Bài 4:

\(\Leftrightarrow n+1\in\left\{1;3\right\}\)

hay \(n\in\left\{0;2\right\}\)

\(\left(n+4\right)⋮\left(n+1\right)\Rightarrow\left(n+1\right)+3⋮\left(n+1\right)\)

\(\Rightarrow\left(n+1\right)\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

Mà \(n\in N\)

\(\Rightarrow n\in\left\{0;2\right\}\)

Bài 4

\(n+4⋮n+1\)

\(n+1+3⋮n+1\)

\(\Rightarrow3⋮n+1\)

\(\Rightarrow n+1\inƯ\left(3\right)\)

\(Ư\left(3\right)=\left\{1;3\right\}\)

\(\Rightarrow n\in\left\{0;2\right\}\)

a.

ĐKXĐ: \(-3\le x\le\dfrac{3}{2}\)

Ta có:

\(4\sqrt{x+3}=2.2\sqrt{x+3}\le2^2+x+3=x+7\)

\(2\sqrt{3-2x}=2.1.\sqrt{3-2x}\le1^2+3-2x=4-2x\)

Do đó:

\(x+4\sqrt{x+3}+2\sqrt{3-2x}\le x+x+7+4-2x=11\)

Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}\sqrt{x+3}=2\\\sqrt{3-2x}=1\end{matrix}\right.\) \(\Leftrightarrow x=1\)

Vậy pt có nghiệm duy nhất \(x=1\)

b.

ĐKXĐ: \(x\ge-\dfrac{3}{2}\)

\(x^2+4x+5-2\sqrt{2x+3}=0\)

\(\Leftrightarrow\left(x^2+2x+1\right)+\left(2x+3-2\sqrt{2x+3}+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)^2+\left(\sqrt{2x+3}-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\\sqrt{2x+3}-1=0\end{matrix}\right.\)

\(\Leftrightarrow x=-1\)

Vậy pt có nghiệm duy nhất \(x=-1\)

Cách 2 câu a:

\(\Leftrightarrow11-x-4\sqrt{x+3}-2\sqrt{3-2x}=0\)

\(\Leftrightarrow\left(x+3-4\sqrt{x+3}+4\right)+\left(3-2x-2\sqrt{3-2x}+1\right)=0\)

\(\Leftrightarrow\left(\sqrt{x+3}-2\right)^2+\left(\sqrt{3-2x}-1\right)^2=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+3}-2=0\\\sqrt{3-2x}-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+3=4\\3-2x=1\end{matrix}\right.\)

\(\Leftrightarrow x=1\)

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10. boat

câu d tìm ra x,y là bao nhiêu

a: \(\left\{{}\begin{matrix}3x+6y=4\\x+4y=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+6y=4\\3x+12y=6\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{1}{3}\\x=2-\dfrac{4}{3}=\dfrac{2}{3}\end{matrix}\right.\)

Bài 4:

\(A=2x^2+3x-10x-15-2x^2+6x+x+7=-8\\ B=x^3-y^3-5+2y^3-x^3-y^3=-5\\ C=x^3-3x^2+3x-1-x^3-3x^2-3x-1-6x^2+6=4\)