a: \(\Leftrightarrow6x+33⋮3x+2\)

\(\Leftrightarrow3x+2\in\left\{1;-1;29;-29\right\}\)

hay \(x\in\left\{-\dfrac{1}{3};-1;9;-\dfrac{31}{3}\right\}\)

b: \(\Leftrightarrow56:x^2+112=126\)

\(\Leftrightarrow56:x^2=14\)

=>x²=4

=>x=2 hoặc x=-2

c: =>2x-14-15+3x=-109

=>5x-29=-109

=>5x=-80

hay x=-16

a: =>2x-6-3x+15=12-4x-18

=>-x+9=-4x-6

=>3x=-15

hay x=-5

b: \(\Leftrightarrow6x+33⋮3x+2\)

\(\Leftrightarrow6x+4+29⋮3x+2\)

\(\Leftrightarrow3x+2\in\left\{1;-1;29;-29\right\}\)

hay \(x\in\left\{-\dfrac{1}{3};-1;9;-\dfrac{31}{3}\right\}\)

c: \(\Leftrightarrow56:x^2+112=126\)

\(\Leftrightarrow56:x^2=14\)

\(\Leftrightarrow x^2=4\)

=>x=2 hoặc x=-2

a: ĐKXĐ: x>=0; x<>1

\(A=\frac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\frac{3\sqrt{x}-2}{1-\sqrt{x}}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\frac{15\sqrt{x}-11}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}-\frac{3\sqrt{x}-2}{\sqrt{x}-1}-\frac{2\sqrt{x}+3}{\sqrt{x}+3}\)

\(=\frac{15\sqrt{x}-11-\left(3\sqrt{x}-2\right)\left(\sqrt{x}+3\right)-\left(2\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{15\sqrt{x}-11-\left(3x+7\sqrt{x}-6\right)-\left(2x+\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{15\sqrt{x}-11-3x-7\sqrt{x}+6-2x-\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{-5x+7\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\)

\(=\frac{-\left(5\sqrt{x}-2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}\)

b: Khi \(x=4-2\sqrt3\) thì \(A=\frac{-5\cdot\sqrt{4-2\sqrt3}+2}{\sqrt{4-2\sqrt3}+3}=\frac{-5\left(\sqrt3-1\right)+2}{\sqrt3-1+3}\)

\(=\frac{-5\sqrt3+5+2}{\sqrt3+2}=\frac{-5\sqrt3+7}{\sqrt3+2}=\left(-5\sqrt3+7\right)\left(2-\sqrt3\right)\)

\(=-10\sqrt3+15+14-7\sqrt3=-17\sqrt3+29\)

c: \(A=\frac12\)

=>\(\frac{-5\sqrt{x}+2}{\sqrt{x}+3}=\frac12\)

=>\(-10\sqrt{x}+4=\sqrt{x}+3\)

=>\(-11\sqrt{x}=-1\)

=>\(\sqrt{x}=\frac{1}{11}\)

=>x=1/121(nhận)

e: \(A+5=\frac{-5\sqrt{x}+2}{\sqrt{x}+3}+5=\frac{-5\sqrt{x}+2+5\sqrt{x}+15}{\sqrt{x}+3}=\frac{17}{\sqrt{x}+3}>0\forall x\) thỏa mãn ĐKXĐ

=>A>-5∀x thỏa mãn ĐKXĐ

huhuhuhu help me cứi tui

a) \(đk:\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne4\end{matrix}\right.\)

b) \(x=3+2\sqrt{2}\Rightarrow\sqrt{x}=\sqrt{3+2\sqrt{2}}=\sqrt{\left(\sqrt{2}+1\right)^2}=\sqrt{2}+1\)

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

c) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}=\dfrac{1}{2}\)

\(\Leftrightarrow4\sqrt{x}-2=\sqrt{x}-2\Leftrightarrow3\sqrt{x}=0\Leftrightarrow x=0\left(tm\right)\)

d) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}>2\)

\(\Leftrightarrow2\sqrt{x}-1>2\sqrt{x}-4\Leftrightarrow-1>-4\left(đúng\forall x\right)\)

e) \(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}-2}=\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}-2}+\dfrac{3}{\sqrt{x}-2}=2+\dfrac{3}{\sqrt{x}-2}\in Z\)

\(\Rightarrow\sqrt{x}-2\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

Do \(x\ge0\)

\(\Rightarrow x\in\left\{1;9;25\right\}\)

`2^(x-1) + 2^x +2^(x+1)  = 112 `

`=> 2^(x-1) + 2^(x-1) xx 2 +2^(x-1) xx 2^2  = 112 `

`=> 2^(x-1) . (1 + 2 + 2^2)   = 112 `

`=> 2^(x-1) . 7   = 112 `

`=> 2^(x-1) = 16`

`=>  2^(x-1) = 2^4`

`=> x - 1 = 4`

`=> x = 5`

đề yêu cầu gì vậy ạ?

Tìm x?

ukm

a)x - 112 = 2³

   x - 112 = 8

       x = 8 + 112

       x = 120

b) Ta có  Ư(35) = {0; 35;70;105;...}

     Vì x\(x\inƯ\left(35\right)̀\)

           và x < 10

     Nên x = 0

c) Để 61* chia hết 3;5 thì x = 5