a: -3/100=-9/300; -2/3=-200/300

=>-3/100>-2/3

b: -3/5=-9/15

-2/3=-10/15

=>-3/5>-2/3

c: -5/4<-1<-3/8

d: -2/3=-8/12; -3/4=-9/12

=>-2/3>-3/4

e: -267/268>-1

-1>-1347/1343

=>-267/268>-1347/1343

a) Ta thấy: \(\frac{-1}{5}< 0\), \(\frac{1}{1000}>0\)

\(\Rightarrow\frac{-1}{5}< \frac{1}{1000}\)

b) Ta có: \(\frac{267}{-268}=\frac{-267}{268}>-1\)

               \(\frac{-1347}{1343}< -1\)

\(\Rightarrow\frac{-1347}{1343}< \frac{-267}{268}\)

(d) qua A(5; 6) : y = mx - 5m + 6 (1) 
(C) : (x - 1)² + (y - 2)² = 1 (2) 
Thay y từ (1) vào (2) ta có phương trình hoành độ giao điểm của (d) và (C) 
(x - 1)² + (mx - 5m + 4)² = 1 
Khai triển ra pt bậc 2 : (m² + 1)x² - 2(5m² - 4m + 1)x + 25m² - 40m + 17 = 0 (*) 
Để (d) tiếp xúc (C) thì (*) phải có nghiệm kép 
∆' = (5m² - 4m + 1)² - (m² + 1)(25m² - 40m + 17) = - 4(3m² - 8m + 4) = 4(m - 2)(2 - 3m) = 0 => m = 3/2; m = 2 
KL : Có 2 đường thẳng cần tìm 
(d1) : y = (3/2)(x - 1) 
(d2) : y = 2x - 4 

∆ ∠ ∡ √ ∛ ∜ x² ⁻¹ ∫ π × ∵ ∴ | | , ⊥,∈∝ ≤ ≥− ± , ÷ ° ≠ → ∞, ≡ , ≅ , ∑,∪,¼ , ½ , ¾ , ≈ , [-b ± √(b² - 4ac) ] / 2a Σ Φ Ω α β γ δ ε η θ λ μ π ρ σ τ φ ω ё й½ ⅓ ⅔ ¼ ⁰ ¹ ² ³ ⁴ ⁵ ⁶ ⁷ ⁸ ⁹ ⁺ ⁻ ⁼ ⁽ ⁾ ⁿ ₁ ₂ ₃₄₅ ₆ ₇ ₈ ₉ ₊ ₋ ₌ ₍ ₎ ∊ ∧ ∏ ∑ ∠ ,∫ ∫ ψ ω Π∮ ∯ ∰ ∇ ∂ • ⇒ ♠ ★

Đặng Trần Anh Thư

a, 

- Ta có : 

  \(\frac{-1}{5}< 1\) ( số âm)

\(\frac{1}{1000}>1\) ( số dương ) 

Dễ thấy \(\frac{-1}{5}< \frac{1}{1000}\)

b) \(\frac{-13}{38}< 1< \frac{29}{-88}\)

\(\frac{-13}{38}< \frac{29}{-88}\)

c)\(\frac{267}{-268}>1>\frac{-1347}{1343}\)

\(\frac{267}{-268}>\frac{-1343}{1343}\)

Bài này chỉ dùng cách so sánh với 1 là nhanh nhất 

\(b,-\frac{.13}{38}\)và \(\frac{19}{-88}\)

\(-\frac{13}{38}< 1< \frac{29}{-88}\)

\(\Rightarrow-\frac{13}{38}< \frac{29}{-88}\)

\(c,\frac{267}{-268}\)và \(-\frac{1347}{1343}\)

\(\frac{267}{-268}>1>-\frac{1347}{1343}\)

\(\frac{\Rightarrow267}{-268}>-\frac{1343}{1343}\)

2. Tìm x

a) \(5-\left|x+\dfrac{1}{2}\right|=1\)

\(\left|x+\dfrac{1}{2}\right|=5-1\)

\(\left|x+\dfrac{1}{2}\right|=4\)

\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}=4\\x+\dfrac{1}{2}=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{2}\\x=\dfrac{-9}{2}\end{matrix}\right.\)

Vậy x=7/2 hoặc -9/2

b) \(\dfrac{4}{3}+\left|2-\dfrac{1}{2}x\right|=7\)

\(\left|2-\dfrac{1}{2}x\right|=7-\dfrac{4}{3}\)

\(\left|2-\dfrac{1}{2}x\right|=\dfrac{17}{3}\)

\(\Rightarrow\left\{{}\begin{matrix}2-\dfrac{1}{2}x=\dfrac{17}{3}\\2-\dfrac{1}{2}x=\dfrac{-17}{3}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{2}x=\dfrac{-11}{3}\\\dfrac{1}{2}x=\dfrac{23}{3}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-22}{3}\\x=\dfrac{46}{3}\end{matrix}\right.\)

Vậy x=-22/3 hoặc x=46/4

1. So sánh:

a. \(\dfrac{-1}{5}\) và \(\dfrac{1}{1000}\)

Ta có:

\(\dfrac{-1}{5}< 0\\ \dfrac{1}{1000}>0\\ \Rightarrow\dfrac{-1}{5}< \dfrac{1}{1000}\)

b.\(\dfrac{267}{-268}\) và \(\dfrac{-1347}{1343}\)

Ta có:

\(\dfrac{267}{-268}>-1\\ \dfrac{-1347}{1343}< -1\\ \Rightarrow\dfrac{267}{-286}>\dfrac{-1347}{1343}\)

c.\(\dfrac{-13}{38}\) và \(\dfrac{29}{-88}\)

Ta có:

\(\dfrac{-13}{38}< \dfrac{-13}{39}=\dfrac{-1}{3}\\ \dfrac{-29}{88}>\dfrac{-29}{87}=\dfrac{-1}{3}\\ \Rightarrow\dfrac{-13}{38}< \dfrac{-29}{88}\)

1, So sánh

\(a,-\dfrac{1}{5}\) và \(\dfrac{1}{1000}\)

Cách 1: \(-\dfrac{1}{5}=-\dfrac{1.200}{5.200}=-\dfrac{200}{1000}\)

Ta có: \(-\dfrac{200}{1000}< \dfrac{1}{1000}\Rightarrow-\dfrac{1}{5}< \dfrac{1}{1000}\)

Cách 2: Ta có:

\(-\dfrac{1}{5}\) là số âm, \(\dfrac{1}{1000}\) là số dương

âm < dương \(\Rightarrow-\dfrac{1}{5}< \dfrac{1}{1000}\)

\(b,\dfrac{267}{-268}\) và \(\dfrac{-1347}{1343}\)

Ta có: \(\dfrac{267}{-268}< -1\left(1\right)\), \(\dfrac{-1347}{1343}>-1\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\)\(\Rightarrow\dfrac{267}{-268}>\dfrac{-1347}{1343}\)

\(c,\dfrac{-13}{38}\) và \(\dfrac{29}{-88}\)

Ta có: \(\dfrac{-13}{38}=-\dfrac{13.44}{38.44}=-\dfrac{572}{1672}\)

\(\dfrac{29}{-88}=-\dfrac{29.19}{88.19}=-\dfrac{551}{1672}\)

\(\Rightarrow-\dfrac{572}{1672}< -\dfrac{551}{1672}\) hay \(-\dfrac{13}{38}< \dfrac{29}{-88}\)

Bài 2: Tìm x:

\(a,5-\left|x+\dfrac{1}{2}\right|=1\)

\(\Rightarrow\left|x+\dfrac{1}{2}\right|=4\)

Ta có \(\left\{{}\begin{matrix}x+\dfrac{1}{2}=4\\x+\dfrac{1}{2}=-4\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{9}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{-\dfrac{9}{2};\dfrac{7}{2}\right\}\)

\(b,\dfrac{4}{3}+\left|2-\dfrac{1}{2}x\right|=7\)

\(\Rightarrow\left|2-\dfrac{1}{2}x\right|=7-\dfrac{4}{3}\)

\(\left|2-\dfrac{1}{2}x\right|=\dfrac{17}{3}\)

Ta có: \(\left\{{}\begin{matrix}2-\dfrac{1}{2}x=\dfrac{17}{3}\\2-\dfrac{1}{2}x=-\dfrac{17}{3}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=\left(2-\dfrac{17}{3}\right):\dfrac{1}{2}=-\dfrac{22}{3}\\x=\left[2-\left(-\dfrac{17}{3}\right)\right]:\dfrac{1}{2}=\dfrac{46}{3}\end{matrix}\right.\)

Vậy \(x\in\left\{-\dfrac{22}{3};\dfrac{46}{3}\right\}\)