Tìm ĐKXĐ
1)\(\frac{4x-9}{8x^3+1}\)
2)\(\frac{8x^2+x+1}{2x^2-7x+5}\)
3)\(\frac{x+4}{x^2-1}\)+\(\frac{2x+3}{x\left(x+2\right)}\)+\(\frac{2}{x^2+4}\)=\(\frac{11}{3}\)
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K
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Những câu hỏi liên quan
AT
24 tháng 3 2020
\(1,\frac{5x-3}{3}-\frac{6x-7}{4}+x=\frac{2x-5}{6}-x+2\)
\(\Leftrightarrow\frac{\left(5x-3\right)4}{12}-\frac{\left(6x-7\right)3}{12}+\frac{12x}{12}=\frac{\left(2x-5\right)2}{12}-\frac{12x}{12}+\frac{24}{12}\)
\(\Leftrightarrow\frac{20x-12}{12}-\frac{18x-21}{12}+\frac{12x}{12}=\frac{4x-10}{12}-\frac{12x}{12}+\frac{24}{12}\)
\(\Rightarrow20x-12-18x+21+12x=4x-10-12x+24\)
\(\Leftrightarrow20x-18x+12x-4x+12x=-10+24+12-21\)
\(\Leftrightarrow22x=5\)
\(\Leftrightarrow x=\frac{5}{22}\)
câu 2 tương tự
24 tháng 3 2020
\(\frac{x-4}{5}+\frac{3x-2}{10}-x=\frac{2x-5}{3}-\frac{7x+2}{6}\)
\(\Leftrightarrow\frac{2x-8}{10}+\frac{3x-2}{10}-x=\frac{4x-10}{6}-\frac{7x+2}{6}\)
\(\Leftrightarrow\frac{5x-10}{10}-x=\frac{-3x-12}{6}\)
\(\Leftrightarrow\frac{2x-2}{5}-x+\frac{x+4}{2}=0\)
\(\Leftrightarrow\frac{4x-4+5x+20-10x}{10}=0\)
\(\Leftrightarrow x=16\)
Vậy tập nghiệm của phương trình là S={16}
IA
23 tháng 3 2020
<=> \(\frac{29}{2}-\frac{2x+6}{5}=\frac{3x}{2}-\frac{2x-14}{3}\)
<=> \(\frac{435}{30}-\frac{12x+36}{30}=\frac{45x}{30}-\frac{20x-140}{30}\)
<=> \(\frac{435-12x-36}{30}=\frac{45x-20x+140}{30}\)
<=> 399 - 12x = 25x + 140
<=> 37x = -259
<=> x = -7
Vậy .....
23 tháng 3 2020
\(\Leftrightarrow\frac{435}{30}-\frac{12\left(x+3\right)}{30}=\frac{45x}{30}-\frac{20\left(x-7\right)}{30}\)
\(\Leftrightarrow\frac{435-12\left(x+3\right)-45x-20\left(x-7\right)}{30}=0\)
\(\Leftrightarrow435-12x-36-45x-20x+140=0\)
\(\Leftrightarrow-77x=-539\)
\(\Leftrightarrow x=7\)
\(Vậyx=7lànghiệmcủaphươngtrình\)
28 tháng 3 2020
a) ĐKXĐ: x khác +2
\(\frac{x-2}{2+x}-\frac{3}{x-2}-\frac{2\left(x-11\right)}{x^2-4}\)
<=> \(\frac{x-2}{2+x}-\frac{3}{x-2}=\frac{2\left(x-11\right)}{\left(x-2\right)\left(x+2\right)}\)
<=> (x - 2)^2 - 3(2 + x) = 2(x - 11)
<=> x^2 - 4x + 4 - 6 - 3x = 2x - 22
<=> x^2 - 7x - 2 = 2x - 22
<=> x^2 - 7x - 2 - 2x + 22 = 0
<=> x^2 - 9x + 20 = 0
<=> (x - 4)(x - 5) = 0
<=> x - 4 = 0 hoặc x - 5 = 0
<=> x = 4 hoặc x = 5
làm nốt đi
KN
18 tháng 8 2020
1. \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
\(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
\(\Leftrightarrow35x-5+60x=96-6x\)
\(\Leftrightarrow95x-5=96-6x\)
\(\Leftrightarrow95x+6x=96+5\)
\(\Leftrightarrow101x=101\)
\(\Leftrightarrow x=1\)
2. \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
\(\Leftrightarrow3\left(10x+3\right)=36+4\left(6+8x\right)\)
\(\Leftrightarrow30x+9=36+24+32x\)
\(\Leftrightarrow30x+9=32x+60\)
\(\Leftrightarrow30x-32x=60-9\)
\(\Leftrightarrow-2x=51\)
\(\Leftrightarrow x=-\frac{51}{2}\)
3. \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow8x-3-2\left(3x-2\right)=2\left(2x-1\right)+x+3\)
\(\Leftrightarrow8x-3-6x+4=4x-2+x+3\)
\(\Leftrightarrow2x+1=5x+1\)
\(\Leftrightarrow2x=5x\)
\(\Leftrightarrow x=0\)
19 tháng 8 2020
4) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
=> \(\frac{9-3x}{8}+\frac{10-2x}{3}=\frac{1-x}{2}-\frac{2}{1}\)
=> \(\frac{3\left(9-3x\right)}{24}+\frac{8\left(10-2x\right)}{24}=\frac{12\left(1-x\right)}{24}-\frac{48}{24}\)
=> \(\frac{27-9x}{24}+\frac{80-16x}{24}=\frac{12-12x}{24}-\frac{48}{24}\)
=> \(\frac{27-9x+80-16x}{24}=\frac{12-12x-48}{24}\)
=> 27 - 9x + 80 - 16x = 12 - 12x - 48
=> 27 - 9x + 80 - 16x - 12 + 12x + 48 = 0
=> (27 + 80 - 12 + 48) + (-9x - 16x + 12x) = 0
=> 143 - 13x = 0
=> 13x = 143
=> x = 11
5) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
=> \(\frac{2x-6}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
=> \(\frac{3\left(2x-6\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)
=> \(\frac{6x-18}{21}+\frac{7x-35}{21}-\frac{13x+4}{21}=0\)
=> \(\frac{6x-18+7x-35-13x-4}{21}=0\)
=> 6x - 18 + 7x - 35 - 13x - 4 = 0
=> (6x + 7x - 13x) + (-18 - 35 - 4) = 0
=> -57 = 0(vô nghiệm)
6) \(\frac{6x+5}{2}-\left(2x+\frac{2x+1}{2}\right)=\frac{10x+3}{4}\)
=> \(\frac{6x+5}{2}-\frac{10x+3}{4}=2x+\frac{2x+1}{2}\)
=> \(\frac{2\left(6x+5\right)}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{2\left(2x+1\right)}{4}\)
=> \(\frac{12x+10}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{4x+2}{4}\)
=> \(\frac{12x+10-\left(10x+3\right)}{4}=\frac{8x+4x+2}{4}\)
=> \(\frac{12x+10-10x-3}{4}=\frac{12x+2}{4}\)
=> \(12x+10-10x-3=12x+2\)
=> \(2x+10-3=12x+2\)
=> 2x + 10 - 3 - 12x - 2 = 0
=> (2x - 12x) + (10 - 3 - 2) = 0
=> -10x + 5 = 0
=> -10x = -5
=> x = 1/2
7) \(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\)
=> \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{x+7}{15}=0\)
=> \(\frac{6x-3}{15}-\frac{5x-10}{15}-\frac{x+7}{15}=0\)
=> \(\frac{6x-3-\left(5x-10\right)-\left(x+7\right)}{15}=0\)
=> 6x - 3 - 5x + 10 - x - 7 = 0
=> (6x - 5x - x) + (-3 + 10 - 7) = 0
=> 0x + 0 = 0
=> 0x = 0
=> x tùy ý
Bài 8 tự làm nhé
1) 8x3 +1 # 0 => x # -1/2
2) 2x2 -7x + 5 # => 2x2 -2x - 5x +5 # 0 => 2x(x-1) -5(x-1) # 0 => (x-1)(2x-5) # 0 => x # 1 và x # 5/2
3) x2 - 1 # 0 => x # 1 và x # -1
x # 0
x + 2 # 0 => x # -2