1)\(\frac{-6}{-4}\) ; \(\frac{-7}{9}\); 0 ;\(\frac{-40}{-50}\); \(\frac{27}{33}\)
2)\(\frac{18}{19}\); \(\frac{4}{3}\); \(\frac{-14}{37}\); \(\frac{17}{20}\); \(\frac{-14}{33}\); 0
HÃY SẮP XẾP THEO THỨ TỰ TĂNG DẦN
( giúp mình nha )
\(A=\frac{\frac{1}{4}+\frac{1}{5}+\frac{1}{6}}{\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)-\frac{1}{4}.\frac{1}{5}.\frac{1}{6}}\)
Ta có \(A=\frac{\frac{1}{4}+\frac{1}{5}+\frac{1}{6}}{\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{6}\right)-\frac{1}{4.5.6}}=\frac{\frac{\left(15+12+10\right)}{60}}{\left(\frac{15+12-10}{60}\right)-\frac{1}{120}}\)\(=\frac{\frac{37}{60}}{\frac{27}{60}-\frac{1}{120}}=\frac{37}{60}:\frac{54}{120}=\frac{37}{60}.\frac{120}{54}=\frac{37.2.60}{60.2.37}=1\)
\(\frac{\frac{1}{4}+\frac{1}{5}+\frac{1}{6}}{(\frac{1}{4}+\frac{1}{5}-\frac{1}{6})-\frac{1}{4}.\frac{1}{5}.\frac{1}{6}}\)=\(\frac{\frac{37}{60}}{\frac{17}{60}-\frac{1}{120}}\)=\(\frac{\frac{37}{60}}{\frac{33}{120}}\)=\(\frac{37}{60}:\frac{33}{120}=\frac{37}{60}.\frac{120}{33}\)=\(\frac{74}{33}\)
Anime Tổng hợp ơi,\((12+15-10)=17\) chứ ko phải =27
So sánh:\(\frac{\frac{\frac{1}{2}}{\frac{3}{4}}}{\frac{\frac{5}{6}}{\frac{7}{8}}}+\frac{\frac{\frac{8}{7}}{\frac{6}{5}}}{\frac{\frac{4}{3}}{\frac{2}{1}}}\) và\(\frac{\frac{\frac{1}{2}}{\frac{3}{4}}+\frac{\frac{8}{7}}{\frac{6}{5}}}{\frac{\frac{5}{6}}{\frac{7}{8}}+\frac{\frac{4}{3}}{\frac{2}{1}}}\)và \(\frac{\frac{\frac{1}{2}+\frac{8}{7}}{\frac{3}{4}+\frac{6}{5}}}{\frac{\frac{5}{6}+\frac{4}{3}}{\frac{7}{8}+\frac{2}{1}}}\)và\(\frac{\frac{\frac{1+8}{2+7}}{\frac{3+6}{4+5}}}{\frac{5+4}{\frac{6+3}{2+1}}}\)
bài 1: tính A:=\(\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+\frac{5}{6}-\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{2}{3}-\frac{1}{2}\)
Bài 2: Cho B=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+.....+\frac{1}{49}-\frac{1}{50}\)
Chứng minh rằng: \(\frac{7}{12}< A< \frac{5}{6}\)
\(\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}-\left(\frac{-5}{6}\right)-\frac{6}{7}-\frac{-7}{8}+\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\)
\(\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}-\left(-\frac{5}{6}\right)-\frac{-7}{8}+\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\)
\(=\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+\frac{5}{6}+\frac{7}{8}+\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{1}{2}\)
\(=\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{2}{3}-\frac{2}{3}\right)+\left(\frac{3}{4}-\frac{3}{4}\right)+\left(\frac{4}{5}-\frac{4}{5}\right)+\left(\frac{5}{6}-\frac{5}{6}\right)+\frac{7}{8}+\frac{6}{7}\)
\(=\frac{7}{8}+\frac{6}{7}=\frac{49}{56}+\frac{48}{56}=\frac{49+48}{56}=\frac{97}{56}\)
cho x,y,z > 0. Cmr: \(\frac{2x}{x^6+y^4}+\frac{2y}{y^6+z^4}+\frac{2z}{z^6+x^4}\le\frac{1}{x^4}+\frac{1}{y^4}+\frac{1}{z^4}\)
\(\frac{1}{x^4}+\frac{1}{y^4}=\frac{x^2}{x^6}+\frac{1}{y^4}\ge\frac{\left(x+1\right)^2}{x^6+y^4}\ge\frac{4x}{x^6+y^4}\)
tương tự
\(\frac{1}{y^4}+\frac{1}{z^4}\ge\frac{4y}{y^6+z^4}\);
\(\frac{1}{z^4}+\frac{1}{x^4}\ge\frac{4z}{z^6+x^4}\);
cộng vế với vế => đpcm
Dấu "=" xảy ra <=> x=y=z=1
Cách khác:
\(x^6+y^4\ge2\sqrt{x^6y^4}=2x^3y^2\)
\(\Rightarrow\frac{2}{x^6+y^4}\le\frac{1}{x^2y^2}\)
CMTT , ta có VT \(\le\frac{1}{x^2y^2}+\frac{1}{y^2z^2}+\frac{1}{z^2x^2}\)
Bổ đề: \(a^2+b^2+c^2\ge ab+bc+ca\) ( luôn đúng)
\(\Rightarrow\frac{1}{x^2y^2}+\frac{1}{y^2z^2}+\frac{1}{z^2x^2}\le\frac{1}{x^4}+\frac{1}{y^4}+\frac{1}{z^4}\)
ĐPCM
Dấu " =" xảy ra khi x=y=z=1
\(3\frac{1}{2}-4\frac{2}{3}+\left[\frac{3}{4}-2\frac{1}{3}\right]-\left(\frac{5}{6}-\frac{7}{4}\right)+5\frac{1}{2}-3\)
\(2\frac{2}{3}-1\frac{2}{5}+1\frac{3}{10}-\left(\frac{2}{5}-\frac{5}{6}\right)+\frac{4}{15}-1\frac{1}{3}\)
\(\left[2\frac{1}{3}-1\frac{4}{3}\right]-\left(\frac{5}{4}-\frac{7}{12}+\frac{-11}{6}\right)+\frac{4}{3}-\frac{3}{4}\)
\(-3\frac{3}{2}+5\frac{4}{3}-\left(\frac{7}{6}-1\frac{3}{4}\right)+\left[\frac{2}{3}-2\frac{1}{4}\right]\)
\(2\frac{2}{3}-\frac{5}{12}-\left(1\frac{3}{4}-2\frac{1}{4}\right)-\left[1-1\frac{1}{6}\right]+\left[\frac{-5}{3}\right]\)
\(1\frac{1}{3}-5\frac{1}{2}-\left[\frac{5}{6}-2\frac{2}{3}\right]+\left[\frac{7}{12}-\frac{5}{6}\right]\)
\(\frac{8}{15}-\left(\frac{2}{5}-3\frac{1}{3}+\left[\frac{-5}{6}\right]\right)+\left[\frac{1}{2}-\frac{4}{5}\right]-\left(\frac{1}{6}-1\frac{1}{3}\right)\)
\(-2\frac{3}{2}+\left[\frac{5}{6}-1\frac{1}{3}\right]-\left(\frac{5}{12}-\frac{7}{6}\right)+\left[\frac{4}{3}-3\frac{1}{4}\right]\)
\(\frac{9}{10}-1\frac{2}{5}-\left(\frac{5}{6}-3\frac{1}{2}\right)-\left[2\frac{1}{4}-5\frac{2}{36}\right]-\left[1-2\frac{1}{15}\right]\)
\(\frac{5}{7}-\frac{5}{21}+1\frac{2}{3}-\left(1\frac{1}{2}-\frac{5}{14}-\frac{1}{3}\right)+\left[\frac{1}{6}-\frac{4}{3}\right]\)
\(\frac{5}{7}-\frac{5}{21}+1\frac{2}{3}-\left(1\frac{1}{2}-\frac{5}{14}-\frac{1}{3}\right)+\left[\frac{1}{6}-\frac{4}{3}\right]\)
\(1\frac{1}{5}-\left(\frac{-9}{10}-2\frac{1}{2}+\frac{3}{4}\right)+\left[\frac{1}{5}-2\frac{1}{2}\right]+\frac{7}{10}-\left(\frac{1}{2}-\frac{1}{4}\right)\)
\(2\frac{1}{3}-\left(5\frac{1}{2}-2\frac{2}{3}\right)+\left[1\frac{1}{6}-2\frac{1}{2}\right]-\frac{5}{12}+\left(\frac{1}{4}-\frac{1}{8}\right)\)
Bai 3\(\frac{1}{2+4}+\frac{1}{2+4+6}+\frac{1}{2+4+6+8}+.....+\frac{1}{2+4+6+8+.....+2\cdot x}=\frac{503}{1007}\)
Tính hợp lý
\(\frac{\frac{5}{3}+\frac{5}{8}-\frac{5}{7}}{\frac{-4}{3}-\frac{-4}{8}+\frac{4}{7}}:\frac{\frac{2}{3}-\frac{1}{6}+\frac{6}{7}}{\frac{-1}{3}+\frac{1}{6}-\frac{1}{7}}\)
\(=\frac{5\left(\frac{1}{3}+\frac{1}{8}-\frac{1}{7}\right)}{-4\left(\frac{1}{3}+\frac{1}{8}-\frac{1}{7}\right)}:\frac{2\left(\frac{1}{3}-\frac{1}{12}+\frac{3}{7}\right)}{ }\)
MÃu thứ hai sao ý
e,\(A=\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+\frac{29}{30}+\frac{41}{42}=\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+\left(1-\frac{1}{12}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{20}\right)+\left(1-\frac{1}{42}\right)\)
\(\Rightarrow A=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+1-\frac{1}{30}+1-\frac{1}{42}=4-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)
\(\Rightarrow A=4-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)=4-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)
\(\Rightarrow A=4-\left(\frac{1}{1}-\frac{1}{7}\right)=4-\frac{6}{7}=3\frac{1}{7}\)
BN mún hỏi j vậy, đây k phải câu hỏi, mà có thì phải là toán lớp 6
\(A=\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}\div\frac{5+\frac{5}{13}+\frac{5}{165}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\times\frac{158158151}{711711711}\)\(B=\frac{-1}{20}-\frac{-1}{30}+\frac{-1}{42}+\frac{-1}{56}+\frac{-1}{72}+\frac{-1}{90}\)
Cau A dat thua so chung la ra
Cau B tach mau thanh h cua 2 thua so lien tiep