1: =>4a^3+4b^3-a^3-3a^2b-3ab^2-b^3>=0

=>a^3-a^2b-ab^2+b^3>=0

=>(a+b)(a^2-ab+b^2)-ab(a+b)>=0

=>(a+b)(a-b)^2>=0(luôn đúng)

2: \(a^4+b^4=\dfrac{a^4}{1}+\dfrac{b^4}{1}>=\dfrac{\left(a^2+b^2\right)^2}{1}=\dfrac{1}{2}\left(\dfrac{a^2}{1}+\dfrac{b^2}{1}\right)^2\)

=>\(a^4+b^4>=\dfrac{1}{2}\left(\dfrac{\left(a+b\right)^2}{2}\right)^2=\dfrac{\left(a+b\right)^4}{8}\)

Bài 1 : chị phân tích ra thừa số nguyên tố, rồi rút gọn đi là ok mak

Bài 2:

\(B=\dfrac{\left(1^4+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)........\left(11^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right)........\left(12^4+\dfrac{1}{4}\right)}\)

\(=\dfrac{\left(1^2+1+\dfrac{1}{2}\right)\left(1^2-1+\dfrac{1}{2}\right).........\left(11^2-11+\dfrac{1}{2}\right)}{\left(2^2+1+\dfrac{1}{2}\right)\left(2^2-2+\dfrac{1}{2}\right).......\left(12^2-12+\dfrac{1}{2}\right)}\)

\(=\dfrac{\dfrac{1}{2}\left(1.2+\dfrac{1}{2}\right)\left(2.3+\dfrac{1}{2}\right).......\left(11.12+\dfrac{1}{2}\right)}{\left(2.3+\dfrac{1}{2}\right)\left(3.4+\dfrac{1}{2}\right)......... \left(12.13+\dfrac{1}{2}\right)}\)

\(=\dfrac{\dfrac{1}{2}}{12.13+\dfrac{1}{2}}\)

\(=\dfrac{1}{313}\)

\(A=\dfrac{35.\left(27^8+2.9^{11}\right)}{15.\left(81^6-12.3^{19}\right)}\)

\(=\dfrac{35.27^8+35.2.9^{11}}{15.81^6-15.12.3^{19}}\)

\(=\dfrac{5.7.\left(3^3\right)^8+5.7.\left(3^2\right)^{11}}{3.5.\left(3^4\right)^6-3.5.3.2^2.3^{19}}\)

\(=\dfrac{5.7.3^{24}+5.7.3^{22}}{5.3^{25}-3^{21}.2^2.5}\)

\(=\dfrac{5.7.3^{22}\left(3^2+1\right)}{5.3^{21}\left(3^4-2^2\right)}\)

\(=\dfrac{7.2.10}{81-4}\)

\(=\dfrac{720}{77}\)

@Phùng Khánh Linh, Nguyễn Nam, Ribi Nkok Ngok, Trần Quốc Lộc, Phạm Hoàng Giang, Thien Tu Borum, Nguyễn Thanh Hằng, Thảo Phương , Trương Hồng Hạnh, ...

a) Ta có: \(\left(\dfrac{-1}{2}\right)^2\cdot\dfrac{7}{4}:\left(\dfrac{5}{8}-1\dfrac{3}{16}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{7}{4}:\left(\dfrac{5}{8}-\dfrac{19}{16}\right)\)

\(=\dfrac{1}{4}\cdot\dfrac{7}{4}:\dfrac{-9}{16}\)

\(=\dfrac{1}{4}\cdot\dfrac{7}{4}\cdot\dfrac{-16}{9}\)

\(=\dfrac{-112}{144}=\dfrac{-7}{9}\)

b) Ta có: \(17\dfrac{6}{11}\cdot\dfrac{4}{27}-8\dfrac{6}{11}:\dfrac{27}{4}+350\%\)

\(=17\dfrac{6}{11}\cdot\dfrac{4}{27}-8\dfrac{6}{11}\cdot\dfrac{4}{27}+350\%\)

\(=\dfrac{4}{27}\left(17+\dfrac{6}{11}-8-\dfrac{6}{11}\right)+\dfrac{7}{2}\)

\(=\dfrac{4}{27}\cdot9+\dfrac{7}{2}\)

\(=\dfrac{4}{3}+\dfrac{7}{2}=\dfrac{8}{6}+\dfrac{21}{6}=\dfrac{29}{6}\)

Bài 1:

(a)

Vì $a,b,c$ là độ dài ba cạnh tam giác nên theo BĐT tam giác ta có:

\(\left\{\begin{matrix} a+b>c\\ b+c>a\\ c+a>b\end{matrix}\right.\Rightarrow \left\{\begin{matrix} c(a+b)>c^2\\ a(b+c)>a^2\\ b(c+a)>b^2\end{matrix}\right.\)

\(\Rightarrow c(a+b)+a(b+c)+b(c+a)> c^2+a^2+b^2\)

\(\Leftrightarrow 2(ab+bc+ac)> a^2+b^2+c^2\)

Ta có đpcm.

(2): Bài này có nhiều cách giải. Nhưng mình xin đưa ra cách làm thuần túy Cô-si nhất.

Đặt

\((a+b-c, b+c-a, c+a-b)=(x,y,z)\Rightarrow (a,b,c)=(\frac{x+z}{2}; \frac{x+y}{2}; \frac{y+z}{2})\)

Khi đó:

\(\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}=\frac{x+z}{2y}+\frac{x+y}{2z}+\frac{y+z}{2x}\)

\(=\frac{x}{2y}+\frac{z}{2y}+\frac{x}{2z}+\frac{y}{2z}+\frac{y}{2x}+\frac{z}{2x}\geq 6\sqrt[6]{\frac{1}{2^6}}=3\) (áp dụng BĐT Cô-si)

Ta có đpcm

Dấu "=" xảy ra khi $x=y=z$ hay $a=b=c$

(c):

Theo BĐT tam giác:

\(b+c>a\Rightarrow 2(b+c)> b+c+a\Rightarrow b+c> \frac{a+b+c}{2}\)

\(\Rightarrow \frac{a}{b+c}< \frac{2a}{a+b+c}\)

Hoàn toàn tương tự với những phân thức còn lại và cộng theo vế:

\(\Rightarrow \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}< \frac{2a}{a+b+c}+\frac{2b}{a+b+c}+\frac{2c}{a+b+c}=2\)

Ta có đpcm.

Bài 2:

Áp dụng BĐT Cô-si cho các số dương:

\(a^2+b^2+c^2+d^2+ab+cd\geq 6\sqrt[6]{a^2.b^2.c^2.d^2.ab.cd}=6\sqrt[6]{(abcd)^3}=6\sqrt[6]{1^3}=6\)

Ta có đpcm

Dấu "=" xảy ra khi \(\left\{\begin{matrix} a^2=b^2=c^2=d^2=ab=cd\\ abcd=1\end{matrix}\right.\Rightarrow a=b=c=d=1\)

Bài 3:

Ta có:

\((x-1)(x-3)(x-4)(x-6)+9\)

\(=[(x-1)(x-6)][(x-3)(x-4)]+9\)

\(=(x^2-7x+6)(x^2-7x+12)+9\)

\(=a(a+6)+9\) (đặt \(x^2-7x+6=a\) )

\(=a^2+6a+9=(a+3)^2\geq 0, \forall a\in\mathbb{R}\)

Ta có đpcm.

`(4^{-2}:(1/3)^2xx1/2)/((-1/6)^2)`
`=((1/16:1/9)xx1/2)/(1/36)`
`=36xx1/2xx(1/16xx9)`
`=18xx9/16`
`=81/8`

Đề sai rồi bạn

\(\dfrac{1}{2}\) là để giống m vuông í

Bài 1:

a) \(3^7:3^5-\left(\dfrac{5}{17}\right)^0=3^{7-5}-1=3^2-1=9-1=8\)

b) \(\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{1}{2}+2\right)^3\)

\(=\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{5}{2}\right)^3\)

\(=\left(\dfrac{5}{2}\right)^{10}\)

c) \(8.\left(\dfrac{1}{4}\right)^3+\left(\dfrac{2}{27}\right)^0-\dfrac{1}{8}\)

\(=8.\dfrac{1}{64}+1-\dfrac{1}{8}\)

\(=\dfrac{1}{8}+1-\dfrac{1}{8}\)

\(=1\)

Bài 2:

a) \(\dfrac{3^4.4^4}{6^4}=\dfrac{3^4.\left(2^2\right)^4}{\left(2.3\right)^4}=\dfrac{3^4.2^8}{2^4.3^4}=\dfrac{2^8}{2^4}=2^4=16\)

b) \(\dfrac{15^3}{10^3}=\dfrac{\left(3.5\right)^3}{ \left(2.5\right)^3}=\dfrac{3^3.5^3}{2^3.5^3}=3^3:2^3=\dfrac{27}{8}\)

c) \(\dfrac{4^2.12^5}{9^2.2^{10}}=\dfrac{\left(2^2\right)^2.\left[3.\left(2^2\right)\right]^5}{\left(3^2\right)^2.2^{10}}=\dfrac{2^4.3^5.2^{10}}{3^4.2^{10}}=2^4.3=16.3=48\)

d) \(\dfrac{6^2+5.2^2+4}{15}=\dfrac{\left(2.3\right)^2+5.2^2+2^2}{15}=\dfrac{2^2.3^2+5.2^2+2^2}{15}=\dfrac{2^2\left(3^2+5+1\right)}{15}=\dfrac{2^2.15}{15}=2^2=4\)

Bài 3:

a) \(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.\left(-1\right)^5}{\left(\dfrac{2}{5}\right)^2.\left(\dfrac{-5}{12}\right)^2}\)

\(=\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.-1}{\left[\dfrac{2}{5}.\left(\dfrac{-5}{12}\right)\right]^2}\)

\(=\dfrac{\left(\dfrac{2}{3}\right)^3. \left(\dfrac{-3}{4}\right)^2.-1}{\left(\dfrac{-1}{6}\right)^2}\)

\(=\left(\dfrac{2}{3}\right)^3.\left[\left(\dfrac{-3}{4}\right).-6\right]^2.-1\)

\(=\left(\dfrac{2}{3}\right)^3.\left(\dfrac{9}{2}\right)^2.-1\)

\(=\left(\dfrac{2}{3}\right)^2.\dfrac{2}{3}.\left(\dfrac{9}{2}\right)^2.-1\)

\(=\left(\dfrac{2}{3}.\dfrac{9}{2}\right)^2.\dfrac{2}{3}.-1\)

\(=9.\dfrac{2}{3}.-1\)

\(=6.-1=-6\)

b) \(\dfrac{6^6+6^3.3^3+3^6}{-73}=\dfrac{\left(2.3\right)^6+\left(2.3\right)^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^6+3^6}{-73}=\dfrac{3^6\left(2^6+2^3+1\right)}{-73}=\dfrac{3^6.73}{-73}=\dfrac{3^6}{-1}=\left(-3\right)^6\)

\(#Wendy.Dang\)

Lần sau bnn gửi từng bài thôi nha, chứ như vầy nhiều quá thì làm không nổi mất. đánh máy nãy giờ lú luôn gòi nè :))

Võ Ngọc Phương

Bài  3b, kết quả -(3)⁶ = - 729 em nhá chứ không phải (-3)⁶

ai giúp mìn vứi ❤

a: \(A=\dfrac{3^6\cdot3^8\cdot5^4-3^{13}\cdot5^{13}\cdot5^{-9}}{3^{12}\cdot5^6+5^6\cdot3^{12}}\)

\(=\dfrac{3^{14}\cdot5^4-3^{13}\cdot5^4}{2\cdot3^{12}\cdot5^6}\)

\(=\dfrac{3^{13}\cdot5^4\cdot\left(3-1\right)}{2\cdot3^{12}\cdot5^6}=\dfrac{3}{5^2}=\dfrac{3}{25}\)

c: \(C=\dfrac{\dfrac{27}{64}+\dfrac{125}{64}-5\cdot\dfrac{16-15}{12}}{\dfrac{25}{64}+\dfrac{4}{9}-\dfrac{5}{6}}\)

\(=\dfrac{47}{24}:\dfrac{1}{576}=47\cdot24=1128\)

Câu a)

Áp dụng BĐT Cauchy-Schwarz:

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}\geq \frac{9}{a+2b}\) (1)

\(\frac{1}{b}+\frac{1}{c}+\frac{1}{c}\geq \frac{9}{b+2c}\)(2)

\(\frac{1}{c}+\frac{1}{a}+\frac{1}{a}\geq \frac{9}{c+2a}\) (3)

Lấy \((1)+2.(2)+3.(3)\) ta có:

\(\Rightarrow \frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{2}{b}+\frac{2}{c}+\frac{2}{c}+\frac{3}{c}+\frac{3}{a}+\frac{3}{a}\geq 9\left(\frac{1}{a+2b}+\frac{1}{b+2c}+\frac{1}{c+2a}\right)\)

\(\Leftrightarrow \frac{7}{a}+\frac{4}{b}+\frac{7}{c}\geq 9\left(\frac{1}{a+2b}+\frac{1}{b+2c}+\frac{1}{c+2a}\right)\)

Ta có đpcm

Dấu bằng xảy ra khi \(a=b=c\)

Câu b)

Áp dụng BĐT Cauchy-Schwarz:

\(\frac{1}{a}+\frac{4}{b}\geq \frac{(1+2)^2}{a+b}=\frac{9}{a+b}\)

\(\Rightarrow \frac{1}{3a}+\frac{4}{3b}\geq \frac{3}{a+b}(1)\)

\(\frac{1}{3b}+\frac{1}{2c}+\frac{1}{2c}\geq \frac{9}{3b+4c}\)

\(\Rightarrow \frac{2}{3b}+\frac{2}{c}\geq \frac{18}{3b+4c}\) (2)

\(\frac{1}{c}+\frac{1}{3a}+\frac{1}{3a}\geq \frac{9}{c+6a}\) (3)

Từ (1); (2); (3) cộng theo vế:

\(\Rightarrow \frac{1}{a}+\frac{2}{b}+\frac{3}{c}\geq \frac{3}{a+b}+\frac{18}{3b+4c}+\frac{9}{c+6a}\)

(đpcm)

Dấu bằng xảy ra khi \(a=\frac{b}{2}=\frac{c}{3}\)

Câu c)

BĐT cần chứng minh tương đương với:
\(\frac{b+c+a}{a}+\frac{2a+c}{b}+\frac{4(a+b)}{a+c}\geq 10\) (*)

Áp dụng BĐT AM-GM:

\(\text{VT}=\frac{b}{a}+\frac{c+a}{2a}+\frac{c+a}{2a}+\frac{a}{b}+\frac{a+c}{2b}+\frac{a+c}{2b}+\frac{a+b}{a+c}+\frac{a+b}{a+c}+\frac{a+b}{a+c}+\frac{a+b}{a+c}\)

\(\geq 10\sqrt[10]{\frac{ba(c+a)^4(a+b)^4}{16a^3b^3(a+c)^4}}=10\sqrt[10]{\frac{(a+b)^4}{16a^2b^2}}\)

Theo AM-GM: \((a+b)^2\geq 4ab\Rightarrow (a+b)^4\geq 16a^2b^2\)

\(\Rightarrow \text{VT}\geq 10\sqrt[10]{\frac{(a+b)^4}{16a^2b^2}}\geq 10\)

Vậy (*) được cm. Ta có đpcm. Dấu bằng xảy ra khi a=b=c