Cho abc= 2016
Tính M = \(\dfrac{2bc-2016}{3c-2bc+2016}-\dfrac{2b}{3-2b+ab}+\dfrac{4032-3ac}{3ac-4032+2016a}\)
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YN
14 tháng 12 2017
- Nhân cả tử và mẫu phân thức thứ nhất với a
- Nhân cả tử và mẫu phân thức thứ 2 với ac
- Thay abc =2016 ta có mẫu số chung là :
3ac - 4032 +2016a
- Rút gọn => đáp án : -1
L
29 tháng 11 2020
\(P=\frac{2bc-2016}{3c-2bc+2016}-\frac{2b}{3-2b+ab}-\frac{4032-3ac}{3ac-4032+2016a}\)
Ta rút gọn từng biểu thức
\(+)\frac{2bc-2016}{3c-2bc+2016}=-1+\frac{3c}{3c-2bc+2016}\)
\(+)\frac{-2b}{3-2b+ab}=\frac{-2bc}{3c-2bc+abc}=\frac{-2bc}{3c-2bc+2016}\)
\(+)\frac{4032-3ac}{3ac-4032+2016a}=-1+\frac{2016a}{3ac-2abc+2016a}\)
\(=-1+\frac{2016}{3c-2bc+2016}\)
\(\Rightarrow P=-1\)
DH
3 tháng 3 2020
Ta có:
\(+)\frac{2bc-2016}{3c-2bc+2016}=-1+\frac{3c}{3c-2bc+2016}\left(1\right)\)
\(+)\frac{-2b}{3-2b+ab}=\frac{-2bc}{3c-2bc+abc}=\frac{-2bc}{3c-2bc+2016}\left(2\right)\)
\(+)\frac{4032-3ac}{3ac-4032+2016a}=-1+\frac{2016a}{3ac-2abc+2016a}=-1+\frac{2016}{3c-2bc+2016}\left(3\right)\)
\(P=\left(1\right)+\left(2\right)+\left(3\right)=-1\)
Vậy .........
BT
26 tháng 12 2017
\(P=\dfrac{2bc-2016}{3c-2bc+2016}-\dfrac{2b}{3-2b+ab}+\dfrac{4032-3ac}{3ac-4032+2016c}\)
\(=\dfrac{2bc-abc}{3c-2bc+abc}-\dfrac{2b}{3-2b+ab}+\dfrac{2abc-3ac}{3ac-2abc+a^2bc}\)
\(=\dfrac{2b-ab}{3-2b+ab}-\dfrac{2b}{3-2b+ab}+\dfrac{2b-3}{3-2b+ab}\)
\(=\dfrac{2b-ab-2b+2b-3}{3-2b+ab}\)
\(=\dfrac{-3+2b-ab}{3-2b+ab}=-1\).
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 8 2021
\(\Leftrightarrow\dfrac{2+3\left(2a+b+2\sqrt{2bc}\right)}{2a+b+2\sqrt{2bc}}\ge\dfrac{16}{\sqrt{2b^2+2\left(a+c\right)^2}+3}\)
\(\Leftrightarrow3+\dfrac{2}{2a+b+2\sqrt{2bc}}\ge\dfrac{16}{\sqrt{2b^2+2\left(a+c\right)^2}+3}\)
Do \(\dfrac{2}{2a+b+2\sqrt{2bc}}\ge\dfrac{2}{2a+b+b+2c}=\dfrac{1}{a+b+c}\)
Và \(2b^2+2\left(a+c\right)^2\ge\left(a+b+c\right)^2\)
Nên ta chỉ cần chứng minh:
\(3+\dfrac{1}{a+b+c}\ge\dfrac{16}{a+b+c+3}\)
Thật vậy, ta có:
\(3+\dfrac{1}{a+b+c}=\dfrac{1}{1}+\dfrac{1}{1}+\dfrac{1}{1}+\dfrac{1}{a+b+c}\ge\dfrac{16}{1+1+1+a+b+c}=\dfrac{16}{a+b+c+3}\) (đpcm)
Dấu "=" xảy ra khi \(a=\dfrac{b}{2}=c=\dfrac{1}{4}\)
19 tháng 2 2017
a. \(4ab.\frac{1}{3}ac-2aca-9a^2.\frac{1}{2}b+10a^2.\frac{1}{5}c+a^2b-a^2bc\)
\(=\left(4.\frac{1}{3}\right)\left(a.a\right).bc-2a^2c-\left(9.\frac{1}{2}\right)a^2b+\left(10.\frac{1}{5}\right)a^2c+a^2b-a^2bc\)
\(=\frac{4}{3}a^2bc-2a^2c-\frac{9}{2}a^2b+2a^2c+a^2b-a^2bc\)
\(=\left(\frac{4}{3}a^2bc-a^2bc\right)+\left(-2a^2c+2a^2c\right)+\left(-\frac{9}{2}a^2b+a^2b\right)\)
\(=\frac{1}{3}a^2bc+\left(-\frac{7}{2}a^2b\right)\)
b. \(2ab-2bc.c+ab+\frac{1}{2}c^2b-4cb^2+2bcb\)
\(=2ab-2bc^2+ab+\frac{1}{2}c^2b-4cb^2+2b^2c\)
\(=\left(2ab+ab\right)+\left(-2bc^2+\frac{1}{2}c^2b\right)+\left(-4cb^2+2b^2c\right)\)
\(=3ab+-\frac{3}{2}bc^2+-2b^2c\)
\(=b\left(3a-\frac{3}{2}c^2-2bc\right)\)
`(2bc-2016)/(3c-2bc+2016)`
`=(-(3c-2bc+2016)+3c)/(3c-2bc+2016)`
`=-1+(3c)/(3c-2bc+2016)`
`(2b)/(3-2b+ab)
`=(2bc)/(3c-2bc+abc)`
`=(2bc)/(3c-2bc+2016)`
`(4032-3ac)/(3ac-4032+2016a)`
`=(-(3ac-4032+2016a)+2016a)/(3ac-4032+2016a)`
`=-1+(2016a)/(3ac-2abc+2016a)`
`=-1+(2016)/(3c-2bc+2016)`
`=>M=-1+(3c)/(3c-2bc+2016)-(2bc)/(3c-2bc+2016)-1+(2016)/(3c-2bc+2016)
`=>M=-2+(3c-2bc+2016)/(3c-2bc+2016)`
`=>M=-2+1`
`=>M=-1`
`(2bc-2016)/(3c-2bc+2016)`
`=(-(3c-2bc+2016)+3c)/(3c-2bc+2016)`
`=-1+(3c)/(3c-2bc+2016)`
`(2b)/(3-2b+ab)`
`=(2bc)/(3c-2bc+abc)`
`=(2bc)/(3c-2bc+2016)`
`(4032-3ac)/(3ac-4032+2016a)`
`=(-(3ac-4032+2016a)+2016a)/(3ac-4032+2016a)`
`=-1+(2016a)/(3ac-2abc+2016a)`
`=-1+(2016)/(3c-2bc+2016)`
`=>M=-1+(3c)/(3c-2bc+2016)-(2bc)/(3c-2bc+2016)-1+(2016)/(3c-2bc+2016)`
`=>M=-2+(3c-2bc+2016)/(3c-2bc+2016)`
`=>M=-2+1`
`=>M=-1`
Nãy thiếu latex ạ sorry~~