Câu 2: 

uses crt;

var a:array[1..100]of integer;

i,n,t:integer;

begin

clrscr;

write('Nhap n='); readln(n);

for i:=1 to n do

begin

write('A[',i,']='); readln(a[i]);

end;

t:=0;

for i:=1 to n do 

if (4<a[i]) and (a[i]<15) then t:=t+a[i];

writeln(t);

readln;

end.

(2015 + 219.830) : (831.220 + 965)

= 183785 : 183785

= 1

Câu 4:

a, \(4P+5O_2\underrightarrow{t^o}2P_2O_5\)

\(4K+O_2\underrightarrow{t^O}2K_2O\)

b, \(CuO+H_2\underrightarrow{t^o}Cu+H_2O\)

c, \(2K+2H_2O\rightarrow2KOH+H_2\)

\(CaO+H_2O\rightarrow Ba\left(OH\right)_2\)

\(SO_3+H_2O\rightarrow H_2SO_4\)

7. Ta có: \(\left(x+\sqrt{x^2+3}\right)\left(\sqrt{x^2+3}-x\right)=x^2+3-x=3\)

\(\Rightarrow\sqrt{x^2+3}-x=y+\sqrt{y^2+3}\Rightarrow x+y=\sqrt{x^2+3}-\sqrt{y^2+3}\left(1\right)\)

Lại có \(\left(y+\sqrt{y^2+3}\right)\left(\sqrt{y^2+3}-y\right)=y^2+3-y=3\)

\(\Rightarrow\sqrt{x^2+3}+x=\sqrt{y^2+3}-y\Rightarrow x+y=\sqrt{y^2+3}-\sqrt{x^2+3}\left(2\right)\)

Lấy \(\left(1\right)+\left(2\right)\Rightarrow2\left(x+y\right)=0\Rightarrow x+y=0\)

9. Ta có: \(\sqrt{55+\sqrt{109}}-\sqrt{55-\sqrt{109}}\)

\(=\sqrt{\dfrac{110+2\sqrt{109}}{2}}-\sqrt{\dfrac{110-2\sqrt{109}}{2}}\)

\(=\sqrt{\dfrac{\left(\sqrt{109}+1\right)^2}{2}}-\sqrt{\dfrac{\left(\sqrt{109}-1\right)^2}{2}}=\dfrac{\sqrt{109}+1}{\sqrt{2}}-\dfrac{\sqrt{109}-1}{\sqrt{2}}\)

\(=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)

Lại có: \(\dfrac{\sqrt{2-\sqrt{4y-y^2}}}{y-2}.\sqrt{4+2\sqrt{4y-y^2}}\)

\(=\dfrac{\sqrt{4-2\sqrt{y\left(4-y\right)}}}{\sqrt{2}\left(y-2\right)}.\sqrt{\left(\sqrt{y}\right)^2+2\sqrt{y\left(4-y\right)}+\left(\sqrt{4-y}\right)^2}\)

\(\dfrac{\sqrt{\left(\sqrt{y}\right)^2-2\sqrt{y\left(4-y\right)}+\left(\sqrt{4-y}\right)^2}}{\sqrt{2}\left(y-2\right)}.\sqrt{\left(\sqrt{y}+\sqrt{4-y}\right)^2}\)

\(=\dfrac{\sqrt{\left(\sqrt{y}-\sqrt{4-y}\right)^2}}{\sqrt{2}\left(y-2\right)}.\left|\sqrt{y}+\sqrt{4-y}\right|=\dfrac{\left|\sqrt{y}-\sqrt{4-y}\right|}{\sqrt{2}\left(y-2\right)}.\left|\sqrt{y}+\sqrt{4-y}\right|\)

Vì \(y>2\Rightarrow\left\{{}\begin{matrix}\sqrt{y}>\sqrt{2}\\\sqrt{4-y}< \sqrt{2}\end{matrix}\right.\Rightarrow\sqrt{y}-\sqrt{4-y}>0\)

\(\Rightarrow\dfrac{\left|\sqrt{y}-\sqrt{4-y}\right|}{\sqrt{2}\left(y-2\right)}.\left|\sqrt{y}+\sqrt{4-y}\right|=\dfrac{\left(\sqrt{y}-\sqrt{4-y}\right)\left(\sqrt{y}+\sqrt{4+y}\right)}{\sqrt{2}\left(y-2\right)}\)

\(=\dfrac{y-\left(4-y\right)}{\sqrt{2}\left(y-2\right)}=\dfrac{2y-4}{\sqrt{2}\left(y-2\right)}=\dfrac{2\left(y-2\right)}{\sqrt{2}\left(y-2\right)}=\sqrt{2}\)

\(\Rightarrow\dfrac{\sqrt{2-\sqrt{4y-y^2}}}{y-2}.\sqrt{4+2\sqrt{4y-y^2}}=\sqrt{55+\sqrt{109}}-\sqrt{55-\sqrt{109}}\)

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this/that dùng cho ng, vật số ít

this: dùng cho ng, vật ở gần

that: dùng cho ng, vật ở xa

these/those dùng cho ng, vật số nhiều

these: ở gần

those: ở xa

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2. This

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4. those

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15. them

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Ban can cau nao nhi?

làm bài nào??

Gọi chiều dài là a(m)

=> Chiều dài là \(\dfrac{5400}{a}\left(m\right)\)

Theo đề bài ta có: \(\dfrac{5400}{a}:a=\dfrac{3}{2}\)

\(\Rightarrow\dfrac{5400}{a^2}=\dfrac{3}{2}\)

\(\Rightarrow a^2=3600\Rightarrow a=60\left(m\right)\)

Vậy chiều rộng là 60m, chiều dài là \(\dfrac{5400}{a}=\dfrac{5400}{60}=90\left(m\right)\)

Chu vi hình chữ nhật là: \(\left(90+60\right).2=300\left(m\right)\)

300m :)