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1mmHg=133,3 N/m\(^3\)

24 tháng 12 2017

Đập tan tham vọng và ý xâm lược Đại Việt của đế chế Mông Nguyên, bảo vệ được độc lập toàn vẹ lãnh thổ và chủ quyền quốc gia.

- Có sự chuẩn bị chu đáo về mặt

-Thể hiện sức mạnh dân tộc đánh bại mọi kẻ thù

- Góp phần xây dựng truyền thống dân tộc đánh bại mọi kẻ thù, xây dựng học thuyết quân sự để lại nhiều bài học cho đời sau.

Mình nghĩ vậy đó chúc bạn học tốt nhé!wink

24 tháng 12 2017

1)Qua 3 lần kháng chiến chống quân xâm lược Mông Nguyên rút ra bài học lịch sử là:

- Đập tan tham vọng và ý xâm lược Đại Việt của đế chế Mông Nguyên, bảo vệ được độc lập toàn vẹ lãnh thổ và chủ quyền quốc gia.

- Có sự chuẩn bị chu đáo về mặt

-Thể hiện sức mạnh dân tộc đánh bại mọi kẻ thù

- Góp phần xây dựng truyền thống dân tộc đánh bại mọi kẻ thù, xây dựng học thuyết quân sự để lại nhiều bài học cho đời sau.

31 tháng 5 2020

*\(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)

\(M=6x^2+9xy-y^2-\left(5x^2-2xy\right)\)

\(M=6x^2+9xy-y^2-5x^2+2xy\)

\(M=\left(6-5\right)x^2+\left(9+2\right)xy-y^2\)

\(M=x^2+11xy-y^2\)

\(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\)

Ta có : \(\hept{\begin{cases}\left(2x-5\right)^{2018}\ge0\forall x\\\left(3y+4\right)^{2020}\ge0\forall y\end{cases}\Rightarrow}\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\ge0\forall x,y\)

Mà đề cho \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\)

=> \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}=0\)

=> \(\hept{\begin{cases}2x-5=0\\3y+4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}}\)

Thay x = 5/2 ; y = -4/3 vào M ta được :

\(M=\left(\frac{5}{2}\right)^2+11\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)-\left(-\frac{4}{3}\right)^2\)

\(M=\frac{25}{4}+\frac{-110}{3}-\frac{16}{9}\)

\(M=\frac{-1159}{36}\)

Vậy giá trị của M = -1159/36 khi x = 5/2 ; y = -4/3

Không chắc nha 

29 tháng 12 2016

gọi V là thể tích của vật

V' là thể tích chìm của vật

D là khối lượng riêng của vật

D' là khối lượng riêng của nước

Trọng lượng của vật là

P = V.d = V.10D

Khi thả vật vào nước thì lực đẩy Ác - si - méc tác dụng vào vật là:

FA = V'.d' = V'.10D'

Khi vật nằm cân bằng trong nước thì

P = FA

V.10D = V'.10D'

=> 560V=1000V'

=> \(\frac{V'}{V}=\frac{560}{1000}=56\%\)

=> V'= 56%V

Vậy vật chìm 56% thể tích của vật

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8 tháng 6 2017

\(a^3\left(a-1\right)+2\left(a-1\right)=\left(a-1\right)\left(a^3+2\right)=0\)\(\Rightarrow\left[{}\begin{matrix}a-1=0\\a^3+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=1\\a^3=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=1\\\sqrt[3]{-2}\end{matrix}\right.\)

8 tháng 6 2017

\(a^3\left(a-1\right)+2\left(a-1\right)=0\\ \Leftrightarrow\left(a-1\right)\left(a^3+2\right)=0\\ \Rightarrow\left[{}\begin{matrix}a-1=0\\a^3+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=1\\a=\sqrt[3]{-2}\end{matrix}\right.\)

uses crt;
var a,b,c,ucln,bcnn,i,min:longint;
begin
clrscr;
write('Nhap a='); readln(a);
write('Nhap b='); readln(b);
write('Nhap c='); readln(c);
min:=a;
if min>b then min:=b;
if min>c then min:=c;
ucln:=1;
for i:=1 to min do
if (a mod i=0) and (b mod i=0) and (c mod i=0) then
begin
if ucln<i then ucln:=i;
end;
bcnn:=a*b*c;
for i:=a*b*c downto 1 do
if (i mod a=0) and (i mod b=0) and (i mod c=0) then
begin
if bcnn>i then bcnn:=i;
end;
writeln('Uoc chung lon nhat: ',ucln);
writeln('Boi chung nho nhat: ',bcnn);
readln;
end.

Ta có: \(\left(2n-1\right)\div\left(3-n\right)\) dư 4

\(\Rightarrow3-n\inƯ\left(4\right)\)

\(\Rightarrow3-n\in\left\{\pm1;\pm2;\pm4\right\}\)

\(\Rightarrow n\in\left\{2;4;1;5;-1;7\right\}\)