Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A = (3,1 – 2,5) – (-2,5 + 3,1) = 3,1 – 2,5 + 2,5 – 3,1 = 0
B = (5,3 – 2,8) – (4 + 5,3) = 5,3 – 2,8 – 4 – 5,3
A=(3,1-2,5)-(-2,5+3,1)=3,1-2,5+2,5-3,1=0
B=(5,3-2,8)-(4+5,3)=5,3-2,8-4-5,3=-6,8
\(\frac{11}{6}-\left(0,04.4,5+0,06.18,\left(6\right)\right)=\frac{11}{6}-0,04.4,5-0,06.\frac{56}{3}=\frac{11}{6}-\frac{9}{50}-\frac{28}{25}=\frac{8}{15}\)
A = \(\left(3,1-2,5\right)-\left(-2,5+3,1\right)\)
\(=3,1-2,5+2,5-3,1=6,2\)
B = \(\left(5,3-2,8\right)-\left(4+5,3\right)\)
\(=5,3-2,8-4-5,3=-6,8\)
C = \(-\left(251.3+281\right)+3.251-\left(1-281\right)\)
\(=-251.3-281+3.251-1+281=-1\)
D = \(-\left(\frac{3}{5}+\frac{3}{4}\right)-\left(-\frac{3}{4}+\frac{2}{5}\right)\)
\(=-\frac{3}{5}-\frac{3}{4}+\frac{3}{4}-\frac{2}{5}=-1\)
(Nhớ k cho mình với nhá!)
\(A=\frac{1}{\left(a-b\right)\left(a-c\right)}+\frac{1}{\left(b-a\right)\left(b-c\right)}+\frac{1}{\left(c-a\right)\left(c-b\right)}\)
\(\frac{b-c-a+c+a-b}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=\frac{0}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)
Ta cần chứng minh \((1+a)(1+b)(1+c) \geq (1+\sqrt[3]{abc})^3\)
\(\Leftrightarrow 1+abc+ab+bc+ca+a+b+c \geq 1+3\sqrt[3]{(abc)^2}+3\sqrt[3]{abc}+abc\)
\(\Leftrightarrow ab+bc+ca+a+b+c \geq 3\sqrt[3]{(abc)^2}+3\sqrt[3]{abc}\)
Đúng theo BĐT AM-GM. Áp dụng vào ta có:
\(\left(1+\frac{1}{a} \right)\left(1+\frac{1}{b} \right)\left(1+\frac{1}{c} \right)=\dfrac{(1+a)(1+b)(1+c)}{abc} \geq \dfrac{(1+\sqrt[3]{abc})^3}{abc} \geq 64\)
Từ \(a+b+c=1 \Rightarrow abc\le \frac{1}{27}\) \(\Rightarrow \dfrac{(1+\sqrt[3]{abc})^3}{abc}=\bigg(\dfrac{1}{\sqrt[3]{abc}}+1\bigg)^3 \geq 64\)
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{3}\)
Ta có : a-b-c=0 \(\Rightarrow\)a-b=c ; a-c=b va b-c=a
Hay : \(\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(b-c\right)\left(b-a\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{a^2}{bc}+\frac{b^2}{ac}+\frac{c^2}{ab}\)
\(=\frac{a^3+b^3+c^3}{abc}\)
\(=\frac{3abc}{abc}\)
=3 (dpcm)
