\(\dfrac{a}{1}\)=\(\dfrac{b}{4}\);\(\dfrac{b}{c}\)=\(\dfrac{3}{4}\) và 4a+b-c=8
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Những câu hỏi liên quan
27 tháng 4 2021
Giải:
a)A=1/56+1/72+1/90+1/110+1/132+1/156
A=1/7.8+1/8.9+1/9.10+1/10.11+1/11.12+1/12.13
A=1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11+1/11-1/12+1/12-1/13
A=1/7-1/13
A=6/91
b)B=4/21+4/77+4/165+4/285+4/437+4/621
B=4/3.7+4/7.11+4/11.15+4/15.19+4/19.23+4/23.27
B=1/3-1/7+1/7-1/11+1/11-1/15+1/15-1/19+1/19-1/23+1/23-1/27
B=1/3-1/27
B=8/27
c) C=1/21+1/77+1/165+1/285+1/437+1/621
C=1/3.7+1/7.11+1/11.15+1/15.19+1/19.23+1/23.27
C=1/4.(4/3.7+4/7.11+4/11.15+4/15.19+4/19.23+4/23.27)
C=1/4.(1/3-1/7+1/7-1/11+1/11-1/15+1/15-1/19+1/19-1/23+1/23-1/27)
C=1/4.(1/3-1/27)
C=1/4.8/27
C=2/27
d) D=1/1.6+1/6.11+1/11.16+1/16.21+1/21.26+1/26.31
D=1/5.(5/1.6+5/6.11+5/11.16+5/16.21+5/21.26+5/26.31)
D=1/5.(1/1-1/6+1/6-1/11+1/11-1/16+1/16-1/21+1/21-1/26+1/26-1/31)
D=1/5.(1/1-1/31)
D=1/5.30/31
D=6/31
Nếu câu d cậu viết thiếu thì làm như vầy nhé!
Chúc bạn học tốt!
27 tháng 4 2021
Nếu như câu d ko chép sai thì làm thế này nha:
d) D=1/1.6+1/6.11+1/11.16+1/16.21+1/26.31
D=1/5.(5/1.6+5/6.11+5/11.16+5/16.21)+1/806
D=1/5.(1/1-1/6+1/6-1/11+1/11-1/16+1/16-1/21)+1/806
D=1/5.(1/1-1/21)+1/806
D=1/5.20/21+1/806
D=4/21+1/806
D=3245/16926
Chúc bạn học tốt!
UK
2 tháng 12 2017
Viết gọn lại, ta cần chứng minh:
\(\sum\left(a+b+\dfrac{1}{4}\right)^2\ge\sum4\left(\dfrac{1}{\dfrac{1}{a}+\dfrac{1}{b}}\right)\)
\(\Leftrightarrow\sum\left(a+b+\dfrac{1}{4}\right)^2\ge\sum4\left(\dfrac{1}{\dfrac{a+b}{ab}}\right)=\sum\dfrac{4ab}{a+b}\)
Thật vậy, ta có:
\(\sum\left(a+b+\dfrac{1}{4}\right)^2\ge\sum\left(2\sqrt{\left(a+b\right).\dfrac{1}{4}}\right)^2=\sum a+b\)
Vậy ta cần chứng minh:
\(\sum a+b\ge\sum\dfrac{4ab}{a+b}\Leftrightarrow\sum\left(a+b\right)^2\ge\sum4ab\Leftrightarrow\sum\left(a-b\right)^2\ge0\)
Vậy ta có đpcm. Đẳng thức xảy ra khi a=b=c
QD
31 tháng 3 2017
a) =
=
b) =
=
=
. ( Với điều kiện b # 1)
c) \(\dfrac{a^{\dfrac{1}{3}}b^{-\dfrac{1}{3}-}a^{-\dfrac{1}{3}}b^{\dfrac{1}{3}}}{\sqrt[3]{a^2}-\sqrt[3]{b^2}}\)= =
=
( với điều kiện a#b).
d) \(\dfrac{a^{\dfrac{1}{3}}\sqrt{b}+b^{\dfrac{1}{3}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}\) = =
=
=
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 3 2022
\(\dfrac{1}{2a-1}+\dfrac{1}{1}\ge\dfrac{4}{2a-1+1}=\dfrac{2}{a}\)
Tương tự: \(\dfrac{1}{2b-1}+1\ge\dfrac{2}{b}\) ; \(\dfrac{1}{2c-1}+1\ge\dfrac{2}{c}\)
Cộng vế:
\(VT\ge\dfrac{2}{a}+\dfrac{2}{b}+\dfrac{2}{c}=\left(\dfrac{1}{a}+\dfrac{1}{b}\right)+\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+\left(\dfrac{1}{c}+\dfrac{1}{a}\right)\ge\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c=1\)
25 tháng 8 2023
a) \(...\dfrac{11}{4}-a+\dfrac{1}{4}=\dfrac{3}{2}\)
\(\dfrac{11}{4}+\dfrac{1}{4}-a=\dfrac{3}{2}\)
\(3-a=\dfrac{3}{2}\)
\(a=3-\dfrac{3}{2}\)
\(a=\dfrac{6}{2}-\dfrac{3}{2}\)
\(a=\dfrac{3}{2}\)
b) \(...\dfrac{13}{4}-a-\dfrac{13}{4}=\dfrac{7}{8}\)
\(\dfrac{13}{4}-\dfrac{13}{4}-a=\dfrac{7}{8}\)
\(0-a=\dfrac{7}{8}\)
\(a=-\dfrac{7}{8}\) (ra số âm lớp 5 chưa học nên bạn xem lại đề)
c) \(...\dfrac{17}{6}-\dfrac{3}{2}-a=\dfrac{1}{6}\)
\(\dfrac{17}{6}-\dfrac{9}{6}-a=\dfrac{1}{6}\)
\(\dfrac{8}{6}-a=\dfrac{1}{6}\)
\(a=\dfrac{8}{6}-\dfrac{1}{6}\)
\(a=\dfrac{7}{6}\)
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
25 tháng 8 2023
a, 2\(\dfrac{3}{4}\) - a + \(\dfrac{1}{4}\) = 1\(\dfrac{1}{2}\)
a = 2 + \(\dfrac{3}{4}\) + \(\dfrac{1}{4}\) - 1 - \(\dfrac{1}{2}\)
a = 2 + 1 - 1 - \(\dfrac{1}{2}\)
a = 2 - \(\dfrac{1}{2}\)
a = \(\dfrac{3}{2}\)
b, 3\(\dfrac{1}{4}\) - a - 3\(\dfrac{1}{4}\) = \(\dfrac{7}{8}\)
(3\(\dfrac{1}{4}\) - 3\(\dfrac{1}{4}\)) - a = \(\dfrac{7}{8}\)
a = - \(\dfrac{7}{8}\)
c, 2\(\dfrac{5}{6}\) - 1\(\dfrac{1}{2}\) - a = \(\dfrac{1}{6}\)
a = 2 + \(\dfrac{5}{6}\) - 1 - \(\dfrac{1}{2}\) - \(\dfrac{1}{6}\)
a = (2-1) + (\(\dfrac{5}{6}\) - \(\dfrac{1}{6}\)) - \(\dfrac{1}{2}\)
a = 1 + \(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)
a = \(\dfrac{7}{6}\)
TA
2 tháng 8 2021
a) `A=a. 1/3 + a. 1/4 - a.1/6 = a. (1/3+1/4 -1/6)=a. 5/12`
Thay `a=-3/5: A=-3/5 . 5/12 =-1/4`
b) `B=b. 5/6+ b. 3/4-b. 1/2=b.(5/6+3/4-1/2)=b. 13/12`
Thay `b=12/13: B=12/13 . 13/12=1`.
2 tháng 8 2021
a) Ta có: \(A=a\cdot\dfrac{1}{3}+a\cdot\dfrac{1}{4}-a\cdot\dfrac{1}{6}\)
\(=a\left(\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{6}\right)\)
\(=a\cdot\left(\dfrac{4}{12}+\dfrac{3}{12}-\dfrac{2}{12}\right)\)
\(=a\cdot\dfrac{5}{12}\)
\(=\dfrac{-3}{5}\cdot\dfrac{5}{12}=\dfrac{-1}{4}\)
b) Ta có: \(B=b\cdot\dfrac{5}{6}+b\cdot\dfrac{3}{4}-b\cdot\dfrac{1}{2}\)
\(=b\left(\dfrac{5}{6}+\dfrac{3}{4}-\dfrac{1}{2}\right)\)
\(=b\cdot\left(\dfrac{10}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right)\)
\(=b\cdot\dfrac{5}{4}\)
\(=\dfrac{12}{13}\cdot\dfrac{5}{4}=\dfrac{60}{52}=\dfrac{15}{13}\)
13 tháng 1 2023
\(M=\dfrac{3^{14}\cdot5^4-3^{12}\cdot5^4}{3^{12}\cdot5^6+7\cdot3^{12}\cdot5^6}=\dfrac{3^{12}\cdot5^4\left(3^2-1\right)}{3^{12}\cdot5^6\left(1+7\right)}=\dfrac{1}{25}\)
3 tháng 8 2023
a,a+1/4=2 3/4-1 1/2
a+1/2=5/4
a=5/4-1/2
a=3/4
b,a-7/4=13/4-7/9
a-7/4=89/36
a= 89/36+7/4
a=152/36
c,3/2-a=17/6-1/6
3/2-a=8/3
a= 3/2-8/3
a= -7/6
MG
1
6 tháng 8 2023
a) \(2\dfrac{3}{4}-a+\dfrac{1}{4}=1\dfrac{1}{2}\)
=> \(\dfrac{11}{4}\) \(-a+\dfrac{1}{4}=\dfrac{3}{2}\)
=> \(\dfrac{11}{4}-a\) = \(\dfrac{3}{2}-\dfrac{1}{4}\)
=> a = \(\dfrac{11}{4}-\dfrac{5}{4}\) =\(\dfrac{3}{2}\)
Vậy a = \(\dfrac{3}{2}\)
b) \(3\dfrac{1}{4}-a-1\dfrac{3}{4}=\dfrac{7}{8}\)
=> \(\dfrac{13}{4}-a-\dfrac{7}{4}=\dfrac{7}{8}\)
=> \(\dfrac{13}{4}-a=\dfrac{21}{8}\)
=> \(a=\dfrac{13}{4}-\dfrac{21}{8}=\dfrac{5}{8}\)
Vậy a = \(\dfrac{5}{8}\)
Từ \(\dfrac{b}{c}=\dfrac{3}{4}\) =>\(\dfrac{b}{3}=\dfrac{c}{4}\) => \(\dfrac{b}{12}=\dfrac{c}{16}\) (1)
Từ \(\dfrac{a}{1}=\dfrac{b}{4}\) =>\(\dfrac{a}{3}=\dfrac{b}{12}\) (2)
Từ (1) và (2) => \(\dfrac{a}{3}=\dfrac{b}{12}=\dfrac{c}{16}=\dfrac{4a}{12}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có :
\(\dfrac{a}{3}=\dfrac{b}{12}=\dfrac{c}{16}=\dfrac{4a}{12}=\dfrac{4a+b-c}{12+12-16}=\dfrac{8}{8}=1\)
=>\(\left\{{}\begin{matrix}a=3\cdot1=3\\b=12\cdot1=12\\c=16\cdot1=16\end{matrix}\right.\)