Cho abc= 2016
Tính M = \(\dfrac{2bc-2016}{3c-2bc+2016}-\dfrac{2b}{3-2b+ab}+\dfrac{4032-3ac}{3ac-4032+2016a}\)
Tính giá trị biểu thức sau, biết abc = 2016
P=2bc-2016/3c-2bc+2016 - 2b/3-2b+ab + 4032-3ac/3ac-4032+2016a
- 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
\(P=\frac{2bc-2016}{3c-2bc+2016}-\frac{2b}{3-2b+ab}-\frac{4032-3ac}{3ac-4032+2016a}\)
Ta rút gọn từng biểu thức
\(+)\frac{2bc-2016}{3c-2bc+2016}=-1+\frac{3c}{3c-2bc+2016}\)
\(+)\frac{-2b}{3-2b+ab}=\frac{-2bc}{3c-2bc+abc}=\frac{-2bc}{3c-2bc+2016}\)
\(+)\frac{4032-3ac}{3ac-4032+2016a}=-1+\frac{2016a}{3ac-2abc+2016a}\)
\(=-1+\frac{2016}{3c-2bc+2016}\)
\(\Rightarrow P=-1\)
Tính giá trị biết thức sau biết abc = 2016
\(P=\frac{2bc-2016}{3c-2bc+2016}-\frac{2b}{3-2b+ab}+\frac{4032-3ac}{3ac-4032+2016a}\)
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 .........
Tính giá trị của biểu thức sau, biết abc = 2016.
P = \(\frac{2bc-2016}{3c-2bc+2016}-\frac{2b}{3-2b+ab}+\frac{4032-3ac}{3ac-4032+2016a}\)
\(P=\dfrac{2bc-2016}{3c-2bc+2016}-\dfrac{2b}{3-2b+ab}+\dfrac{4032-3ac}{3ac-4032+2016c}\)
\(=\dfrac{2bc-abc}{3c-2bc+abc}-\dfrac{2b}{3-2b+ab}+\dfrac{2abc-3ac}{3ac-2abc+a^2bc}\)
\(=\dfrac{2b-ab}{3-2b+ab}-\dfrac{2b}{3-2b+ab}+\dfrac{2b-3}{3-2b+ab}\)
\(=\dfrac{2b-ab-2b+2b-3}{3-2b+ab}\)
\(=\dfrac{-3+2b-ab}{3-2b+ab}=-1\).
+) Cho các số dương a,b,c thỏa mãn: a+2b+3c=3
CM: \(\sqrt{\dfrac{2ab}{2ab+9c}}+\sqrt{\dfrac{2bc}{2bc+a}}+\sqrt{\dfrac{ac}{ac+2b}}\le\dfrac{3}{2}\)
+) Cho a,b,c >0 và a+b+c≤3
Tìm min P\(=\dfrac{1}{a^2+b^2}+\dfrac{1}{b^2+c^2}+\dfrac{1}{c^2+a^2}\)
Thu gọn các biểu thức sau
a, 4ab.1/3 - 2aca - 9a^2 . 1/2b + 10a^2 . 1/5c +a^2b . a^2bc
b, 2ab - 2bc . c +ab + 1/2c^2b - 4cb^2 + 2bcb
c, x/3 + x/6 + 3x/2 - 4/3mn^2 + 0,2mn^2 - 1/1/3mn^2
d, (2/3ac)^2 . c^2 - 2/5a(c.c)^2 + 2/3ac^3 . c - 1/4ac^4
giúp mình nhanh nha
Bài 1:
a)
ta có: \(\dfrac{50}{100}=\dfrac{1}{2};\dfrac{-\dfrac{4}{13}}{-\dfrac{8}{13}}=\dfrac{1}{2};\dfrac{\dfrac{2}{15}}{\dfrac{4}{15}}=\dfrac{1}{2};\dfrac{-\dfrac{2}{17}}{-\dfrac{4}{17}}=\dfrac{1}{2}\)
\(\dfrac{50}{100}=\dfrac{\dfrac{4}{13}}{\dfrac{8}{13}}=\dfrac{\dfrac{2}{15}}{\dfrac{4}{15}}=\dfrac{\dfrac{2}{17}}{\dfrac{4}{17}}=\dfrac{50-\dfrac{4}{13}+\dfrac{2}{15}-\dfrac{2}{17}}{100-\dfrac{8}{13}+\dfrac{4}{15}-\dfrac{4}{17}}=\dfrac{1}{2}\)
vậy \(A=\dfrac{1}{2}\)
b)
\(B=\dfrac{1}{19}+\dfrac{9}{19.29}+\dfrac{9}{29.39}+...+\dfrac{9}{1999.2009}\\ B=\dfrac{1}{19}-\dfrac{1}{19}+\dfrac{2}{29}-\dfrac{2}{29}+\dfrac{3}{39}-...-\dfrac{199}{1999}+\dfrac{200}{2009}\\ B=\dfrac{200}{2009}\)
Bài 2:
\(\dfrac{a}{b}=\dfrac{b}{3c}=\dfrac{c}{9a}=\dfrac{b+c}{3c+9a}\)
suy ra: \(b=\dfrac{3c\left(b+c\right)}{3c+9a}=\dfrac{3cb+3c^2}{3c+9a}=\dfrac{bc+c^2}{c+3a}\)
\(c=\dfrac{9a\left(b+c\right)}{3c+9a}=\dfrac{9ab+9ac}{3c+9a}=\dfrac{3ab+3ac}{c+3a}\)
giả sử b=c là đúng thì :\(\dfrac{bc+c^2}{c+3a}=\dfrac{3ab+3ac}{c+3a}\)
hay \(bc+c^2=3ab+3ac\\ \Leftrightarrow c^2+bc-3ab-3ac=0\)
\(\Leftrightarrow\left(b+c\right)\left(c-3a\right)=0\Rightarrow c-3a=0\Rightarrow c=3a\)
b) \(\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{2013.2015}+\dfrac{1}{2014.2016}\\ =\dfrac{1}{2}\left(\dfrac{2}{1.3}+\dfrac{2}{2.4}+\dfrac{2}{3.5}+...+\dfrac{2}{2013.2015}+\dfrac{2}{2014.2016}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{2016}\right)=\dfrac{2015}{4032}< 1\)
mà \(1< \dfrac{4}{3}\) nên \(\dfrac{2015}{4032}< \dfrac{4}{3}\)
hay \(\dfrac{1}{1.3}+\dfrac{1}{2.4}+\dfrac{1}{3.5}+...+\dfrac{1}{2013.2015}+\dfrac{1}{2014.2016}< \dfrac{4}{3}\)
bài 3:
a)\(\left(x-y\right)\left(x+y\right)=x^2-y^2-xy+xy=x^2-y^2\) (đpcm)
b) áp dụng BĐT tam giác, ta có:
\(a+b>c\Rightarrow a+b-c>0\\ b+c>a\Rightarrow b+c-a< 0\\ a+c>b\Rightarrow a-b+c>0\)
suy ra: \(\left(a+b-c\right)\left(b+c-a\right)\left(a-b+c\right)< 0\: \: \: \: \: \: \)
đồng thời \(abc>0\) với mọi a, b, c dương.
nên \(\left(a+b-c\right)\left(b+c-a\right)\left(a-b+c\right)< abc\)
ko tìm dc dấu bằng xảy ra.
Cho ba số thực dương a,b,c . Chứng minh : \(\dfrac{2+6a+3b+6\sqrt{2bc}}{2a+b+2\sqrt{2bc}}\) ≥ \(\dfrac{16}{\sqrt{2b^2+2\left(a+c\right)^2}+3}\)
\(\Leftrightarrow\dfrac{2+3\left(2a+b+2\sqrt{2bc}\right)}{2a+b+2\sqrt{2bc}}\ge\dfrac{16}{\sqrt{2b^2+2\left(a+c\right)^2}+3}\)
\(\Leftrightarrow3+\dfrac{2}{2a+b+2\sqrt{2bc}}\ge\dfrac{16}{\sqrt{2b^2+2\left(a+c\right)^2}+3}\)
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 \(a=\dfrac{b}{2}=c=\dfrac{1}{4}\)
Thu gọn các biểu thức sau
a , 4ab . 1/3ac - 2aca - 9a^2 . 1/2b + 10a^2 . 1/5c + a^2b - a^2bc
b, 2ab - 2bc . c + ab + 1/2c^2b - 4cb^2 + 2bcb
c, x/3 + x/6 + 3x/2 - 4/3mn^2+ 0,2mn^2 - 1/1/3mn^2
d, (2/3ac)^2 . c^2 - 2/5a(c.c)^2 + 2/3ac^3 . c - 1/4ac^4
Giúp mình nha mình cần câu trả lời lắm lun
kí hiệu :
dấu chấm là dấu nhân
kí hiêu này ^ là mũ
kí hiệu nay / là phần nha VD: 1 phần 3 mình viết là 1/3 nha
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. \(2ab-2bc.c+ab+\frac{1}{2}c^2b-4cb^2+2bcb\)
\(=2ab-2bc^2+ab+\frac{1}{2}c^2b-4cb^2+2b^2c\)
\(=\left(2ab+ab\right)+\left(-2bc^2+\frac{1}{2}c^2b\right)+\left(-4cb^2+2b^2c\right)\)
\(=3ab+-\frac{3}{2}bc^2+-2b^2c\)
\(=b\left(3a-\frac{3}{2}c^2-2bc\right)\)
1) Cho △ABC. Khẳng định nào đúng?
\(A.S_{\Delta ABC}=\dfrac{1}{2}a.b.c\)
\(B.\dfrac{a}{\sin A}=R\)
\(C.\cos B=\dfrac{b^2+c^2-a^2}{2bc}\)
\(D.m_c^2=\dfrac{2b^2+2a^2-c^2}{4}\)