Diện tích xung quanh căn phòng là: 

`2 . h . (a+b) = 2 . 3 . (8 + 4) = 72 (m^2)`

Đổi `72m^2 = 7200dm^2`

Cần số viên gạch ống là: 

`7200 : 1,2 = 6000` (viên)

Đáp số: `6000` viên

\(5^{n+3}-5^{n+1}=5^{12}.120\)

\(\Leftrightarrow5^n.5^3-5^n.5=5^{12}.120\Leftrightarrow5^n\left(5^3-5\right)=5^{12}.120\Leftrightarrow5^n.120=5^{12}.120\Leftrightarrow5^n=5^{12}\Rightarrow n=12\)

Sửa đề:

 \(3^{2n+3}-3^{2n+2}=2\cdot3^{10}\\ \Rightarrow3^{2n+2}\cdot3-3^{2n+2}=2\cdot3^{10}\\ \Rightarrow3^{2n+2}\cdot\left(3-1\right)=2\cdot3^{10}\\ \Rightarrow2\cdot3^{2n+2}=2\cdot3^{10}\\ \Rightarrow3^{2n+2}=3^{10}\\ \Rightarrow2n+2=10\\ \Rightarrow2n=8\\ \Rightarrow n=4\)

Vậy...
 

1: (2x-1)*x>0

TH1: \(\left\{{}\begin{matrix}2x-1>0\\x>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>\dfrac{1}{2}\\x>0\end{matrix}\right.\Leftrightarrow x>\dfrac{1}{2}\)

TH2: \(\left\{{}\begin{matrix}2x-1< 0\\x< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< \dfrac{1}{2}\\x< 0\end{matrix}\right.\)

=>x<0

2: 
ĐKXĐ: x<>1

\(\dfrac{x+3}{x-1}< 0\)

TH1: \(\left\{{}\begin{matrix}x+3>0\\x-1< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>-3\\x< 1\end{matrix}\right.\)

=>-3<x<1

TH2: \(\left\{{}\begin{matrix}x+3< 0\\x-1>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x< -3\\x>1\end{matrix}\right.\)

=>Loại

c: 

ĐKXĐ: x<>0

\(\dfrac{x^2-2}{5x}< 0\)

TH1: \(\left\{{}\begin{matrix}x^2-2< 0\\5x>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x^2< 2\\x>0\end{matrix}\right.\Leftrightarrow0< x< \sqrt{2}\)

TH2: \(\left\{{}\begin{matrix}x^2-2>0\\5x< 0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x^2>2\\x< 0\end{matrix}\right.\Leftrightarrow-\sqrt{2}< x< 0\)

d: (x-3)(x+7)>0

TH1: \(\left\{{}\begin{matrix}x-3>0\\x+7>0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x>3\\x>-7\end{matrix}\right.\Leftrightarrow x>3\)

TH2: \(\left\{{}\begin{matrix}x-3< 0\\x+7< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 3\\x< -7\end{matrix}\right.\)

=>x<-7

\(\left(7x+2\right)^{-1}=3^{-2}\)

=>\(\dfrac{1}{7x+2}=\dfrac{1}{3^2}=\dfrac{1}{9}\)

=>7x+2=9

=>7x=7

=>x=1

\(\dfrac{120^5}{40^5}+\dfrac{8^{13}}{4^{10}}-\dfrac{390^4}{130^4}\)

\(=3^5+\dfrac{2^{39}}{2^{20}}-3^4\)

\(=243+2^{19}-81=524450\)

làm ơn giúp mình với

Let's break down the problem step by step:

Step 1:

We are given a right triangle ABC at vertex A, with altitude AH and median AD. We also know that I and J are the points where the medians of triangles ABH and ACH intersect with each other.

Step 2:

Since triangle ABC is a right triangle, we know that angle A is a right angle (90°). Therefore, we can conclude that triangle ABE is also a right triangle (with angle ABE being a right angle).

Step 3:

Now, let's focus on triangle ABH. Since I is the point where the median of triangle ABH intersects with the line segment AB, we know that AI = IB (by definition of median). Similarly, since J is the point where the median of triangle ACH intersects with the line segment AC, we know that AJ = JC (by definition of median).

Step 4:

Using the fact that I and J are on opposite sides of angle ABE, we can write:

AI + IB = AJ + JC

Since AI = IB and AJ = JC, we can simplify this equation to:

2IB = 2JC

Step 5:

Now, let's look at the triangles ABE and ACE. Since they share side AE and angle E is common to both triangles, we can say that:

∠EAB = ∠ECA (common angles)

Using this fact, we can conclude that:

AE = EB (since opposite sides of equal angles are equal)

Step 6:

Now we have:

AE = EB and IB = JC

Using these two equations, we can write:

IJ = IB - JC = AE - AE = 0

So, IJ is a zero-length line segment!

Conclusion:

Since IJ is a zero-length line segment, it means that I and J coincide with each other. This implies that:

IJ ⊥ AD (I and J are collinear with AD)

Therefore, we have shown that triangle ABE is a right triangle and IJ is perpendicular to AD.

Answer:

a. Tam giác ABE vuông b) IJ vuông góc với AD

\(\left(-13\cdot\dfrac{2}{5}+\dfrac{2}{9}:2\dfrac{1}{2}+\dfrac{2}{5}\cdot\dfrac{11}{9}\right)\cdot2\dfrac{1}{2}\)

\(=\left(-\dfrac{26}{5}+\dfrac{2}{9}:\dfrac{5}{2}+\dfrac{22}{45}\right)\cdot\dfrac{5}{2}\)

\(=\left(-\dfrac{26}{5}+\dfrac{2}{9}\cdot\dfrac{2}{5}+\dfrac{22}{45}\right)\cdot\dfrac{5}{2}\)

\(=\left(-\dfrac{234}{45}+\dfrac{4}{45}+\dfrac{22}{45}\right)\cdot\dfrac{5}{2}\)

\(=\dfrac{-208}{45}\cdot\dfrac{5}{2}=-\dfrac{104}{9}\)

Chữ"và"trong ngoặc có phải dấu k bạn