2: Để hàm số \(y=\frac{2x+1}{\sqrt{\left(m-1\right)x^2+2\left(m+2\right)x+2m-1}}\) xác định với mọi x thì

\(\left(m-1\right)x^2+2\left(m+2\right)x+2m-1>0\) ∀ x(1)

TH1: m=1

(1) sẽ trở thành \(\left(1-1\right)\cdot x^2+2\left(1+2\right)x+2\cdot1-1>0\)

=>6x+1>0

=>6x>-1

=>x>-1/6

=>Loại

TH2: m<>1

\(\Delta=\left\lbrack2\left(m+2\right)\right\rbrack^2-4\left(m-1\right)\left(2m-1\right)\)

\(=4\left(m^2+4m+4\right)-4\left(2m^2-3m+1\right)=4\left(m^2+4m+4-2m^2+3m-1\right)=4\left(-m^2+7m+3\right)\)

Để (1) luôn đúng thì Δ<0 và a>0

=>m-1>0 và \(4\left(-m^2+7m+3\right)<0\)

=>m>1 và \(m^2-7m-3>0\)

=>m>1 và \(m^2-7m+\frac{49}{4}-\frac{61}{4}>0\)

=>m>1 và \(\left(m-\frac72\right)^2>\frac{61}{4}\)

=>m>1 và \(\left[\begin{array}{l}m-\frac72>\frac{\sqrt{61}}{2}\\ m-\frac72<-\frac{\sqrt{61}}{2}\end{array}\right.\Rightarrow\left[\begin{array}{l}m>\frac{\sqrt{61}+7}{2}\\ m<\frac{-\sqrt{61}+7}{2}\end{array}\right.\)

=>\(m>\frac{\sqrt{61}+7}{2}\)

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 \(n\in N\)

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

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 \(x=-1\)

1. illegal

2. traffic jam

3. seatbelt

4. safely

5. railway station

6. safety

7. traffic signs

8. helicopter

9. tricycle

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:

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

Bài 4:

a: \(\sqrt{1,6}\cdot\sqrt{250}+\sqrt{19,6}:\sqrt{4,9}\)

\(=\sqrt{1,6\cdot250}+\sqrt4\)

\(=\sqrt{400}+2=20+2=22\)

b: \(\sqrt{1\frac34\cdot2\frac27\cdot5\frac49}\)

\(=\sqrt{\frac74\cdot\frac{16}{7}\cdot\frac{49}{9}}=\sqrt{\frac{16}{4}\cdot\frac{49}{9}}=2\cdot\frac73=\frac{14}{3}\)

c: \(\left(20\sqrt{300}-15\sqrt{675}+5\sqrt{75}\right):\sqrt{15}\)

\(=20\sqrt{20}-15\sqrt{45}+5\sqrt5\)

\(=40\sqrt5-45\sqrt5+5\sqrt5=0\)

d: \(\left(\sqrt{325}-\sqrt{117}+2\sqrt{208}\right):\sqrt{13}\)

\(=\sqrt{25}-\sqrt9+2\cdot\sqrt{16}\)

\(=5-3+2\cdot4\)

=2+8

=10

e: \(\frac{2\sqrt8-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\frac{\sqrt5+\sqrt{27}}{\sqrt{30}+\sqrt{162}}\)

\(=\frac{4\sqrt2-2\sqrt3}{\sqrt6\left(\sqrt3-2\sqrt2\right)}-\frac{\sqrt5+\sqrt{27}}{\sqrt6\left(\sqrt5+\sqrt{27}\right)}\)

\(=\frac{2\left(2\sqrt2-\sqrt3\right)}{-\sqrt6\left(2\sqrt2-\sqrt3\right)}-\frac{1}{\sqrt6}=-\frac{2}{\sqrt6}-\frac{1}{\sqrt6}=-\frac{3}{\sqrt6}=\frac{-3\sqrt6}{6}=-\frac{\sqrt6}{2}\)

f: \(\frac{3+2\sqrt3}{\sqrt3}+\frac{2+\sqrt2}{\sqrt2+1}-\left(\sqrt2+\sqrt3\right)\)

\(=2+\sqrt3+\frac{\sqrt2\left(\sqrt2+1\right)}{\sqrt2+1}-\sqrt2-\sqrt3\)

\(=2-\sqrt2+\sqrt2\)

=2