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a)|x+0,573|=2
=>x+0,573=2 hoặc -2
Xét x+0,573=2
=>x=1,427
Xét x+0,573=-2
=>x=-2,573
a) | x + 0,573 | = 2
\(\Rightarrow\)x + 0,573 = 2 hoặc x + 0,573 = -2
+) x + 0,573 = 2\(\Rightarrow\)x = 1,427
+) x + 0,573 = -2\(\Rightarrow\)x = -2,573
Vậy x = 1,427 hoặc -2,573
b) \(\left|x+\frac{1}{3}\right|-4=-1\)
\(\Rightarrow\left|x+\frac{1}{3}\right|=3\)
\(\Rightarrow x+\frac{1}{3}=3\) hoặc \(x+\frac{1}{3}=-3\)
+) \(x+\frac{1}{3}=3\Rightarrow x=\frac{8}{3}\)
+) \(x+\frac{1}{3}=-3\Rightarrow x=\frac{-10}{3}\)
Vậy \(x=\frac{8}{3}\) hoặc \(x=\frac{-10}{3}\)
Các phần khác làm tương tự nhé bạn
Ta có: \(\frac{3}{\left(x+2\right)\left(x+5\right)}=\frac{1}{x+2}-\frac{1}{x+5}\); \(\frac{5}{\left(x+5\right)\left(x+10\right)}=\frac{1}{x+5}-\frac{1}{x+10}\)
\(\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{1}{x+10}-\frac{1}{x+17}\);
=> Phương trình tương đương:
\(\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)<=> \(\frac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
<=> \(\frac{15}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
=> x=15
Đáp số: x=15
=> \(\frac{\left(x+5\right)-\left(x+3\right)}{\left(x+2\right)\left(x+5\right)}+\frac{\left(x+10\right)-\left(x+5\right)}{\left(x+5\right)\left(x+10\right)}+\frac{\left(x+17\right)-\left(x+10\right)}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
=> \(\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
=> \(\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\) => \(\frac{15}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\) => x = 15
\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\frac{\left(x+5\right)-\left(x+2\right)}{\left(x+2\right)\left(x+5\right)}+\frac{\left(x+10\right)-\left(x+5\right)}{\left(x+5\right)\left(x+10\right)}+\frac{\left(x+17\right)-\left(x+10\right)}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\frac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\frac{15}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow x=15\)
Theo đề ta có :
\(\frac{3}{\left(x+2\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Leftrightarrow\frac{\left(x+5\right)-\left(x+2\right)}{\left(x+2\right)\left(x+5\right)}+\frac{\left(x+10\right)-\left(x+5\right)}{\left(x+5\right)\left(x+10\right)}+\frac{\left(x+17\right)-\left(x+10\right)}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\frac{1}{x+2}-\frac{1}{x+17}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\frac{\left(x+17\right)-\left(x+2\right)}{\left(x+2\right)\left(x+17\right)}=\frac{x}{\left(x+2\right)\left(x+17\right)}\)
\(\Rightarrow\left(x+17\right)-\left(x+2\right)=x\)
\(\Rightarrow x=15\)
\(\left|x+\frac{1}{3}\right|+\frac{4}{5}=\left|-3,2+\frac{2}{5}\right|+\left(27-\frac{3}{5}\right)\left(27-\frac{3^2}{6}\right)...\left(27-\frac{3^5}{9}\right)...\left(27-\frac{3^{2010}}{2014}\right)\)
\(\Leftrightarrow\left|x+\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}+\left(27-\frac{3^2}{6}\right)\left(27-\frac{3^3}{7}\right)...\left(27-27\right)...\left(27-\frac{3^{2010}}{2014}\right)\)
\(\Leftrightarrow\left|x+\frac{1}{3}\right|+\frac{4}{5}=\frac{14}{5}\)
\(\Leftrightarrow\left|x+\frac{1}{3}\right|=2\)
\(\Rightarrow\hept{\begin{cases}x+\frac{1}{3}=2\\x+\frac{1}{3}=-2\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{5}{3}\\x=-\frac{7}{3}\end{cases}}}\)
bạn ơi, có một chỗ chưa chuẩn .bạn kiểm tra lại giú mình. chỗ vế trái bạn thiếu \(\left(27-\frac{3}{5}\right)\). bạn bổ sung vào cho đúng nhé. dù sao vẫn cảm ơn bạn.
tìm x biết:
\(\left(\frac{1}{4}x-1\right)-\left(\frac{5}{6}x+2\right)-\left(1-\frac{5}{8}x\right)=3\)
\(\left(\frac{1}{4}x-1\right)-\left(\frac{5}{6}x+2\right)-\left(1-\frac{5}{8}x\right)=3\)
=> \(\frac{1}{4}x-1-\frac{5}{6}x-2-1+\frac{5}{8}x=3\)
=> \(\left(\frac{1}{4}x-\frac{5}{6}x+\frac{5}{8}x\right)+\left(-1-2-1\right)=3\)
=> \(\frac{1}{24}x-4=3\)
=> \(\frac{1}{24}x=7\)
=> \(x=7:\frac{1}{24}=7\cdot24=168\)
Vậy x = 168
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: \(\frac{1}{x}-\frac{1}{x+k}=\frac{x+k}{x\left(x+k\right)}-\frac{x}{x\left(x+k\right)}=\frac{x+k-x}{x\left(x+k\right)}=\frac{k}{x\left(x+k\right)}\)
\(x=\frac{1,56\left(5\right)-2,65\left(6\right)}{3,6\left(5\right)}\)
\(x=\frac{1,56+0,00\left(5\right)-2,65-0,00\left(6\right)}{3,6+0,0\left(5\right)}\)
\(x=\frac{\frac{39}{25}+\frac{0,\left(5\right)}{100}-\frac{53}{20}-\frac{0,\left(6\right)}{100}}{\frac{18}{5}+\frac{0,\left(5\right)}{10}}\)
\(x=\frac{\frac{39}{25}+\frac{5}{9}:100-\frac{53}{20}-\frac{6}{9}:100}{\frac{18}{5}+\frac{5}{9}:10}\)
\(x=\frac{\frac{39}{25}+\frac{1}{180}-\frac{53}{20}-\frac{1}{150}}{\frac{18}{5}+\frac{1}{18}}\)
\(x=\frac{\frac{1404}{300}+\frac{5}{900}-\frac{2385}{900}-\frac{6}{900}}{\frac{324}{90}+\frac{5}{90}}\)
\(x=\frac{-982}{900}:\frac{329}{90}\)
\(x=\frac{-491}{450}.\frac{90}{329}=\frac{-491}{1645}\)