pt đã cho \(\Leftrightarrow\left(x-y\right)\left(x+3y\right)=2x+6y-x+y\)

\(\Leftrightarrow\left(x-y\right)\left(x+3y\right)=2\left(x+3y\right)-\left(x-y\right)\)

Đặt \(\left\{{}\begin{matrix}x-y=a\\x+3y=b\end{matrix}\right.\) với \(a,b\inℤ\) và \(b\ge4\)

pt thành \(ab=2a-b\)

\(\Leftrightarrow ab-2a+b-2=-2\)

\(\Leftrightarrow\left(a+1\right)\left(b-2\right)=-2\) (*)

 Vì \(b\ge4\Leftrightarrow b-2\ge2\). Do đó (*) \(\Rightarrow\) \(b-2=2\) hay \(b=4\), nghĩa là dấu "=" phải xảy ra \(\Leftrightarrow x=y=1\). Thử lại, ta thấy không thỏa mãn.

 Vậy pt đã cho không có nghiệm nguyên dương.

a) ĐK: \(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\\4-x^2\ne0\\x^2-4\ne0\end{matrix}\right.\Leftrightarrow x\ne\pm2\)

\(A=\left(\dfrac{x+2}{x-2}-\dfrac{1}{x+2}-\dfrac{x-4}{4-x^2}\right):\dfrac{1}{x^2-4}\)

\(A=\left[\dfrac{x+2}{x-2}-\dfrac{1}{x+2}+\dfrac{x-4}{\left(x+2\right)\left(x-2\right)}\right]\cdot\left(x^2-4\right)\)

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

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

\(A=x^2+4x+2\)

b) \(A=14\)

\(\Leftrightarrow x^2+4x+2=14\)

\(\Leftrightarrow x^2+4x-12=0\)

\(\Leftrightarrow x^2-2x+6x-12=0\)

\(\Leftrightarrow x\left(x-2\right)+6\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(ktm\right)\\x=-6\left(tm\right)\end{matrix}\right.\)

a; (\(x+y\))² - 4.(\(x-y\))²

   = \(x^2+2xy+y^2\) - 4\(x^2+8xy-4y^2\)

  =  (\(x^2-4x^2\))  + (2\(xy+8xy\)) + (y² - 4y²)

= - 3\(x^2\) + 10\(xy\)  - 3y²

b; (\(x+y\))³ - 2\(x^3\) + (\(x-y\))³

  = \(x^3+3x^2y+3xy^2+y^3\) - 2\(x^3\) + \(x^3-3x^2y+3xy^2-y^3\) 

= (\(x^3\) + \(x^3\)- 2\(x^3\)) + (3\(x^2y-3xy^2\)) + (3\(xy^2\) + 3\(xy^2\)) + (y³-y³)

= 0 + 0 + 6\(xy^2\) + 0

= 6\(xy^2\) 

Lời giải:

Gọi đa thức dư khi chia $f(x)$ cho $(x-2))(x^2+1)$ là $ax^2+bx+c$

Ta có:

$f(x)=(x-2)(x^2+1)Q(x)+ax^2+bx+c$

$f(2) = 4a+2b+c=7(1)$

$f(x) = (x-2)(x^2+1)Q(x)+a(x^2+1)+bx+(c-a)$

$=(x^2+1)[(x-2)Q(x)+a]+bx+(c-a)$

$\Rightarrow bx+(c-a)=3x+5$

$\Rightarrow b=3; c-a=5(2)$
Từ $(1); (2)\Rightarrow a=\frac{-4}{5}; b=3; c=\frac{21}{5}$

Vậy đa thức dư là $\frac{-4}{5}x^2+3x+\frac{21}{5}$

\(Đkxđ:\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)

\(M=\left(\dfrac{1}{x-2}-\dfrac{1}{x+2}\right):\dfrac{2}{x+2}\\ =\left(\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{x+2}{2}\\ =\dfrac{x+2-x+2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{2}\\ =\dfrac{4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{2}\\ =\dfrac{2}{x-2}\)

Để `M=1` Thì

\(\dfrac{2}{x-2}=1\\ \Leftrightarrow\dfrac{2}{x-2}-1=0\\ \Leftrightarrow\dfrac{2}{x-2}-\dfrac{x-2}{x-2}=0\\ \Leftrightarrow2-x+2=0\\ \Leftrightarrow4-x=0\\ \Leftrightarrow x=4\)

a) Để \(M\) xác định thì \(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\\\dfrac{2}{x+2}\ne0\end{matrix}\right.\Rightarrow x\ne\pm2\)

Khi đó: \(M=\left(\dfrac{1}{x-2}-\dfrac{1}{x+2}\right):\dfrac{2}{x+2}\)

\(=\left[\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\cdot\dfrac{x+2}{2}\)

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

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

Bạn nên gõ đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người hiểu đề và hỗ trợ tốt hơn bạn nhé.

Để biểu thức trên xác định thì: \(\begin{cases} x+2\ne0\\ x-2\ne0\\ x^2-4\ne0\\ \dfrac{6}{x+3}\ne0\\ x+3\ne0 \end{cases} \Leftrightarrow \begin{cases} x\ne\pm2\\ x\ne-3 \end{cases} \)

Khi đó: \(\left(\dfrac{1}{x+2}-\dfrac{5}{x-2}+\dfrac{4}{x^2-4}\right):\dfrac{6}{x+3}\)

\(=\left[\dfrac{x-2}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{4}{\left(x-2\right)\left(x+2\right)}\right]\cdot\dfrac{x+3}{6}\)

\(=\dfrac{x-2-5x-10+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+3}{6}\)

\(=\dfrac{\left(-4x-8\right)\left(x+3\right)}{6\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{-4\left(x+2\right)\left(x+3\right)}{6\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{-2\left(x+3\right)}{3\left(x-2\right)}=\dfrac{-2x-6}{3x-6}\)

a, Để \(A\) xác định thì: \(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\\9-x^2\ne0\\\dfrac{x-1}{x+3}\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm3\\x\ne1\end{matrix}\right.\)

Với \(x\ne\pm3;x\ne1\) ta có:

\(A=\left(\dfrac{2x}{x-3}+\dfrac{x}{x+3}+\dfrac{2x^2+3x+1}{9-x^2}\right):\dfrac{x-1}{x+3}\)

\(=\left[\dfrac{2x}{x-3}+\dfrac{x}{x+3}-\dfrac{2x^2+3x+1}{x^2-9}\right]\cdot\dfrac{x+3}{x-1}\)

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

\(=\dfrac{2x^2+6x+x^2-3x-2x^2-3x-1}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x-1}\)

\(=\dfrac{x^2-1}{x-3}\cdot\dfrac{1}{x-1}\)

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

Vậy \(A=\dfrac{x+1}{x-3}\) với \(x\ne\pm3;x\ne1\).

b, Với \(x\ne\pm3;x\ne1\):

Để \(A=3\) thì \(\dfrac{x+1}{x-3}=3\)

\(\Rightarrow x+1=3x-9\)

\(\Leftrightarrow3x-x=1+9\)

\(\Leftrightarrow2x=10\)

\(\Leftrightarrow x=5\left(tmdk\right)\)

Vây \(A=3\) khi \(x=5\).

c. Để \(A< 1\) thì \(\dfrac{x+1}{x-3}< 1\)

\(\Leftrightarrow\dfrac{x+1}{x-3}-1< 0\)

\(\Leftrightarrow\dfrac{x+1-\left(x-3\right)}{x-3}< 0\)

\(\Leftrightarrow\dfrac{4}{x-3}< 0\)

\(\Rightarrow x-3< 0\) (vì \(4>0\))

\(\Leftrightarrow x< 3\)

Kết hợp với ĐKXĐ của \(x\), ta được: \(x< 3;x\ne-3;x\ne1\)

Vậy \(A< 1\) khi \(x< 3;x\ne-3;x\ne1\).

\(Toru\)

Lời giải:
a. Các đơn thức: $\frac{4\pi r^3}{3}; \frac{p}{2\pi}; 0; \frac{1}{\sqrt{2}}$

b. Đa thức:

$\frac{4\pi r^3}{3}$ có 1 hạng tử

$\frac{p}{2\pi}$ có 1 hạng tử

$0$ có 1 hạng tử

$\frac{1}{\sqrt{2}}$ có 1 hạng tử 

$ab-\pi r^2$ có 2 hạng tử

 $x^3-x+1$ có 3 hạng tử