-4)=0
=>(2x-3)(x-4)(x-1)=0
=>\(x\in\left\{1;4;\dfrac{3}{2}\right\}\)
Ta có:
\(\left|x-3\right|=x-3\Leftrightarrow x-3\ge0\Leftrightarrow x\ge3\)
\(\left|x-3\right|=-x+3\Leftrightarrow x-3< 0\Leftrightarrow x< 3\)
Với \(x\ge3\Rightarrow x-3=9-2x\)
\(\Leftrightarrow3x=12\)
\(\Leftrightarrow x=4\) (Thoả mãn)
Với \(x< 3\Rightarrow-x+3=9-2x\)
\(\Leftrightarrow x=6\) (Loại)