Giari phương trình
\(\frac{x+3}{x-3}+\frac{x-3}{x+3}=\frac{9}{x^2-9}\)
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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`
\(< =>\left[{}\begin{matrix}x=0\\2x-9=0\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{9}{2}\left(tm\right)\end{matrix}\right.\)
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
22 tháng 3 2020
a, \(2+\frac{3}{x-5}=1\Leftrightarrow\frac{3}{x-5}=-1\)
\(\Leftrightarrow x-5=\frac{3}{-1}=-3\Leftrightarrow x=2\)
Vậy .............
22 tháng 3 2020
b, ....................
\(\Leftrightarrow\frac{x-9}{x^2-3^2}-\frac{2}{x+3}=\frac{1}{x-3}\)
\(\Leftrightarrow\frac{x-9}{\left(x-3\right)\left(x+3\right)}-\frac{2x-6}{\left(x-3\right)\left(x+3\right)}-\frac{x+3}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{x-9-2x+6-x+3}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\frac{-2x}{\left(x-3\right)\left(x+3\right)}=0\Rightarrow-2x=0\Rightarrow x=0\)
Vậy .............
BV
Giải phương trình: \(\frac{x^2}{9}+\frac{16}{x^2}=\frac{10}{3}\left(\frac{x}{3}-\frac{4}{x}\right)\)
1
9 tháng 4 2018
Điều kiện:\(x\ne0\)
Đặt \(\frac{x}{3}-\frac{4}{x}=t\).Ta có:\(t^2=\left(\frac{x}{3}-\frac{4}{x}\right)^2=\frac{x^2}{9}-2.\frac{x}{3}.\frac{4}{x}+\frac{16}{x^2}=\frac{x^2}{9}+\frac{16}{x^2}-\frac{8}{3}\)
\(\Rightarrow\frac{x^2}{9}+\frac{16}{x^2}=t^2+\frac{8}{3}\).Thay vào pt ta có:\(t^2+\frac{8}{3}=\frac{10}{3}.t\)
\(\Leftrightarrow3t^2-10t+8=0\)\(\Leftrightarrow3t^2-4t-6t+8=0\)
\(\Leftrightarrow t\left(3t-4\right)-2\left(3t-4\right)=0\)
\(\Leftrightarrow\left(t-2\right)\left(3t-4\right)=0\Rightarrow\orbr{\begin{cases}t=2\\t=\frac{4}{3}\end{cases}}\)
Với \(t=2\) thì \(\frac{x^2-12}{3x}=2\Leftrightarrow x^2-12-6x=0\)\(\Rightarrow x^2-6x+9-21=0\)
\(\Leftrightarrow\left(x-3\right)^2=21\Rightarrow\orbr{\begin{cases}x-3=\sqrt{21}\\x-3=-\sqrt{21}\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\sqrt{21}+3\\x=3-\sqrt{21}\end{cases}}\)
Với \(t=\frac{4}{3}\) thì \(\frac{x^2-12}{3x}=\frac{4}{3}\Leftrightarrow x^2-4x-12=0\Leftrightarrow\left(x+2\right)\left(x-6\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-2\\x=6\end{cases}}\)
Tập nghiệm của pt S=\(\left\{\sqrt{21}+3;3-\sqrt{21};-2;6\right\}\)
NN
Nguyễn Ngọc Anh Minh
CTVHS
VIP
13 tháng 7 2017
Ta thấy \(\left(x-3\right)\left(2x+3\right)=2x^2-3x-9.\)
\(\left(1\right)\Leftrightarrow\frac{x}{x-3}-\frac{2x^2+9}{\left(x-3\right)\left(2x+3\right)}=\frac{1}{2x+3}\)
ĐK: \(x\ne3\)và \(x\ne-\frac{3}{2}\)
\(\Rightarrow x\left(2x+3\right)-2x^2-9=x-3\)
\(\Leftrightarrow2x^2+3x-2x^2-9=x-3\Leftrightarrow2x=6\Leftrightarrow x=2\)
Thỏa mãn ĐK
Các trường hợp khác làm tương tự
IT
1
26 tháng 9 2016
\(\frac{x^2}{9}+\frac{16}{x^2}=\frac{10}{3}\left(\frac{x}{3}-\frac{4}{x}\right)\)
\(\Leftrightarrow\frac{x^2}{9}-\frac{10x}{9}+\frac{40}{3x}+\frac{16}{x^2}=0\)
\(\Leftrightarrow\frac{x^4-10x^3+120x+144}{9x^2}=0\)
\(\Leftrightarrow x^4-10x^3+120x+144=0\)
\(\Leftrightarrow x^4-6x^3-12x^2-4x^3+24x^2+48x-12x^2+72x+144=0\)
\(\Leftrightarrow x^2\left(x^2-6x-12\right)-4x\left(x^2-6x-12\right)-12\left(x^2-6x-12\right)=0\)
\(\Leftrightarrow\left(x^2-4x-12\right)\left(x^2-6x-12\right)=0\)
\(\Leftrightarrow\left(x^2+2x-6x-12\right)\left(x^2-6x-12\right)=0\)
\(\Leftrightarrow\left[x\left(x+2\right)-6\left(x+2\right)\right]\left(x^2-6x-12\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+2\right)\left(x^2-6x-12\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-6=0\\x+2=0\\x^2-6x-12=0\left(1\right)\end{array}\right.\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x=6\\x=-2\end{array}\right.\)(tm)
\(\Delta_{\left(1\right)}=\left(-6\right)^2-\left(-4\left(1.12\right)\right)=84\)
\(\Rightarrow\)\(x_{1,2}=\frac{6\pm\sqrt{84}}{2}\) (tm)
Vậy pt có nghiệm là \(x=-2;x=6\)và \(x=\frac{6\pm\sqrt{84}}{2}\)
6 tháng 11 2017
Đề \(\Leftrightarrow\frac{\left(x-2\right)\left(x+3\right)+\left(1+x\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{x+31}{x^2-9}\)\(\left(ĐKXĐ:x\ne3,-3\right)\)
\(\Leftrightarrow\frac{x^2+3x-2x-6+x-3+x^2-3x}{x^2-9}=\frac{x+31}{x^2-9}\)
\(\Leftrightarrow2x^2-x-9=x+31\Leftrightarrow2x^2-2x-40=0\Leftrightarrow x^2-x-20=0\)
\(\Leftrightarrow x^2-5x+4x-20=0\Leftrightarrow x\left(x-5\right)+4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+4\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x-5=0\\x+4=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=5\left(chọn\right)\\x=-4\left(chọn\right)\end{cases}}}\)
Vậy \(S=\left\{5;-4\right\}\)
NN
7 tháng 4 2019
\(\frac{x+3}{x-3}-\frac{x-3}{x+3}=\frac{9}{x^2-9}\)
\(\Leftrightarrow\)\(\frac{\left(x+3\right).\left(x+3\right)}{\left(x-3\right).\left(x+3\right)}\)\(-\)\(\frac{\left(x-3\right).\left(x-3\right)}{\left(x+3\right).\left(x-3\right)}\)\(=\frac{9}{\left(x+3\right).\left(x-3\right)}\)
\(\Leftrightarrow\) \(\left(x+3\right)^2-\left(x-3\right)^2=9\)
\(\Leftrightarrow\)\(x^2+6x+9-x^2+6x-9=9\)
\(\Leftrightarrow\) \(12x=9\)
\(\Leftrightarrow\)\(x=\frac{3}{4}\)
Vậy phương trình có nghiệm là: \(x=\frac{3}{4}\)
\(\frac{x+3}{x-3}+\frac{x-3}{x+3}=\frac{9}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\)
<=> \(\frac{x+3}{x-3}+\frac{x-3}{x+3}=\frac{9}{\left(x-3\right)\left(x+3\right)}\)
<=> \(\frac{\left(x+3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-3\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{9}{\left(x-3\right)\left(x+3\right)}\)
=> \(\left(x+3\right)\left(x+3\right)+\left(x-3\right)\left(x-3\right)=9\)
<=> \(x^2+3x+3x+9+x^2-3x-3x+9=9\)
<=> \(2x^2+18=9\)
<=> \(2x^2=-9\)
<=> \(x^2=-\frac{9}{2}\)
Vì \(x^2\ge0,-\frac{9}{2}< 0\)nên không tìm được x
Vậy PT vô nghiệm
<=> \(S=\varnothing\)
ĐKXĐ : x ≠ ±3
<=> \(\frac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\frac{9}{\left(x-3\right)\left(x+3\right)}\)
=> 2x2 + 18 = 9
<=> x2 = -9/2 ( vô lí )
Vậy phương trình vô nghiệm