\(a,\dfrac{x}{8}=\dfrac{7}{-2}\\ \Rightarrow x=-28\\ b,\dfrac{1-2x}{6}=\dfrac{-1}{2}\\ \Leftrightarrow2-4x=-6\\ \Leftrightarrow4x=8\\ \Leftrightarrow x=2\\ c,\dfrac{x+2}{3}=\dfrac{x+3}{4}\\ \Leftrightarrow4x+8=3x+9\\ \Leftrightarrow x=1\\ d,\dfrac{10}{2-x}=2\\ \Leftrightarrow4-2x=10\\ \Leftrightarrow2x=-6\\ \Leftrightarrow x=-3\)

tách đi bạn

a) (2x - 3)(6 - 2x) = 0

=> \(\left[{}\begin{matrix}2x-3=0\\6-2x=0\end{matrix}\right.=>\left[{}\begin{matrix}2x=3\\2x=6\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=3\end{matrix}\right.\)

b) \(5\dfrac{4}{7}:x=13=>\dfrac{39}{7}:x=13=>x=\dfrac{39}{7}:13=>x=\dfrac{3}{7}\)

c) \(2x-\dfrac{3}{7}=6\dfrac{2}{7}=>2x-\dfrac{3}{7}=\dfrac{44}{7}=>2x=\dfrac{47}{7}=>x=\dfrac{47}{14}\)

d) \(\dfrac{x}{5}+\dfrac{1}{2}=\dfrac{6}{10}=>\dfrac{x}{5}=\dfrac{6}{10}-\dfrac{1}{2}=>\dfrac{x}{5}=\dfrac{1}{10}=>x.10=5=>x=\dfrac{1}{2}\)

e) \(\dfrac{x+3}{15}=\dfrac{1}{3}=>\left(x+3\right).3=15=>x+3=5=>x=2\)

f)\(\dfrac{x-12}{4}=\dfrac{1}{2}=\dfrac{x-12}{4}=\dfrac{2}{4}\)

⇒\(x-12=2\)

   \(x=2+12\)

  x = 14

g)2\(\dfrac{1}{4}.\left(x-7\dfrac{1}{3}\right)=1,5\)

\(\dfrac{9}{4}.\left(x-\dfrac{22}{3}\right)=1,5\) 

      \(\left(x-\dfrac{22}{3}\right)=\dfrac{3}{2}:\dfrac{9}{4}\)

       \(x-\dfrac{22}{3}=\dfrac{2}{3}\)

       \(x=\dfrac{2}{3}+\dfrac{22}{3}\) 

      \(x=8\)

a) Ta có: \(\dfrac{4}{5}-3\left|x\right|=\dfrac{1}{5}\)

\(\Leftrightarrow3\left|x\right|=\dfrac{4}{5}-\dfrac{1}{5}=\dfrac{3}{5}\)

\(\Leftrightarrow\left|x\right|=\dfrac{1}{5}\)

hay \(x\in\left\{\dfrac{1}{5};-\dfrac{1}{5}\right\}\)

b) Ta có: \(4x-\dfrac{1}{2}x+\dfrac{3}{5}x=\dfrac{4}{5}\)

nên \(\dfrac{41}{10}x=\dfrac{4}{5}\)

hay \(x=\dfrac{8}{41}\)

c) Ta có: \(\left(2x-8\right)\left(10-5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-8=0\\10-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=8\\5x=10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)

d) Ta có: \(\dfrac{3}{4}+\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}\)

\(\Leftrightarrow\dfrac{1}{4}\left|2x-1\right|=\dfrac{7}{2}-\dfrac{3}{4}=\dfrac{14}{4}-\dfrac{3}{4}=\dfrac{11}{4}\)

\(\Leftrightarrow\left|2x-1\right|=11\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=11\\2x-1=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=12\\2x=-10\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\)

a) ĐKXĐ: \(x\notin\left\{0;-5\right\}\)

Ta có: \(B=\dfrac{x^2+2x}{2x+10}+\dfrac{x-5}{x}-\dfrac{5x-50}{2x^2+10x}\)

\(=\dfrac{x^2+2x}{2\left(x+5\right)}+\dfrac{x-5}{x}-\dfrac{5x-50}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+2x^2}{2x\left(x+5\right)}+\dfrac{2\left(x+5\right)\left(x-5\right)}{2x\left(x+5\right)}-\dfrac{5x-50}{2x\left(x+5\right)}\)

\(=\dfrac{x^3+2x^2+2x^2-50-5x+50}{2x\left(x+5\right)}\)

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

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

\(=\dfrac{x^2+5x-x-5}{2\left(x+5\right)}\)

\(=\dfrac{x\left(x+5\right)-\left(x+5\right)}{2\left(x+5\right)}\)

\(=\dfrac{\left(x+5\right)\left(x-1\right)}{2\left(x+5\right)}\)

\(=\dfrac{x-1}{2}\)

b) Để B=0 thì \(\dfrac{x-1}{2}=0\)

\(\Leftrightarrow x-1=0\)

hay x=1(nhận)

Vậy: Để B=0 thì x=1

Để \(B=\dfrac{1}{4}\) thì \(\dfrac{x-1}{2}=\dfrac{1}{4}\)

\(\Leftrightarrow4\left(x-1\right)=2\)

\(\Leftrightarrow4x-4=2\)

\(\Leftrightarrow4x=6\)

hay \(x=\dfrac{3}{2}\)(nhận)

Vậy: Để \(B=\dfrac{1}{4}\) thì \(x=\dfrac{3}{2}\)

c) Thay x=3 vào biểu thức \(B=\dfrac{x-1}{2}\), ta được:

\(B=\dfrac{3-1}{2}=\dfrac{2}{2}=1\)

Vậy: Khi x=3 thì B=1

d) Để B<0 thì \(\dfrac{x-1}{2}< 0\)

\(\Leftrightarrow x-1< 0\)

\(\Leftrightarrow x< 1\)

Kết hợp ĐKXĐ, ta được: 

\(\left\{{}\begin{matrix}x< 1\\x\notin\left\{0;-5\right\}\end{matrix}\right.\)

Vậy: Để B<0 thì \(\left\{{}\begin{matrix}x< 1\\x\notin\left\{0;-5\right\}\end{matrix}\right.\)

Để B>0 thì \(\dfrac{x-1}{2}>0\)

\(\Leftrightarrow x-1>0\)

hay x>1

Kết hợp ĐKXĐ, ta được: x>1

Vậy: Để B>0 thì x>1

Bổ sung phần c và d luôn:

c, C = \(\dfrac{2}{5}\)

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

\(\Leftrightarrow\) 5(x² - 1) = 2(2x² + 3)

\(\Leftrightarrow\) 5x² - 5 = 4x² + 6

\(\Leftrightarrow\) x² = 11

\(\Leftrightarrow\) x² - 11 = 0

\(\Leftrightarrow\) (x - \(\sqrt{11}\))(x + \(\sqrt{11}\)) = 0

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x-\sqrt{11}=0\\x+\sqrt{11}=0\end{matrix}\right.\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=\sqrt{11}\left(TM\right)\\x=-\sqrt{11}\left(TM\right)\end{matrix}\right.\)

d, Ta có: \(\dfrac{x^2-1}{2x^2+3}\) = \(\dfrac{x^2+\dfrac{3}{2}-\dfrac{5}{2}}{2\left(x^2+\dfrac{3}{2}\right)}\) = \(\dfrac{1}{2}\) - \(\dfrac{5}{4\left(x^2+\dfrac{3}{2}\right)}\)

C nguyên \(\Leftrightarrow\) \(\dfrac{5}{4\left(x^2+\dfrac{3}{2}\right)}\) nguyên \(\Leftrightarrow\) 5 \(⋮\) 4(x² + \(\dfrac{3}{2}\))

\(\Leftrightarrow\) 4(x² + \(\dfrac{3}{2}\)) \(\in\) Ư(5)

Xét các TH:

4(x² + \(\dfrac{3}{2}\)) = 5 \(\Leftrightarrow\) x² = \(\dfrac{-1}{4}\) \(\Leftrightarrow\) x² + \(\dfrac{1}{4}\) = 0 (Vô nghiệm)

4(x² + \(\dfrac{3}{2}\)) = -5 \(\Leftrightarrow\) x² = \(\dfrac{-11}{4}\) \(\Leftrightarrow\) x² + \(\dfrac{11}{4}\) = 0 (Vô nghiệm)

4(x² + \(\dfrac{3}{2}\)) = 1 \(\Leftrightarrow\) x² = \(\dfrac{-5}{4}\) \(\Leftrightarrow\) x² + \(\dfrac{5}{4}\) = 0 (Vô nghiệm)

4(x² + \(\dfrac{3}{2}\)) = -1 \(\Leftrightarrow\) x² = \(\dfrac{-7}{4}\) \(\Leftrightarrow\) x² + \(\dfrac{7}{4}\) = 0 (Vô nghiệm)

Vậy không có giá trị nào của x \(\in\) Z thỏa mãn C \(\in\) Z

Chúc bn học tốt! (Ko bt đề sai hay ko nữa :v)

a, đk x khác 0

<=> x^2 = 16 <=> x = 4 ; x = -4 (tm)

b, <=> 36x +252 = -360 <=> x = -17 

c. đk x khác -1 

<=> (x+1)^2 = 16 

TH1 : x + 1 = 4 <=> x = 3 (tm)

TH2 : x + 1 = -4 <=> x = -5 (tm) 

d, đk x khác 1/2 

<=> (2x-1)^2 = 81 

TH1 : 2x - 1 = 9 <=> x = 5 (tm) 

TH2 : 2x - 1 = -9 <=> x = -4 (tm) 

a: \(\Leftrightarrow x^2=16\)

hay \(x\in\left\{4;-4\right\}\)

b: =>x+7/15=-2/3

=>x+7=-10

hay x=-17

c: \(\Leftrightarrow\left(x+1\right)^2=16\)

\(\Leftrightarrow x+1\in\left\{4;-4\right\}\)

hay \(x\in\left\{3;-5\right\}\)

a) \(\dfrac{x}{4}=\dfrac{4}{x}\)=>x²=4.4=16 =>x²=4²

=>x=2 hay x=-2.

b) \(\dfrac{x+7}{15}=-\dfrac{24}{36}\)=>\(\dfrac{x+7}{15}=-\dfrac{2}{3}\)=>x+7=-\(\dfrac{2}{3}.15\)=-10 =>x=-17

c)\(\dfrac{x+1}{8}=\dfrac{2}{x+1}\)=>(x+1)²=2.8=16=4²

=>x+1=4 hay x+1=-4

=>x=3 hay x=-5.

d) \(\dfrac{2x-1}{\left(-3\right)^2}=\dfrac{\left(-3\right)^2}{2x-1}\)=>\(\dfrac{2x-1}{9}=\dfrac{9}{2x-1}\)=>(2x-1)²=9²

=>2x-1=9 hay 2x-1=-9

=>x=5 hay x=-4.

\(a,3-x=x+1,8\)

\(\Rightarrow-x-x=1,8-3\)

\(\Rightarrow-2x=-1,2\)

\(\Rightarrow x=0,6\)

\(b,2x-5=7x+35\)

\(\Rightarrow2x-7x=35+5\)

\(\Rightarrow-5x=40\)

\(\Rightarrow x=-8\)

\(c,2\left(x+10\right)=3\left(x-6\right)\)

\(\Rightarrow2x+20=3x-18\)

\(\Rightarrow2x-3x=-18-20\)

\(\Rightarrow-x=-38\)

\(\Rightarrow x=38\)

\(d,8\left(x-\dfrac{3}{8}\right)+1=6\left(\dfrac{1}{6}+x\right)+x\)

\(\Rightarrow8x-3+1=1+6x+x\)

\(\Rightarrow8x-3=7x\)

\(\Rightarrow8x-7x=3\)

\(\Rightarrow x=3\)

\(e,\dfrac{2}{9}-3x=\dfrac{4}{3}-x\)

\(\Rightarrow-3x+x=\dfrac{4}{3}-\dfrac{2}{9}\)

\(\Rightarrow-2x=\dfrac{10}{9}\)

\(\Rightarrow x=-\dfrac{5}{9}\)

\(g,\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{3}{4}x-\dfrac{1}{2}\)

\(\Rightarrow\dfrac{1}{2}x-\dfrac{3}{4}x=-\dfrac{1}{2}-\dfrac{5}{6}\)

\(\Rightarrow-\dfrac{1}{4}x=-\dfrac{4}{3}\)

\(\Rightarrow x=\dfrac{16}{3}\)

\(h,x-4=\dfrac{5}{6}\left(6-\dfrac{6}{5}x\right)\)

\(\Rightarrow x-4=5-x\)

\(\Rightarrow x+x=5+4\)

\(\Rightarrow2x=9\)

\(\Rightarrow x=\dfrac{9}{2}\)

\(k,7x^2-11=6x^2-2\)

\(\Rightarrow7x^2-6x^2=-2+11\)

\(\Rightarrow x^2=9\Rightarrow\left[{}\begin{matrix}x=3\\x=-3\end{matrix}\right.\)

\(m,5\left(x+3\cdot2^3\right)=10^2\)

\(\Rightarrow5\left(x+24\right)=100\)

\(\Rightarrow x+24=20\)

\(\Rightarrow x=-4\)

\(n,\dfrac{4}{9}-\left(\dfrac{1}{6^2}\right)=\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{4}{9}-\dfrac{1}{36}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2+\dfrac{5}{12}=\dfrac{5}{12}\)

\(\Rightarrow\dfrac{2}{3}\left(x-\dfrac{2}{3}\right)^2=0\)

\(\Rightarrow x-\dfrac{2}{3}=0\Rightarrow x=\dfrac{2}{3}\)

#\(Urushi\text{☕}\)

a: 3-x=x+1,8

=>-2x=-1,2

=>x=0,6

b: 2x-5=7x+35

=>-5x=40

=>x=-8

c: 2(x+10)=3(x-6)

=>3x-18=2x+20

=>x=38

d; 8(x-3/8)+1=6(1/6+x)+x

=>8x-3+1=1+6x+x

=>8x-2=7x+1

=>x=3

e: =>-3x+x=4/3-2/9

=>-2x=12/9-2/9=10/9

=>x=-5/9

g: =>3/4x-1/2x=5/6+1/2

=>1/4x=5/6+3/6=8/6=4/3

=>x=4/3*4=16/3

h: =>x-4=-x+5

=>2x=9

=>x=9/2

a, \(\dfrac{6}{2x+1}\Rightarrow2x+1\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)

c, \(\dfrac{x-3}{x-1}=\dfrac{x-1-2}{x-1}=1-\dfrac{2}{x-1}\Rightarrow x-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

tương tự .... 

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