`a,` Với `x=3`

\(B=\dfrac{x^2-x}{2x+1}\\ \Rightarrow\dfrac{3^2-3}{2\cdot3+1}\\ =\dfrac{9-3}{6+1}\\ =\dfrac{6}{7}\)

`b,` Ta có `M=A*B`

\(M=\left(\dfrac{1}{x-1}+\dfrac{x}{x^2-1}\right)\cdot\dfrac{x^2-x}{2x+1}\\ =\left(\dfrac{1}{x-1}+\dfrac{x}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{x\left(x-1\right)}{2x+\text{ }1}\\ =\left(\dfrac{x+1}{\left(x-1\right)\left(x+1\right)}+\dfrac{x}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{x\left(x-1\right)}{2x+1}\\ =\dfrac{x+1+x}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x\left(x-1\right)}{2x+1}\\ =\dfrac{2x+1}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x\left(x-1\right)}{2x+1}\\ =\dfrac{x}{x+1}\)

`c,` Để `M=1/2`

`=> x/(x+1)=1/3`

`<=> (3x)/(3(x+1))= (x+1)/(3(x+1))`

`<=> 3x=x+1`

`<=>3x-x=1`

`<=>2x=1`

`<=>x=1/2`

các học bá đâu rùi

a: Khi x=9 thì A=(9-2)/(3+2)=7/5

b: \(B=\dfrac{x-\sqrt{x}+2\sqrt{x}+2-4}{x-1}=\dfrac{x+\sqrt{x}-2}{x-1}\)

\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)

c: P=A*B

\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\cdot\dfrac{x-2}{\sqrt{x}+2}=\dfrac{x-2}{\sqrt{x}+1}\)

P=7/4

=>(x-2)/(căn x+1)=7/4

=>4x-8=7căn 7+7

=>4x-7căn x-15=0

=>căn x=3(nhận) hoặc căn x=-5/4(loại)

=>x=9

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)đk:x>=0,x ne 25`

`A=9=>A=(3+2)/(3-5)=-5/2`

`B)B=(3sqrtx-15+20-2sqrtx)/(x-25)`

`=(sqrtx+5)/(x-25)`

`=1/(sqrtx-5)`

`A=B.|x-4|`

`<=>A/B=|x-4|`

`<=>\sqrtx+2=|x-4|`

`<=>\sqrtx+2=(sqrtx+2)|sqrtx-2|`

`<=>|sqrtx-2|=1`

`+)sqrtx-2=1<=>x=9(tm)`

`+)sqrtx-2=-1<=>x=1(tm)`

Vậy `S={1,9}`

a, Thay x=9 vào biểu thức A ta có

\(A=\dfrac{\sqrt{9}+2}{\sqrt{9}-5}\)

\(A=\dfrac{3+2}{3-5}=\dfrac{5}{-2}=-2,5\)

Vậy A =-2,5 khi x=9

a. A=\(\dfrac{\sqrt{x}+2}{\sqrt{x}-5}\)

=\(\dfrac{\sqrt{9}+2}{\sqrt{9}-5}=\dfrac{-5}{2}\)

ĐKXĐ : \(x\ne0;x\ne\pm1\)

a) Bạn ghi lại rõ đề.

b) \(B=\dfrac{x-1}{x+1}+\dfrac{3x-x^2}{x^2-1}=\dfrac{x-1}{x+1}+\dfrac{3x-x^2}{\left(x-1\right).\left(x+1\right)}\)

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

c) \(P=A.B=\dfrac{x^2+x-2}{x.\left(x-1\right)}=\dfrac{\left(x-1\right).\left(x+2\right)}{x\left(x-1\right)}=\dfrac{x+2}{x}=1+\dfrac{2}{x}\)

Không tồn tại Min P \(\forall x\inℝ\)

b: \(B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{x^2-9}=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)

b: \(B=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{3x-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)

\(a,B=\dfrac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{1+\sqrt{x}}-\sqrt{x}\\ B=x-\sqrt{x}+1-\sqrt{x}=\left(\sqrt{x}-1\right)^2\)

Mà \(x=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)

\(\Rightarrow B=\left(\sqrt{3}-1-1\right)^2=\left(\sqrt{3}-2\right)^2=7-4\sqrt{3}\)

\(b,P=AB=\dfrac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\left(\sqrt{x}-1\right)^2\\ P=\dfrac{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{x+\sqrt{x}+1}=\sqrt{x}-1\\ c,Q=\sqrt{x}+\dfrac{1}{P}=\sqrt{x}+\dfrac{1}{\sqrt{x}-1}\\ Q=\sqrt{x}-1+\dfrac{1}{\sqrt{x}-1}+1\ge2\sqrt{1}+1=3\\ Q_{min}=3\Leftrightarrow\left(\sqrt{x}-1\right)^2=1\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}-1=1\\1-\sqrt{x}=1\end{matrix}\right.\Leftrightarrow\sqrt{x}=2\left(x>1\Leftrightarrow\right)x=4\left(tm\right)\)

a: \(B=\left(\sqrt{x}-1\right)^2=\left(\sqrt{3}-2\right)^2=7-4\sqrt{3}\)

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

`a)` Thay `x=2` vào `B` có: `B=[-10]/[2-4]=5`

`b)` Với `x ne -1;x ne -5` có:

`A=[(x+2)(x+1)-5x-1-(x+5)]/[(x+1)(x+5)]`

`A=[x^2+x+2x+2-5x-1-x-5]/[(x+1)(x+5)]`

`A=[x^2-3x-4]/[(x+1)(x+5)]`

`A=[(x+1)(x-4)]/[(x+1)(x+5)]`

`A=[x-4]/[x+5]`

`c)` Với `x ne -5; x ne -1; x ne 4` có:

`P=A.B=[x-4]/[x+5].[-10]/[x-4]`

           `=[-10]/[x+5]`

Để `P` nguyên `<=>[-10]/[x+5] in ZZ`

    `=>x+5 in Ư_{-10}`

Mà `Ư_{-10}={+-1;+-2;+-5;+-10}`

`=>x={-4;-6;-3;-7;0;-10;5;-15}` (t/m đk)