Giải phương trình: 1 3 x − 1 + 1 2 x + 4 = 1 9 x − 2 + 1 5 − 4 x
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Những câu hỏi liên quan
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
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
LT
4
6 tháng 7 2020
a, \(\frac{9}{x^2-4}=\frac{x-1}{x+2}+\frac{3}{x-2}\left(ĐKXĐ:x\ne\pm2\right)\)
\(\frac{9}{\left(x-2\right)\left(x+2\right)}=\frac{x-1}{x+2}+\frac{3}{x-2}\)
\(\frac{9}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
Khử mẫu : \(9=\left(x-1\right)\left(x-2\right)+3\left(x+2\right)\)
Đến đây nhường bn, rất dễ =))
b, \(\frac{1}{x-5}-\frac{3}{x^2-6x+5}=\frac{5}{x-1}\)
\(\frac{1}{x-5}-\frac{3}{\left(x-5\right)\left(x-1\right)}=\frac{5}{\left(x-1\right)}\)
\(\frac{\left(x-1\right)}{x-5}-\frac{3}{\left(x-5\right)\left(x-1\right)}=\frac{5\left(x-5\right)}{\left(x-1\right)\left(x-5\right)}\)
Khử mẫu \(x-1-3=5\left(x-5\right)\)
Tự lm nốt mà cho mk hỏi, đề bài có bpt mà bpt đâu
PN
6 tháng 7 2020
\(\frac{9}{x^2-4}=\frac{x-1}{x+2}+\frac{3}{x-2}\left(ĐKXĐ:x\ne2;-2\right)\)
\(< =>\frac{9}{x^2-2^2}=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(< =>\frac{9}{\left(x-2\right)\left(x+2\right)}=\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{3x+6}{\left(x+2\right)\left(x-2\right)}\)
\(< =>9=x^2-2x-x+2+3x+6\)
\(< =>x^2-\left(2x+x-3x\right)+\left(2+6-9\right)=0\)
\(< =>x^2-2=0\)\(< =>x^2=2\)
\(< =>x=\pm\sqrt{2}\left(tmđk\right)\)
Vậy tập nghiệm của phương trình trên là \(\pm\sqrt{2}\)
BN
0
HP
2 tháng 2 2021
1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
HP
2 tháng 2 2021
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
19 tháng 8 2021
1: Ta có: \(\dfrac{-3}{x-4}-\dfrac{3-5x}{x^2-16}=\dfrac{1}{x+4}\)
Suy ra: \(-3\left(x+4\right)-3+5x=x-4\)
\(\Leftrightarrow-3x-12-3+5x-x+4=0\)
\(\Leftrightarrow x=11\left(nhận\right)\)
AH
Akai Haruma
Giáo viên
19 tháng 8 2021
2. ĐKXĐ: $x\neq \pm 2$
PT \(\Leftrightarrow \frac{3(x-2)}{(2+x)(x-2)}-\frac{x-1}{(x-2)(x+2)}=\frac{2(x+2)}{(x-2)(x+2)}\)
\(\Leftrightarrow \frac{3(x-2)-(x-1)}{(x-2)(x+2)}=\frac{2(x+2)}{(x-2)(x+2)}\)
\(\Rightarrow 3(x-2)-(x-1)=2(x+2)\)
\(\Leftrightarrow 2x-5=2x+4\Leftrightarrow 9=0\) (vô lý)
Vậy pt vô nghiệm
16 tháng 1 2021
\(a,\left(2x-3\right)^2=\left(x+1\right)^2\\ \Leftrightarrow\left(2x-3\right)^2-\left(x+1\right)^2=0\\ \Leftrightarrow\left(2x-3+x+1\right)\left(2x-3-x-1\right)=0\\ \Leftrightarrow\left(3x-2\right)\left(x-4\right)\\ \Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\x=4\end{matrix}\right. \\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=4\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{2}{3};4\right\}\)
16 tháng 1 2021
\(b,x^2-6x+9=9\left(x-1\right)^2\\ \Leftrightarrow\left(x-3\right)^2=9\left(x-1\right)^2\\ \Leftrightarrow\left(x-3\right)^2-9\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-3\right)^2-3^2\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-3\right)^2-\left[3\left(x-1\right)\right]^2=0\\ \Leftrightarrow\left(x-3\right)^2-\left(3x-3\right)^2=0\\ \Leftrightarrow\left(x-3+3x-3\right)\left(x-3-3x+3\right)=0\\ \Leftrightarrow-2x\left(4x-6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-2x=0\\4x-6=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\4x=6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{0;\dfrac{3}{2}\right\}\)
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
=>\(\left[{}\begin{matrix}x+4=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-4\left(nhận\right)\end{matrix}\right.\)
14 tháng 2 2022
1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\)
hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)
2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)
hay \(x\in\left\{1;5\right\}\)
3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)\left(2x+1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)
4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{-4;3;-3\right\}\)
5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)
\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)
\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)
hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)
14 tháng 2 2022
1.
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(5x-2\right)\)
\(\Leftrightarrow x+3=5x-2\)
\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)
2.
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)
\(\Leftrightarrow x^2+x+1=x^2-2x+16\)
\(\Leftrightarrow3x=15\Leftrightarrow x=5\)
3.
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2};x=-\dfrac{1}{2}\end{matrix}\right.\)
1 3 x − 1 + 1 2 x + 4 = 1 9 x − 2 + 1 5 − 4 x Đ K : x ≠ 1 3 , x ≠ − 2 , x ≠ 2 9 , x ≠ 5 4
Ta có pt: 5 x + 3 ( 3 x − 1 ) ( 2 x + 4 ) = 5 x + 3 ( 9 x − 2 ) ( 5 − 4 x )
< = > x = − 3 5 ( 3 x − 1 ) ( 2 x + 4 ) = ( 9 x − 2 ) ( 5 − 4 x ) < = > x = − 3 5 6 x 2 + 12 x − 2 x − 4 = − 36 x 2 + 45 x + 8 x − 10 < = > x = − 3 5 ( T M ) x = 6 7 ( T M ) x = 1 6 ( T M )
Vậy phương trình đã có có 3 nghiệm phân biệt như trên.