bài 1 câu a)

x/5=5/x suy ra x.x=5.5

x^2=25=5^2

x=5

1a) \(\frac{x}{5}\) = \(\frac{5}{x}\)

Vì\(\frac{x}{5}\) = \(\frac{5}{x}\) nên x. x= 5. 5

x\(^2\) = 25

x\(^2\) = 5\(^2\)

-> x\(^2\) = 5\(^2\) hoặc x= -5\(^2\)

=> x= 5 hoặc x= -5

Chúc bn học tốt!

\(a.\)

Ta có : \(\frac{x}{5}=\frac{5}{x}\)

\(\Rightarrow x^2=5^2\)

\(\Rightarrow x^2=25\)

\(\Rightarrow x\in\left\{\pm5\right\}\)

a: \(\Leftrightarrow-x^2-3x+x+3+x^2-6x=11\)

=>-8x+3=11

=>-8x=8

hay x=-1

b: \(\Leftrightarrow3x^2-15x+x-5-3x^2+3x=5\)

=>-11x=10

hay x=-10/11

bài 1:

a) (x+1)^2-(x-1)^2-3(x+1)(x-1)

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

=2x^2-3x^2-3

=-x^2-3

Bài 1: 

a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)

\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)

\(\Leftrightarrow-12x^2+14x+13=0\)

\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)

b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)

hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)

ai giúp mik vs

Bài 3: 

a) Ta có: \(2x-3=x+\dfrac{1}{2}\)

\(\Leftrightarrow2x-x=\dfrac{1}{2}+3\)

\(\Leftrightarrow x=\dfrac{7}{2}\)

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

\(\Leftrightarrow3x-\dfrac{1}{2}-2x+\dfrac{1}{2}-5=0\)

\(\Leftrightarrow x=5\)

\(a,A=\dfrac{9-3x+x^2+10x+25-x^2+1}{\left(x-1\right)\left(x+5\right)}\\ A=\dfrac{7x+35}{\left(x-1\right)\left(x+5\right)}=\dfrac{7\left(x+5\right)}{\left(x-1\right)\left(x+5\right)}=\dfrac{7}{x-1}\\ b,A\in Z\\ \Leftrightarrow x-1\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\\ \Leftrightarrow x\in\left\{-6;0;2;8\right\}\left(tm\right)\\ b,A< 0\Leftrightarrow x-1< 0\left(7>0\right)\\ \Leftrightarrow x< 1;x\ne-5\\ c,\left|A\right|=3\Leftrightarrow\dfrac{7}{\left|x-1\right|}=3\Leftrightarrow\left|x-1\right|=\dfrac{7}{3}\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}+1=\dfrac{10}{3}\left(tm\right)\\x=-\dfrac{7}{3}+1=-\dfrac{4}{3}\left(tm\right)\end{matrix}\right.\)

a, (x + 9) + (x - 2) + (x + 7) + (x - 4) + (x + 5) + (x - 6) + (x + 3) + (x - 8) + (x + 1) = 95

    x + 9 + x - 2 + x + 7 + x - 4 + x + 5 + x - 6 + x + 3 + x - 8 + x + 1 = 95

    9x + 9 - 2 + 7 - 4 + 5 - 6 + 3 - 8 + 1 = 95

    9x + 5 = 95

     9x       = 95 - 5

     9x       = 90

       x       = 90 : 9

       x       = 10

Vậy x = 10

b, {[(x : 2) - 1/2 ] : 4 - 1/4 } : 5 = 5 + 1/5

    {[(x : 2) - 1/2 ] : 4 - 1/4 } : 5 = 26/5

    {[(x : 2) - 1/2 ] : 4 - 1/4 }      = 26/5 x 5

    [(x : 2) - 1/2 ] : 4 - 1/4         = 26

    [(x : 2) - 1/2 ] : 4                 = 26 + 1/4

    [(x : 2) - 1/2 ] : 4                 = 105/4

     (x : 2) - 1/2                        = 105/4 x 4

     (x : 2) - 1/2                        = 105

      x : 2                                 = 105 + 1/2

      x : 2                                 = 211/2

      x                                      = 211/2 x 2

      x                                      = 211

Vậy x = 211

a)P=9x+5=95=>9x=90=>x=90:9=10                                                                                                                                                    vay x = 10

mk chả hỉu mô tê j cả làm chi tiết đi!!

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a)B=x+5 +x +x-5/x(x-5)=3x/x(x-5)=3/x-5

        b)đkxđ   x khác 5

a)B=x+5 +x +x-5/x(x-5)=3x/x(x-5)=3/x-5

        b)đkxđ   x khác 5

1, đk x khác 0 ; -5 

\(B=\dfrac{x+5+x+x-5}{x\left(x+5\right)}=\dfrac{3}{x+5}\)

2, B = 3/(x+5) > 0 => x + 5 > 0 <=> x > -5 

3, \(\Rightarrow x+5\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)