x=-1/6            

x + x² = 0

=> x(1 + x) = 0

=> x = 0 hoặc x + 1 = 0

=> x = 0 hoặc x = -1

vậy_

mk biến đổi về pt tích, sau đó bạn tính nốt nhé:

b) \(x+1-\left(x+1\right)^2=0\)

<=> \(\left(x+1\right)\left(1-x-1\right)=0\)

<=> \(-x\left(x+1\right)=0\)

c) \(15y\left(4y-9\right)-3\left(4y-9\right)=0\)

<=> \(3\left(4y-9\right)\left(5y-1\right)=0\)

d) \(8\left(25z+7\right)-27z\left(25z+7\right)=0\)

<=> \(\left(25z+7\right)\left(8-27z\right)=0\)

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

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

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

c) \(15y\left(4y-9\right)-3\left(4y-9\right)=0\Leftrightarrow\left(15y-3\right)\left(4y-9\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{15}=\dfrac{1}{5}\\x=\dfrac{9}{4}\end{matrix}\right.\)

d) \(8\left(25z+7\right)-27z\left(25z+7\right)=0\Leftrightarrow\left(8-27z\right)\left(25z+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}z=\dfrac{8}{27}\\z=\dfrac{-7}{25}\end{matrix}\right.\)

Ta có : 

\(\left|x-\frac{2}{3}\right|< \frac{1}{3}\)

\(\Leftrightarrow\)\(\frac{-1}{3}< x-\frac{2}{3}< \frac{1}{3}\)

\(\Leftrightarrow\)\(\frac{-1}{3}+\frac{2}{3}< x-\frac{2}{3}+\frac{2}{3}< \frac{1}{3}+\frac{2}{3}\) ( cộng 3 vế cho \(\frac{2}{3}\) ) 

\(\Leftrightarrow\)\(\frac{1}{3}< x< 1\)

Vậy \(\frac{1}{3}< x< 1\)

Chúc bạn học tốt ~ 

Với các giá trị nguyên của \(x\ne-1\), để A nguyên thì \(\left(x^5+1\right)⋮\left(x^3+1\right)\)

\(\Leftrightarrow\left(x^5+x^2-\left(x^2-1\right)\right)⋮\left(x^3+1\right)\)

\(\Leftrightarrow\left(x^2\left(x^3+1\right)-\left(x^2-1\right)\right)⋮\left(x^3+1\right)\)

\(\Leftrightarrow\left(x^2-1\right)⋮\left(x^3+1\right)\)

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

\(\Rightarrow x\left(x-1\right)⋮\left(x^2-x+1\right)\)

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

\(\Leftrightarrow1⋮\left(x^2-x+1\right)\)

\(\Rightarrow\left[{}\begin{matrix}x^2-x+1=1\\x^2-x+1=-1\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x\left(x-1\right)=0\\\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}=0\left(vn\right)\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

ĐKXĐ: \(x\ge0\)

Đặt \(\sqrt{x}=a\)

\(\Rightarrow a^2-2a-1=0\)

\(\Rightarrow\left(a-1\right)^2=2\)

\(\Rightarrow\orbr{\begin{cases}a-1=\sqrt{2}\\a-1=-\sqrt{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}a=\sqrt{2}+1\\a=-\sqrt{2}+1\end{cases}\Leftrightarrow}\orbr{\begin{cases}\sqrt{x}=\sqrt{2}+1\\\sqrt{x}=-\sqrt{2}+1< 0\left(v\text{ô}l\text{ý}\right)\end{cases}}}\Leftrightarrow x=\left(\sqrt{2}+1\right)^2=3+2.\sqrt{2}\)Vậy \(x=3+2.\sqrt{2}\)

P/S: Không chắc lắm

1. \(B=\left(\frac{21}{\left(x-3\right)\left(x+3\right)}+\frac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\frac{\left(x-3\right)\left(x-1\right)}{\left(x-3\right)\left(x+3\right)}\right):\frac{x+3-1}{x+3}\)\(=\frac{3x+6}{\left(x-3\right)\left(x+3\right)}.\frac{x+3}{x+2}=\frac{3\left(x+2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)\left(x+2\right)}=\frac{3}{x-3}\)
2. Điều kiện \(x\ne3\) \(\Rightarrow\frac{-3}{5}=\frac{3}{x-3}\Leftrightarrow x-3=-5\Leftrightarrow x=-2\)
3. \(B=\frac{3}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow x< 3\)

a) B=(\(\frac{21}{x^2-9}\)-\(\frac{x-4}{3-x}\)-\(\frac{x-1}{3+x}\)) : (1-\(\frac{1}{x+3}\)) (ĐK: x khác +-3)

=(\(\frac{21}{\left(x-3\right).\left(x+3\right)}\)+\(\frac{x-4}{x-3}\)-\(\frac{x-1}{x+3}\)) : (1-\(\frac{1}{x+3}\))

=(\(\frac{21+\left(x+4\right).\left(x+3\right)-\left(x-1\right).\left(x-3\right)}{\left(x-3\right).\left(x+3\right)}\):(\(\frac{x+3-1}{x+3}\))

=(\(\frac{3x+6}{\left(x-3\right).\left(x+3\right)}\)) . (\(\frac{x+3}{x+2}\))

=(\(\frac{3.\left(x+2\right)}{\left(x-3\right).\left(x+3\right)}\). \(\frac{x+3}{x+2}\)

=\(\frac{3}{x-3}\)

b) B=\(\frac{3}{x-3}\)=\(\frac{-3}{5}\)

(=) \(\frac{3.5}{x-3}\)=-3

(=) -3.(x-3) = 15

(=) -3x=6

(=) x=-2

vậy x=2 thì B=\(\frac{-3}{5}\)

c) B=\(\frac{3}{x-3}\)<0

(=) 3 < x - 3

(=) -x < - 3 - 3

(=) x > 6

Vậy với x > 6 thì B < 0

BẠN VÀO CÂU HỎI TƯƠNG TỰ RỒI TÌM .

segerheth