a)Vì |4x - 2| = 6 <=> 4x - 2 ϵ {6,-6} <=> x ϵ {2,-1}

Thay x = 2, ta có B không tồn tại

Thay x = -1, ta có B = \(\dfrac{1}{3}\)

b)ĐKXĐ:x ≠ 2,-2

Ta có \(A=\dfrac{5}{x+2}+\dfrac{3}{2-x}-\dfrac{15-x}{4-x^2}=\dfrac{10-5x+3x+6}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{16-2x}{\left(x+2\right)\left(2-x\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{\left(x+2\right)\left(x-2\right)}-\dfrac{15-x}{4-x^2}=\dfrac{2x-16}{x^2-4}+\dfrac{15-x}{x^2-4}=\dfrac{x-1}{x^2-4}\)c)Từ câu b, ta có \(A=\dfrac{x-1}{x^2-4}\)\(\Rightarrow\dfrac{2A}{B}=\dfrac{\dfrac{\dfrac{2x-2}{x^2-4}}{2x+1}}{x^2-4}=\dfrac{2x-2}{2x+1}< 1\) với mọi x

Do đó không tồn tại x thỏa mãn đề bài

a: Sửa đề: \(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\)

ĐKXĐ: x∉{0;2;-2;3}

Ta có: \(A=\left(\frac{2+x}{2-x}-\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\left(\frac{x^2-3x}{2x^2-x^3}\right)\)

\(=\left\lbrack\frac{-\left(x+2\right)}{x-2}-\frac{4x^2}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{x+2}\right\rbrack:\frac{x\left(x-3\right)}{x^2\cdot\left(2-x\right)}\)

\(=\frac{-\left(x+2\right)^2-4x^2+\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}:\frac{x-3}{x\left(2-x\right)}\)

\(=\frac{-x^2-4x-4-4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}\cdot\frac{-x\left(x-2\right)}{x-3}\)

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

b: Để A>0 thì \(\frac{4x^2}{x-3}>0\)

=>x-3>0

=>x>3

c: |x-7|=4

=>\(\left[\begin{array}{l}x-7=4\\ x-7=-4\end{array}\right.\Rightarrow\left[\begin{array}{l}x=11\left(nhận\right)\\ x=3\left(loại\right)\end{array}\right.\)

Thay x=11 vào A, ta được:

\(A=\frac{4\cdot11^2}{11-3}=\frac{4\cdot121}{8}=\frac{121}{2}\)

giúp mình với . mình đang cần gấp nhé!

`a)|x-2|=2<=>[(x=4(ko t//m)),(x=0(t//m)):}`

Thay `x=0` vào `A` có: `A=[2\sqrt{0}-3]/[\sqrt{0}-2]=3/2`

`b)` Với `x >= 0,x ne 4` có:

`B=[2(\sqrt{x}-3)+\sqrt{x}(\sqrt{x}+3)-4\sqrt{x}]/[(\sqrt{x}+3)(\sqrt{x}-3)]`

`B=[2\sqrt{x}-6+x+3\sqrt{x}-4\sqrt{x}]/[(\sqrt{x}+3)(\sqrt{x}-3)]`

`B=[x+\sqrt{x}-6]/[(\sqrt{x}+3)(\sqrt{x}-3)]`

`B=[(\sqrt{x}+3)(\sqrt{x}-2)]/[(\sqrt{x}+3)(\sqrt{x}-3)]`

`B=[\sqrt{x}-2]/[\sqrt{x}-3]`

`c)` Với `x >= 0,x ne 4` có:

`C=A.B=[2\sqrt{x}-3]/[\sqrt{x}-2].[\sqrt{x}-2]/[\sqrt{x}-3]=[2\sqrt{x}-3]/[\sqrt{x}-3]`

Có: `C >= 1`

`<=>[2\sqrt{x}-3]/[\sqrt{x}-3] >= 1`

`<=>[2\sqrt{x}-3-\sqrt{x}+3]/[\sqrt{x}-3] >= 0`

`<=>[\sqrt{x}]/[\sqrt{x}-3] >= 0`

  Vì `x >= 0=>\sqrt{x} >= 0`

  `=>\sqrt{x}-3 > 0`

`<=>x > 9` (t/m đk)

\(a,\\ A=25x^2-10x+11\\ =\left(5x\right)^2-2.5x.1+1^2+10\\ =\left(5x+1\right)^2+10\ge10\forall x\in R\\ Vậy:min_A=10.khi.5x+1=0\Leftrightarrow x=-\dfrac{1}{5}\\ B=\left(x-3\right)^2+\left(11-x\right)^2\\ =\left(x^2-6x+9\right)+\left(121-22x+x^2\right)\\ =x^2+x^2-6x-22x+9+121=2x^2-28x+130\\ =2\left(x^2-14x+49\right)+32\\ =2\left(x-7\right)^2+32\\ Vì:2\left(x-7\right)^2\ge0\forall x\in R\\ Nên:2\left(x-7\right)^2+32\ge32\forall x\in R\\ Vậy:min_B=32.khi.\left(x-7\right)=0\Leftrightarrow x=7\\Tương.tự.cho.biểu.thức.C\)

b:

\(D=-25x^2+10x-1-10\)

\(=-\left(25x^2-10x+1\right)-10\)

\(=-\left(5x-1\right)^2-10< =-10\)

Dấu = xảy ra khi x=1/5

\(E=-9x^2-6x-1+20\)

\(=-\left(9x^2+6x+1\right)+20\)

\(=-\left(3x+1\right)^2+20< =20\)

Dấu = xảy ra khi x=-1/3

\(F=-x^2+2x-1+1\)

\(=-\left(x^2-2x+1\right)+1=-\left(x-1\right)^2+1< =1\)

Dấu = xảy ra khi x=1