Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1.\(\frac{456}{461}va\frac{123}{128}\)
Ta có: \(\frac{456}{461}+\frac{5}{461}=1\)
\(\frac{123}{128}+\frac{5}{128}=1\)
vì \(\frac{5}{461}< \frac{5}{128}\)nên \(\frac{456}{461}>\frac{123}{128}\)
2.\(\frac{53}{57}va\frac{531}{571}\)
Vì \(\frac{53}{57}< 1\)
\(\Rightarrow\frac{53}{57}=\frac{530}{570}< \frac{530+1}{570+1}=\frac{531}{571}\)
\(\Rightarrow\frac{53}{57}< \frac{531}{571}\)
a) ta có: \(1-\frac{88}{89}=\frac{1}{89}\)
\(1-\frac{95}{96}=\frac{1}{96}\)
Vì 89 < 96 nên \(\frac{1}{89}>\frac{1}{96}\) nên \(\frac{88}{89}>\frac{95}{96}\)
\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)
=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)
=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)
Vậy \(x\in\left\{\frac{9}{20}\right\}\)
\(b,x+\frac{1}{4}=\frac{4}{3}\)
=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)
Vậy \(x\in\left\{\frac{13}{12}\right\}\)
\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)
=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)
Vậy \(x\in\left\{\frac{25}{42}\right\}\)
\(d,\left|x+5\right|-6=9\)
=> \(\left|x+5\right|=9+6=15\)
=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)
Vậy \(x\in\left\{10;-20\right\}\)
\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)
=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)
\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{6}\)
=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)
\(g,x^2=16\)
=> \(\left|x\right|=\sqrt{16}=4\)
=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
vậy \(x\in\left\{4;-4\right\}\)
\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)
=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)
Vậy \(x\in\left\{\frac{5}{6}\right\}\)
\(i,3^3.x=3^6\)
\(x=3^6:3^3=3^3=27\)
Vậy \(x\in\left\{27\right\}\)
\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)
=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)
Vậy \(x\in\left\{\frac{5}{27}\right\}\)
\(k,1\frac{2}{3}:x=6:0,3\)
=> \(\frac{5}{3}:x=20\)
=> \(x=\frac{5}{3}:20=\frac{1}{12}\)
Vậy \(x\in\left\{\frac{1}{12}\right\}\)
a) Ta có: \(\frac{179}{197}< 1\)và \(\frac{971}{917}>1\)
\(\Rightarrow\frac{179}{197}< 1< \frac{971}{917}\)
\(\Rightarrow\frac{179}{197}< \frac{971}{917}\)
b)Ta có: \(\frac{64}{85}< \frac{64}{81}\) mà \(\frac{73}{81}>\frac{64}{81}\)
\(\Rightarrow\frac{64}{85}< \frac{64}{81}< \frac{73}{81}\)
\(\Rightarrow\frac{64}{85}< \frac{73}{81}\)
Tự làm tiếp nha ~
a) Vì 179/197 < 1 ; 971/917 > 1
=> 179/197 < 971/917
b) Vì 64/85 < 64/81 < 73/81
=> 64/85 < 73/81
c) Ta có: 456/461 = 1 - 5/461 ; 123/128 = 1 - 5/128
Vì 5/461 < 5/128 => 1 - 5/461 > 1 - 5/128
=> 456/461 > 123/128
d) 53/57 = 530/570
Áp dụng a/b < 1 => a/b < a+m/b+m (a,b,m thuộc N*)
Do 530/570 < 1 => 530/570 < 530+1/570+1
=> 53/57 < 531/571
e) Ta có: 58/89 > 58/87 = 2/3
36/53 < 36/54 = 2/3
=> 58/89 > 36/53
g) Ta có: 11/32 > 11/33 = 1/3
16/49 < 16/48 = 1/3
=> 11/32 > 16/49