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Ta có : \(\left|x+\frac{13}{14}\right|=-\left|x-\frac{3}{7}\right|\)
\(\Rightarrow\left|x+\frac{13}{14}\right|+\left|x-\frac{3}{7}\right|=0\)
Mà : \(\left|x+\frac{13}{14}\right|\ge0\forall x\)
\(\left|x-\frac{3}{7}\right|\ge0\forall x\)
Nên : \(\orbr{\begin{cases}\left|x+\frac{13}{14}\right|=0\\\left|x-\frac{3}{7}\right|=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{13}{14}=0\\x-\frac{3}{7}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{13}{14}\\x=\frac{3}{7}\end{cases}}\)
\(16\frac{2}{7}:\left(\frac{-3}{5}\right)+28\frac{2}{7}:\frac{3}{5}\)
=\(\frac{114}{7}:\left(\frac{-3}{5}\right)+\frac{198}{7}:\frac{3}{5}\)
=\(-\frac{190}{7}+\frac{330}{7}\)
=\(\frac{-190+330}{7}=\frac{140}{7}=20\)
a, A lớn nhất khi 7x la nguyên dương nho nhất
\(\Rightarrow7x=1\)
\(\Rightarrow x=\frac{1}{7}\)
\(b,B=\frac{10+4-x}{4-x}\)
\(B=\frac{10}{4-x}+1\)
b lon nhat khi 4-xla nguyen duong nho nhat
\(\Rightarrow4-x=1\)
\(\Rightarrow x=4-1=3\)
\(c,C=\frac{27-2x}{12-x}=\frac{3+24-2x}{12-x}=\frac{3}{12-x}+2\)
c lon nhat khi 12-x la nguyen duong nho nhat
\(\Rightarrow12-x=1\Rightarrow x=11\)
Bài 1:
\(A=\left(\frac{-5}{11}+\frac{7}{22}-\frac{4}{33}-\frac{5}{44}\right):\left(38\frac{1}{122}-39\frac{7}{22}\right)\)
\(=\frac{-49}{132}:\left(-\frac{879}{671}\right)=\frac{2989}{105408}\)
Bài 2:
\(\frac{4}{5}-\left(\frac{-1}{8}\right)=\frac{7}{8}-x\)
<=> \(\frac{7}{8}-x=\frac{27}{40}\)
<=> \(x=\frac{7}{8}-\frac{27}{40}=\frac{1}{5}\)
Vậy...
Ta có 1 - a2 = 1 - a + a - a2 = 1 - a + a(1 - a) = (1 - a)(1 + a)
Khi đó \(\left(\frac{1}{2^2}-1\right)\left(\frac{1}{3^2}-1\right)....\left(\frac{1}{100^2}-1\right)=\frac{1-2^2}{2^2}.\frac{1-3^2}{3^2}...\frac{1-100^2}{100^2}\)
= \(\frac{\left(1-2\right)\left(1+2\right)}{2^2}.\frac{\left(1-3\right)\left(1+3\right)}{3^2}...\frac{\left(1-100\right)\left(1+100\right)}{100^2}\)
= \(-\frac{\left(2-1\right)\left(2+1\right).\left(3-1\right)\left(3+1\right)...\left(100-1\right)\left(100+1\right)}{2^2.3^2.4^2....100^2}\)
\(=-\frac{1.3.2.4...99.101}{2.2.3.3.4.4...100.100}=-\frac{\left(1.2.3...99\right).\left(3.4.5...101\right)}{\left(2.3.4...100\right).\left(2.3.4...100\right)}=-\frac{1.101}{100.2}=-\frac{101}{200}\)
A=(1-1/1)+(1-1/4)+(1-1/9)+(1/16)+..........+(1-1/100)
=>1-99/100
\(\left(\frac{1}{7}\right)^7.7^7\)=\(\frac{1^7}{7^7}.7^7=1^7=1\)