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Bài 1: Tìm \( x \)
\[
x - \frac{25\%}{100}x = \frac{1}{2}
\]
Để giải phương trình này, trước hết chúng ta phải chuyển đổi phần trăm thành dạng thập phân:
\[
\frac{25\%}{100} = 0.25
\]
Phương trình ban đầu trở thành:
\[
x - 0.25x = \frac{1}{2}
\]
Tổng hợp các hạng tử giống nhau:
\[
1x - 0.25x = \frac{1}{2}
\]
\[
0.75x = \frac{1}{2}
\]
Giải phương trình ta được:
\[
x = \frac{\frac{1}{2}}{0.75} = \frac{2}{3}
\]
Vậy, \( x = \frac{2}{3} \)
Bài 2: Tính hợp lý
a) \[
\frac{5}{-4} + \frac{3}{4} + \frac{4}{-5} + \frac{14}{5} - \frac{7}{3}
\]
Chúng ta cần tìm một mẫu số chung cho tất cả các phân số. Mẫu số chung nhỏ nhất là 60.
\[
= \frac{75}{-60} + \frac{45}{60} + \frac{-48}{60} + \frac{168}{60} - \frac{140}{60}
\]
\[
= \frac{75 + 45 - 48 + 168 - 140}{60}
\]
\[
= \frac{100}{60} = \frac{5}{3}
\]
b) \[
\frac{8}{3} \times \frac{2}{5} \times \frac{3}{10} \times \frac{10}{92} \times \frac{19}{92}
\]
Tích của các phân số là:
\[
= \frac{8 \times 2 \times 3 \times 10 \times 19}{3 \times 5 \times 10 \times 92 \times 92}
\]
\[
= \frac{9120}{4131600} = \frac{57}{25825}
\]
c) \[
\frac{5}{7} \times \frac{2}{11} + \frac{5}{7} \times \frac{9}{14} + \frac{1}{5}
\]
Tích của các phân số là:
\[
= \frac{10}{77} + \frac{45}{98} + \frac{1}{5}
\]
\[
= \frac{980}{7546} + \frac{3485}{7546} + \frac{15092}{75460}
\]
\[
= \frac{2507}{7546}
\]
Bài 2:
\(M=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{2008.2009}\)
\(\Rightarrow M=\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{2009-2008}{2008.2009}\)
\(\Rightarrow M=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{2008}-\frac{1}{2009}=\frac{1}{2}-\frac{1}{2009}\)
Bài 1:
Ta có: \(\frac{x}{2}+\frac{x}{3}=x\left(\frac{1}{2}+\frac{1}{3}\right)=\frac{5}{6}x=\frac{1}{4}\Rightarrow x=\frac{3}{10}\)
\(\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}\right):x=\frac{3}{1.2}+\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{15.16}\)
\(\left(\frac{8}{16}+\frac{4}{16}+\frac{2}{16}+\frac{1}{16}\right).\frac{1}{x}=3.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{15.16}\right)\)
\(\frac{8+4+2+1}{16}.\frac{1}{x}=3.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{15}-\frac{1}{16}\right)\)
\(\frac{15}{16}.\frac{1}{x}=3.\left(1-\frac{1}{16}\right)\)
\(\frac{15}{16}.\frac{1}{x}=3.\frac{15}{16}\)
=> \(\frac{1}{x}=3\)
=> \(x=\frac{1}{3}\)
a)Xét 1/2-1/3-1/6=3/6-2/6-1/6=0
=> (1+2+3+...+2018).(3/1.2+3/2.3+...+3/2018.2019).(1/2-1/3-1/6)=(1+2+3+...+2018).(3/1.2+3/2.3+...+3/2018.2019).0=0
b) 4A=1.2.3.4+2.3.4.4+..+x(x+1)(x+2)4
=1.2.3.4+2.3.4.5-1.2.3.4+...+x(x+1)(x+2)(x+3)-x(x+1)(x+2)(x-1)
= (x-1)x(x+1)(x+2)
=> A=x(x+1)(x+2)(x-1)/4
a) \(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(=1-\frac{1}{100}< 1\)
\(\Rightarrow A< 1\)
b) \(B=\frac{1}{3}+\left(\frac{1}{3}\right)^2+...+\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow3B=1+\frac{1}{3}+...+\left(\frac{1}{3}\right)^{99}\)
\(\Rightarrow3B-B=1-\left(\frac{1}{3}\right)^{100}\)
\(\Rightarrow2B=1-\left(\frac{1}{3}\right)^{100}< 1\)
\(\Rightarrow2B< 1\)
\(\Rightarrow B< \frac{1}{2}\)
\(\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\right)\cdot100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ \left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\right)\cdot100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ \left(1-\frac{1}{10}\right)\cdot100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ \frac{9}{10}\cdot100-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ 90-\left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=89\\ \left[\frac{5}{2}:\left(X+\frac{206}{100}\right)\right]:\frac{1}{2}=1\\ \frac{5}{2}:\left(X+\frac{206}{100}\right)=\frac{1}{2}\\ X+\frac{206}{100}=5\\ X=\frac{500}{100}-\frac{206}{100}\\ X=\frac{294}{100}=\frac{147}{50}\)
Vậy \(X=\frac{147}{50}\)
( 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ......+ 1/9 - 1/10) . 100 - [ 5/2 : ( x + 103/50 ) ] = 89 . 1/2
( 1 - 1/10) . 100 - [ 5/2 : ( x + 103/50 ) ] = 89/2
90 - 5/2 : ( x + 103/50 ) = 89/2
5/2 : ( x + 103/50 ) = 90 - 89/2
5/2 : ( x + 103/50 ) = 91/2
x + 103/50 = 5/2 : 91/2
x + 103/50 = 5/91
x = 5/91 - 103/50
x = -9,123/4550
b) \(\left(\sqrt{2x+3}-3\right)+\left(\sqrt{x+1}-2\right)+5=3x+2\left(\sqrt{2x^2+5x+3}-6\right)+12-16\)
\(\Leftrightarrow\left(\sqrt{2x+3}-3\right)+\left(\sqrt{x+1}-2\right)=3\left(x-3\right)+2\left(\sqrt{2x^2+5x+3}-6\right)\)
\(\Leftrightarrow\frac{2\left(x-3\right)}{\sqrt{2x+3}+3}+\frac{x-3}{\sqrt{x+1}+2}-3\left(x-3\right)-\frac{2\left(x-3\right)\left(2x+11\right)}{\sqrt{2x^2+5x+3}+6}=0\Leftrightarrow x-3=0\Leftrightarrow x=3.\)
=>\(-B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2012}\right)\)
=\(\frac{1}{2}.\frac{2}{3}...\frac{2011}{2012}=\frac{1}{2012}\)
mong là trước ngày mai
`Answer:`
Bài 1:
a. \(\frac{1}{2}-\left(\frac{2}{3}x-\frac{1}{3}\right)=\frac{2}{3}\)
\(\Leftrightarrow\frac{1}{2}-\frac{2}{3}x+\frac{1}{3}=\frac{2}{3}\)
\(\Leftrightarrow\frac{5}{6}-\frac{2}{3}x=\frac{2}{3}\)
\(\Leftrightarrow-\frac{2}{3}=\frac{2}{3}-\frac{5}{6}\)
\(\Leftrightarrow-\frac{2}{3}x=-\frac{1}{6}\)
\(\Leftrightarrow x=-\frac{1}{6}:-\frac{2}{3}\)
\(\Leftrightarrow x=\frac{1}{4}\)
b. \(\frac{3}{x+5}=15\%\left(ĐKXĐ:x\ne-5\right)\)
\(\Leftrightarrow\frac{3}{x+5}=\frac{3}{20}\)
\(\Leftrightarrow\frac{60}{20\left(x+5\right)}=\frac{3\left(x+5\right)}{20\left(x+5\right)}\)
\(\Leftrightarrow60x=3x+15\)
\(\Leftrightarrow-3x=-45\)
\(\Leftrightarrow x=15\)
Bài 2:
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=1-\frac{1}{50}\)
\(=\frac{49}{50}\)