GPT
\(\frac{3}{3x^2-4x+1}+\frac{13}{3x^2+2x+1}=\frac{6}{x}\)
Những câu hỏi liên quan
a,frac{3}{x}+frac{1}{x+3}+frac{3}{x+6}+frac{1}{x+7}frac{1}{1-x}b, frac{1}{x-5}+frac{1}{x-2}+frac{1}{x-1}+frac{1}{x}+frac{1}{x+3}frac{3x-3}{4}c,frac{1}{x-3}+frac{1}{3x+1}+frac{10x-13}{4x-6}frac{1}{x+1}+frac{1}{2x-1}+frac{1}{3x+7}d,frac{x^2+x+1}{2x-1}left(frac{3x^2-x+5}{4x-2}-3right)8e,frac{2x^2-3}{3x-1}left(2x-frac{7+4x}{3x-1}right)2f,frac{xleft(3x-1right)left(3x^2+1right)left(6x^2-3x-1right)}{left(x+1right)^3}frac{1}{2}g, xleft(x^2+2right)left(x^2+2x+8+frac{12}{x-2}right)3left(x-2right)
Đọc tiếp
a,\(\frac{3}{x}+\frac{1}{x+3}+\frac{3}{x+6}+\frac{1}{x+7}=\frac{1}{1-x}\)
b, \(\frac{1}{x-5}+\frac{1}{x-2}+\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+3}=\frac{3x-3}{4}\)
c,\(\frac{1}{x-3}+\frac{1}{3x+1}+\frac{10x-13}{4x-6}=\frac{1}{x+1}+\frac{1}{2x-1}+\frac{1}{3x+7}\)
d,\(\frac{x^2+x+1}{2x-1}\left(\frac{3x^2-x+5}{4x-2}-3\right)=8\)
e,\(\frac{2x^2-3}{3x-1}\left(2x-\frac{7+4x}{3x-1}\right)=2\)
f,\(\frac{x\left(3x-1\right)\left(3x^2+1\right)\left(6x^2-3x-1\right)}{\left(x+1\right)^3}=\frac{1}{2}\)
g, \(x\left(x^2+2\right)\left(x^2+2x+8+\frac{12}{x-2}\right)=3\left(x-2\right)\)
GPT: \(\frac{12x}{x^2+4x+2}-\frac{3x}{x^2+2x+2}=1\)
ĐKXĐ: \(x^2+4x+2\ne0\)
Nhận thấy \(x=0\) không phải nghiệm, phương trình đã cho tương đương:
\(\frac{12}{x+\frac{2}{x}+4}-\frac{3}{x+\frac{2}{x}+2}=1\)
Đặt \(x+\frac{2}{x}+2=a\Rightarrow x+\frac{2}{x}+4=a+2\) ta được:
\(\frac{12}{a+2}-\frac{3}{a}=1\Leftrightarrow12a-3\left(a+2\right)=a\left(a+2\right)\)
\(\Leftrightarrow a^2-7a+6=0\Rightarrow\left[{}\begin{matrix}a=1\\a=6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{2}{x}+2=1\\x+\frac{2}{x}+2=6\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2+x+2=0\\x^2-4x+2=0\end{matrix}\right.\) \(\Rightarrow...\)
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Bài 1. Giải các phương trình sau
1) frac{3x+2}{2}-frac{3x+1}{6}frac{5}{3}-2x
2) frac{x-3}{5}6-frac{1-2x}{3}
3) 2left(x+frac{3}{5}right)5-left(frac{13}{5}+xright)
4) frac{2x+3}{3}frac{5-4x}{2}
5) frac{5x+3}{12}frac{1+2x}{9}
6) x-frac{x+1}{3}frac{2x+1}{5}
7) frac{3left(x-3right)}{4}+frac{4x-10,5}{10}frac{3left(x+1right)}{5}+6
8) frac{2left(3x+1right)+1}{4}-5frac{2
left(3x-1right)}{5}-frac{3x+2}{10}
9) frac{x+1}{3}+frac{3left(2x+1right)}{4}frac{2x+3left(x+1right)}{6}+frac{7+12x}{12}
10...
Đọc tiếp
Bài 1. Giải các phương trình sau
1) \(\frac{3x+2}{2}-\frac{3x+1}{6}=\frac{5}{3}-2x\)
2) \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)
3) \(2\left(x+\frac{3}{5}\right)=5-\left(\frac{13}{5}+x\right)\)
4) \(\frac{2x+3}{3}=\frac{5-4x}{2}\)
5) \(\frac{5x+3}{12}=\frac{1+2x}{9}\)
6) \(x-\frac{x+1}{3}=\frac{2x+1}{5}\)
7) \(\frac{3\left(x-3\right)}{4}+\frac{4x-10,5}{10}=\frac{3\left(x+1\right)}{5}+6\)
8) \(\frac{2\left(3x+1\right)+1}{4}-5=\frac{2 \left(3x-1\right)}{5}-\frac{3x+2}{10}\)
9) \(\frac{x+1}{3}+\frac{3\left(2x+1\right)}{4}=\frac{2x+3\left(x+1\right)}{6}+\frac{7+12x}{12}\)
10) \(\frac{2x-1}{3}-\frac{5x+2}{7}=x+13\)
Giải pt: \(\frac{2}{3x^2-4x+1}+\frac{13}{3x^2+2x+1}=\frac{6}{x}\)
GPT: \(\frac{2}{x^2-3x+12}+\frac{6}{x^2+2x+12}=\frac{1}{x}\)
ĐKXĐ: ...
\(\Leftrightarrow\frac{2x}{x^2-3x+12}+\frac{6x}{x^2+2x+12}=1\)
\(\Leftrightarrow\frac{2}{x+\frac{12}{x}-3}+\frac{6}{x+\frac{12}{x}+2}=1\)
Đặt \(x+\frac{12}{x}-3=t\)
\(\Rightarrow\frac{2}{t}+\frac{6}{t+5}=1\Leftrightarrow2\left(t+5\right)+6t=t\left(t+5\right)\)
\(\Leftrightarrow t^2-3t-10=0\Rightarrow\left[{}\begin{matrix}t=5\\t=-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x+\frac{12}{x}-3=-2\\x+\frac{12}{x}-3=5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-x+12=0\\x^2-8x+12=0\end{matrix}\right.\) (casio)
1 tháng 1 2020 lúc 22:36
gpt : a) \(\frac{5x}{\sqrt{4-x^2}}+\frac{8}{x^2}+\frac{2x}{4-x^2}+\frac{5\sqrt{4-x^2}}{x}+4=0\)
b) \(\frac{2x}{\sqrt{8x^2+25}}+\frac{125}{x^2}-14=0\)
c) \(\left(x^3-3x+2\right)\sqrt{3x-2}-2x^3+6x^2-4x=0\)
d) \(\sqrt{x^2-x+6}+\frac{4}{x-1}=x^2+x\)
Akai Haruma, No choice teen, Arakawa Whiter, HISINOMA KINIMADO, tth, Nguyễn Việt Lâm, Phạm Hoàng Lê Nguyên, @Nguyễn Thị Ngọc Thơ
Mn giúp em vs ạ! Thanks trước!
\(c,\left(x^3-3x+2\right)\sqrt{3x-2}-2x^3+6x^2-4x=0\)
\(\Rightarrow\left(x+2\right)\left(x-1\right)^2\sqrt{3x-2}-2x\left(x^2-3x+2\right)=0\)
\(\Rightarrow\left(x+2\right)\left(x-1\right)^2\sqrt{3x-2}-2x\left(x-1\right)\left(x-2\right)=0\)
\(\Rightarrow x=1\)
Hoặc là: \(\Rightarrow\left(x+2\right)\left(x-1\right)\sqrt{3x-2}-2x\left(x-2\right)=0\)
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Còn cần nữa không, hôm bữa chị giải ra câu a mà quên béng mất, mấy hôm lại bận làm thuyết trình Tiếng Anh nên bỏ dở.
Giờ mà cần chị cũng chỉ làm được câu a thôi '-'
GPT: \(\frac{2}{x^2-3x+12}+\frac{6}{x^2+2x+12}=\frac{1}{x}\)
\(pt\Leftrightarrow\frac{2x}{x^2-3x+12}+\frac{6x}{x^2+2x+12}=1\)
\(\Leftrightarrow\frac{2}{x-3+\frac{12}{x}}+\frac{6}{x+2+\frac{12}{x}}=1\)
Đặt \(x+\frac{12}{x}=t\)
Khi đó:
\(pt\Leftrightarrow\frac{2}{t-3}+\frac{6}{t+2}=1\Leftrightarrow2t+4+6t-18=t^2-t-6\)
\(\Leftrightarrow t^2-t-6=8t-14\)
\(\Leftrightarrow t^2-9t+8=0\)
\(\Leftrightarrow\left(t-8\right)\left(t-1\right)=0\)
\(\Leftrightarrow x+\frac{12}{x}=8;x+\frac{12}{x}=1\)
Thôi,bí rồi
Bài 4: Giải các phương trình sau
a) 4(x+5)(x+6)(x+10)(x+12)=\(3x^2\)
b) \(\frac{1}{x^2-3x+3}+\frac{2}{x^2-3x+4}=\frac{6}{x^2-3x+5}\)
c) \(\frac{4x}{4x^2-8x+7}+\frac{3x}{4x^2-10x+7}=1\)
d) \(\frac{2x}{2x^2-5x+3}+\frac{13x}{2x^2+x+3}\)
a) 4 ( x + 5 )( x + 6 )( x + 10 )( x + 12 ) = 3x2
Do x = 0 không là nghiệm pt nên chia 2 vế pt cho \(x^2\ne0\), ta được :
\(\frac{4}{x^2}\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3\)
\(\Leftrightarrow4\left(x+\frac{60}{x}+17\right)\left(x+\frac{60}{x}+16\right)=3\)
Đến đây ta đặt \(x+\frac{60}{x}+16=t\left(1\right)\)
Ta được :
\(4t\left(t+1\right)=3\Leftrightarrow4t^2+4t-3=0\Leftrightarrow\left(2t+3\right)\left(2t-1\right)=0\)
Từ đó ta lắp vào ( 1 ) tính được x
giải phương trình
a) \(\frac{4x-8}{2x^2+1}=0\)
b)\(\frac{x^2-x-6}{x-3}=0\)
c)\(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}\)
d)\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)
a.ĐK: 2x2+1\(\ne0\) \(\forall x\)
Để phương trình bằng 0 thì 4x-8=0 ( Vì 2x2+1 >0 với mọi x)
\(\Leftrightarrow x=2\) (TM)
Vậy ...
b.ĐK: x-3\(\ne0\) \(\Leftrightarrow x\ne3\)
Để phương trình bằng 0 thì x2-x-6=0 (Vì x-3\(\ne0\))
\(\Leftrightarrow\left[{}\begin{matrix}x=2\:\left(TM\right)\\x=-3\:\left(TM\right)\end{matrix}\right.\)
Vậy ...
c. ĐK: x\(\ne\)2
\(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}\Leftrightarrow\frac{x+5}{3\left(x-2\right)}-\frac{1}{2}=\frac{2x-3}{2\left(x-2\right)}\)
\(\Leftrightarrow\frac{2\left(x+5\right)-3\left(x-2\right)}{6\left(x-2\right)}=\frac{3\left(2x-3\right)}{6\left(x-2\right)}\)
\(\Leftrightarrow2x+10-3x+6=6x-9\) (x\(\ne\)2)
\(\Leftrightarrow x=\frac{25}{7}\left(TM\right)\)
Vậy ...
d. ĐK: \(x\ne\pm\frac{1}{3}\)
\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)
\(\Leftrightarrow\frac{12}{1-9x^2}=\frac{\left(1-3x\right)^2-\left(1+3x\right)^2}{1-9x^2}\)
\(\Leftrightarrow12=1-6x+9x^2-1-6x-9x^2\) (\(x\ne\pm\frac{1}{3}\))
\(\Leftrightarrow x=-2\:\left(TM\right)\)
Vậy...