giải các phương trình sau :
1) 3x - 2( x - 3 ) = 6
2) ( x - 1 )2 = 9 ( x + 1 )2
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Những câu hỏi liên quan
26 tháng 5 2021
1) \(3x-2x+6=6\Leftrightarrow x=0\)
2) \(4\left(2x-1\right)-12x-12=3\left(x+2\right)\)
\(\Leftrightarrow8x-4-12x-12-3x-6=0\)
\(\Leftrightarrow7x=-22\Leftrightarrow x=\dfrac{-22}{7}\)
TB
26 tháng 5 2021
3, \(\left(x-1\right)2=9\left(x+1\right)2\)
\(\Leftrightarrow2x-2\) \(=18x+18\)
\(\Leftrightarrow2x-18x=18+2\)
\(\Leftrightarrow-16x\) \(=20\)
\(\Leftrightarrow x\) \(=\dfrac{-5}{4}\)
Vậy pt đã cho có tập nghiệm là S= \(\left\{\dfrac{-5}{4}\right\}\)
4, \(\dfrac{x-4}{x-1}+\dfrac{x+4}{x+1}=2\) ( ĐKXĐ : \(x\ne\pm1\) )
\(\Leftrightarrow\dfrac{\left(x-4\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{\left(x+4\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\dfrac{2\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(\Rightarrow x^2-3x-4+x^2+3x-4=2x^2-2\)
\(\Leftrightarrow2x^2-8-2x^2+2=0\)
\(\Leftrightarrow0\) \(=6\) ( Vô lí )
Vậy pt đã cho vô nghiệm
2 tháng 2 2024
a: 11x+4=-3/2
=>\(11x=-\dfrac{3}{2}-4=-\dfrac{11}{2}\)
=>\(x=-\dfrac{1}{2}\)
b: \(x^2-9+2\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(x+3\right)+2\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(x+3+2\right)=0\)
=>(x-3)(x+5)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
c: \(\dfrac{x-3}{5}+\dfrac{1+2x}{3}=6\)
=>\(\dfrac{3\left(x-3\right)+5\left(2x+1\right)}{15}=6\)
=>\(3x-9+10x+5=90\)
=>13x-4=90
=>13x=94
=>\(x=\dfrac{94}{13}\)
d: \(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\)(ĐKXĐ: \(x\notin\left\{-1;2\right\}\))
=>\(\dfrac{2\left(x-2\right)-\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{3x-11}{\left(x-2\right)\left(x+1\right)}\)
=>3x-11=2x-4-x-1
=>3x-11=x-5
=>2x=6
=>x=3(nhận)
9 tháng 1 2021
Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
PN
16 tháng 5 2021
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
23 tháng 11 2023
1: \(2^x=64\)
=>\(x=log_264=6\)
2: \(2^x\cdot3^x\cdot5^x=7\)
=>\(\left(2\cdot3\cdot5\right)^x=7\)
=>\(30^x=7\)
=>\(x=log_{30}7\)
3: \(4^x+2\cdot2^x-3=0\)
=>\(\left(2^x\right)^2+2\cdot2^x-3=0\)
=>\(\left(2^x\right)^2+3\cdot2^x-2^x-3=0\)
=>\(\left(2^x+3\right)\left(2^x-1\right)=0\)
=>\(2^x-1=0\)
=>\(2^x=1\)
=>x=0
4: \(9^x-4\cdot3^x+3=0\)
=>\(\left(3^x\right)^2-4\cdot3^x+3=0\)
Đặt \(a=3^x\left(a>0\right)\)
Phương trình sẽ trở thành:
\(a^2-4a+3=0\)
=>(a-1)(a-3)=0
=>\(\left[{}\begin{matrix}a-1=0\\a-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=1\left(nhận\right)\\a=3\left(nhận\right)\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}3^x=1\\3^x=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=0\end{matrix}\right.\)
5: \(3^{2\left(x+1\right)}+3^{x+1}=6\)
=>\(\left[3^{x+1}\right]^2+3^{x+1}-6=0\)
=>\(\left(3^{x+1}\right)^2+3\cdot3^{x+1}-2\cdot3^{x+1}-6=0\)
=>\(3^{x+1}\left(3^{x+1}+3\right)-2\left(3^{x+1}+3\right)=0\)
=>\(\left(3^{x+1}+3\right)\left(3^{x+1}-2\right)=0\)
=>\(3^{x+1}-2=0\)
=>\(3^{x+1}=2\)
=>\(x+1=log_32\)
=>\(x=-1+log_32\)
6: \(\left(2-\sqrt{3}\right)^x+\left(2+\sqrt{3}\right)^x=2\)
=>\(\left(\dfrac{1}{2+\sqrt{3}}\right)^x+\left(2+\sqrt{3}\right)^x=2\)
=>\(\dfrac{1}{\left(2+\sqrt{3}\right)^x}+\left(2+\sqrt{3}\right)^x=2\)
Đặt \(b=\left(2+\sqrt{3}\right)^x\left(b>0\right)\)
Phương trình sẽ trở thành:
\(\dfrac{1}{b}+b=2\)
=>\(b^2+1=2b\)
=>\(b^2-2b+1=0\)
=>(b-1)2=0
=>b-1=0
=>b=1
=>\(\left(2+\sqrt{3}\right)^x=1\)
=>x=0
7: ĐKXĐ: \(x^2+3x>0\)
=>x(x+3)>0
=>\(\left[{}\begin{matrix}x>0\\x< -3\end{matrix}\right.\)
\(log_4\left(x^2+3x\right)=1\)
=>\(x^2+3x=4^1=4\)
=>\(x^2+3x-4=0\)
=>(x+4)(x-1)=0
=>\(\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
VD
18 tháng 3 2022
\(a,4+3x=25-4x\\ \Leftrightarrow7x=21\\ \Leftrightarrow x=3\\ b,\left(x-1\right)^2+\left(x-1\right)\left(x+3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-1+x+3\right)=0\\ \Leftrightarrow\left(x-1\right)\left(2x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
c, ĐKXĐ:\(x\ne-1,x\ne2\)
\(\dfrac{1}{x+1}+\dfrac{3}{x-2}=\dfrac{9}{\left(x+1\right)\left(x-2\right)}\\ \Leftrightarrow\dfrac{x-2}{\left(x+1\right)\left(x-2\right)}+\dfrac{3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{9}{\left(x+1\right)\left(x-2\right)}=0\\ \Leftrightarrow\dfrac{x-2+3x+3-9}{\left(x+1\right)\left(x-2\right)}=0\\ \Rightarrow4x-8=0\\ \Leftrightarrow x=2\left(ktm\right)\)
28 tháng 8 2021
d: Ta có: \(\dfrac{x}{x+3}-\dfrac{2x}{x-3}-\dfrac{3x}{9-x^2}=0\)
\(\Leftrightarrow x^2-3x-2x^2-6x+3x=0\)
\(\Leftrightarrow-x^2-6x=0\)
\(\Leftrightarrow-x\left(x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-6\left(nhận\right)\end{matrix}\right.\)
19 tháng 1 2024
a: 5-3x=6x+7
=>-3x-6x=7-5
=>-9x=2
=>\(x=-\dfrac{2}{9}\)
b: \(\dfrac{3x-2}{6}-5=3-\dfrac{2\left(x+7\right)}{4}\)
=>\(\dfrac{3x-2}{6}+\dfrac{x+7}{2}=8\)
=>\(\dfrac{3x-2+3\left(x+7\right)}{6}=8\)
=>3x-2+3x+14=48
=>6x+12=48
=>6x=36
=>\(x=\dfrac{36}{6}=6\)
c: \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)
=>\(\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0\)
=>(x-1)(5x+3-3x+8)=0
=>(x-1)(2x+11)=0
=>\(\left[{}\begin{matrix}x-1=0\\2x+11=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{11}{2}\end{matrix}\right.\)
d: \(\left(2x-1\right)^2-\left(x+3\right)^2=0\)
=>\(\left(2x-1-x-3\right)\left(2x-1+x+3\right)=0\)
=>\(\left(x-4\right)\left(3x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{3}\end{matrix}\right.\)
H
20 tháng 4 2023
\(\dfrac{2x-1}{3}+\dfrac{3x-2}{4}=\dfrac{4x-3}{5}\)
\(\Leftrightarrow\dfrac{20\left(2x-1\right)}{60}+\dfrac{15\left(3x-2\right)}{60}=\dfrac{12\left(4x-3\right)}{60}\)
`<=> 20(2x-1) +15(3x-2) =12(4x-3)`
`<=> 40x - 20 + 45x - 30 = 48x - 36`
`<=> 85x -50 = 48x - 36`
`<=> 85x-48x = -36+50`
`<=> 37x =14`
`<=> x= 14/37`
Vậy phương trình có nghiệm `x=14/37`
__
\(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{x^2-9}\)
\(\Leftrightarrow\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{\left(x-3\right)\left(x+3\right)}\)
ĐKXĐ : \(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
Ta có : \(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-6}{\left(x-3\right)\left(x+3\right)}\)
`=> 5x + 15 + 4x -12=x-6`
`<=> 9x + 3=x-6`
`<=> 9x-x=-6-3`
`<=> 8x = -9`
`<=>x=-9/8(tm)`
Vậy phương trình có nghiệm `x=-9/8`
` @ yngoc`
1) 3x- 2(x - 3 ) =6
<=> 3x - 2x+ 6 =6
<=>x + 6 = 6
=>x= 6 - 6 =0
vậy pt vô nghiệm
1)3x-2(x+3)=6
3x-2x-6-6=0
x=0
không có vô nghiệm nhé
2)(x-1)2=9(x+1)2
x2-2x+1-9(x2+2x+1)=0
-8x2+16x-8=0
-8(x2-2x+1)=0
x2-2x+\(\frac{1}{4}-\frac{1}{4}\)+1=0
(x+1/2)2=-3/4 (vô lý)(vì n2 luôn luôn lớn hơn hoặc bằng hai
=> x thuộc rỗng