\(\frac{5}{2x-4}+\frac{7}{x+2}-\frac{10}{x^2-4}\)
ai giải nhanh giùm mình với
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\(ĐKXĐ:x\ne\pm5\)
\(\frac{3}{4\left(x-5\right)}+\frac{15}{50-2x^2}=\frac{-7}{6\left(x+5\right)}\)
\(\Rightarrow\frac{3\left(x+5\right)}{4\left(x-5\right)\left(x+5\right)}+\frac{30}{4\left(25-x^2\right)}=\frac{-7\left(x-5\right)}{6\left(x+5\right)\left(x-5\right)}\)
\(\Rightarrow\frac{3x+15}{4\left(x-5\right)\left(x+5\right)}+\frac{-30}{4\left(x-5\right)\left(x+5\right)}=\frac{-7\left(x-5\right)}{6\left(x+5\right)\left(x-5\right)}\)
\(\Rightarrow\frac{3x+15-30}{4\left(x-5\right)\left(x+5\right)}=\frac{-7\left(x-5\right)}{6\left(x+5\right)\left(x-5\right)}\)
\(\Rightarrow\frac{3x-15}{4\left(x-5\right)\left(x+5\right)}=\frac{-7\left(x-5\right)}{6\left(x+5\right)\left(x-5\right)}\)
\(\Rightarrow\frac{3\left(x-5\right)}{4\left(x-5\right)\left(x+5\right)}=\frac{-7\left(x-5\right)}{6\left(x+5\right)\left(x-5\right)}\)
\(\Rightarrow\frac{3}{4\left(x+5\right)}=\frac{-7}{6\left(x+5\right)}\)
\(\Rightarrow18\left(x+5\right)=-28\left(x+5\right)\)
\(\Rightarrow18\left(x+5\right)+28\left(x+5\right)=0\)
\(\Rightarrow46\left(x+5\right)=0\Leftrightarrow x+5=0\Leftrightarrow x=-5\)(ktm)
Vậy pt vô nghiệm
\(\frac{2}{5}:x=-\frac{1}{4}\)
\(x=\frac{2}{5}:-\frac{1}{4}\)
\(x=\frac{2}{5}\cdot-\frac{4}{1}\)
\(x=-\frac{8}{5}\)
Vậy \(x=-\frac{8}{5}\)
\(\frac{4}{7}\cdot x-\frac{2}{3}=\frac{1}{5}
\)
\(\frac{4}{7}\cdot x=\frac{1}{5}+\frac{2}{3}\)
\(\frac{4}{7}\cdot x=\frac{3}{15}+\frac{10}{15}\)
\(\frac{4}{7}\cdot x=\frac{13}{15}\)
\(x=\frac{13}{15}:\frac{4}{7}\)
\(x=\frac{13}{15}\cdot\frac{7}{4}\)
\(x=\frac{91}{60}\)
Vậy \(x=\frac{91}{60}\)
2/5 : x = -1/4
=> x = 2/5 : -1/4
=> x = -8/5
4/7.x - 2/3 = 1/5
=> 4/7x = 1/5-2/3
=> 4/7x = -7/15
=> x = -49/60
\(A=\frac{2-\frac{2}{19}+\frac{2}{43}-\frac{2}{1995}}{3-\frac{3}{19}+\frac{3}{43}-\frac{3}{1995}}:\frac{4-\frac{4}{29}+\frac{4}{41}-\frac{4}{2941}}{5-\frac{5}{29}+\frac{5}{41}-\frac{5}{2941}}\)
\(\Rightarrow A=\frac{2\left(1-\frac{1}{19}+\frac{1}{43}-\frac{1}{1995}\right)}{3\left(1-\frac{1}{19}+\frac{1}{43}-\frac{1}{1995}\right)}:\frac{4\left(1-\frac{1}{29}+\frac{1}{41}-\frac{1}{2941}\right)}{5\left(1-\frac{1}{29}+\frac{1}{41}-\frac{1}{2941}\right)}\)
\(\Rightarrow A=\frac{2}{3}:\frac{4}{5}\)
\(\Rightarrow A=\frac{2}{3}.\frac{5}{4}\)
\(\Rightarrow A=\frac{10}{12}=\frac{5}{6}\)
Theo đề bài cho => \(4-\frac{5}{6-\frac{7}{8-\frac{9}{10}}}=4-\frac{5}{x}\)
=> \(x=6-\frac{7}{8-\frac{9}{10}}\)
Vậy x=...
Nếu thế mk đúng thì ủng hộ nha
Ta có:\(2-\frac{3}{4-\frac{5}{6-\frac{7}{8-\frac{9}{10}}}}=2-\frac{3}{4-\frac{5}{6-\frac{7}{\frac{71}{10}}}}=2-\frac{3}{4-\frac{5}{6-\frac{70}{71}}}\)
\(=2-\frac{3}{4-\frac{5}{\frac{356}{71}}}\)
=>x=\(\frac{356}{71}\)
1) \(\frac{x-3}{2}+\frac{4x+1}{3}=\frac{2x-7}{6}\)
<=> 3(x - 3) + 2(4x + 1) = 2x - 7
<=> 3x - 9 + 8x + 2 = 2x - 7
<=> 11x - 7 = 2x - 7
<=> 11x - 7 - 2x = -7
<=> 9x - 7 = -7
<=> 9x = -7 + 7
<=> 9x = 0
<=> x = 0
\(x^2-2x+3=t\left(t\ge0\right)\)
\(pt\Leftrightarrow\frac{1}{t-1}+\frac{1}{t}=\frac{9}{2\left(t+1\right)}\)
\(\Leftrightarrow\frac{2t\left(t+1\right)}{2t\left(t^2-1\right)}+\frac{2\left(t^2-1\right)}{2t\left(t^2-1\right)}-\frac{9t\left(t-1\right)}{2t\left(t^2-1\right)}=0\)
\(\Leftrightarrow-5t^2+11t-2=0\)
\(\Leftrightarrow\left(5x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=2\end{cases}}\)
1.
\(\frac{2x+3}{4}-\frac{5x+3}{6}=\frac{3-4x}{12}\)
\(MC:12\)
Quy đồng :
\(\Rightarrow\frac{3.\left(2x+3\right)}{12}-\left(\frac{2.\left(5x+3\right)}{12}\right)=\frac{3x-4}{12}\)
\(\frac{6x+9}{12}-\left(\frac{10x+6}{12}\right)=\frac{3x-4}{12}\)
\(\Leftrightarrow6x+9-\left(10x+6\right)=3x-4\)
\(\Leftrightarrow6x+9-3x=-4-9+16\)
\(\Leftrightarrow-7x=3\)
\(\Leftrightarrow x=\frac{-3}{7}\)
2.\(\frac{3.\left(2x+1\right)}{4}-1=\frac{15x-1}{10}\)
\(MC:20\)
Quy đồng :
\(\frac{15.\left(2x+1\right)}{20}-\frac{20}{20}=\frac{2.\left(15x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-20=2\left(15x-1\right)\)
\(\Leftrightarrow30x+15-20=15x-2\)
\(\Leftrightarrow15x=3\)
\(\Leftrightarrow x=\frac{3}{15}=\frac{1}{5}\)
Bài 2
a. \(-1\frac{2}{3}-|2x-1|:\frac{3}{5}=-2\)
\(|2x-1|:\frac{3}{5}=\frac{5}{3}-2\)
\(|2x-1|:\frac{3}{5}=-\frac{1}{3}\)
\(|2x-1|=-\frac{1}{5}\)
Vì giá trị tuyệt đối luôn \(\ge0\)với mọi x
mà \(-\frac{1}{5}< 0\)
=> \(x\in\varnothing\)
Tìm x hay đề yêu cầu làm gì?