-9/25 . 17 và 2/3 - -3/5 ^2 . 22/3
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\(\dfrac{3}{7}\cdot\left(-\dfrac{2}{5}\right)\cdot2\dfrac{1}{2}\cdot20\cdot\dfrac{19}{72}\)
\(=\dfrac{3}{7}\cdot\left(-\dfrac{2}{5}\right)\cdot\dfrac{5}{2}\cdot20\cdot\dfrac{19}{72}\)
\(=\left(\dfrac{3}{7}\cdot\dfrac{19}{72}\right)\cdot\left(-\dfrac{2}{5}\cdot\dfrac{5}{2}\right)\cdot20\)
\(=\dfrac{19}{168}\cdot-1\cdot20\)
\(=\dfrac{19}{168}\cdot-20\)
\(=\dfrac{19\cdot-5}{42}\)
\(=\dfrac{-95}{42}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(1+\dfrac{x^3-x}{x^2+1}\cdot\left(\dfrac{1}{1-x}-\dfrac{1}{1-x^2}\right)\)
\(=1+\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\left(\dfrac{-1}{x-1}+\dfrac{1}{\left(x-1\right)\left(x+1\right)}\right)\)
\(=1+\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\dfrac{-x-1+1}{\left(x-1\right)\left(x+1\right)}\)
\(=1+\dfrac{x\left(x-1\right)\left(x+1\right)}{x^2+1}\cdot\dfrac{-x}{\left(x-1\right)\left(x+1\right)}\)
\(=1-\dfrac{x^2}{x^2+1}=\dfrac{1}{x^2+1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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 \(\dfrac{AD}{DC}=\dfrac{HK}{KA}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Từ 100 đến 1000 có 10 số tròn trăm
b) Từ 100 đến 200 có 11 số tròn chục
c) ??? đề chưa rõ
![](https://rs.olm.vn/images/avt/0.png?1311)
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: ...
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt: \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\)
\(2A=2\cdot\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\right)\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2004}}\)
\(2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2004}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2005}}\right)\)
\(A=1-\dfrac{1}{2^{2005}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(-\dfrac{9}{25}\cdot17\dfrac{2}{3}-\left(-\dfrac{3}{5}\right)^2\cdot\dfrac{22}{3}\)
\(=-\dfrac{9}{25}\cdot\dfrac{53}{3}-\dfrac{9}{25}\cdot\dfrac{22}{3}\)
\(=-\dfrac{9}{25}\left(\dfrac{53}{3}+\dfrac{22}{3}\right)=-\dfrac{9}{25}\cdot25=-9\)