bài viết dài thì bạn nên tự làm ấy, chứ cop nguyên của ngkh nó kì lắm

dùng trí tưởng tượng -> nghĩ ra cốt truyện -> dùng gg dịch hoặc cambridge dic -> xong bài :)))

Bài 2 :

\(a,\left(x+2\right)\left(x^2+3x-2\right)=2\left(x+2\right)x^2\)

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

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

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

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

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

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

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

Vậy \(S=\left\{-2;2;1\right\}\)

\(b,9x^2-\left(6x+2\right)\left(x-5\right)=1\)

\(\Leftrightarrow9x^2-\left(6x^2-30x+2x-10\right)-1=0\)

\(\Leftrightarrow9x^2-6x^2+30x-2x+10-1=0\)

\(\Leftrightarrow3x^2+28x+9=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-9\end{matrix}\right.\)

Vậy \(S=\left\{-\dfrac{1}{3};-9\right\}\)

\(c,\dfrac{x}{3x-2}-\dfrac{x}{2+3x}=\dfrac{6x^2}{9x^2-4}\left(dkxd:x\ne\pm\dfrac{2}{3}\right)\)

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

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

\(\Leftrightarrow3x^2+2x-3x^2+2x-6x^2=0\)

\(\Leftrightarrow4x-6x^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(tmdk\right)\\x=\dfrac{2}{3}\left(ktmdk\right)\end{matrix}\right.\)

Vậy \(S=\left\{0\right\}\)

Bài 1 :

\(a,P=\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}-\dfrac{1}{1-x}+\dfrac{2-x^2}{x^2-x}\right)\left(dkxd:x\ne0,x\ne\pm1\right)\)

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

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

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

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

\(b,P=-\dfrac{1}{2}\Rightarrow\dfrac{x^2}{x-1}=-\dfrac{1}{2}\)

\(\Rightarrow2x^2=-\left(x-1\right)\)

\(\Rightarrow2x^2=-x+1\)

\(\Rightarrow2x^2+x-1=0\)

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

Vậy \(P=-\dfrac{1}{2}\) thì \(x=\dfrac{1}{2};x=-1\)

\(c,\) Để P nhận giá trị nguyên dương thì \(P\ge0\)

\(\Leftrightarrow\dfrac{x^2}{x-1}\ge0\Leftrightarrow x\ge0\)

1:

a: =>x^3+3x^2-2x+2x^2+6x-4=2x^3+4x^2

=>2x^3+4x^2=x^3+5x^2+4x-4

=>x^3-x^2-4x+4=0

=>(x-1)(x-2)(x+2)=0

=>x=1;x=2;x=-2

b: =>9x^2-6x^2+30x-2x+10=1

=>3x^2+28x+9=0

=>(x+9)(3x+1)=0

=>x=-1/3;x=-9

c: =>x(3x+2)-x(3x-2)=6x^2

=>6x^2=3x^2+2x-3x^2+2x=4x

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

=>x=2/3(loại); x=0(nhận)

Sorry, mình không biết làm phần Task 1 nhaaa :<<<

Task 2: Complete the sentences using PAST SIMPLE TENSE.

1. didn't like

2. phoned

3. went

4. Did they leave

5. Did you like

6. didn't finish

7. Did she go

8. stayed / got

9. went

10. Did you travel

Chúc bạn học tốt!! ^^

thank

a) \(=3\left(xy-4\right)\)

b) \(=x^2\left(x-y\right)+4\left(x-y\right)=\left(x-y\right)\left(x^2+4\right)\)

c) \(=x^2-\left(y^2-12y+36\right)=x^2-\left(y-6\right)^2=\left(x-y+6\right)\left(x+y-6\right)\)

d) \(=\left(4p^2-36p+81\right)-25=\left(2p-9\right)^2-25=\left(2p-9-5\right)\left(2p-9+5\right)=4\left(p-7\right)\left(p-2\right)\)

b: \(\Leftrightarrow x^2-4-x-x=0\)

hay x=-4

1 saw - didn't see - was looking

2 met - went - went - had - were waiting

3 was cycling - stepped - was going - managed - didn't hit

4 was waiting - arrived

5 were you doing

6 Did you go

7 wore

8 were you driving - happened

9 took - wasn't looking 

10 were - didn't know

`#3107.101107`

c)

Ta có:

`2x = 3y`

`=> x/3 = y/2 => x/15 = y/10`

`4y = 5z`

`=> y/5 = z/4 => y/10 = z/8`

`=> x/15 = y/10 = z/8`

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

`x/15 = y/10 = z/8 = (x - y - z)/(15 - 10 - 8) = (-27)/(-3) = 27/3 = 9`

`=> x/15 = y/10 = z/8 = 9`

`=>`\(x=15\cdot9=135\\ y=9\cdot10=90\\ z=8\cdot9=72\)

Vậy, giá trị của `x; y; z` lần lượt là `135; 90; 72`

d)

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

`(x - 5)/3 = (y - 4)/2 = (x - 5 + y - 4)/(3 + 2) = (x + y - 5 - 4)/5 = (-1-9)/5 = (-10)/5 = -2`

`=> (x - 5)/3 = (y - 4)/2 = -2`

`=>` 

\(x=\left(-2\cdot3\right)+5=-1\)

\(y=\left(-2\cdot2\right)+4=0\)

Vậy, giá trị của `x; y` lần lượt là `-1; 0`

e)

`(x - 1)/2 = (y - 2)/3 = (z - 3)/4`

`=> (x - 1)/2 = (2y - 4)/6 = (3z - 9)/12`

Áp dụng tính chất dãy tỉ số bằng nhau ta có:

`(x - 1)/2 = (2y - 4)/6 = (3z - 9)/12 = (x - 1 - (2y - 4) + 3z - 9)/(2 - 6 + 12) =`\(\dfrac{x-1-2y+4+3z-9}{8}=\dfrac{x-2y+3z+\left(-1+4-9\right)}{8}\\ =\dfrac{-10-6}{8}=\dfrac{-16}{8}=-2\)

`=> (x - 1)/2 = (y - 2)/3 = (z - 3)/4 = -2`

`=>`\(x=\left(-2\cdot2\right)+1=-3\); \(y=\left(-2\cdot3\right)+2=-4\); \(z=\left(-2\cdot4\right)+3=-5\)

Vậy, giá trị của `x; y; z` lần lượt là `-3; -4; -5.`