1 . => Loan visited mother last week

2 . => we are going to see a movie H2 at 8 o' clock this morning

I - viết lại câu sao nghĩa k thay đổi

1, I used to own a dog .The dog never barked at them

dog I own never barked at them

2,My friend Bill has decided to buy a motor- bike .His car was stolen last week

My friend Bill , whose car was stolen last week , has decided to buy a motor - bike

3,I bought a new car . It cost me a lot of money

The new car is bought , which cost me a lot of money.

4,I just talked to girl .I can never remember her name

I just talked to the girl whose name I can never remember

5,The librarian wants to speak to a man.We borrwed his car yesterday

The librarian wants to speak to a man whose car we borrwed yesterday

6,Carol is a vegetarian.I cooked a meal for her last week

Carol , whom I cooked a meal last week is a vegetarian

7, I was sitting next to a boy in the exam .He told me the answer

was sitting next to in the exam told me the answer.

============

Câu cuối á , tại vì phần viết lại ko có who nên chắc bỏ who đi

thông cảm sai thì đừng ném gạch nhé

a , PTHH :

Mg + H₂SO₄ -> MgSO₄ + H₂

x ...........x...........................x (mol)

Zn + H₂SO₄ -> ZnSO₄ + H₂

y............y..........................y (mol)

n_H2 = \(\dfrac{4,48}{22,4}=0,2\left(mol\right)\)

Gọi số mol Mg , Zn trong hỗn hợp lần lượt là x,y(mol)

=> x+ y = 0,2 (mol) (*)

Theo bài ra ta có : 24x+65y = 8,9 (g) (**)

Từ (*) và (**) ta có hệ :

\(\left\{{}\begin{matrix}x+y=0,2\\24x+65y=8,9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0,1\left(mol\right)\\y=0,1\left(mol\right)\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}m_{Mg}=2,4\left(g\right)\\m_{Zn}=6,5\left(g\right)\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\%m_{Mg}\approx26,97\%\\\%m_{Zn}=73,03\%\end{matrix}\right.\)

b , Theo PTHH , \(\Sigma n_{H2SO4}=0,2\left(mol\right)\)

=> V = \(\dfrac{n}{C_M}=\dfrac{0,2}{0,2}=1\left(l\right)\)

Bài 1 :

a , \(\sqrt{25^2-24^2}=\sqrt{\left(25-24\right)\left(25+24\right)}=\sqrt{49}=7\)

b , \(\sqrt{21.8^2-18.2^2}=\) \(\sqrt{21.4^2.2^3-18.2^2}=\sqrt{2^2\left(21.4^2-18\right)}=2\sqrt{6\left(7.8-3\right)}=2\sqrt{6.53}=2\sqrt{318}\)

Bai 2 : \(\sqrt{\sqrt{3}+\sqrt{2}}.\sqrt{\sqrt{3}-\sqrt{2}}=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}=\sqrt{3-2}=\sqrt{1}=1\)

Ta có : \(3^{x+2}+3^x=10\)

<=> \(3^x.9+3^x=10\Leftrightarrow3^x\left(9+1\right)=10\)

<=> \(3^x=1\Rightarrow x=0\)

a , \(\left(5\sqrt{2}-2\sqrt{5}\right)\sqrt{5}\) - \(\sqrt{250}\)

= \(5\sqrt{10}-2.5-5\sqrt{10}=-2.5=-10\)

b , \(6\sqrt{\dfrac{1}{3}}+\dfrac{9}{\sqrt{3}}-\dfrac{2}{\sqrt{3}-1}=\dfrac{15}{\sqrt{3}}-\dfrac{2}{\sqrt{3}-1}\)

= \(\dfrac{5\sqrt{3}\left(\sqrt{3}-1\right)-2}{\sqrt{3}-1}=\dfrac{13-5\sqrt{3}}{\sqrt{3}-1}\)

Vế A

Ta có : A =

2A = \(2^{101}-2^{100}+2^{99}-2^{98}+...+2^3-2^2\)

=> 2A + A = 3A = \(2^{100}-2\Rightarrow A=\dfrac{2^{100}-2}{3}\)

=================

B làm tương tự , nhân 3 lên rồi cộng lại là ra

dựa vào câu hỏi tương tự

b , Ta có : \(\dfrac{x\sqrt{x}-y\sqrt{y}}{\sqrt{x}-\sqrt{y}}=\dfrac{\left(\sqrt{x}\right)^3-\left(\sqrt{y}\right)^3}{\left(\sqrt{x}-\sqrt{y}\right)}\) = \(\dfrac{\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)}{\sqrt{x}-\sqrt{y}}=x+\sqrt{xy}+y\)

d , Ta có : \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2=\left(\dfrac{1-a\sqrt{a}+\sqrt{a}-a}{1-\sqrt{a}}\right)\dfrac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)= \(\dfrac{\left(1-a\right)+\sqrt{a}\left(1-a\right)}{1-\sqrt{a}}.\dfrac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)

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

= \(\dfrac{\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)}{\left(1-a\right)}=\dfrac{\left(1-a\right)}{\left(1-a\right)}=1\)