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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 \(S=\left\{50\right\}\)
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+1) . 5 = (2x + 5) . 3
5x + 5 = 6x + 15
5x + 6x = 15-5
11x =10
x = 11\10
câu a
\(\left(\dfrac{2x}{3x+1}-1\right):\left(1-\dfrac{8x^2}{9x^2-1}\right)\\ =\left(\dfrac{2x}{3x+1}-\dfrac{3x+1}{3x+1}\right):\left(\dfrac{9x^2-1}{9x^2-1}-\dfrac{8x^2}{9x^2-1}\right)\\ =\left(\dfrac{2x}{3x+1}-\dfrac{3x+1}{3x+1}\right):\left(\dfrac{9x^2-1}{\left(3x-1\right)\left(3x+1\right)}-\dfrac{8x^2}{\left(3x-1\right)\left(3x+1\right)}\right)\\ =\left(\dfrac{2x-3x-1}{3x+1}\right):\left(\dfrac{9x^2-1-8x^2}{\left(3x-1\right)\left(3x+1\right)}\right)\)
\(=\left(\dfrac{-x-1}{3x+1}\right):\left(\dfrac{x^2-1}{\left(3x-1\right)\left(3x+1\right)}\right)\\ =\dfrac{-x-1}{3x+1}\cdot\dfrac{\left(3x-1\right)\left(3x+1\right)}{x^2-1}\)
\(=\dfrac{-\left(x+1\right)\cdot\left(3x-1\right)\cdot\left(3x+1\right)}{\left(3x+1\right)\cdot\left(x-1\right)\cdot\left(x+1\right)}\\ =\dfrac{-3x+1}{x-1}\)
câu b
thay \(x=2\) vào P ta được
\(\dfrac{-3\cdot2+1}{2-1}=\dfrac{-6+1}{1}=-5\)
vậy \(P=5\) khi \(x=2\)
câu a)
\(\dfrac{2y-1}{y}-\dfrac{2x+1}{x}\\ =\dfrac{2xy-x}{xy}-\dfrac{2xy+y}{xy}\\ =\dfrac{2xy-x-2xy-y}{xy}\\ =\dfrac{-x-y}{xy}\)
câu b)
\(\dfrac{2x}{3}:\dfrac{5}{6x^2}\\ =\dfrac{2x}{3}\cdot\dfrac{6x^2}{5}\\ =\dfrac{2x\cdot6x^2}{3\cdot5}\\ =\dfrac{12x^3}{15}=\dfrac{4x^3}{5}\)
\(A=4x^2+16x-9=4\left(x^2+4x+4\right)-25=4\left(x+2\right)^2-25\ge-25\)
\(A_{min}=-25\) khi \(x=-2\)
\(B=-5x^2-29x-20=-5\left(x^2+\dfrac{29}{5}x+\dfrac{841}{100}\right)+\dfrac{441}{20}\)
\(B=-5\left(x+\dfrac{29}{10}\right)^2+\dfrac{441}{20}\le\dfrac{441}{20}\)
\(B_{max}=\dfrac{441}{20}\) khi \(x=-\dfrac{29}{10}\)
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