\(=4\left(\frac{4}{20.24}+\frac{4}{24.28}+...+\frac{4}{76.80}\right)\)

\(=4\left(\frac{1}{20}-\frac{1}{24}+\frac{1}{24}-\frac{1}{28}+...+\frac{1}{76}-\frac{1}{80}\right)\)

\(=4\left(\frac{1}{20}-\frac{1}{80}\right)\)

\(=4\times\frac{3}{80}\)

\(=\frac{12}{80}=\frac{3}{20}<1\)

đpcm

frtji

giúp mình với mai phải nộp rồi

\(C=9\cdot\left(\frac{\frac{26299}{2023}-\frac{3757}{2023}-\frac{91}{2023}}{\frac{8092}{2023}-\frac{1156}{2023}-\frac{28}{2023}}\right):\left(\frac{\frac{840255}{21545}+\frac{168051}{21545}+\frac{6045}{21545}+\frac{27105}{21545}}{\frac{344720}{21545}+\frac{68944}{21545}+\frac{2480}{21545}+\frac{11120}{21545}}\right)\cdot\frac{158158}{164164}\)

\(C=9\cdot\left(\frac{\frac{22451}{2023}}{\frac{6908}{2023}}\div\frac{\frac{1041456}{21545}}{\frac{427264}{21545}}\right)\cdot\frac{158158}{164164}\)

\(C=9\cdot\left(\frac{13}{4}\div\frac{39}{16}\right)\cdot\frac{158158}{164164}\)

\(C=9\cdot\frac{4}{3}\cdot\frac{2\cdot79\cdot2\cdot79}{2\cdot82\cdot2\cdot82}=9\cdot\frac{4}{3}\cdot\frac{79}{82}\)

\(C=\frac{474}{41}\)

Trả lời

Sửa đề câu a 1 tí

a)39/16:5/8-17/16:5/8+4/5

=(39/16-17/16):5/8+4/5

=11/8:5/8+4/5

=11/5+4/5

=15/5=3

b)16/17.12/19+16/17.7/19-2/3

=16/17.(12/19+7/19)-2/3

=16/17-2/3

=48/56-14/56

=34/56=17/28

Học tốt !

\(3\frac{14}{19}+\frac{13}{17}+\frac{35}{43}+6\)

\(=\frac{71}{19}+\frac{13}{17}+\frac{35}{43}+6\)

\(=\frac{1454}{323}+\frac{35}{43}+6\)

\(=5,...+6\)

\(=11,...\)

\(Bai2a\)\(A=\frac{\sqrt{3}-\sqrt{6}}{1-\sqrt{2}}-\frac{2+\sqrt{8}}{1+\sqrt{2}}\)

\(=\frac{\sqrt{3}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{2\left(1+\sqrt{2}\right)}{1+\sqrt{2}}\)

\(=\sqrt{3}-2\) 

\(VayA=\sqrt{3}-2\)

\(\frac{5-\frac{5}{3}+\frac{5}{9}-\frac{5}{27}}{8-\frac{8}{3}+\frac{8}{9}-\frac{8}{27}}:\frac{15-\frac{15}{11}+\frac{15}{121}}{16-\frac{16}{11}+\frac{16}{121}}\)

\(=\frac{5\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}{8\left(1-\frac{1}{3}+\frac{1}{9}-\frac{1}{27}\right)}:\frac{15\left(1-\frac{1}{11}+\frac{1}{121}\right)}{16\left(1-\frac{1}{11}+\frac{1}{121}\right)}\)

\(=\frac{5}{8}:\frac{15}{16}\)

\(=\frac{2}{3}\)

2/3

Tk mk nha

Cac ban giup mik voi, mik k cho! ( 10 k cho 3 nguoi dau tien tra loi cau hoi cua mik)

tự tính  nhé dễ lắm bạn 

\(=\frac{\frac{5}{12}+\frac{9}{12}-\frac{12}{12}}{\frac{36}{12}-\frac{10}{12}+\frac{8}{12}}+\frac{16\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\right).}{17\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}\right).}\)

\(=\frac{1}{\frac{6}{\frac{17}{6}}}+\frac{16}{17}\)\(=\frac{1}{17}+\frac{16}{17}=1\)

Ta có :

\(\frac{a}{b^3+16}=\frac{a}{16}-\frac{ab^3}{16\left(b^3+16\right)}\ge\frac{a+b+c}{16}-\frac{ab^2+bc^2+ca^2}{192}.\)(1)

Không mất tính tổng quát, giả sử \(a\ge b\ge c\)ta có:

\(\text{a(a−b)(b−c)≥0 ⇔abc+a^2b≥ab^2+ca^2}\)

Ta có: \(ab^2+bc^2+ca^2+abc\le bc^2+2abc+a^2b=b(a+c)^2\le\frac{4\left(a+b+c\right)^3}{27}=4\)(2)

Từ (1) và (2) suy ra dpcm

Dấu ''='' xảy ra khi (a,b,c)=(0,1,2)(a,b,c)=(0,1,2) cùng các hoán vị.

Gỉa sử \(a\ge b\ge c\)

Ta có:

\(b\le\frac{a+b+c}{3}\)(1)

\(\left(a+c\right)^2\le\left(\frac{2\left(a+b+c\right)}{3}\right)^2=\frac{4\left(a+b+c\right)^2}{9}\)(2)

nhân theo vế (1)(2) suy ra dpcm

tại sao b=<(a+b+c)/3

Ta có \(\cos 2a = 2{\cos ^2}a - 1 = 2.{\left( {\frac{1}{4}} \right)^2} - 1 = \frac{{ - 7}}{8}\)

Chọn B