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

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

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

h

Bài 6:

a:

ĐKXĐ: x∉{0;-2;1}

\(\frac{1-x^2}{x^2+2x}:\frac{2-2x}{3x}\)

\(=\frac{\left(1-x\right)\left(1+x\right)}{x\left(x+2\right)}\cdot\frac{3x}{2\left(1-x\right)}\)

\(=\frac{3\left(1+x\right)}{2\left(x+2\right)}=\frac{3x+3}{2x+4}\)

b: ĐKXĐ: x<>1

\(\frac{x^3+1}{x-1}:\left(x^2-x+1\right)\)

\(=\frac{\left(x+1\right)\left(x^2-x+1\right)}{\left(x-1\right)\left(x^2-x+1\right)}\)

\(=\frac{x+1}{x-1}\)

c: ĐKXĐ: x∉{-1;-2;0;2}

Ta có: \(\frac{x^2-x-2}{x^2+3x+2}:\frac{x^2-4x+4}{x^2+2x}\)

\(=\frac{\left(x-2\right)\left(x+1\right)}{\left(x+1\right)\left(x+2\right)}\cdot\frac{x\left(x+2\right)}{\left(x-2\right)^2}\)

\(=\frac{x}{x-2}\)

d: ĐKXĐ: x<>2y;x<>-y

Ta có: \(\frac{x-2y}{x^2-xy+y^2}:\frac{x^2-4xy+4y^2}{x^3+y^3}\)

\(=\frac{x-2y}{x^2-xy+y^2}\cdot\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{\left(x-2y\right)^2}=\frac{x+y}{x-2y}\)

Bài 5:

a: \(\frac{x^2+3x}{x-4}:P=\frac{x^2-9}{x^2-4x}\)

=>\(P=\frac{x^2+3x}{x-4}:\frac{x^2-9}{x^2-4x}\)

=>\(P=\frac{x\left(x+3\right)}{x-4}\cdot\frac{x\left(x-4\right)}{\left(x-3\right)\left(x+3\right)}=\frac{x^2}{x-3}\)

b: \(Q:\frac{x-2}{2x+3}=\frac{4x^2+12x+9}{x^2-4}\)

=>\(Q:\frac{x-2}{2x+3}=\frac{\left(2x+3\right)^2}{\left(x-2\right)\left(x+2\right)}\)

=>\(Q=\frac{\left(2x+3\right)^2}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x-2}{2x+3}=\frac{2x+3}{x+2}\)

Bài 4:

a: \(\frac{x+4}{x+5}:\frac{x+5}{x+6}:\frac{x+6}{x+4}\)

\(=\frac{x+4}{x+5}\cdot\frac{x+6}{x+5}\cdot\frac{x+4}{x+6}=\frac{\left(x+4\right)^2}{\left(x+5\right)^2}\)

b: \(\frac{x-7}{x+8}:\left(\frac{x-7}{x-9}:\frac{x+8}{x-9}\right)\)

\(=\frac{x-7}{x+8}:\left(\frac{x-7}{x-9}\cdot\frac{x-9}{x+8}\right)\)

\(=\frac{x-7}{x+8}:\frac{x-7}{x+8}\)

=1

Bài 3:

a: ĐKXĐ: x∉{3;-3}

\(\left(x^2-9\right):\frac{2x+6}{x-3}\)

\(=\left(x-3\right)\left(x+3\right)\cdot\frac{x-3}{2\left(x+3\right)}=\frac{\left(x-3\right)^2}{2}\)

b: ĐKXĐ: x<>3/2; x<>0; y<>0

\(\frac{xy}{2x-3}:\frac{x^2y^2}{6-4x}\)

\(=\frac{xy}{2x-3}\cdot\frac{-2\left(2x-3\right)}{x^2y^2}=\frac{-2}{xy}\)

c: ĐKXĐ: x∉{0;1;-2;2}

\(\frac{x^2+2x}{x^2-2x+1}:\frac{x^2-4}{x^2-x}\)

\(=\frac{x\left(x+2\right)}{\left(x-1\right)^2}\cdot\frac{x\left(x-1\right)}{\left(x-2\right)\left(x+2\right)}\)

\(=\frac{x^2}{\left(x-2\right)\left(x-1\right)}\)

d: ĐKXĐ: x<>-3y/2; x<>2

\(\frac{2x+3y}{2-x}:\frac{4x^2+12xy+9y^2}{x^3-8}\)

\(=\frac{2x+3y}{2-x}\cdot\frac{\left(x-2\right)\left(x^2+2x+4\right)}{\left(2x+3y\right)^2}\)

\(=\frac{-\left(x^2+2x+4\right)}{2x+3y}\)

pt đã cho \(\Leftrightarrow\dfrac{2x-50}{50}-1+\dfrac{2x-51}{49}-1+\dfrac{2x-52}{48}-1+\dfrac{2x-53}{47}-1+\dfrac{2x-200}{25}+4=0\)

\(\Leftrightarrow\dfrac{2x-50-50}{50}+\dfrac{2x-51-49}{49}+\dfrac{2x-52-48}{48}+\dfrac{2x-53-47}{47}+\dfrac{2x-200+100}{25}=0\)

\(\Leftrightarrow\dfrac{2x-100}{50}+\dfrac{2x-100}{49}+\dfrac{2x-100}{48}+\dfrac{2x-100}{47}+\dfrac{2x-100}{25}=0\)

\(\Leftrightarrow\left(2x-100\right)\left(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}\right)=0\)

\(\Leftrightarrow2x-100=0\) (vì \(\dfrac{1}{50}+\dfrac{1}{49}+\dfrac{1}{48}+\dfrac{1}{47}+\dfrac{1}{25}>0\))

\(\Leftrightarrow x=50\)

Vậy pt đã cho có tập nghiệm \(S=\left\{50\right\}\)

x= 1phần 2

Do AB // DE (gt)

Theo hệ quả của định lý Thalès, ta có:

AB/DE = BC/CD

x = BC = AB.CD : DE

x = BC = 5.7,2 : 15 = 2,4

Do AB // DE (gt)

Theo hệ quả của định lý Thalès, ta có:

AB/DE = AC/CE

y = CE = AC.DE : AB

= 3.15 : 7,2

= 6,25

x= 2,5.y= 7,2