1 Tìm x
\(a,\frac{5}{7}-\left|3x-2\right|=0\)
\(b,\frac{13}{15}-\frac{13}{21}+x.\frac{7}{12}=\frac{7}{10}\)
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\(\frac{x}{2}-\left(\frac{3}{5}x-\frac{13}{5}\right)=-\left(\frac{7}{5}+\frac{7}{10}x\right)\)
\(\Rightarrow\frac{5x-6x+26+14+7x}{10}=0\Rightarrow6x+40=0\Rightarrow x=-\frac{20}{3}\)
A) \(\frac{7}{\left(x+3\right)\left(x+10\right)}+\frac{11}{\left(x+10\right)\left(x+21\right)}+\frac{13}{\left(x+21\right)\left(x+34\right)}\)
\(=\frac{\left(x+10\right)-\left(x+3\right)}{\left(x+3\right)\left(x+10\right)}+\frac{\left(x+21\right)-\left(x+10\right)}{\left(x+10\right)\left(x+21\right)}+\frac{\left(x+34\right)-\left(x+21\right)}{\left(x+21\right)\left(x+34\right)}\)
\(=\frac{1}{x+3}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+21}+\frac{1}{x+21}-\frac{1}{x+34}\)
\(=\frac{1}{x+3}-\frac{1}{x+34}\)
\(=\frac{\left(x+34\right)-\left(x+3\right)}{\left(x+3\right)\left(x+34\right)}\)\(=\frac{x}{\left(x+3\right)\left(x+34\right)}\)
\(\Rightarrow\left(x+34\right)-\left(x+3\right)=x\)
\(\Rightarrow x=31\)
Vậy, x = 31
Bạn áp dụng: \(\frac{k}{x\cdot\left(x+k\right)}=\frac{1}{x}-\frac{1}{x+k}\) với \(x,k\inℝ;x\ne0;x\ne-k\)
Chứng minh: \(\frac{1}{x}-\frac{1}{x+k}=\frac{x+k}{x\left(x+k\right)}-\frac{x}{x\left(x+k\right)}=\frac{x+k-x}{x\left(x+k\right)}=\frac{k}{x\left(x+k\right)}\)
\(\left(\frac{13}{21}+x\right).\frac{7}{12}=\frac{13}{15}-\frac{7}{10}\)
\(\left(\frac{13}{21}+x\right).\frac{7}{12}=\frac{26}{30}-\frac{21}{30}\)
\(\left(\frac{13}{21}+x\right).\frac{7}{12}=\frac{1}{6}\)
\(\frac{13}{21}+x=\frac{1}{6}:\frac{7}{12}\)
\(\frac{13}{21}+x=\frac{1.12}{6.7}\)
\(\frac{13}{21}+x=\frac{2}{7}\)
\(x=\frac{2}{7}-\frac{13}{21}\)
\(x=\frac{6}{21}-\frac{13}{21}\)
\(x=\frac{-1}{3}\)
\(\frac{13}{15}-\left(\frac{13}{21}+x\right).\frac{7}{12}=\frac{7}{10}\)
\(\left(\frac{13}{21}+x\right).\frac{7}{12}=\frac{13}{15}-\frac{7}{10}\)
\(\left(\frac{13}{21}+x\right).\frac{7}{12}=\frac{1}{6}\)
\(\frac{13}{21}+x=\frac{1}{6}:\frac{7}{12}\)
\(\frac{13}{21}+x=\frac{2}{7}\)
\(x=\frac{2}{7}-\frac{13}{21}\)
\(x=-\frac{1}{3}\)
\(\frac{1}{2}+\frac{5}{6}-\frac{3}{8}\)
\(=\frac{12}{24}+\frac{20}{24}-\frac{9}{24}\)
\(=\frac{23}{24}\)
\(\frac{10}{15}\cdot\frac{7}{4}+\frac{7}{4}\cdot\frac{9}{15}-\frac{4}{15}\cdot\frac{7}{4}\)
\(=\frac{7}{4}\cdot\left(\frac{10}{15}+\frac{9}{15}-\frac{4}{15}\right)\)
\(=\frac{7}{4}\cdot1=\frac{7}{4}\)
\(\left(\frac{4}{5}+\frac{1}{2}\right)\left(\frac{3}{13}-\frac{8}{13}\right)\)
\(=\left(\frac{8}{10}+\frac{5}{10}\right)\cdot\left(-\frac{5}{13}\right)\)
\(=\frac{13}{10}\cdot\left(-\frac{5}{13}\right)=-\frac{1}{2}\)
a, 3x-12 = 30
=> 3x = 30 + 12
=> 3x = 42
=> x = 42 : 3 = 14
Vậy x = 14
b, \(\frac{2}{3}x+\frac{1}{4}=\frac{7}{12}\)
\(\Rightarrow\frac{2}{3}x=\frac{7}{12}-\frac{1}{4}\)
\(\Rightarrow\frac{2}{3}x=\frac{7}{12}-\frac{3}{12}\)
\(\Rightarrow\frac{2}{3}x=\frac{1}{3}\)
\(\Rightarrow x=\frac{1}{3}\div\frac{2}{3}\Rightarrow\frac{1}{3}\cdot\frac{3}{2}=\frac{1}{2}\)
Vậy x = \(\frac{1}{2}\)
c, 2x2 = 32
=> x2 = 32 : 2
=> x2 = 16
=> x2 = 42
=> x = 4
Vậy x = 4
\(a.\)\(\Rightarrow\) \(3x-2=\frac{5}{7}\)hoặc \(3x-2=-\frac{5}{7}\)
\(\Rightarrow x=\frac{19}{21}\) hoặc \(x=\frac{3}{7}\)
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