a: \(A=\dfrac{x-2\sqrt{x}}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x\sqrt{x}+x+\sqrt{x}}+\dfrac{1+2x-2\sqrt{x}}{x^2-\sqrt{x}}\)

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

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

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

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

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

b: \(A\in Z\)

=>\(\sqrt{x}+1⋮x+\sqrt{x}+1\)

=>\(\sqrt{x}\left(\sqrt{x}+1\right)+1-1⋮x+\sqrt{x}+1\)

=>\(-1⋮x+\sqrt{x}+1\)

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

=>\(x+\sqrt{x}+1=1\)

=>\(x+\sqrt{x}=0\)

=>x=0(loại)

a: \(\Delta=\left[-\left(m+5\right)\right]^2-4\cdot1\cdot\left(3m+6\right)\)

\(=m^2+10m+25-12m-24\)

\(=m^2-2m+1=\left(m-1\right)^2>=0\forall m\)

=>Phương trình luôn có nghiệm với mọi m

b: Theo Vi-et, ta có:

\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=m+5\\x_1x_2=\dfrac{c}{a}=3m+6\end{matrix}\right.\)

Để x1,x2 là độ dài hai cạnh của một tam giác vuông có độ dài cạnh huyền bằng 5 thì \(x_1^2+x_2^2=25\)

=>\(\left(x_1+x_2\right)^2-2x_1x_2=25\)

=>\(\left(m+5\right)^2-2\left(3m+6\right)-25=0\)

=>\(m^2+10m+25-6m-12-25=0\)

=>\(m^2+4m-12=0\)

=>(m+6)(m-2)=0

=>\(\left[{}\begin{matrix}m=-6\\m=2\end{matrix}\right.\)

Câu 2:

a: \(B=\dfrac{2\left(x+4\right)}{x-3\sqrt{x}-4}+\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{8}{\sqrt{x}-4}\)

\(=\dfrac{2\left(x+4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{8}{\sqrt{x}-4}\)

\(=\dfrac{2\left(x+4\right)+\sqrt{x}\left(\sqrt{x}-4\right)-8\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{2x+8+x-4\sqrt{x}-8\sqrt{x}-8}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}=\dfrac{3x-12\sqrt{x}}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}\)

\(=\dfrac{3\sqrt{x}\left(\sqrt{x}-4\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+1}\)

b: Để B là số nguyên thì \(3\sqrt{x}⋮\sqrt{x}+1\)

=>\(3\sqrt{x}+3-3⋮\sqrt{x}+1\)

=>\(-3⋮\sqrt{x}+1\)

=>\(\sqrt{x}+1\in\left\{1;3\right\}\)

=>\(\sqrt{x}\in\left\{0;2\right\}\)

=>\(x\in\left\{0;4\right\}\)

1:

b: \(B=\dfrac{3}{2+\sqrt{3}}-\dfrac{13}{4-\sqrt{3}}+\dfrac{6}{\sqrt{3}}\)

\(=\dfrac{3\left(2-\sqrt{3}\right)}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}-\dfrac{13\left(4+\sqrt{3}\right)}{16-3}+2\sqrt{3}\)

\(=6-3\sqrt{3}-4-\sqrt{3}+2\sqrt{3}=2-2\sqrt{3}\)

a: \(A=\sqrt{12}-\sqrt{27}-\sqrt{48}-\sqrt{108}\)

\(=2\sqrt{3}-3\sqrt{3}-4\sqrt{3}-6\sqrt{3}\)

\(=-11\sqrt{3}\)

dạ em cảm ơn anh Thịnh nhiều

a: \(x^2-3x^2+4=0\)

=>\(4-2x^2=0\)

=>\(2-x^2=0\)

=>\(x^2=2\)

=>\(x=\pm\sqrt{2}\)

b: \(2x^2+5x-7=0\)

=>\(2x^2+7x-2x-7=0\)

=>(2x+7)(x-1)=0

=>\(\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=1\end{matrix}\right.\)

c: \(8x-4=0\)

=>8x=4

=>\(x=\dfrac{4}{8}=\dfrac{1}{2}\)

d: 4-5x=0

=>5x=4

=>\(x=\dfrac{4}{5}\)

e: \(3x^4+7x^2-10=0\)

=>\(3x^4+10x^2-3x^2-10=0\)

=>\(\left(3x^2+10\right)\left(x^2-1\right)=0\)

=>\(x^2-1=0\)

=>\(x^2=1\)

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

Để hệ có nghiệm duy nhất thì \(\dfrac{1}{m}\ne\dfrac{m}{-1}\)

=>\(m^2\ne-1\)(luôn đúng)

=>Hệ luôn có nghiệm duy nhất

\(\left\{{}\begin{matrix}x+my=1\\mx-y=-m\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-my\\m\left(1-my\right)-y=-m\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=1-my\\m-m^2y-y=-m\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=1-my\\y\left(m^2+1\right)=2m\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=\dfrac{2m}{m^2+1}\\x=1-my=1-\dfrac{2m^2}{m^2+1}=\dfrac{m^2+1-2m^2}{m^2+1}=\dfrac{-m^2+1}{m^2+1}\end{matrix}\right.\)

\(S=x+y=\dfrac{2m}{m^2+1}+\dfrac{-m^2+1}{m^2+1}=\dfrac{-m^2+2m-1+2}{m^2+1}\)

\(=\dfrac{-\left(m-1\right)^2+2}{m^2+1}< =\dfrac{2}{1}=2\forall m\)

Dấu '=' xảy ra khi m=0

S lại là <= 2/1 v ạ??

Vs lại s dấu = xra khi m = 0 v a

Em nên gõ đề bài bằng công thức toán học biểu tượng \(\Sigma\) góc trái màn hình. 

\(6x^4+7x^3+5x^2-x-2=0\)

=>\(6x^4-3x^3+10x^3-5x^2+10x^2-5x+4x-2=0\)

=>\(\left(2x-1\right)\left(3x^3+5x^2+5x+2\right)=0\)

=>\(\left(2x-1\right)\left(3x^3+2x^2+3x^2+2x+3x+2\right)=0\)

=>\(\left(2x-1\right)\left(3x+2\right)\left(x^2+x+1\right)=0\)

mà \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\)

nên (2x-1)(3x+2)=0

=>\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-\dfrac{2}{3}\end{matrix}\right.\)

ĐKXĐ: m ≠ 2

' = (-m)² - (m - 2)(m + 2)

= m² - m² + 4

= 4 > 0

Vậy phương trình luôn có hai nghiệm với mọi m ∈ R

x₁ = (m + 2)/(m - 2)

x₂ = (m - 2)/(m - 2) = 1 ∈ Z

Xét x₁ = (m + 2)/(m - 2)

= (m - 2 + 4)/(m - 2)

= 1 + 4/(m - 2)

Để x₁ ∈ Z thì 4 ⋮ (m - 2)

⇔ m - 2 ∈ Ư(4) = {-4; -2; -1; 1; 2; 4}

⇔ m ∈ {-2; 0; 1; 3; 4; 6}

Vậy m ∈ {-2; 0; 1; 3; 4; 6} thì phương trình đã cho có hai nghiệm x₁, x₂ ∈ Z