giải phương trình sau:
\(\frac{x+2}{x^2+2x+4}\)-\(\frac{x-2}{x^2-2x+4}\)=\(\frac{16\left(x^2-x\right)}{x^6-8^2}\)
K
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MN
3 tháng 2 2020
\(ĐKXĐ:x\ne2\)
\(\frac{2\left(x^2+x+6\right)}{x^2-8}+\frac{2}{2-x}=\frac{3}{x^2+2x+4}\)
\(\Leftrightarrow\frac{2x^2+2x+12}{\left(x-2\right)\left(x^2+2x+4\right)}-\frac{2}{x-2}-\frac{3}{x^2+2x+4}=0\)
\(\Leftrightarrow\frac{2x^2+2x+12-2\left(x^2+2x+4\right)-3\left(x-2\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=0\)
\(\Leftrightarrow2x^2+2x+12-2x^2-4x-8-3x+6=0\)
\(\Leftrightarrow-5x+10=0\)
\(\Leftrightarrow x=2\left(ktm\right)\)
Vậy phương trình vô nghiệm
MN
3 tháng 2 2020
Nhầm :)) Dòng đầu tiên cậu sửa cho tớ thành như này nhé :
\(\frac{2\left(x^2+x+6\right)}{x^3-8}+\frac{2}{2-x}=\frac{3}{x^2+2x+4}\)
Đề bài của cậu bị sai rồi nhé :D
9 tháng 3 2020
tớ ko bt lm abc , tớ lm d thôi nha , thứ lỗi
\(\frac{5}{2x-3}-\frac{1}{x+2}=\frac{5}{x-6}-\frac{7}{2x-1}\)
\(\frac{3x+13}{2x^2+x-6}=\frac{5}{x-6}+\frac{7}{1-2x}\)
\(\frac{3x+13}{\left(x+2\right)\left(2x-3\right)}=\frac{3x+37}{\left(x-6\right)\left(2x-1\right)}\)
\(\frac{10-9x}{-4x^3+32x^2-51x+18}=0\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{10}{9}\end{cases}}\)
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
HN
15 tháng 1 2019
a, \(\frac{\left(x-2\right)^2}{3}-\frac{\left(2x-3\right).\left(2x+3\right)}{8}+\frac{\left(x-4\right)^2}{6}=0\)
\(\Leftrightarrow\frac{x^2-4x+4}{3}+\frac{9-4x^2}{8}+\frac{x^2-8x+16}{6}=0\)
\(\Leftrightarrow\frac{8\left(x^2-4x+4\right)+3\left(9-4x^2\right)+4\left(x^2-8x+16\right)}{24}=0\)
\(\Leftrightarrow\frac{8x^2-32x+32+27-12x^2+4x^2-32x+64}{24}=0\)
\(\Leftrightarrow\frac{123-64x}{24}=0\Leftrightarrow123-64x=0\Leftrightarrow x=\frac{123}{64}\)