Ta có: \(a+b+c=0\Rightarrow a+b=-c\Rightarrow\left(a+b\right)^3=\left(-c\right)^3\)

\(\Rightarrow a^3+3a^2b+3ab^2+b^3=-c^3\Rightarrow a^3+b^3+3ab\left(a+b\right)=-c^3\)

\(\Rightarrow a^3+b^3+c^3+3ab.\left(-c\right)=0\Rightarrow a^3+b^3+c^3-3abc=0\)

\(\Rightarrow a^3+b^3+c^3=3abc\) => đpcm

Gọi S là tổng chiều dài quãng đường ô tô chuyển động trong 3 chặng.

Thời gian ô tô chuyển động trên chặng đầu là: \(t_1=\dfrac{S}{3}:v_1=\dfrac{S}{3}:12=\dfrac{S}{36}\) (giây)

Thời gian ô tô chuyển động trên chặng sau là: \(t_2=\dfrac{S}{3}:v_2=\dfrac{S}{3}:8=\dfrac{S}{24}\) (giây)

Thời gian ô tô chuyển động trên chặng cuối là: \(t_3=\dfrac{S}{3}:v_3=\dfrac{S}{3}:16=\dfrac{S}{48}\) (giây)

Vận tốc trung bình của ô tô trên cả 3 chặng đường là:

\(v_{tb}=\dfrac{S}{t_1+t_2+t_3}=S:\left(\dfrac{S}{36}+\dfrac{S}{24}+\dfrac{S}{48}\right)=S:\left(\dfrac{4S}{144}+\dfrac{6S}{24}+\dfrac{3S}{48}\right)\)

\(=S:\dfrac{13S}{144}=\dfrac{144}{13}\approx11,08\) (m/s)

Vậy \(v_{tb}\) của ô tô trên cả 3 chặng là \(\approx11,08\) (m/s)

a, \(7A-2B=7.\left(5x+2y\right)-2.\left(9x+7y\right)\)

\(=35x+14y-18x+14y=17x\)

Vậy 7A-2B=17x

b, Ta có: \(5x+2y⋮17\Rightarrow5.\left(5x+2y\right)⋮17\Rightarrow25x+10y⋮17\) (1)

Mà \(\left(2x+10y\right)+\left(9x+7y\right)=25x+10y+9x+7y\)

\(=34x+17y=17.\left(2x+y\right)⋮17\) (2)

Từ (1) và (2) \(\Rightarrow9x+7y⋮17\) => đpcm

Đặt \(A=\dfrac{\left(1^4+\dfrac{1}{4}\right)\left(3^4+\dfrac{1}{4}\right)...\left(19^4+\dfrac{1}{4}\right)}{\left(2^4+\dfrac{1}{4}\right)\left(4^4+\dfrac{1}{4}\right)...\left(20^4+\dfrac{1}{4}\right)}\)

\(=\dfrac{\left[\left(1^4+\dfrac{1}{4}\right).2^4\right]\left[\left(3^4+\dfrac{1}{4}\right).2^4\right]...\left[\left(19^4+\dfrac{1}{4}\right).2^4\right]}{\left[\left(2^4+\dfrac{1}{4}\right).2^4\right]\left[\left(4^4+\dfrac{1}{4}\right).2^4\right]...\left[\left(20^4+\dfrac{1}{4}\right).2^4\right]}\)

\(=\dfrac{\left(2^4+4\right)\left(6^4+4\right)...\left(38^4+4\right)}{\left(4^4+4\right)\left(8^4+4\right)...\left(40^4+4\right)}\)

Lưu ý: \(a^4+4=\left(a^4+4a^2+4\right)-4a^2=\left(a^2+2\right)^2-\left(2a\right)^2\)

\(=\left(a^2-2a+2\right)\left(a^2+2a+2\right)\)

Áp dụng vào biểu thức A, ta có:

\(A=\dfrac{\left(2^4+4\right)\left(6^4+4\right)...\left(38^4+4\right)}{\left(4^4+4\right)\left(8^4+4\right)...\left(40^4+4\right)}\)

\(=\dfrac{\left(2^2-2.2+2\right)\left(2^2+2.2+2\right)...\left(38^2-38.2+2\right)\left(38^2+38.2+2\right)}{\left(4^2-2.4+2\right)\left(4^2+2.4+2\right)...\left(40^2-2.40+2\right)\left(40^2+2.40+2\right)}\)

\(=\dfrac{2.10.26..1370.1522}{10.26.50...1522.1682}=\dfrac{2}{1682}=\dfrac{1}{841}\)

Vậy \(A=\dfrac{1}{841}\)

a)Ta có:

8⁵=(2³)⁵=2¹⁵

=>8⁵+2¹¹=2¹⁵+2¹¹=2¹¹.(2⁴+1)=2¹¹.17 chia hết cho 17

=>8⁵+2¹¹ chia hết cho 17

b)Ta có:

8⁷=(2³)⁷=2²¹

=>8⁷-2¹⁹=2²¹-2¹⁹=2¹⁹.(2²-1)=2¹⁹.3 ????????????

\(3x^2-18x+27=3.\left(x^2-6x+9\right)=3.\left(x-3\right)^2\)

a, \(A=1999.2001=\left(2000-1\right)\left(2000+1\right)=2000^2-1< 2000^2=B\)

Vậy A<B

b, \(A=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)=\left(2^{16}-1\right)\left(2^{16}+1\right)\)

\(=2^{32}-1< 2^{32}=B\)

Vậy A<B

Ta có: \(x^2-2x+y^2-4y+6=\left(x^2-2x+1\right)+\left(y^2-4y+4\right)+1\)

\(=\left(x-1\right)^2+\left(y-2\right)^2+1\)

Vì \(\left(x-1\right)^2\ge0;\left(y-2\right)^2\ge0\Rightarrow\left(x-1\right)^2+\left(y-2\right)^2+1\ge1\)

hay \(x^2-2x+y^2-4y+6\ge1\)

Dấu ''='' xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\left(y-2\right)^2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x-1=0\\y-2=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

Vậy \(min_{\left(x^2-2x+y^2-4x+6\right)}=1\) khi x=1; y=2

Vì 2>a,b,c>0 => a(2-b); b(2-c); c(2-a) là các số thực dương.

Áp dụng bất đẳng thức Cauchy cho 6 số, ta có:

\(\dfrac{a+\left(2-b\right)+b+\left(2-c\right)+c+\left(2-a\right)}{6}\ge\)

\(\sqrt[6]{a.\left(2-b\right).b.\left(2-c\right).c.\left(2-a\right)}\)

\(\Rightarrow\dfrac{a+b+c-a-b-c+2+2+2}{6}\ge\sqrt[6]{a.\left(2-b\right).b.\left(2-c\right).c.\left(2-a\right)}\)

\(\Rightarrow1\ge\sqrt[6]{a.\left(2-b\right).b.\left(2-c\right).c.\left(2-a\right)}\)

\(\Rightarrow1^6\ge a.\left(2-b\right).b.\left(2-c\right).c.\left(2-a\right)\Rightarrow1\ge a.\left(2-b\right).b.\left(2-c\right).c.\left(2-a\right)\)

=> a(2-b); b(2-c); c(2-a) không đồng thời lớn hơn 1

=> đpcm

a, Vì \(\left|x+1\right|\ge0;-\left|y+3\right|\le0\Rightarrow\left|x+1\right|=-\left|y+3\right|\)

\(\Leftrightarrow\left|x+1\right|=-\left|y+3\right|=0\Rightarrow\left\{{}\begin{matrix}\left|x+1\right|=0\\-\left|y+3\right|=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+1=0\\y+3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\)

Vậy x=-1; y=-3

b, Do \(\left|3x-1\right|\ge0\Rightarrow\left|3x-1\right|+1>0\). Mà \(-\left|2y-5\right|\le0\)

=> không tồn tại giá trị của x, y để \(\left|3x-1\right|+1=-\left|2y-5\right|\)

Vậy không tồn tại giá trị của x, y thỏa mãn yêu cầu bài toán