Bài 5:

a) \(x^2-xy+x-y\)

\(=\left(x^2-xy\right)+\left(x-y\right)\)

\(=x\left(x-y\right)+\left(x-y\right)\)

\(=\left(x+1\right)\left(x-y\right)\)

b) \(xz+yz+4x+4y\)

\(=\left(xz+yz\right)+\left(4x+4y\right)\)

\(=z\left(x+y\right)+4\left(x+y\right)\)

\(=\left(z+4\right)\left(x+y\right)\)

c) \(x^2-x-y^2+y\)

\(=\left(x^2-y^2\right)-\left(x-y\right)\)

\(=\left(x+y\right)\left(x-y\right)-\left(x-y\right)\)

\(=\left(x-y\right)\left(x+y-1\right)\)

d) \(x^2+2x+2z-z^2\)

\(=\left(x^2-z^2\right)+\left(2x+2z\right)\)

\(=\left(x+z\right)\left(x-z\right)+2\left(x+z\right)\)

\(=\left(x+z\right)\left(x-z+2\right)\)

18, \(\frac{x}{2}+\frac{x^2}{8}=0\Leftrightarrow4x+x^2=0\Leftrightarrow x\left(x+4\right)=0\Leftrightarrow x=-4;x=0\)

19, \(4-x=2\left(x-4\right)^2\Leftrightarrow\left(4-x\right)-2\left(4-x\right)^2=0\)

\(\Leftrightarrow\left(4-x\right)\left[1-2\left(4-x\right)\right]=0\Leftrightarrow\left(4-x\right)\left(-7+2x\right)=0\Leftrightarrow x=4;x=\frac{7}{2}\)

20, \(\left(x^2+1\right)\left(x-2\right)+2x-4=0\Leftrightarrow\left(x^2+1\right)\left(x-2\right)+2\left(x-2\right)=0\)

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

21, \(x^4-16x^2=0\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\Leftrightarrow x=0;x=\pm4\)

22, \(\left(x-5\right)^3-x+5=0\Leftrightarrow\left(x-5\right)^3-\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left[\left(x-5\right)^2-1\right]=0\Leftrightarrow\left(x-5\right)\left(x-6\right)\left(x-4\right)=0\Leftrightarrow x=4;x=5;x=6\)

23, \(5\left(x-2\right)-x^2+4=0\Leftrightarrow5\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(5-x-2\right)=0\Leftrightarrow x=2;x=3\)

đề là tìm x ạ

\(PT\Leftrightarrow\left(3x+5\right)\dfrac{x+1}{2}+\left(3x+5\right)\dfrac{2x-2}{3}-\left(3x+5\right)=0\)

\(\Leftrightarrow\left(3x+5\right)\left(\dfrac{x+1}{2}+\dfrac{2x-2}{3}-1\right)=0\)

\(\Leftrightarrow\left(3x+5\right)\dfrac{3\left(x+1\right)+2\left(2x-2\right)-6}{6}=0\)

\(\Leftrightarrow\left(3x+5\right)\dfrac{7x-7}{6}=0\)

\(\Leftrightarrow\left(3x+5\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5}{3}\\x=1\end{matrix}\right.\)

KL: Phương trình có tập nghiệm \(S=\left\{\dfrac{-5}{3};1\right\}\)

g, cộng 1 vào cả 2 vế của pt , ta được :

<=>\(\dfrac{x+1}{7}+1+\dfrac{x+2}{6}=\dfrac{x+3}{5}+1-\dfrac{x+4}{4}+1=0\)

<=>\(\dfrac{x+8}{7}+\dfrac{x+8}{6}-\dfrac{x+8}{5}+\dfrac{x+8}{4}=0\)

<=>\(\left(x+8\right)\left(\dfrac{1}{7}+\dfrac{1}{6}-\dfrac{1}{5}+\dfrac{1}{4}\right)=0\)

<=> x+8 =0 

<=> x=-8

h, 

\(\dfrac{3x-2}{3}\left(x-1\right)+\dfrac{3x-2}{4}\left(2x-2\right)=0\)

<=>\(\dfrac{3x-2}{3}\left(x-1\right)+2.\dfrac{3x-2}{4}\left(x-1\right)=0\)

<=>\(\dfrac{3x-2}{3}\left(x-1\right)+\dfrac{6x-4}{4}\left(x-1\right)=0\)

<=>\(\left(x-1\right).\left(\dfrac{3x-2}{3}+\dfrac{6x-4}{4}\right)=0\)

<=>x-1 =0 

<=> x=1

\(A=-\left(x^2-4x+4\right)-\left(y^2+4y+4\right)+10\\ A=-\left(x-2\right)^2-\left(y+2\right)^2+10\le10\\ A_{max}=10\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=-2\end{matrix}\right.\)

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

- 27/1=81/3 (Ngược lại)

- 3/9=27/81 (Ngược lại)

- 27/9=3/1 (Ngược lại)

- 81/9=27/3 (Ngược lại)

- 1/27=3/81 (Ngược lại)

\(\dfrac{1}{1}\) = \(\dfrac{3}{3}\) = \(\dfrac{9}{9}\) = \(\dfrac{27}{27}\) = \(\dfrac{81}{81}\)

a) Pt : \(CuO+2HCl\rightarrow CuCl_2+H_2O|\)

              1           2              1            1

              a           2a

             \(Fe_2O_3+6HCl\rightarrow2FeCl_3+3H_2O\)

                   1            6              2              3

                   b           6b

b) Gọi a là số mol của CuO

           b là số mol của Fe₂O₃

\(m_{CuO}+m_{Fe2O3}=25\left(g\right)\)

⇒ \(n_{CuO}.M_{CuO}+n_{Fe2O3}.M_{Fe2O3}=25g\)

 ⇒ 80a + 160b = 25g (1) 

Ta có : 200ml = 0,2l

\(n_{HCl}=3,5.0,2=0,7\left(mol\right)\)

⇒ 2a + 6b = 0,7(2)

Từ (1),(2) ta có hệ phương trình :

      80a + 160b = 25g

       2a + 6b = 0,7

    ⇒ \(\left\{{}\begin{matrix}a=0,2375\\b=0,0375\end{matrix}\right.\)

\(m_{CuO}=0,2375.80=19\left(g\right)\)

\(m_{Fe2O3}=0,0375.160=6\left(g\right)\)

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