Bài 1:

Theo đề ta có : A₁=G₁; T₁=3A₁; X₁ = 2T₁

=>\(\left\{{}\begin{matrix}X_1=6A_1\\T_1=3A_1\\G_1=A_1\end{matrix}\right.\)=>\(\left\{{}\begin{matrix}A=4A_1\\G=7A_1\end{matrix}\right.\)

Lại có: 2A + 3G= 5800

=> 29A₁ = 5800 => A₁=200

=> G = 1400

4. \(x^2-3x+xy-3y=0\)

\(\Leftrightarrow x\left(x-3\right)+y\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+y\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-y\end{matrix}\right.\)

5. \(x^2-8x-3x+24=0\)

\(\Leftrightarrow x\left(x-8\right)-3\left(x-8\right)=0\)

\(\Leftrightarrow\left(x-8\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=3\end{matrix}\right.\)

6. \(\left(x-2\right)^2-5\left(2-x\right)=0\)

\(\Leftrightarrow\left(x-2\right)^2+5\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-2+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-2+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

7. \(3x\left(x-1\right)-x^2+2x-1=0\)

\(\Leftrightarrow3x\left(x-1\right)-\left(x^2-2x+1\right)=0\)

\(\Leftrightarrow3x\left(x-1\right)-\left(x-1\right)^2=0\)

\(\Leftrightarrow\left(x-1\right)\left[3x-\left(x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)

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

8. \(x^2\left(x-3\right)+18-6x=0\)

\(\Leftrightarrow x^2\left(x-3\right)-6\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x^2-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x^2-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\pm\sqrt{6}\end{matrix}\right.\)

10. \(\left(x-5\right)^2-\left(x-2\right)^2=0\)

\(\Leftrightarrow\left[\left(x-5\right)-\left(x-2\right)\right]\left[\left(x-5\right)+\left(x-2\right)\right]=0\)

\(\Leftrightarrow\left(x-5-x+2\right)\left(x-5+x-2\right)=0\)

\(\Leftrightarrow-3\left(2x-7\right)=0\)

\(\Leftrightarrow2x-7=0\)

\(\Leftrightarrow x=\dfrac{7}{2}\)

12. \(x^2\left(x-3\right)-4x+12=0\)

\(\Leftrightarrow x^2\left(x-3\right)-4\left(x-3\right)=0\)

\(\Leftrightarrow\left(x^2-4\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\\x=3\end{matrix}\right.\)

14. \(3x^2-7x-10=0\)

\(\Leftrightarrow3x^2+3x-10x-10=0\)

\(\Leftrightarrow3x\left(x+1\right)-10\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(3x-10\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\3x-10=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{10}{3}\end{matrix}\right.\)

#Urushi

4: x^2-3x+xy-3y=0

=>x(x-3)+y(x-3)=0

=>(x-3)(x+y)=0

=>x=3 và x+y=0

=>x=3 và y=-3

6: (x-2)^2-5(2-x)=0

=>(x-2)^2+5(x-2)=0

=>(x-2)(x-2+5)=0

=>(x-2)(x+3)=0

=>x=-3 hoặc x=2

8: x^2(x-3)+18-6x=0

=>x^2(x-3)-6(x-3)=0

=>(x-3)(x^2-6)=0

=>x=3 hoặc \(x=\pm\sqrt{6}\)

10: (x-5)^2-(x-2)^2=0

=>(x-5-x+2)(x-5+x-2)=0

=>-3(2x-7)=0

=>2x-7=0

=>x=7/2

12: x^2(x-3)-4x+12=0

=>x^2(x-3)-4(x-3)=0

=>(x-3)(x^2-4)=0

=>(x-3)(x-2)(x+2)=0

=>\(x\in\left\{3;2;-2\right\}\)

14: 3x^2-7x-10=0

=>3x^2-10x+3x-10=0

=>(3x-10)(x+1)=0

=>x=10/3 hoặc x=-1

Câu 2

\((1) MnO_2 + 4HCl \to MnCl_2 + Cl_2 + 2H_2O\\ (2) Cl_2 + H_2 \xrightarrow{as} 2HCl\\ (3) 3Cl_2 + 2Fe \xrightarrow{t^o} 2FeCl_3\\ (4) 2FeCl_3 + Fe \to 3FeCl_2\\ (5) 2NaOH + Cl_2 \to NaCl + NaClO + H_2O\)

\((1) 4Al + 3O_2 \xrightarrow{t^o} 2Al_2O_3\\ (2) 2Fe + 3Cl_2 \xrightarrow{t^o} 2FeCl_3\\ (3) C + O_2 \xrightarrow{t^o} CO_2\\ (4) 2KMnO_4 \xrightarrow{t^o} K_2MnO_4 + MnO_2 + O_2\\ (5) 4P + 5O_2 \xrightarrow{t^o} 2P_2O_5\\ (6) 2KClO_3 \xrightarrow{t^o} 2KCl + 3O_2\\ (7) Fe + H_2SO_4 \to FeSO_4 + H_2\\ (8) Cu + 2H_2SO_4 \to CuSO_4 + SO_2 + 2H_2O\\ (9) 2Fe + 6H_2SO_4 \to Fe_2(SO_4)_3 + 3SO_2 + 6H_2O\\ (10) 2Al + 6H_2SO_4 \to Al_2(SO_4)_3 + 3SO_2 + 6H_2O\)

1 bỏ so

-In order to V(inf): Để làm gì

3 -So as to V(inf): Để làm gì

6 me->her

-Could/can/Will+V(inf)

-Help+O+V/to V

11 ..... me where the nearest post office is?

-Can/could+S+V+wh-questions+S+V?

14 -Shall+S+V(inf)?

17 ........ going to help him revise his lessons

-"be" going to V(inf): Sẽ làm gì( mang tính chắc chắn)

18 -Would+S+love/like+to V/N?

19 -Let's+V(inf)

= Shall+we+V(inf)?

20 -Trong ngữ cảnh này "to V" được dùng với nghĩa "để làm gì"

*Inf: Infinitive

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