\(\dfrac{3}{7}\cdot\left(-\dfrac{2}{5}\right)\cdot2\dfrac{1}{2}\cdot20\cdot\dfrac{19}{72}\)

\(=\dfrac{3}{7}\cdot\left(-\dfrac{2}{5}\right)\cdot\dfrac{5}{2}\cdot20\cdot\dfrac{19}{72}\)

\(=\left(\dfrac{3}{7}\cdot\dfrac{19}{72}\right)\cdot\left(-\dfrac{2}{5}\cdot\dfrac{5}{2}\right)\cdot20\)

\(=\dfrac{19}{168}\cdot-1\cdot20\)

\(=\dfrac{19}{168}\cdot-20\)

\(=\dfrac{19\cdot-5}{42}\)

\(=\dfrac{-95}{42}\)

=-6/5

Bài 2:

a) ĐKXĐ: \(\left\{{}\begin{matrix}x+3\ne0\\x-3\ne0\\9-x^2\ne0\end{matrix}\right.\Leftrightarrow x\ne\pm3\)

b) \(A=\dfrac{3}{x+3}+\dfrac{1}{x-3}-\dfrac{18}{9-x^2}\)

\(A=\dfrac{3}{x+3}+\dfrac{1}{x-3}+\dfrac{18}{x^2-9}\)

\(A=\dfrac{3\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{x+3}{\left(x+3\right)\left(x-3\right)}+\dfrac{18}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{3x-9+x+3+18}{\left(x+3\right)\left(x-3\right)}\)

\(A=\dfrac{4x+12}{\left(x+3\right)\left(x-3\right)}\)

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

\(A=\dfrac{4}{x-3}\) 

c) Thay `x=-1` vào A ta có:

\(A=\dfrac{4}{-1-3}=\dfrac{4}{-4}=-1\)

d) `A=-4` khi: \(\dfrac{4}{x-3}=-4\)

\(\Leftrightarrow x-3=-1\)

\(\Leftrightarrow x=2\left(tm\right)\)

Bài 1:

a: ĐKXĐ: x<>3

\(\dfrac{9}{x-3}+\dfrac{3x}{3-x}\)

\(=\dfrac{9}{x-3}-\dfrac{3x}{x-3}=\dfrac{9-3x}{x-3}\)

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

b: \(\dfrac{5}{x+5}+\dfrac{-4}{x+4}\)

\(=\dfrac{5\left(x+4\right)-4\left(x+5\right)}{\left(x+5\right)\left(x+4\right)}\)

\(=\dfrac{5x+20-4x-20}{\left(x+5\right)\left(x+4\right)}=\dfrac{x}{\left(x+5\right)\left(x+4\right)}\)

c: \(\dfrac{x+5}{2x-3}-\dfrac{2x-7}{3-2x}-\dfrac{x+4}{3-2x}\)

\(=\dfrac{x+5}{2x-3}+\dfrac{2x-7}{2x-3}+\dfrac{x+4}{2x-3}\)

\(=\dfrac{x+5+2x-7+x+4}{2x-3}\)

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

d: \(\dfrac{x^2-y^2}{10x^3y}:\dfrac{x-y}{5xy}\)

\(=\dfrac{\left(x-y\right)\left(x+y\right)}{10x^3y}\cdot\dfrac{5xy}{x-y}\)

\(=\dfrac{x+y}{1}\cdot\dfrac{5xy}{10x^3y}\)

\(=\dfrac{x+y}{2x^2}\)

e: \(\dfrac{2x^2-20x+50}{3x+3}\cdot\dfrac{x^2-1}{4\left(x-5\right)^3}\)

\(=\dfrac{2\left(x^2-10x+25\right)}{3\left(x+1\right)}\cdot\dfrac{\left(x+1\right)\left(x-1\right)}{4\left(x-5\right)^3}\)

\(=\dfrac{2\left(x-5\right)^2}{4\left(x-5\right)^3}\cdot\dfrac{x-1}{3}\)

\(=\dfrac{x-1}{3\cdot2\left(x-5\right)}=\dfrac{x-1}{6x-30}\)

f: \(\dfrac{x-2}{x+1}:\dfrac{x^2-5x+6}{x^2-2x-3}\)

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

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

g: \(\dfrac{x}{x-2y}+\dfrac{x}{x+2y}+\dfrac{4xy}{4y^2-x^2}\)

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

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

\(=\dfrac{2x^2-4xy}{\left(x-2y\right)\left(x+2y\right)}=\dfrac{2x\left(x-2y\right)}{\left(x-2y\right)\left(x+2y\right)}=\dfrac{2x}{x+2y}\)

h: \(\dfrac{1}{x-y}+\dfrac{3xy}{y^3-x^3}+\dfrac{x-y}{x^2+xy+y^2}\)

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

\(=\dfrac{x^2+xy+y^2-3xy+\left(x-y\right)^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)

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

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

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

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

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

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

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

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

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

Bài 5:

a: Xét ΔABC vuông tại A và ΔHBA vuông tại H có

\(\widehat{ABC}\) chung

Do đó: ΔABC~ΔHBA

=>\(\dfrac{AC}{HA}=\dfrac{BC}{BA}\)

=>\(AC\cdot AB=AH\cdot BC\)

b: ΔABC vuông tại A

=>\(AB^2+AC^2=BC^2\)

=>\(AC^2=7,5^2-4,5^2=36=6^2\)

=>AC=6(cm)

=>\(AH=\dfrac{4.5\cdot6}{7,5}=\dfrac{27}{7,5}=3,6\left(cm\right)\)

ΔAHB vuông tại H

=>\(AH^2+HB^2=AB^2\)

=>\(HB^2=4,5^2-3,6^2=2,7^2\)

=>HB=2,7(cm)

HB+HC=BC

=>HC+2,7=7,5

=>HC=4,8(cm)

c: Xét ΔBAH có BK là phân giác

nên \(\dfrac{KH}{KA}=\dfrac{BH}{BA}\left(1\right)\)

Xét ΔBAC có BD là phân giác

nên \(\dfrac{AD}{DC}=\dfrac{BA}{BC}\left(2\right)\)

Ta có: ΔBAH~ΔBCA

=>\(\dfrac{BH}{BA}=\dfrac{BA}{BC}\left(3\right)\)

Từ (1),(2),(3) suy ra \(\dfrac{AD}{DC}=\dfrac{HK}{KA}\)

a) Từ 100 đến 1000 có 10 số tròn trăm

b) Từ 100 đến 200 có 11 số tròn chục

c) ??? đề chưa rõ 

ai làm được mik tick cho nha

a) Đổi: 1,2m = 12dm; 0,8m = 8dm 

Diện tích kính làm bể là:

\(\left(12+8\right)\times2\times6+12\times8=336\left(dm^2\right)\)

b) Đổi: 40cm = 4dm 

Thể tích nước đang có trong bể là:

\(4\times12\times8=384\left(dm^3\right)\) 

c) Thể tích của bể là:

\(6\times12\times8=576\left(dm^3\right)=576\left(l\right)\)

Đổi: \(384\left(dm^3\right)=384\left(l\right)\)

Cần đổ thêm số lít nước là:
\(576-384=192\left(l\right)\) 

d) Đổi: 55cm = 5,5dm 

Thể tích cả nước và đá trong bể là:

\(5,5\times12\times8=528\left(dm^3\right)\)

Thể tích của hòn đá là:

\(528-384=144\left(dm^3\right)\)

ĐS: ... 

Đặt: \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\)

\(2A=2\cdot\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\right)\)

\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2004}}\)

\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2004}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\right)\)

\(A=1-\dfrac{1}{2^{2005}}\) 

\(\sqrt[2]{5+6}=\sqrt{11}\)

bằng: 3.31662479

\(B=x\left(y^2-1\right)+y\left(x^2-1\right)\)

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

\(=xy\left(x+y\right)-\left(x+y\right)\)

=5xy-5