245 m² là diện tích toàn phần hay diện tích xung quanh em ơi? 

Ta có: AB//DC

=>\(\widehat{A_1}=\widehat{D_4}\)(hai góc so le trong)

=>\(\widehat{D_4}=110^0\)
Ta có: \(\widehat{D_1}=\widehat{D_4}\)(hai góc đối đỉnh)

mà \(\widehat{D_4}=110^0\)

nên \(\widehat{D_1}=110^0\)

Ta có: AB//DC

=>\(\widehat{C_3}=\widehat{B_2}\)(hai góc so le trong)

=>\(\widehat{B_2}=135^0\)

Ta có: \(\widehat{B_1}+\widehat{B_2}=180^0\)(hai góc kề bù)

=>\(\widehat{B_1}=180^0-135^0=45^0\)

1. He would have solved the puzzle had he watched the news.
2. If I had had a mobile, I could have phoned you.
3. Were I to ask you to lend me your dictionary, would you do it?
4. The money will only be paid should a new contract be signed.
5. I’d appreciate it if you would reply at your earliest convenience.
6. If only I had known you earlier.
7. But for your absent-mindedness then, the soup would have tasted excellent.
8. They would have paid less if they had booked the tickets yesterday.
9.If you like, you can stay for two days.
10.If the parents were to buy the cat, their children would be very happy.

Bài 6:

a: \(\left|x+\dfrac{1}{2}\right|>=0\forall x;\left|y-\dfrac{3}{4}\right|>=0\forall y;\left|z-1\right|>=0\forall z\)

Do đó: \(\left|x+\dfrac{1}{2}\right|+\left|y-\dfrac{3}{4}\right|+\left|z-1\right|>=0\forall x,y,z\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x+\dfrac{1}{2}=0\\y-\dfrac{3}{4}=0\\z-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=\dfrac{3}{4}\\z=1\end{matrix}\right.\)

b: \(\left|x-\dfrac{3}{4}\right|>=0\forall x;\left|\dfrac{2}{5}-y\right|>=0\forall y;\left|x-y+z\right|>=0\forall x,y,z\)

Do đó: \(\left|x-\dfrac{3}{4}\right|+\left|\dfrac{2}{5}-y\right|+\left|x-y+z\right|>=0\forall x,y,z\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{3}{4}=0\\\dfrac{2}{5}-y=0\\x-y+z=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=\dfrac{2}{5}\\z=-x+y=-\dfrac{3}{4}+\dfrac{2}{5}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{3}{4}\\y=\dfrac{2}{5}\\z=-\dfrac{7}{20}\end{matrix}\right.\)

c: \(\left|x-\dfrac{2}{3}\right|>=0\forall x;\left|x+y+\dfrac{3}{4}\right|>=0\forall x,y;\left|y-z-\dfrac{5}{6}\right|>=0\forall y,z\)

Do đó: \(\left|x-\dfrac{2}{3}\right|+\left|x+y+\dfrac{3}{4}\right|+\left|y-z-\dfrac{5}{6}\right|>=0\forall x,y,z\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-\dfrac{2}{3}=0\\x+y+\dfrac{3}{4}=0\\y-z-\dfrac{5}{6}=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-x-\dfrac{3}{4}=-\dfrac{2}{3}-\dfrac{3}{4}\\z=y-\dfrac{5}{6}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=\dfrac{2}{3}\\y=-\dfrac{17}{12}\\z=-\dfrac{17}{12}-\dfrac{10}{12}=-\dfrac{27}{12}=-\dfrac{9}{4}\end{matrix}\right.\)

Bài 5:

a: \(\left|-\dfrac{3}{5}+\dfrac{1}{2}\right|-\left(\dfrac{3}{4}-\dfrac{5}{8}\right)+\left|-\dfrac{3}{2}\right|\)

\(=\left|-\dfrac{6}{10}+\dfrac{5}{10}\right|-\dfrac{1}{8}+\dfrac{3}{2}\)

\(=\dfrac{1}{10}-\dfrac{1}{8}+\dfrac{3}{2}=\dfrac{4}{40}-\dfrac{5}{40}+\dfrac{60}{40}=\dfrac{59}{40}\)

b: \(\dfrac{2}{3}-\left|-\dfrac{7}{3}+\dfrac{3}{4}\right|-\left|-\dfrac{5}{2}+1\right|\)

\(=\dfrac{2}{3}-\left|-\dfrac{28}{12}+\dfrac{9}{12}\right|-\left|-\dfrac{5}{2}+\dfrac{2}{2}\right|\)

\(=\dfrac{2}{3}-\dfrac{19}{12}-\dfrac{3}{2}=\dfrac{8}{12}-\dfrac{19}{12}-\dfrac{18}{12}\)

\(=-\dfrac{29}{12}\)

c: \(\dfrac{1}{5}-\left(\dfrac{3}{10}-\dfrac{-3}{5}\right)-\left|\dfrac{1}{4}-\dfrac{2}{5}\right|\)

\(=\dfrac{1}{5}-\dfrac{3}{10}-\dfrac{3}{5}-\left|\dfrac{5}{20}-\dfrac{8}{20}\right|\)

\(=-\dfrac{7}{10}-\left|\dfrac{-3}{20}\right|=-\dfrac{7}{10}-\dfrac{3}{20}=-\dfrac{17}{20}\)

d: \(\left|-\dfrac{5}{2}+\dfrac{3}{4}-\dfrac{1}{3}\right|-\left(-\dfrac{3}{4}+\dfrac{-5}{3}\right)\)

\(=\left|-\dfrac{30}{12}+\dfrac{9}{12}-\dfrac{4}{12}\right|+\dfrac{3}{4}+\dfrac{5}{3}\)

\(=\dfrac{25}{12}+\dfrac{9}{12}+\dfrac{20}{12}=\dfrac{54}{12}=\dfrac{9}{2}\)

n=2

Giải chi tiết giúp e

Số tiền lãi là:

\(7500000\times15\%=1125000\) (đồng)

Đáp số: \(1125000\) đồng

Để tính số tiền lãi, ta có thể sử dụng công thức: Số tiền lãi = Số tiền vốn * Tỷ lệ lãi suất. Với số tiền vốn là 7,500,000 và tỷ lệ lãi suất là 15%, ta có: Số tiền lãi = 7,500,000 * 0.15 = 1,125,000 đồng. Vậy số tiền lãi mà cửa hàng đó đã kiếm được là 1,125,000 đồng.

a: \(2\sqrt{27}-3\sqrt{54}-\dfrac{1}{3}\sqrt{48}\)

\(=2\cdot3\sqrt{3}-3\cdot3\sqrt{6}-\dfrac{1}{3}\cdot4\sqrt{3}\)

\(=6\sqrt{3}-9\sqrt{6}-\dfrac{4}{3}\sqrt{3}=\dfrac{14}{3}\sqrt[]{3}-9\sqrt{6}\)

b: \(-\dfrac{1}{2}\sqrt{108}+\dfrac{1}{15}\cdot\sqrt{75}-\dfrac{1}{3}\cdot\sqrt{363}\)

\(=-\dfrac{1}{2}\cdot6\sqrt{3}+\dfrac{1}{15}\cdot5\sqrt{3}-\dfrac{1}{3}\cdot11\sqrt{3}\)

\(=-3\sqrt{3}+\dfrac{1}{3}\sqrt{3}-\dfrac{11}{3}\sqrt{3}=-\dfrac{19}{3}\sqrt{3}\)

c: \(\dfrac{5}{8}\sqrt{48}-\dfrac{1}{33}\cdot\sqrt{363}+\dfrac{3}{14}\cdot\sqrt{147}\)

\(=\dfrac{5}{8}\cdot4\sqrt{3}-\dfrac{1}{33}\cdot11\sqrt{3}+\dfrac{3}{14}\cdot7\sqrt{3}\)

\(=\dfrac{5}{2}\sqrt{3}-\dfrac{1}{3}\sqrt{3}+\dfrac{3}{2}\sqrt{3}=\dfrac{11}{3}\sqrt{3}\)

d:

ĐKXĐ: x>=0; x<>9

 Sửa đề:\(\dfrac{x-9}{x-3\sqrt{x}}-\dfrac{x-4}{\sqrt{x}+2}\)

\(=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\)

\(=\dfrac{\sqrt{x}+3}{\sqrt{x}}-\left(\sqrt{x}-2\right)=\dfrac{\sqrt{x}+3-x+2\sqrt{x}}{\sqrt{x}}\)

\(=\dfrac{-x+3\sqrt{x}+3}{\sqrt{x}}\)

e: ĐKXĐ: x>=0; x<>4

\(\dfrac{x+2\sqrt{x}+1}{\sqrt{x}+1}-\dfrac{x-4\sqrt{x}+4}{\sqrt{x}-2}\)

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

\(=\sqrt{x}+1-\sqrt{x}+2=3\)

\(\left(3x-1\right)^3=125\)

=>\(\left(3x-1\right)^3=5^3\)

=>3x-1=5

=>3x=6

=>\(x=\dfrac{6}{3}=2\)

\(\left(3x-1\right)^3=125\\ \Rightarrow\left(3x-1\right)^3=5^3\\ \Rightarrow3x-1=5\\ \Rightarrow3x=5+1\\ \Rightarrow3x=6\\ \Rightarrow x=6:3\\ \Rightarrow x=2\)