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Bài 1:
a, \(\frac{1}{-16}-\frac{3}{45}=\frac{-1}{16}-\frac{1}{15}\)
\(=\frac{-15}{240}-\frac{16}{240}\)
\(=\frac{-31}{240}\)
b, \(=\frac{-10}{12}-\frac{-12}{12}\)
\(=\frac{2}{12}=\frac{1}{6}\)
c, \(=\frac{-30}{6}-\frac{1}{6}\)
\(=\frac{-31}{6}\)
Bài 2:
a, \(x=-\frac{1}{2}-\frac{3}{4}\)
\(x=-\frac{1}{4}\)
b, \(\frac{1}{2}+x=-\frac{11}{2}\)
\(x=-\frac{11}{2}-\frac{1}{2}\)
\(x=-6\)
Bạn nhớ k đúng và chọn câu trả lời này nhé!!!! Mình giải đúng và chính xác hết ^_^
\(B=-\frac{1}{3}+\frac{2}{5}-\frac{2}{3}-\frac{3}{5}+\frac{1}{5}\)
\(=\left(-\frac{1}{3}-\frac{2}{3}\right)+\left(\frac{2}{5}-\frac{3}{5}+\frac{1}{5}\right)\)
\(=-\frac{3}{3}+0\)
\(=-1\)
=.= hk tốt!!
B=\(\frac{-4}{12}+\frac{18}{45}+\frac{-6}{9}+\frac{-21}{35}+\frac{6}{30}\)
=\(\frac{-4}{4\cdot3}+\frac{2\cdot9}{5\cdot9}+\frac{\left(-2\right)\cdot3}{3\cdot3}+\frac{\left(-4\right)\cdot7}{5\cdot7}+\frac{6}{5\cdot6}\)
=\(\frac{-1}{3}+\frac{2}{5}+\frac{-2}{3}+\frac{-4}{5}+\frac{1}{5}\)
= \(\left(\frac{-1}{3}+\frac{-2}{3}\right)+\left(\frac{-4}{5}+\frac{2}{5}+\frac{1}{5}\right)\)
=\(\frac{-3}{3}+\frac{-1}{5}\)
= \(-1+\frac{-1}{5}\)=\(\frac{-5-1}{5}=\frac{-6}{5}\)
\(a,\frac{1}{6}-\frac{2}{3}\)
\(=\frac{1}{6}-\frac{4}{6}=-\frac{3}{6}=-\frac{1}{2}\)
Vì \(B=\frac{2014^{11}+2}{2014^{12}+2}<1\)
\(\Rightarrow B=\frac{2014^{11}+2}{2014^{12}+2}<\frac{2014^{11}+2+4026}{2014^{12}+2+4026}=\frac{2014^{11}+4028}{2014^{12}+4028}=\frac{2014.\left(2014^{10}+2\right)}{2014\left(2014^{11}+2\right)}=\frac{2014^{10}+2}{2014^{11}+2}=A\)
Vậy B<A hay A<B
ta chứng minh bài toán phụ:
nếu ta có b<d \(\frac{a}{b}\)>\(\frac{c}{d}\) thì ad>bc
dễ thây \(\frac{ad}{bd}>\frac{cb}{bd}\)
=> ad>bd
áp dụng:
dat 2014=a ta co
\(A=\frac{a^{10}+2}{a^{11+2}}\)
\(B=\frac{a^{11}+2}{a^{12}+2}\)
ta có
\(A=\frac{a^{10}+2.a^{12}+2}{a^{11}+2.a^{12}+2}\)
\(B=\frac{a^{11}+2.a^{11}+2}{a^{12}+2.a^{11}+2}\)=\(\frac{a^{10}+2a^{12}+2}{a^{12}+2a^{11}+2}\)
=> A=B
mk hok chắc đâu nha
2/7 + 7/-5 + -2/32 = -41/35
4/3 + | 5\-6 | + | -7/-6 | = 4/3 + 5/6 + 7/6 = 10/3
hok tốt
.............
\(\frac{2}{7}+\frac{7}{-5}+\frac{-2}{35}\)
\(=\frac{2}{7}+\frac{-7}{5}+\frac{-2}{35}\)
\(=\frac{10}{35}+\frac{-49}{35}+\frac{-2}{35}\)
\(=\frac{-41}{35}\)
\(\frac{4}{3}+\left|\frac{5}{6}\right|+\left|\frac{-7}{-6}\right|\)
\(=\frac{4}{3}+\frac{5}{6}+\frac{7}{6}\)
\(=\frac{8}{6}+\frac{5}{6}+\frac{7}{6}\)
\(=\frac{20}{6}=\frac{10}{3}\)
a)(1,5.x+3/7):1,5-1,5=1 b)(2+x+4/7+3/7):1,4-5/7=-2
1,5.x:1,5+3/7:3/2=1+1,5 (2+x+1):1,4-5/7=-2
x+ 3/7x 2/3=2,5 (3+x):1,4=-2+5
x+2/7=2,5 (3+x):1,4=3
x+2/7=5/2 3+x=3x1,4
x=5/2-2/7 3+x=4,2
x=31/14 x=4,2-3=1,2
\(A>\frac{196}{197+198}+\frac{197}{198+197}=\frac{196+197}{198+197}=B\)
\(\Leftrightarrow A>B\)
\(\frac{x+5}{4}=\frac{x-3}{5}\)
\(\Rightarrow x+5=\left(x-3\right).\frac{4}{5}\)
\(\Rightarrow x+5=\frac{4}{5}x-\frac{12}{5}\)
\(\Rightarrow\frac{1}{5}x=-\frac{37}{5}\)
\(\Rightarrow x=-37\)
Vậy x = -37
\(\frac{x+5}{4}=\frac{x-3}{5}\Leftrightarrow5\left(x+5\right)=4\left(x-3\right)\Leftrightarrow5x+25=4x-13\Leftrightarrow x=-38\)
a) \(\frac{2^{10}\left(2+3\right)}{\left(2^2\right)^5.5.2}=\frac{2^{10}.5}{2^{10}.5.2}=\frac{1}{2}\); b) \(=\frac{\left(3^2\right)^4.2-3^6}{3^6.34.3}=\frac{3^6\left(2.3^2-1\right)}{3^6.34.3}=\frac{3^6.17}{3^6.17.2.3}=\frac{1}{6}\)