\(1,15x-3-x^2+2x+x^2-13x=7\\ 15x+2x-13x-x^2+x^2=7+3\\ 4x=10\\ x=\dfrac{5}{2}\\ 2,4x+8-14x+7+27x-36=30\\ 4x-14x+27x=30-8-7+36\\ 17x=51\\ x=3\)

\(3,48x^2-20x-12x+5+3x-7-48x^2+112x=81\\ 48x^2-48x^2-20x-12x+3x+112x=81+7-5\\ 83x=83\\ x=1\)

\(4,2\left(6x^2-2x+15x-5\right)-6\left(2x^2-x+4x-2\right)=-6\\ 12x^2-4x+30x-10-12x^2+6x-24x+12=-6\\ 12x^2-12x^2-4x+30x+6x-24x=-6-12+10\\ 8x=-8\\ x=-1\)

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

b - d - d - c - c

II

1d , 2c , 3d , 4a , 5b , 6(cái này chịu, thấy nó đúng ngữ pháp hết nên thử xem trong sách có câu nào vậy ko)  , 7c , 8c , 9b , 10b

III

wants 

is

boils

does stacy practice

i love cold drinks but i have a sore throat

he usually goes climbing mountain in the summer

people volunteer because they want to improve their community

we encouraged people to donate books and clothes to street children

pho is one of the most popular dishes in vietnam 

1c

2a

3b

4c

3, ta có:

\(B=\dfrac{\sqrt{x}-3+2\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x-3}\right)}\cdot\dfrac{2\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\\ =\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{2\left(\sqrt{x}+3\right)}{\sqrt{x}-1}\\ =\dfrac{6}{\sqrt{x}-3}\)

để B=3 thì ta có:

\(\dfrac{6}{\sqrt{x}-3}=3\\ \Leftrightarrow\dfrac{6}{\sqrt{x}-3}=\dfrac{3\sqrt{x}-9}{\sqrt{x}-3}\\ \Leftrightarrow6=3\sqrt{x}-9\\ \Leftrightarrow3\sqrt{x}=15\\ \Leftrightarrow\sqrt{x}=5\\ \Leftrightarrow x=25\)

vậy để B=3 thì x=25

\(=\sqrt{6+1-2\sqrt{6}}-\dfrac{\sqrt{5}\left(\sqrt{10}-\sqrt{5}\right)}{\sqrt{10}-\sqrt{5}}+\sqrt{\left(11-\sqrt{120}\right)\left(11+\sqrt{120}\right)^2}\\ =\sqrt{\left(\sqrt{6}-\sqrt{1}\right)^2}-\sqrt{5}+\sqrt{\left(11^2-120\right)\left(11+2\sqrt{30}\right)}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{1\left(6+5+2\sqrt{6\cdot5}\right)}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{\left(\sqrt{6}+\sqrt{5}\right)^2}\\ =\sqrt{6}-\sqrt{1}-\sqrt{5}+\sqrt{6}+\sqrt{5}=2\sqrt{6}-\sqrt{1}\)

1. A={x∈ N /x≤6}

A={0;1;2;3;4;5;6}

\(2\\ a,A=\left\{13;14;15\right\}\\ b,B=\left\{1;2;3;4;5\right\}\\ c,C=\left\{13;14;15;16\right\}\\ 3.\\ \)

M={x∈ N* / 3<x<10 }

\(M=\left\{4;5;6;7;8;9\right\}\)

∈    ;   ∉

\(3^x\left(1+3\right)=108\\ 3^x\cdot4=108\\ 3^x=27\\ x=3\)