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)

Refer: 

123 Cau Giay Rd, Hanoi

Nov 20, 2020

Hello Anna,

How are you doing? Are you interested in an art exhibition? Let’s join an Exhibition of Modern at Arts Centre with me this weekend. It is held at 9 o’clock this Saturday morning. I think we will meet at 8.15 at the bus stop near the Arts Center and then we’ll go to the exhibition together.

Write to me soon. I’m sure it will be fun.

Love,

Linh

          Ngồi buồn nhớ mẹ ta xưa

Miếng nhai cơm búng , lưỡi lừa cá xương

        Bé thì nhờ mẹ , nhờ cha

    Lớn lên nhờ vợ lúc già nhờ con

          Lên non với biết non cao 

Nuôi con mới biết công lao mẹ , thầy 

       Ơn cha nặng lắm ai ơi

Nghĩa mẹ bằng trời , chín tháng cưu mang

       Con người có tổ , có tông 

Như cây có cuội , như sông có nguồn

k mik nha !

=>\(25\cdot\dfrac{\sqrt{a-3}}{5}-7\cdot\dfrac{2}{3}\cdot\sqrt{a-3}-7\sqrt{a^2-9}+18\cdot\dfrac{1}{3}\sqrt{a^2-9}=0\)

=>\(\sqrt{a-3}\cdot\dfrac{1}{3}-\sqrt{a^2-9}=0\)

=>\(\sqrt{a-3}\left(\dfrac{1}{3}-\sqrt{a+3}\right)=0\)

=>a-3=0 hoặc a+3=1/9

=>a=3 hoặc a=-26/9

\(\frac{119}{153}=\frac{7}{9}\)    

\(\frac{322}{345}=\frac{14}{15}\)

Học tốt 

\(P=sin^22x-\left[2sin\dfrac{x}{2}cos\dfrac{x}{2}\left(cos^4\dfrac{x}{2}-sin^4\dfrac{x}{2}\right)\right]^2\)

\(=sin^22x-\left[sinx\left(cos^2\dfrac{x}{2}-sin^2\dfrac{x}{2}\right)\left(cos^2\dfrac{x}{2}+sin^2\dfrac{x}{2}\right)\right]^2\)

\(=sin^22x-\left[sinx.cosx.1\right]^2\)

\(=sin^22x-\left[\dfrac{1}{2}sin2x\right]^2\)

\(=\dfrac{3}{4}sin^22x=\dfrac{3}{4}\left(1-cos^22x\right)=\dfrac{3}{4}\left(1-\dfrac{1}{4}\right)=\dfrac{9}{16}\)

`#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.`