(1- 1/2 )x(1-1/3)x(1-1/4)x(1-1/5)x......x(1-1/10) =
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Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\dfrac{1}{2}+x=\dfrac{5}{6}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{1}{2}=\dfrac{5}{6}-\dfrac{3}{6}=\dfrac{2}{6}=\dfrac{1}{3}\)
b) \(x+\dfrac{1}{4}=\dfrac{3}{4}\)
\(\Rightarrow x=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{2}{4}=\dfrac{1}{2}\)
c) \(x-\dfrac{1}{5}=\dfrac{3}{10}\)
\(\Rightarrow x=\dfrac{3}{10}+\dfrac{1}{5}=\dfrac{3}{10}+\dfrac{2}{10}=\dfrac{5}{10}=\dfrac{1}{2}\)
d) \(\dfrac{5}{6}-x=\dfrac{1}{3}\)
\(\Rightarrow x=\dfrac{5}{6}-\dfrac{1}{3}=\dfrac{5}{6}-\dfrac{2}{6}=\dfrac{3}{6}=\dfrac{1}{2}\)
e) \(\dfrac{3}{10}+x=\dfrac{1}{2}\)
\(\Rightarrow x=\dfrac{1}{2}-\dfrac{3}{10}=\dfrac{5}{10}-\dfrac{3}{10}=\dfrac{2}{10}=\dfrac{1}{5}\)
g) \(x+\dfrac{1}{4}=\dfrac{3}{8}\)
\(\Rightarrow x=\dfrac{3}{8}-\dfrac{1}{4}=\dfrac{3}{8}-\dfrac{2}{8}=\dfrac{1}{8}\)
a) 12+x=5612+x=56
⇒x=56−12=56−36=26=13⇒x=56−12=56−36=26=13
b) x+14=34x+14=34
⇒x=34−14=24=12⇒x=34−14=24=12
c) x−15=310x−15=310
⇒x=310+15=310+210=510=12⇒x=310+15=310+210=510=12
d) 56−x=1356−x=13
⇒x=56−13=56−26=36=12⇒x=56−13=56−26=36=12
e) 310+x=12310+x=12
⇒x=12−310=510−310=210=15⇒x=12−310=510−310=210=15
g) x+14=38x+14=38
⇒x=38−14=38−28=18⇒x=38−14=38−28=18
Đọc tiếp
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{1}{\left(x-1\right)\left(x-2\right)}+\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-2}+\dfrac{1}{x-2}-\dfrac{1}{x-3}+\dfrac{1}{x-3}-....+\dfrac{1}{x-5}-\dfrac{1}{x-6}=\dfrac{1}{10}\)
\(\Leftrightarrow\dfrac{1}{x-1}-\dfrac{1}{x-6}=\dfrac{1}{10}\Leftrightarrow\dfrac{x-6-x+1}{\left(x-1\right)\left(x-6\right)}=\dfrac{1}{10}\)
\(\Leftrightarrow x^2-7x+56=0\Leftrightarrow x^2-2.\dfrac{7}{2}x+\dfrac{49}{4}+\dfrac{175}{4}=\left(x-\dfrac{7}{2}\right)^2+\dfrac{175}{4}>0\)
Vậy phương trình vô nghiệm
![](https://rs.olm.vn/images/avt/0.png?1311)
`@` ` \text {Ans}`
`\downarrow`
`a,`
`1/4+3/4*x=3/2-x`
`=> 1/4 + 3/4x - 3/2 + x = 0`
`=> (1/4 - 3/2) + (3/4x + x) = 0`
`=> -5/4 + 7/4x = 0`
`=> 7/4x = 5/4`
`=> x = 5/4 \div 7/4`
`=> x = 5/7`
Vậy, `x=5/7`
`b,`
`3/5*x-1/4=1/10*x-1/2`
`=> 3/5x - 1/4 - 1/10x + 1/2 = 0`
`=> (3/5x - 1/10x) + (-1/4 + 1/2)=0`
`=> 1/2x + 1/4 = 0`
`=> 1/2x = -1/4`
`=> x = -1/4 \div 1/2`
`=> x = -1/2`
Vậy, `x=-1/2`
`c,`
`3x-3/5=x-1/4`
`=> 3x - 3/5 - x + 1/4 = 0`
`=> (3x - x) - (3/5 - 1/4) = 0`
`=> 2x - 7/20 = 0`
`=> 2x = 0,35`
`=> x = 0,35 \div 2`
`=> x = 7/40`
Vậy, `x=7/40`
`d,`
`3/2*x-2/5=1/3*x-1/4`
`=> 3/2x - 2/5 - 1/3x + 1/4 = 0`
`=> (3/2x - 1/3x) - (2/5 - 1/4) = 0`
`=> 7/6x - 3/20 = 0`
`=> 7/6x = 3/20`
`=> x = 3/20 \div 7/6`
`=> x = 9/70`
Vậy, `x=9/70`
`@` `\text {Kaizuu lv uuu}`
![](https://rs.olm.vn/images/avt/0.png?1311)
1) 1/3 x 1/2 x 3/7 = 1/6 x 3/7 = 1/14
2) 5/4 x 1/3 + 1/7 = 5/12 + 1/7 = 47/84
3) 8 x (8/9 - 2/3) = 8 x 2/9 = 16/9
4) 5/6 x 48/20 x 1/2 = 2 x 1/2 = 1
5) (2/5 + 3/4) x 8 = 23/20 x 8 = 46/5
6) 10 x (1/2 - 1/5) = 10 x 3/10 = 3
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{1}{n\left(n+1\right)}=\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n}-\frac{1}{n+1}\)
Lại Có: ĐKXĐ: x≠1,x≠2,x≠3,x≠4,x≠5,x≠6
\(\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-6\right)}=\frac{1}{10}\)<=>\(\frac{1}{\left(x-6\right)\left(x-5\right)}+\frac{1}{\left(x-5\right)\left(x-4\right)}+\frac{1}{\left(x-4\right)\left(x-3\right)}+\frac{1}{\left(x-3\right)\left(x-2\right)}+\frac{1}{\left(x-2\right)\left(x-1\right)}=\frac{1}{10}\)
<=>\(\frac{1}{x-6}-\frac{1}{x-5}+\frac{1}{x-5}-\frac{1}{x-4}+\frac{1}{x-4}-\frac{1}{x-3}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-2}-\frac{1}{x-1}=\frac{1}{10}\)
<=> \(\frac{1}{x-6}-\frac{1}{x-1}=\frac{1}{10}\)
<=> \(\frac{x-1-x+6}{\left(x-6\right)\left(x-1\right)}=\frac{1}{10}\)
<=> \(\frac{5}{\left(x-6\right)\left(x-1\right)}=\frac{1}{10}\)
<=>(x-6)(x-1)=50
<=>x2-7x+6-50=0
<=>x2+4x-11x-44=0
<=>x(x+4)-11(x+4)=0
<=>(x+4)(x-11)=0
<=>\(\left[{}\begin{matrix}x+4=0\\x-11=0\end{matrix}\right.\)
<=>\(\left[{}\begin{matrix}x=-4\\x=11\end{matrix}\right.\)(Thỏa mãn)
Vậy phương trình thuộc tập nghiệm S={-4;11}
![](https://rs.olm.vn/images/avt/0.png?1311)
Nguyễn Trà My
Phần a)
\(3\times\left(\frac{1}{2}-x\right)+\frac{1}{3}=\frac{7}{6}-x\)
\(32-3x+13=76-x\)
\(116-3x=76-x\)
\(116-76=3x-x\)
\(46=2x\)
\(x=46\div2\)
\(x=13\)
\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times...\times\left(1-\frac{1}{10}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{9}{10}\)
\(=\frac{1}{10}\)
\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{10}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{9}{10}\)
\(=\frac{1.2.3....9}{2.3.4...10}=\frac{1}{10}\)
Ủng hộ mk nha !!! ^_^