Giải phương trình sau
3/x^2+5x+4+2/x^2+10x+24+4/3+9/x^2+3x-18
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Khách
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Những câu hỏi liên quan
30 tháng 12 2018
a) \(x^3-2x^2-5x+6=0\)
\(x^3-x^2-x^2+x-6x+6=0\)
\(x^2\left(x-1\right)-x\left(x-1\right)-6\left(x-1\right)=0\)
\(\left(x-1\right)\left(x^2-x-6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x^2-x-6=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x^2-2x+3x-6=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\\left(x+3\right)\left(x-2\right)=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=1\\x=\left\{2;-3\right\}\end{cases}}\)
30 tháng 12 2018
\(a,x^3-2x^2-5x+6=0\)
\(\Leftrightarrow\left(x^3-x^2\right)-\left(x^2-x\right)-\left(6x-6\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-x\left(x-1\right)-6\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-x-6\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[\left(x^2-3x\right)+\left(2x-6\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left[x\left(x-3\right)+2\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x-3\right)=0\)
\(\Leftrightarrow x-1=0\left(h\right)x+2=0\left(h\right)x-3=0\)
\(\Leftrightarrow x=1\left(h\right)x=-2\left(h\right)x=3\)
Vậy \(x\in\left\{-2;1;3\right\}\)
P/S: (h) là hoặc nhé
11 tháng 5 2018
\(\text{a) }\dfrac{3}{x^2+5x+4}+\dfrac{2}{x^2+10x+24}=\dfrac{4}{3}+\dfrac{9}{x^2+3x-18}\\ ĐKXĐ:x\ne-1;x\ne-3;x\ne-4;x\ne-6\\ \Rightarrow\dfrac{3}{x^2+4x+x+4}+\dfrac{2}{x^2+6x+4x+24}=\dfrac{4}{3}+\dfrac{9}{x^2+6x-3x-18}\\ \Rightarrow\dfrac{3}{x\left(x+4\right)+\left(x+4\right)}+\dfrac{2}{x\left(x+6\right)+4\left(x+6\right)}=\dfrac{4}{3}+\dfrac{9}{x\left(x+6\right)-3\left(x+6\right)}\\ \Rightarrow\dfrac{3}{\left(x+1\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}=\dfrac{4}{3}+\dfrac{9}{\left(x-3\right)\left(x+6\right)}\)\(\Rightarrow\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}=\dfrac{4}{3}+\dfrac{1}{x-3}-\dfrac{1}{x+6}\\ \Rightarrow\dfrac{1}{x+1}-\dfrac{1}{x+6}-\dfrac{1}{x-3}+\dfrac{1}{x+6}=\dfrac{4}{3}\\ \Rightarrow\dfrac{1}{x+1}-\dfrac{1}{x-3}=\dfrac{4}{3}\\ \Rightarrow\dfrac{3\left(x-3\right)}{3\left(x+1\right)\left(x-3\right)}-\dfrac{3\left(x+1\right)}{3\left(x+1\right)\left(x-3\right)}=\dfrac{4\left(x+1\right)\left(x-3\right)}{3\left(x+1\right)\left(x-3\right)}\\ \Rightarrow3x-9-3x-3=4\left(x^2-2x-3\right)\\ \Leftrightarrow4x^2-8x-12=-12\\ \Leftrightarrow4x^2-8x=0\\ \Leftrightarrow4x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}4x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)Vậy phương trình có tập nghiệm \(S=\left\{0;2\right\}\)
TS
2
TA
21 tháng 2 2018
\(\dfrac{3}{\left(x+1\right)\left(x+4\right)}+\dfrac{2}{\left(x+4\right)\left(x+6\right)}=\dfrac{9}{\left(x-3\right)\left(x+6\right)}=\dfrac{4}{3}\)
=> \(\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}=\dfrac{1}{x-3}-\dfrac{1}{x+6}=\dfrac{4}{3}\)
=> \(\dfrac{1}{x+1}-\dfrac{1}{x+6}-\dfrac{1}{x-3}+\dfrac{1}{x+6}=0\)
=> \(\dfrac{1}{x+1}-\dfrac{1}{x-3}=0\)
Ma \(\dfrac{1}{x+1}-\dfrac{1}{x-3}=\dfrac{4}{3}\)
=> pt vo nghiem
TA
21 tháng 2 2018
\(\dfrac{1}{x+1}-\dfrac{1}{x+4}+\dfrac{1}{x+4}-\dfrac{1}{x+6}-\dfrac{1}{x+3}+\dfrac{1}{x+6}=\dfrac{4}{3}\)
=> \(\dfrac{1}{x+1}-\dfrac{1}{x+3}=\dfrac{4}{3}\)
=> \(\dfrac{2}{\left(x+1\right)\left(x+3\right)}=\dfrac{4}{3}\)
=> 4(x+1)(x+3)=6
=> 4(x2+4x+3)=6
=> 4x2+16x+6=0
=> (4x2+16x+16)-10=0
=> (2x+4)2=10
=> \(\left[{}\begin{matrix}2x+4=\sqrt{10}\\2x+4=-\sqrt{10}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\dfrac{\sqrt{10}-4}{2}\\x=\dfrac{-\sqrt{10}-4}{2}\end{matrix}\right.\)
9 tháng 1 2023
a. 3(x-2)-10=5(2x + 1)
<=> 3x - 6 - 10 = 10x + 5
<=> 3x - 10x = 5 + 6 + 10
<=> -7x = 21
<=> x = -3
b. 3x + 2=8 -2(x-7)
<=> 3x + 2 = 8 - 2x + 14
<=> 3x + 2x = 8 + 14 - 2
<=> 5x = 20
<=> x = 4
c. 2x-(2+5x)= 4(x + 3)
<=> 2x - 2 - 5x = 4x + 12
<=> 2x - 5x - 4x = 12 + 2
<=> -7x = 14
<=> x = -2
d. 5-(x +8)=3x + 3(x-9)
<=> 5 - x - 8 = 3x + 3x - 27
<=> -x - 3x - 3x = -27 + 8 - 5
<=> -7x = -24
<=> x = 24/7
e. 3x - 18 + x= 12-(5x + 3)
<=> 3x - 18 + x = 12 - 5x - 3
<=> 3x + x - 5x = 12 - 3 + 18
<=> -x = 27
<=> x = - 27
9 tháng 1 2023
a. 3(x-2)-10=5(2x + 1)
<=> 3x - 6 - 10 = 10x + 5
<=> 3x - 10x = 5 + 6 + 10
<=> -7x = 21
<=> x = -3
b. 3x + 2=8 -2(x-7)
<=> 3x + 2 = 8 - 2x + 14
<=> 3x + 2x = 8 + 14 - 2
<=> 5x = 20
<=> x = 4
c. 2x-(2+5x)= 4(x + 3)
<=> 2x - 2 - 5x = 4x + 12
<=> 2x - 5x - 4x = 12 + 2
<=> -7x = 14
<=> x = -2
d. 5-(x +8)=3x + 3(x-9)
<=> 5 - x - 8 = 3x + 3x - 27
<=> -x - 3x - 3x = -27 + 8 - 5
<=> -7x = -24
<=> x = 24/7
e. 3x - 18 + x= 12-(5x + 3)
<=> 3x - 18 + x = 12 - 5x - 3
<=> 3x + x - 5x = 12 - 3 + 18
<=> -x = 27
<=> x = - 27
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 7 2021
a.
\(\Leftrightarrow\left\{{}\begin{matrix}3x-2\ge0\\3x^2-17x+4=\left(3x-2\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{2}{3}\\3x^2-17x+4=9x^2-12x+4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{2}{3}\\6x^2+5x=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\dfrac{2}{3}\\\left[{}\begin{matrix}x=0< \dfrac{2}{3}\left(loại\right)\\x=-\dfrac{5}{6}< \dfrac{2}{3}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
Vậy pt đã cho vô nghiệm
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 7 2021
b.
ĐKXĐ: \(\left[{}\begin{matrix}x\ge4\\x\le1\end{matrix}\right.\)
Đặt \(\sqrt{x^2-5x+4}=t\ge0\Leftrightarrow x^2-5x=t^2-4\)
\(\Rightarrow2x^2-10x=2t^2-8\)
Phương trình trở thành:
\(2t^2-8-3t+6=0\)
\(\Leftrightarrow2t^2-3t-2=0\Rightarrow\left[{}\begin{matrix}t=2\\t=-\dfrac{1}{2}< 0\left(loại\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2-5x+4}=2\)
\(\Leftrightarrow x^2-5x=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)
PN
17 tháng 8 2020
a, \(12-2\left(1-x\right)^2=\left(3x-2\right)\left(2x-3\right)\)
\(< =>12-2\left(1-2x+x^2\right)=6x^2-9x-4x+6\)
\(< =>12-2+4x-2x^2=6x^2-13x+6\)
\(< =>10+4x-2x^2-6x^2+13x-6=0\)
\(< =>-8x^2+17x+4=0< =>\orbr{\begin{cases}x=\frac{17-\sqrt{417}}{16}\\x=\frac{17+\sqrt{417}}{16}\end{cases}}\)
b, \(10x+3-5x=4x+12< =>5x+3-4x-12=0\)
\(< =>x-9=0< =>x=9\)
c, \(11x+42-2x=100-9x-22< =>9x+42-100+9x+22=0\)
\(< =>18x+64-100=0< =>18x-36=0< =>x=\frac{36}{18}=2\)
d, \(2x-\left(3-5x\right)=4\left(x+3\right)< =>2x-3+5x=4x+12\)
\(< =>7x-3-4x-12=0< =>3x-15=0< =>x=\frac{15}{3}=5\)
e, \(2\left(x-3\right)+5x\left(x-1\right)=5x^2< =>2x-6+5x^2-5=5x^2\)
\(< =>2x-11+5x^2-5x^2=0< =>2x-11=0< =>x=\frac{11}{2}\)
f, \(-6\left(1,5-2x\right)=3\left(-15+2x\right)< =>-6\left(\frac{3}{2}-2x\right)=3\left(2x-15\right)\)
\(< =>-9+12x-6x+45=0< =>6x+36=0< =>x=-6\)
g, \(14x-\left(2x+7\right)=3x+12x-13< =>14x-2x-7=15x-13\)
\(< =>12x-7-15x+13=0< =>-3x+6=0< =>x=-2\)
h, \(\left(x-4\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)
\(< =>x^2-16-6x+4=x^2-8x+16\)
\(< =>x^2-6x-12-x^2+8x-16=0\)
\(< =>2x-28=0< =>x=\frac{28}{2}=14\)
q, \(4\left(x-2\right)-\left(x-3\right)\left(2x-5\right)=?\)thiếu đề
Sửa đề:\(\frac{3}{x^2+5x+4}+\frac{2}{x^2+10x+24}=\frac{4}{3}=\frac{9}{x^2+3x-18}\)
\(\Leftrightarrow\frac{3}{\left(x+1\right)\left(x+4\right)}+\frac{2}{\left(x+4\right)\left(x+6\right)}=\frac{9}{\left(x-3\right)\left(x+6\right)}=\frac{4}{3}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+6}=\frac{1}{x-3}-\frac{1}{x+6}=\frac{4}{3}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x-6}=\frac{1}{x-3}-\frac{1}{x+6}=\frac{4}{3}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+6}-\frac{1}{x-3}+\frac{1}{x+6}=\frac{4}{3}\)
\(\Leftrightarrow\frac{1}{x+1}-\frac{1}{x+3}=\frac{4}{3}\)
Tự giải tiếp
Quyên sai rồi, tử là 1 mới đc tách kiểu đó, mà 2 pt đó bằng 4/3 thì xét 1 pt thôi được rồi, bước 3 từ dưới lên sai bét