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

\(\Leftrightarrow2x+2=0\left(\text{vì }x^2+1\ne0\right)\)

\(\Leftrightarrow2x=-2\text{ }\Leftrightarrow x=-1\)

\(\text{Vậy S}=\left\{-1\right\}\)

mong các bạn k đúng cho mình

hehe :)))) tam thức bậc 2 anh êi

P=9xy+10yz+11zx=9xy+z(10y+11x)=9xy(1-x-y)(10y+11x)

khai triển và rút gọn ta được :

\(P=-11x^2-10y^2+11x+10y-12xy\)

tương đương với :

 \(11x^2+\left(12y-11\right)x+10y^2-10y+P\ge0\)(1)

Coi đây là tam thức bậc 2 ẩn x do đk của x => (1) phải có nghiệm  hay

\(\Delta-\left(12y-11\right)^2-44\left(10y^2-10y+P\right)\ge0\)

Hay \(-296y^2+176y+121-44P\ge0\)

tương đương với

\(P\le-\frac{74}{11}\left(y^2-\frac{22}{37}y-\frac{121}{296}\right)\)

dùng phép tách thành bình phương ; ta dễ thấy :

\(y^2-\frac{22}{37}y-\frac{121}{296}\ge-\frac{5445}{10952}\)

=> \(P\le\left(\frac{74}{-11}\right).\left(-\frac{5445}{10952}\right)-\frac{495}{148}\)

vậy \(MaxP=\frac{495}{148}\)đạt được khi \(y=\frac{11}{37};x=\frac{25}{74};z=\frac{27}{74}\)

(2x + 1)(3x + 3) = 0

<=> 2x + 1 = 0 hoặc 3x + 3 = 0

<=> x = -1/2 hoặc x = -1

\(\left(2x+1\right)\left(3x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\3x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=-1\\3x=-3\end{cases}\Leftrightarrow}}\orbr{\begin{cases}x=\frac{-1}{2}\\x=-1\end{cases}}\)

Vậy ...

\(\frac{x^3+x^2-x-1}{x^3+2x-5}\)

\(\Leftrightarrow\frac{x^3+x^2-x-1}{x^3+2x-5}=0\)

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

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

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

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

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

\(\Leftrightarrow x=\pm1\)

Vậy \(x\in\left\{\pm1\right\}\)

\(\frac{x^3+x^2-x-1}{x^3+2x-5}=\frac{x^2\left(x+1\right)-\left(x+1\right)}{x^3+2x-5}\)

\(=\frac{\left(x+1\right)\left(x^2-1\right)}{x^3+2x-5}\)

Để \(\frac{x^3+x^2-x-1}{x^3+2x-5}=0\Leftrightarrow\left(x-1\right)\left(x^2-1\right)=0\left(x^3+2x-5\ne0\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x^2-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x^2=1\end{cases}\Leftrightarrow x=\pm}\)

Vậy x={-1;1}

1) \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)

<=> \(\frac{21x}{24}-\frac{100\left(x-9\right)}{24}=\frac{80x+6}{24}\)

<=> 21x - 100x + 900 = 80x + 6

<=> -79x - 80x = 6 - 900

<=> -159x = -894

<=> x = 258/53

Vậy S = {258/53}

2) \(\frac{\left(2x+1\right)^2}{5}-\frac{\left(x+1\right)^2}{3}=\frac{7x^2-14x-5}{15}\)

<=> \(\frac{3\left(4x^2+4x+1\right)}{15}-\frac{5\left(x^2+2x+1\right)}{15}=\frac{7x^2-14x-5}{15}\)

<=> 12x² + 12x + 3 - 5x² - 10x - 5 = 7x² - 14x - 5

<=> 7x² + 2x - 7x² + 14x = -5 + 2

<=> 16x = 3

<=> x = 3/16

Vậy S  = {3/16}

3) 4(3x - 2) - 3(x - 4) = 7x+  10

<=> 12x - 8 - 3x + 12 = 7x + 10

<=> 9x - 7x = 10 - 4

<=> 2x = 6

<=> x = 3

Vậy S = {3}

4) \(\frac{\left(x+10\right)\left(x+4\right)}{12}-\frac{\left(x+4\right)\left(2-x\right)}{4}=\frac{\left(x+10\right)\left(x-2\right)}{3}\)

<=> \(\frac{x^2+14x+40}{12}+\frac{3\left(x^2+2x-8\right)}{12}=\frac{4\left(x^2+8x-20\right)}{12}\)

<=> x² + 14x + 40 + 3x² + 6x - 24 = 4x² + 32x - 80

<=> 4x² + 20x - 4x² - 32x = -80 - 16

<=> -12x = -96

<=> x = 8

Vậy S = {8}

tớ đang cần gấp 

Đề như sai rồi bạn ơi. 

Câu 2: 6x² + 7x - 3

= 6x² + 9x - 2x - 3

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

= (3x - 1)(2x + 3)