Nhìn sơ qua thì thấy bài 3, b thay -2 vào x rồi giải bình thường tìm m

Bài 2:

a) \(x+x^2=0\)

\(\Leftrightarrow x\left(x+1\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x=0\\x+1=0\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=0\\x=0-1\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}x=0\\x=-1\end{cases}}\)

b) \(0x-3=0\)

\(\Leftrightarrow0x=3\)

\(\Rightarrow vonghiem\)

c) \(3y=0\)

\(\Leftrightarrow y=0\)

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

\(\Rightarrow\frac{3}{x-5}=-1\)

\(\Rightarrow3=-x+5\)

\(\Leftrightarrow x+3=5\)

\(\Rightarrow x=2\)

Bài 1"
a) \(x^2-4x+3\ge0\)

\(\Leftrightarrow x^2-x-3x+3\ge0\)

\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)\ge0\)

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\ge0\)

\(\Leftrightarrow\begin{cases}x-1\ge0\\x-3\ge0\end{cases}\) hoặc \(\begin{cases}x-1\le0\\x-3\le0\end{cases}\)

\(\Leftrightarrow\begin{cases}x\ge1\\x\ge3\end{cases}\) hoặc \(\begin{cases}x\le1\\x\le3\end{cases}\)

\(\Leftrightarrow x\ge3\) hoặc \(x\le1\)

Sao có mỗi một câu vậy bạn giúp mình nốt đi !! <3

a) 2x^2 + 3 = 2x(x + 4) - 7

<=> 2x^2 + 3 = 2x^2 + 8x - 7

<=> 2x^2 - 2x^2 - 8x = - 7 - 3

<=> -8x = -10

<=> x = -10/-8 = 5/4

b) 4x^2 - 12x + 5 = 0

<=> 4x^2 - 2x - 10x + 5 = 0

<=> 2x(2x - 1) - 5(2x - 1) = 0

<=> (2x - 5)(2x - 1) = 0

<=> 2x - 5 = 0 hoặc 2x - 1 = 0

<=> x = 5/2 hoặc x = 1/2

c) |5 - 2x| = 1 - x
<=> \(\hept{\begin{cases}5-2x\text{ nếu }5-2x\ge0\Leftrightarrow x\ge\frac{5}{2}\\-\left(5-2x\right)\text{ nếu }5-2x< 0\Leftrightarrow x< \frac{5}{2}\end{cases}}\)

+) nếu x >= 5/2, ta có:

5 - 2x = 1 - x

<=> -2x + 1 = 1 - 5

<=> -x = -4

<=> x = 4 (tm)

+) nếu x < 5/2, ta có:

-(5 - 2x) = 1 - x

<=> -5 + 2x = 1 - x

<=> 2x + 1 = 1 + 5

<=> 3x = 6

<=> x = 2 (ktm)

d) \(\frac{2}{x-1}=\frac{\left(2x-1\right)\left(2x+1\right)}{x^3-1}-\frac{2x+3}{x^2+x+1}\) ; ĐKXĐ: x # 1 

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

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

<=> 2(x^2 + x + 1) = (2x - 1)(2x + 1) - (2x + 3)(x - 1)

<=> 2x^2 + 2x + 2 = 2x^2 - x + 2

<=> 2x^2 - 2x^2 + 2x - x = 2 - 2

<=> x = 0

mạn phép vô đây để kiếm câu trả lời 

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

\(< =>2x^2+3=2x.x+4.2x-7\)

\(< =>2x^2+3=2x^2+8x-7\)

\(< =>2x^2+3-2x^2=8x-7\)

\(< =>\left(2x^2-2x^2\right)-8x=-7-3\)

\(< =>-8x=-10< =>8x=10\)

\(< =>x=10:8=\frac{10}{8}=\frac{5}{4}\)

\(P=\left(\frac{x-1}{x+3}+\frac{2}{x-3}+\frac{x^2+3}{9-x^2}\right):\left(\frac{2x-1}{2x+1}-1\right)\)\(\left(đkcđ:x\ne\pm3;x\ne-\frac{1}{2}\right)\)

\(=\left(\frac{\left(x-1\right).\left(x-3\right)+2.\left(x+3\right)-\left(x^2+3\right)}{x^2-9}\right):\left(\frac{2x-1-\left(2x+1\right)}{2x+1}\right)\)

\(=\frac{x^2-4x+3+2x+6-x^2-3}{x^2-9}:\frac{-2}{2x+1}\)

\(=\frac{-2x-6}{x^2-9}.\frac{2x+1}{-2}\)

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

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

b)\(\left|x+1\right|=\frac{1}{2}\Leftrightarrow\orbr{\begin{cases}x+1=\frac{1}{2}\\x+1=-\frac{1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\left(koTMđkxđ\right)\\x=-\frac{3}{2}\left(TMđkxđ\right)\end{cases}}}\)

thay \(x=-\frac{3}{2}\)  vào P tâ đc:   \(P=\frac{2x+1}{x-3}=\frac{2.\left(-\frac{3}{2}\right)+1}{-\frac{3}{2}-3}=\frac{4}{9}\)

c)ta có:\(P=\frac{x}{2}\Leftrightarrow\frac{2x+1}{x-3}=\frac{x}{2}\)

\(\Rightarrow2.\left(2x+1\right)=x.\left(x-3\right)\)

\(\Leftrightarrow4x+2=x^2-3x\)

\(\Leftrightarrow x^2-7x-2=0\)

\(\Leftrightarrow x^2-2.\frac{7}{2}+\frac{49}{4}-\frac{57}{4}=0\)

\(\Leftrightarrow\left(x-\frac{7}{2}\right)^2-\frac{57}{4}=0\)

\(\Leftrightarrow\left(x-\frac{7}{2}-\frac{\sqrt{57}}{2}\right).\left(x-\frac{7}{2}+\frac{\sqrt{57}}{2}\right)\)

bạn tự giải nốt nhé!!

d)\(x\in Z;P\in Z\Leftrightarrow\frac{2x+1}{x-3}\in Z\Leftrightarrow\frac{2x-6+7}{x-3}=2+\frac{7}{x-3}\in Z\)

\(2\in Z\Rightarrow\frac{7}{x-3}\in Z\Leftrightarrow x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)

bạn tự làm nốt nhé

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

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

b, \(\left|x+1\right|=\dfrac{1}{2}\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}-1\\x=-\dfrac{1}{2}-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\left(ktmđk\right)\\x=-\dfrac{3}{2}\end{matrix}\right.\)

Thay x = -3/2 ta được \(\dfrac{2\left(-\dfrac{3}{2}\right)+1}{-\dfrac{3}{2}+3}=\dfrac{-2}{\dfrac{3}{2}}=-\dfrac{4}{3}\)

a) $0,25x+1,5=0$

$\mathrm{=>\; x\; =\; (0\; -\; 1,5)\; :\; 0,25\; =\; -1,5\; :\; 0,25\; =\; -6}$

Vậy x = -6.

b) $6,36-5,3x=0$

=> x = (0 + 6,36) : 5,3 = 6,36 : 5,3 =\(\dfrac{6}{5}=1,2\)
Vậy x = 1,2.

c) $\frac{4}{3}x-\frac{5}{6}=\frac{1}{2}$

Vậy x = 1.

d) $-\frac{5}{9}x+1=\frac{2}{3}x-10$

$\mathrm{=>\; \backslash (\backslash dfrac\{-5\}\{9\}x-\backslash dfrac\{2\}\{3\}x=\backslash dfrac\{-11\}\{9\}x=-10-1=-11\backslash )}$

$\mathrm{=>\; \backslash (x=-11:\backslash dfrac\{-11\}\{9\}=9\backslash )}$

Vậy x = 9.