Lời giải:

a.

$|2x-5|=12-3x$

Nếu $x\geq \frac{5}{2}$ thì $2x-5=12-3x$

$\Leftrightarrow x=3,4$ (thỏa mãn)

Nếu $x< \frac{5}{2}$ thì: $5-2x=12-3x$

$\Leftrightarrow x=7$ (loại)

Vậy......

b.

$4x=|x+1|+|x+2|+|x+3|\geq 0$

$\Rightarrow x\geq 0$

Do đó: $|x+1|+|x+2|+|x+3|=(x+1)+(x+2)+(x+3)=3x+6$

Vậy: $3x+6=4x$

$\Leftrightarrow x=6$ (thỏa mãn)

c.

$|x^2+|x+2||=x^2+3$

$\Leftrightarrow x^2+|x+2|=x^2+3$
$\Leftrightarrow |x+2|=3$

$\Leftrightarrow x+2=3$ hoặc $x+2=-3$

$\Leftrightarrow x=1$ hoặc $x=-5$

d.

$|x^2-3|=6$

$\Leftrightarrow x^2-3=6$ hoặc $x^2-3=-6$

$\Leftrightarrow x^2=9$ (chọn) hoặc $x^2=-3< 0$ (loại)

$\Leftrightarrow x=\pm 3$

Lời giải:

1.

$4x+9=0$

$4x=-9$

$x=\frac{-9}{4}$
2.

$-5x+6=0$

$-5x=-6$

$x=\frac{6}{5}$

3.

$x^2-1=0$

$x^2=1=1^2=(-1)^2$

$x=\pm 1$

4.

$x^2-9=0$

$x^2=9=3^2=(-3)^2$

$x=\pm 3$

5.

$x^2-x=0$

$x(x-1)=0$

$x=0$ hoặc $x-1=0$

$x=0$ hoặc $x=1$

6.

$x^2-2x=0$

$x(x-2)=0$

$x=0$ hoặc $x-2=0$

$x=0$ hoặc $x=2$

7.

$x^2-3x=0$

$x(x-3)=0$

$x=0$ hoặc $x-3=0$ 

$x=0$ hoặc $x=3$

8.

$3x^2-4x=0$

$x(3x-4)=0$

$x=0$ hoặc $3x-4=0$

$x=0$ hoặc $x=\frac{4}{3}$

a: \(=\dfrac{x^2+3x+2-x^2+2x+8}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}=\dfrac{5}{x-2}\)

b: \(=\dfrac{x^2-4x+3-x^2-3x-2+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x-1}\)

c: \(=\dfrac{x+2}{x\left(x-2\right)}+\dfrac{2}{x\left(x+2\right)}+\dfrac{3x+2}{\left(x+2\right)\left(x-2\right)}\)

\(=\dfrac{x^2+2x+2x-4+3x+2}{x\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+7x-2}{x\left(x-2\right)\left(x+2\right)}\)

a,

\(\dfrac{x+1}{x-2}-\dfrac{x}{x+2}+\dfrac{8}{x^2-4}\\ =\dfrac{x^2+3x+2-x^2+2x+8}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}=\dfrac{5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{5}{x-2}\)

b,

\(\dfrac{x-3}{x+1}-\dfrac{x+2}{x-1}+\dfrac{8x}{x^2-1}\\ =\dfrac{x^2-4x+3-x^2-3x-2+8x}{\left(x-1\right)\left(x+1\right)}=\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{1}{x-1}\)

Bài 1.

a: \(\Leftrightarrow x-3=-13\)

hay x=-10

Dễ

 Thế

Mà

Cũnhoir

Dc

Ạ

Chịu

Chắc

Phải

Ngu 

Lamqs

Mới

Hỏi

Câu

Này

6: \(-x^2y\left(xy^2-\dfrac{1}{2}xy+\dfrac{3}{4}x^2y^2\right)\)

\(=-x^3y^3+\dfrac{1}{2}x^3y^2-\dfrac{3}{4}x^4y^3\)

7: \(\dfrac{2}{3}x^2y\cdot\left(3xy-x^2+y\right)\)

\(=2x^3y^2-\dfrac{2}{3}x^4y+\dfrac{2}{3}x^2y^2\)

8: \(-\dfrac{1}{2}xy\left(4x^3-5xy+2x\right)\)

\(=-2x^4y+\dfrac{5}{2}x^2y^2-x^2y\)

9: \(2x^2\left(x^2+3x+\dfrac{1}{2}\right)=2x^4+6x^3+x^2\)

10: \(-\dfrac{3}{2}x^4y^2\left(6x^4-\dfrac{10}{9}x^2y^3-y^5\right)\)

\(=-9x^8y^2+\dfrac{5}{3}x^6y^5+\dfrac{3}{2}x^4y^7\)

11: \(\dfrac{2}{3}x^3\left(x+x^2-\dfrac{3}{4}x^5\right)=\dfrac{2}{3}x^3+\dfrac{2}{3}x^5-\dfrac{1}{2}x^8\)

12: \(2xy^2\left(xy+3x^2y-\dfrac{2}{3}xy^3\right)=2x^2y^3+6x^3y^3-\dfrac{4}{3}x^2y^5\)

13: \(3x\left(2x^3-\dfrac{1}{3}x^2-4x\right)=6x^4-x^3-12x^2\)

\(a,\left(-31\right).\left(x+7\right)=0\\ \Rightarrow x+7=0\\ \Rightarrow x=-7\\ b,\left(8-x\right).\left(x+13\right)=0\\ \Rightarrow\left[{}\begin{matrix}8-x=0\\x+13=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=8\\x=-13\end{matrix}\right.\\ c,\left(x^2-25\right)\left(3-x\right)=0\\ \Rightarrow\left(x-5\right)\left(x+5\right)\left(3-x\right)=0\\\Rightarrow \left[{}\begin{matrix}x-5=0\\x+5=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=-5\\x=3\end{matrix}\right.\\ d,\left(x-3\right)\left(x^2+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x^2+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x^2=-4\left(loại\right)\end{matrix}\right.\\ \Rightarrow x=3\)

a, (–31) . (x +7)=0

<=>  x +7 = 0

<=> x = -7

Vậy x \(\in\left\{-7\right\}\)

b, (8 – x) . (x + 13) = 0

<=> \(\left[{}\begin{matrix}8-x=0\\x+13=0\end{matrix}\right.\)   <=> \(\left[{}\begin{matrix}x=8\\x=-13\end{matrix}\right.\)

Vậy x \(\in\left\{8;-13\right\}\)

c,(x²– 25) . (3– x )=0

<=> (x - 5) (x + 5) (3 - x) = 0

<=> \(\left[{}\begin{matrix}x-5=0\\x+5=0\\3-x=0\end{matrix}\right.\)   <=> \(\left[{}\begin{matrix}x=5\\x=-5\\x=3\end{matrix}\right.\)

Vậy x \(\in\left\{5;-5;3\right\}\)

d, ( x - 3 ) (x² + 4) = 0

<=> \(\left[{}\begin{matrix}x-3=0\\x^2+4=0\end{matrix}\right.\)   <=> \(\left[{}\begin{matrix}x=3\\x^2=-4\end{matrix}\right.\)_{(vô lý)}

Vậy x \(\in\left\{3\right\}\)