\(\frac{x^2+x+1}{x^2+2x+1}+\frac{x^2+3x+1}{x^2+4x+1}=\frac{19}{12}\)
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, \(x\left(x^2+2\right)\left(x^2+2x+8+\frac{12}{x-2}\right)=3\left(x-2\right)\)
GIẢI PHƯƠNG TRÌNH
a)\(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
b)\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
c)\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)
d)\(\frac{x+4}{x^2-3x+2}+\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)
a) \(\frac{1-x}{x+1}+3=\frac{2x+3}{x+1}\)
<=> 1 - x + 3(x + 1) = 2x + 3
<=> 1 - x + 3x + 3 = 2x + 3
<=> 1 - x + 3x + 3 - 2x = 3
<=> 4 = 3 (vô lý)
=> pt vô nghiệm
b) ĐKXĐ: \(x\ne1;x\ne2\)
\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
<=> (x - 2)(2 - x) - 5(x + 1)(2 - x) = 15(x - 2)
<=> 2x - x2 - 4 + 2x - 5x - 5x2 + 10 = 15x - 30
<=> -x + 4x2 - 14 = 15x - 30
<=> x - 4x2 + 14 = 15x - 30
<=> x - 4x2 + 14 + 15x - 30 = 0
<=> 16x - 4x2 - 16 = 0
<=> 4(4x - x2 - 4) = 0
<=> -x2 + 4x - 4 = 0
<=> x2 - 4x + 4 = 0
<=> (x - 2)2 = 0
<=> x - 2 = 0
<=> x = 2 (ktm)
=> pt vô nghiệm
c) xem bài 4 ở đây: Câu hỏi của gjfkm
d) ĐKXĐ: \(x\ne1;x\ne2;x\ne3\)
\(\frac{x+4}{x^2-3x+2}+\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)
<=> \(\frac{x+4}{\left(x-1\right)\left(x-2\right)}+\frac{x+1}{\left(x-1\right)\left(x-3\right)}=\frac{2x+5}{\left(x-1\right)\left(x-3\right)}\)
<=> (x + 4)(x - 3) + (x + 1)(x - 2) = (2x + 5)(x - 2)
<=> x2 - 3x + 4x - 12 + x2 - 2x + x - 2 = 2x2 - 4x + 5x - 10
<=> 2x2 - 14 = 2x2 + x - 10
<=> 2x2 - 14 - 2x2 = x - 10
<=> -14 = x - 10
<=> -14 + 10 = x
<=> -4 = x
<=> x = -4
bài 1:giải các pt sau:
a/frac{1-x}{x+1}+3frac{2x+3}{x+1}
b/frac{left(x+2right)^2}{2x-3}-1frac{x^2+10}{2x-3}
c/frac{1}{x+1}-frac{5}{x-2}frac{15}{left(x+1right)left(2-xright)}
d/frac{1-6x}{x-2}+frac{9x+4}{x+2}frac{xleft(3x-2right)+1}{x^2-4}
e/frac{12}{1-9x^2}frac{1-3x}{1+3x}-frac{1+3x}{1-3x}
ffrac{x+4}{x^2-3x+2}+frac{x+1}{x^2-4x+3}frac{2x+5}{x^2-4x+3}
Đọc tiếp
bài 1:giải các pt sau:
a/\(\frac{1-x}{x+1}\)+3=\(\frac{2x+3}{x+1}\)
b/\(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2+10}{2x-3}\)
c/\(\frac{1}{x+1}-\frac{5}{x-2}=\frac{15}{\left(x+1\right)\left(2-x\right)}\)
d/\(\frac{1-6x}{x-2}+\frac{9x+4}{x+2}=\frac{x\left(3x-2\right)+1}{x^2-4}\)
e/\(\frac{12}{1-9x^2}=\frac{1-3x}{1+3x}-\frac{1+3x}{1-3x}\)
f\(\frac{x+4}{x^2-3x+2}+\frac{x+1}{x^2-4x+3}=\frac{2x+5}{x^2-4x+3}\)
Gpt
a) \(x^2+\sqrt{x+5}=5\)
b)\(\sqrt{x-\frac{x}{1}}-\sqrt{1-\frac{1}{x}}=1-\frac{1}{x}\)
c) \(\frac{x^2-x+1}{x^2-2x+1}+\frac{x^2+3x+1}{x^2+4x+1}=\frac{19}{12}\)
d) \(\sqrt{1-2x}+\sqrt{1+2x}=2-x^2\)
a) \(x^2-5+\sqrt{x+5}=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+5\right)+\sqrt{x+5}=0\)(tự làm tiếp)
b) Đề hơi sai sai
c) Mik chưa nghĩ ra
d) \(\left(\sqrt{1-2x}-1\right)+\left(\sqrt{1+2x}-1\right)+x^2=0\)
\(\frac{-2x}{\sqrt{1-2x}+1}+\frac{2x}{\sqrt{1+2x}+1}+x^2=0\)(tự lm tiếp)
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Giải pt:
a) \(\frac{x^2+2x-16}{x^2-x-12}+1=\frac{2x+1}{x+3}+\frac{3x-8}{x-4}\)
b) \(\frac{2x-1}{x+2}+\frac{7x+9}{\left(x+2\right)\left(x-1\right)}=\frac{3x-1}{x-1}\)
c) \(\frac{x+1}{20}+\frac{x+2}{19}+\frac{x+3}{18}=\frac{x+20}{1}+\frac{x+19}{2}+\frac{x+18}{3}\)
Giải giúp mình với ạ :((
\(\frac{x-1}{12}+\frac{2x-12}{14}=\frac{3x-14}{25}+\frac{4x-29}{27}\)\(\frac{3}{x+1}+\frac{4}{x-2}=/frac{5}{x-3}+\frac{6}{x-4}\)\(\frac{2x-1}{2}+\frac{2x+1}{3}=/frac{2x+7}{6}+\frac{2x+9}{7}\)
Xem chi tiết
1)2x(25x-4)-(5x-2)(5x+1)8 / 5)2left(x-2right)-3left(3x-1right)left(x-3right)
2)x(4x-3)-(2x-2)(2x-1)5 / 6)frac{2}{x+1}-frac{1}{x-2}frac{3x-11}{left(x+1right)left(x-2right)}
3)frac{5}{2x+3}+frac{3}{9-x^2}frac{8}{7left(x3right)} / 7)frac{5x-2}{6}+frac{3-4x}{2}2-frac{x+7}{3}
4)frac{2}{3left(x-2right)}+frac{5}{12-3x^2}frac{3}{4left(x+2right)} /...
Đọc tiếp
1)2x(25x-4)-(5x-2)(5x+1)=8 / 5)\(2\left(x-2\right)-3\left(3x-1\right)=\left(x-3\right)\)
2)x(4x-3)-(2x-2)(2x-1)=5 / 6)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
3)\(\frac{5}{2x+3}+\frac{3}{9-x^2}=\frac{8}{7\left(x=3\right)}\) / 7)\(\frac{5x-2}{6}+\frac{3-4x}{2}=2-\frac{x+7}{3}\)
4)\(\frac{2}{3\left(x-2\right)}+\frac{5}{12-3x^2}=\frac{3}{4\left(x+2\right)}\) / 8)\(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right)\left(x-2\right)}\)
Đây là lớp 8 nha các b giúp mk với
Do mk viết nhầm
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Giải pt:
a) \(\frac{^{x^2+2x-16}}{x^2-x-12}+1=\frac{2x+1}{x+3}+\frac{3x-8}{x-4}\)
b) \(\frac{2x-1}{x+2}+\frac{7x+9}{\left(x+2\right)\left(x-1\right)}=\frac{3x-1}{x-1}\)
c) \(\frac{x+1}{20}+\frac{x+2}{19}+\frac{x+3}{18}=\frac{x+20}{1}+\frac{x+19}{2}+\frac{x+18}{3}\)
a) \(\frac{7}{8x}+\frac{5-x}{4x^2-8x}=\frac{x-1}{2x^2-4x}+\frac{1}{8x-16}\)
b) 3x.|x+1|−2x|x+2|=12
<=> \(\frac{7}{8x}+\frac{5-x}{4x\left(x-2\right)}=\frac{x-1}{2x\left(x-2\right)}+\frac{1}{8\left(x-2\right)}\)(DK: x khác 0 và 2)
<=>\(\frac{7x\left(x-2\right)}{8x\left(x-2\right)}+\frac{10-2x}{8x\left(x-2\right)}=\frac{4x-4}{8x\left(x-2\right)}=\frac{x}{8x\left(x-2\right)}\)
<=>\(7x^2-14x+10-2x=4x-4+x\)
<=>\(7x^2-14x-2x-4x-x=-4-10\)
<=>\(7x^2-21x+14=0\)
<=>\(7\left(x^2-3x+2\right)=0\)
<=>\(x^2-3x+2=0\)
<=>\(x^2-x-2x+2=0\)
<=>\(x\left(x-1\right)-2\left(x-1\right)=0\)
<=>\(\left(x-1\right)\left(x-2\right)=0\)
<=>\(\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\left(TMDK\right)\\x=2\left(KTMDK\right)\end{cases}}\)
Vậy: x=1
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