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a. Ta có : 0,(123) = 0, 123123123.......
Vì 1= 1 → 0,123123... = 0,123
hay 0,(123) = 0,123
b) Ta có : -0,4(32) = -0,4323232.............
Vì 432= 432 → -0,432=-0,4(32)
Bài 1:
Theo đề, ta có:
\(\dfrac{-13}{2}< \dfrac{11}{a}< \dfrac{-13}{3}\)
\(\Leftrightarrow\dfrac{-143}{26}< \dfrac{-143}{-13a}< \dfrac{-143}{33}\)
\(\Leftrightarrow\dfrac{143}{26}>\dfrac{143}{-13a}>\dfrac{143}{33}\)
hay \(a\in\varnothing\)
\(\frac{2a+13b}{3a-7b}=\frac{2c+13d}{3c-7d}\)
<=> (2a+13b)(3c-7d)=(2c+13d)(3a-7b)
<=> 6ac-14ad+39bc-91bd=6ac-14bc+39ad-91bd
<=>14ad-39bc=14bc-39ad
<=>53ad=53bc<=> ad=bc<=>a/b=c/d
=> ĐPCM
Bài 1:
\(\frac{\frac{5}{131}+\frac{5}{141}-\frac{5}{191}-\frac{5}{4011}}{\frac{7}{131}+\frac{7}{141}+\frac{7}{-191}-\frac{7}{4011}}=\frac{5\left(\frac{1}{131}+\frac{1}{141}-\frac{1}{191}-\frac{1}{4011}\right)}{7\left(\frac{1}{131}+\frac{1}{141}-\frac{1}{191}-\frac{1}{4011}\right)}=\frac{5}{7}\)
Bài 2:
a) \(\frac{x}{7}+\left(\frac{-3}{7}\right)^2=\frac{2}{7}:\frac{4}{3}\)
\(\Rightarrow\frac{x}{7}+\frac{9}{49}=\frac{3}{14}\)
\(\Rightarrow\frac{x}{7}=\frac{3}{98}\)
\(\Rightarrow98x=21\)
\(\Rightarrow x=\frac{3}{14}\)
Vậy \(x=\frac{3}{14}\)
b) \(\left(x-1\right)^{x+6}=\left(x-1\right)^{x+4}\)
\(\Rightarrow\left(x-1\right)^{x+6}-\left(x-1\right)^{x+4}=0\)
\(\Rightarrow\left(x-1\right)^{x+4}.\left[\left(x-1\right)^2-1\right]=0\)
\(\Rightarrow\left(x-1\right)^{x+1}=0\) hoặc \(\left(x-1\right)^2-1=0\)
+) \(\left(x-1\right)^{x+1}=0\Rightarrow x-1=0\Rightarrow x=1\)
+) \(\left(x-1\right)^2-1=0\)
\(\Rightarrow\left(x-1\right)^2=1\)
\(\Rightarrow\left(x-1\right)=\pm1\)
+ \(x-1=1\Rightarrow x=2\)
+ \(x-1=-1\Rightarrow x=0\)
Vậy \(x\in\left\{0;2;1\right\}\)
1)
\(\frac{\frac{5}{131}+\frac{5}{141}-\frac{5}{191}-\frac{5}{4011}}{\frac{7}{131}+\frac{7}{141}+\frac{7}{-191}-\frac{7}{4011}}\)
\(=\frac{5\left(\frac{1}{131}+\frac{1}{141}-\frac{1}{191}-\frac{1}{4011}\right)}{7\left(\frac{1}{131}+\frac{1}{141}-\frac{1}{191}-\frac{1}{4011}\right)}\)
\(=\frac{5}{7}\)
2) \(\frac{x}{7}+\left(-\frac{3}{7}\right)^2=\frac{2}{7}:\frac{4}{3}\)
\(=\frac{x}{7}+\frac{9}{49}=\frac{3}{14}\)
\(=\frac{x}{7}=\frac{3}{14}-\frac{9}{49}=\frac{3}{98}\)
\(\Rightarrow98x=21\)
\(\Rightarrow x=\frac{3}{14}\)
\(\Leftrightarrow\left(2a+13b\right)\left(3c-7d\right)=\left(2c+13d\right)\left(3a-7b\right)\)
\(\Leftrightarrow6ac-14ad+39bc-91bd=6ac-14bc+39ad-91bd\)
\(\Leftrightarrow-14ad+14bc=39ad-39bc\)
\(\Leftrightarrow-14\left(ad-bc\right)=39\left(ad-bc\right)\)
=>ad-bc=0
=>ad=bc
hay a/b=c/d
a.\(\frac{13}{17}\)=1-\(\frac{4}{17}\); \(\frac{46}{50}\)=1-\(\frac{4}{50}\)
Vì \(\frac{4}{17}\)>\(\frac{4}{50}\)=> 1-\(\frac{4}{17}\)<1-\(\frac{4}{50}\)
Vậy\(\frac{13}{17}\)<\(\frac{46}{50}\)
a)2,(13) = 2,(131)
b)-0,123 < 0,123
c)1,(472) < 2,(472)
d)0,428571 > 3/7