\(\sqrt[3]{64}+\sqrt[3]{6859}+\sqrt[3]{729}\)

\(=\sqrt[3]{4^3}+\sqrt[3]{19^3}+\sqrt[3]{9^3}\)

\(=4+19+9\)

\(=32\)

Ta có:

+ 3√512=3√83=8;5123=833=8;

+ 3√−729=3√(−9)3=−9;−7293=(−9)33=−9;

+ 3√0,064=3√0,43=0,4;0,0643=0,433=0,4;

+ 3√−0,216=3√(−0,6)3=−0,6;−0,2163=(−0,6)33=−0,6;

+ 3√−0,008=3√(−0,2)3=−0,2.

Đáp án:

( lần lượt như trên nhé!!! Ko viết lại đề)

8 ; - 9 ; 0,4 ; - 0,6 ; - 0,2

Kết quả lần lượt là $8$ ; $-9$ ; $0,4$ ; $-0,6$ ; $-0,2$

Ta có: \(A=\left[6.\left(\frac{-1}{3}\right)^2-\left(-\frac{1}{3}\right)+1\right]:\left(\frac{-1}{3}-1\right)\)

\(\Rightarrow A=\left[6.\frac{1}{9}+\frac{1}{3}+1\right]:\left(\frac{-1}{3}-\frac{3}{3}\right)\)

\(\Rightarrow A=\left[\frac{2}{3}+\frac{1}{3}+1\right]:\frac{-4}{3}\)

\(\Rightarrow A=\left[1+1\right].\frac{-3}{4}=2.\frac{-3}{4}=\frac{-3}{2}\)

Mà \(B=\left(729-1^3\right)\left(729-2^3\right)\left(729-3^3\right)...\left(729-125^3\right)\)

\(=\left(729-1^3\right)\left(729-2^3\right)...\left(729-9^3\right)...\left(729-125^3\right)\)

\(=\left(729-1^3\right)\left(729-2^3\right)...0...\left(729-125^3\right)=0\)

Vì \(\frac{-3}{2}< 0\)nên A < B

\(729:3^4=9\)

\(729:3^3:9=3\)

Phân tích số dưới dấu căn ra thừa số nguyên tố hoặc đổi thành phân số.

3\(\sqrt{ }\)512 = 3\(\sqrt{ }\)29 = 3\(\sqrt{ }\)(23)3= 23 = 8

3\(\sqrt{ }\)-729 = – 3\(\sqrt{ }\)729 = – 3\(\sqrt{ }\)36=- 3\(\sqrt{ }\)(32)3 = – (32)= -9

3\(\sqrt{ }\)-216 = -3/5

3\(\sqrt{ }\)-0,008 = -1/5

-597871

nhầm nha

viết sai rồi 

câu 2 này ms làm tức thì nà

đầu tiên t c/m câu phụ \(\left(a-b\right)\left(b-c\right)\left(c-a\right)\le\dfrac{3\sqrt{3}}{2}\)

đặt P =VT ta có \(P\le\left|P\right|=\sqrt{P^2}\)

vậy ta c/m \(P^2\le\dfrac{27}{4}\)

<=> \(\left(a-b\right)^2\left(b-c\right)^2\left(c-a\right)^2\le\dfrac{27}{4}\)

không mất tính tổng wat giả sử \(a\ge b\ge c\) (2)

dễ thấy \(\left(b-c\right)^2\le b^2;\left(c-a\right)^2\le a^2\)

=> c/m :\(a^2b^2\left(a-b\right)^2\le\dfrac{27}{4}\Leftrightarrow4a^2b^2\left(a-b\right)^2\le\dfrac{27}{4}\)

áp dụng AM-GM ta có

\(4a^2b^2\left(a-b\right)^2=\left(2ab\right)\left(2ab\right)\left(a^2-2ab+b^2\right)\le\left[\dfrac{2\left(2ab\right)+\left(a^2-2ab+b^2\right)}{3}\right]^3=\left(\dfrac{a^2+2ab+b^2}{3}\right)^3=\dfrac{\left(a+b\right)^6}{27}\)

mặt khác từ (2) ta có \(a+b\le a+b+c=3\)

=>dpcm

@quay trở lại bài toán áp dụng câu phụ mik vừa ns c2 <=> c/m

\(\left(a^3+b^3+c^3\right)\left(a+b\right)\left(b+c\right)\left(c+a\right)\le\dfrac{243}{4}\)

nhân 3 cho 2 vế r áp dụng AM-GM

\(\left(a^3+b^3+c^3\right)3\left(a+b\right)\left(a+c\right)\left(c+b\right)\)\(\le\dfrac{\left[a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\right]^2}{4}=\dfrac{\left(a+b+c\right)^6}{4}=\dfrac{729}{4}\)

=> dpcm

giúp jum t @Neet;@Ace Legona (có cách khác AM-GM thì qá tốt nha!!)

áp dụng BĐT \(\sqrt[3]{\dfrac{a^3+b^3+c^3}{3}}\ge\dfrac{a+b+c}{3}\) và \(\sqrt[3]{\dfrac{a^3+b^3}{2}}\ge\dfrac{a+b}{2}\) (c/m dưới dạng tổng quát)

\(\sqrt[3]{a^2+3}=\sqrt[3]{4}.\sqrt[3]{\dfrac{\dfrac{a^2+1}{2}+1}{2}}\ge\sqrt[3]{4}.\dfrac{\sqrt[3]{\dfrac{a^2+1}{2}}+1}{2}\)

\(\sqrt[3]{b^2+3}=\sqrt[3]{7}.\sqrt[3]{\dfrac{5.\dfrac{b^2+1}{5}+1+1}{7}}\ge\sqrt[3]{7}.\dfrac{5\sqrt[3]{\dfrac{b^2+1}{5}}+1+1}{ }\)

\(\sqrt[3]{c^2+3}=\sqrt[3]{12}.\sqrt[3]{\dfrac{5.\dfrac{c^2+1}{10}+1}{6}}\ge\sqrt[3]{12}.\dfrac{5\sqrt[3]{\dfrac{c^2+1}{10}}+1}{6}\)

đặt P = VT của dpcm,ta đc

\(P\ge\dfrac{1}{\sqrt[3]{2}}\left(\sqrt[3]{\dfrac{a^2+1}{2}}+1\right)+\dfrac{1}{5\sqrt[3]{2}}\left(5\sqrt[3]{\dfrac{b^2+1}{5}}+2\right)+\dfrac{1}{5\sqrt[3]{2}}\left(\sqrt[3]{\dfrac{c^2+1}{10}}+1\right)=\left(\sqrt[3]{\dfrac{a^2+1}{4}+\sqrt[3]{\dfrac{b^2+1}{10}}+\sqrt[3]{\dfrac{c^2+1}{20}}}\right)+\dfrac{8}{5\sqrt[3]{2}}\)

AM-GM bộ 3 số ta được

\(\sqrt[3]{\dfrac{a^2+1}{4}}+\sqrt[3]{\dfrac{b^2+1}{10}}+\sqrt[3]{\dfrac{c^2+1}{20}}\ge3\sqrt[9]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{800}}\)

we c/m \(3\sqrt[9]{\dfrac{\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)}{800}}+\dfrac{8}{5\sqrt[3]{2}}\ge\dfrac{23}{5\sqrt[3]{2}}\)

<=>\(\left(a^2+1\right)\left(b^2+1\right)\left(c^2+1\right)\ge100\)

cắn bút bín đổi ta đc \(\left(a^2+1\right)\left[\left(b+c\right)^2+\left(bc-1\right)^2\right]\ge100\)

áp dụng BĐT cauchy- gì gì đó

\(\left(a^2+1\right)\left[\left(b+c\right)^2+\left(bc-1\right)^2\right]\ge\left[a\left(b+c\right)+\left(bc-1\right)\right]^2=\left(ab+bc+ca-1\right)^2\ge10^2=100\)=> dpcm

dấu = xảy ra <=> a=1,b=2,c=3

p/s:có j sai ns t nha cách làm của t khá rườm rà @@

\(=\dfrac{3\cdot7\cdot3^4\cdot3^6+3^6\cdot3^4\cdot3^3}{3^2\cdot3^4\cdot2\cdot3^{12}\cdot13+3^2\cdot2\cdot3^3\cdot2\cdot3^4\cdot2\cdot3^2+723\cdot729}\)

\(=\dfrac{3^{11}\cdot7+3^{13}}{3^{18}\cdot26+3^{11}\cdot8+3^7\cdot241}\)

\(=\dfrac{3^{11}\left(7+9\right)}{3^7\left(3^{11}\cdot26+3^4\cdot8+241\right)}=\dfrac{3^7\cdot16}{17\cdot101\cdot2683}\)