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NN
0
N
19 tháng 9 2017
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20 tháng 8 2020
Ta có: \(\left|x+\frac{8}{139}\right|\ge0\forall x\)
\(\Rightarrow-\left|x+\frac{8}{139}\right|\le0\forall x\)
\(\Rightarrow-\left|x+\frac{8}{139}\right|+\frac{141}{272}\le\frac{141}{272}\forall x\)
Dấu '=' xảy ra khi \(x+\frac{8}{139}=0\)
hay \(x=-\frac{8}{139}\)
Vậy: Giá trị lớn nhất của biểu thức \(A=-\left|x+\frac{8}{139}\right|+\frac{141}{272}\) là \(\frac{141}{272}\) khi \(x=-\frac{8}{139}\)
12 tháng 8 2016
x.(x + 4) + 4.(x + 4) = 36
=> (x + 4).(x + 4) = 36
=> (x + 4)2 = 36
=> \(\orbr{\begin{cases}x+4=6\\x+4=-6\end{cases}}\)=> \(\orbr{\begin{cases}x=2\\x=-10\end{cases}}\)
6x-2 = 36
=> 6x-2 = 62
=> x - 2 = 2
=> x = 2 + 2
=> x = 4
Ta có:
\(\frac{139}{140}=1-\frac{1}{140};\frac{140}{141}=1-\frac{1}{141}\)
Vì \(\frac{1}{140}>\frac{1}{141}\)=> \(1-\frac{1}{140}< 1-\frac{1}{141}\)
=> \(\frac{139}{140}< \frac{140}{141}\)
23 tháng 2 2017
M = (3+ 1/417).1/762 - 1/139(4 +761/762) - 4/417.762 + 5/139
= 3/762 + 1/417.762 - 4/139 - 761/139.762 - 4/417.762 + 5/139
= 3/762 - 3/417.762 + 1/139 - 761/139.762 =
= 1179/ 139.762 - 761/139.762 - 3/417.762 =
= 418/139.762 - 3/417.762 =418/139.762 - 1/139.762 = 417/139.762 = 3/762 = 1/254
31 tháng 12 2016
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 \(x\in\left\{0;2;1\right\}\)
HN
31 tháng 12 2016
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) \(\frac{x}{7}+\left(-\frac{3}{7}\right)^2=\frac{2}{7}:\frac{4}{3}\)
\(=\frac{x}{7}+\frac{9}{49}=\frac{3}{14}\)
\(=\frac{x}{7}=\frac{3}{14}-\frac{9}{49}=\frac{3}{98}\)
\(\Rightarrow98x=21\)
\(\Rightarrow x=\frac{3}{14}\)
AP
12 tháng 7 2020
\(M\)\(=\)\(313\) \(\times\) \(\frac{4}{417}\) \(\times\) \(\frac{1}{762}\) \(-\) \(\frac{4}{417}\) \(\times\) \(\frac{1}{762}\) \(-\) \(\frac{1}{139}\) \(\times4\frac{761}{762}\)\(+\frac{1}{139}\times5\)
\(M=\)\(\frac{4}{417}\times\frac{1}{762}\times312\)\(-\frac{1}{139}\left(4\frac{761}{762}-5\right)\)
\(M=\frac{-416}{139}\times\frac{-1}{762}\)\(-\frac{1}{139}\times\frac{-1}{762}\)
\(M=\frac{-1}{762}\left(\frac{-416}{139}-\frac{1}{139}\right)\) \(=\frac{415}{105918}\)