11.D

12.B

8. A

9. B

10. B

8.A

9.B

10.B

Câu 6: B

Câu 7: C

Câu 4: C

Câu 5: B

ai giúp với

a: \(6\left(x-1\right)-5=3x\)

=>\(6x-6-5-3x=0\)

=>3x-11=0

=>3x=11

=>\(x=\dfrac{11}{3}\)

b: \(\left(2x+5\right)^2-\left(x-2\right)^2=0\)

=>\(\left(2x+5+x-2\right)\left(2x+5-x+2\right)=0\)

=>(3x+3)(x+7)=0

=>3(x+1)(x+7)=0

=>(x+1)(x+7)=0

=>\(\left[{}\begin{matrix}x+1=0\\x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-7\end{matrix}\right.\)

c: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

\(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{5x-2}{4-x^2}\)

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

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

=>\(x^2-3x+2-x^2-2x=-5x+2\)

=>-5x+2=-5x+2

=>0x=0(luôn đúng)

Vậy: S={x|\(x\notin\left\{2;-2\right\}\)}

Câu 1: A

Câu 2: D

Câu 3: A

a: \(M=\left(\dfrac{x}{x+2}+\dfrac{x^3-8}{x^3+8}\cdot\dfrac{x^2-2x+4}{4-x^2}\right):\dfrac{2}{x+2}\)

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

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

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

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

b: \(x^2+2x=0\)

=>x(x+2)=0

=>\(\left[{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-2\left(loại\right)\end{matrix}\right.\)

Thay x=0 vào M, ta được:

\(M=\dfrac{-2}{0+2}=\dfrac{-2}{2}=-1\)

c: Để M*x nguyên thì \(-2x⋮x+2\)

=>\(-2x-4+4⋮x+2\)

=>\(4⋮x+2\)

=>\(x+2\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(x\in\left\{-1;-3;0;-4;2;-6\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{-1;-3;0;-4;-6\right\}\)

\(Tacó:\) \(x+y=3\)

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

Áp dụng BĐT Cô si cho 3 số, ta có:

\(3=\left(\dfrac{x}{2}+\dfrac{x}{2}+y\right)\ge3.^3\sqrt{\dfrac{x}{2}.\dfrac{x}{2}.y}=3.^3\sqrt{\dfrac{x^2y}{4}}\)

⇒\(1\ge^3\sqrt{\dfrac{x^2y}{4}}\)

⇒\(1\ge\dfrac{x^2y}{4}\Leftrightarrow x^2y\le4\)

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