1+ 1 x 3 + 1
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a)(1-1/2)xx(1-1/3)xx(1-1/4)xx(1-1/5)`
`=1/2xx2/3xx3/4xx4/5`
`=[1xx2xx3xx4]/[2xx3xx4xx5]`
`=1/5`
`b)(1-3/4)xx(1-3/7)xx(1-3/10)xx(1-3/13)xx .... xx(1-3/97)xx(1-3/100)`
`=1/4xx4/7xx7/10xx10/13xx .... xx94/97xx97/100`
`=[1xx4xx7xx10xx...xx94xx97]/[4xx7xx10xx13xx....xx97xx100]`
`=1/100`
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=2x^3+3x^2-3-5x^2-5x=2x^3-2x^2-5x-3\\ B=125-150x+60x^2-8x^3-25+9x^2=-8x^3+69x^2-150x+100\\ C=\left(3x+1-2x+1\right)\left(3x+1+2x-1\right)=5x\left(x+2\right)=5x^2+10x\\ D=\left(2x+1-3+x\right)^2=\left(3x-2\right)^2=9x^2-12x+4\\ E=x^3-6x^2+12x-8-x^3+x+6x^2-18x=-5x-8\\ F=x^3-3x^2+3x-1-3+3x^2-x^3+1-3x=-3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Giải:
a) \(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{4}\right).\left(1-\dfrac{1}{5}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}\)
\(=\dfrac{1.2.3.4}{2.3.4.5}\)
\(=\dfrac{1}{5}\)
b) \(\left(1-\dfrac{3}{4}\right).\left(1-\dfrac{3}{7}\right).\left(1-\dfrac{3}{10}\right).\left(1-\dfrac{3}{13}\right).....\left(1-\dfrac{3}{97}\right).\left(1-\dfrac{3}{100}\right)\)
\(=\dfrac{1}{4}.\dfrac{4}{7}.\dfrac{7}{10}.\dfrac{10}{13}.....\dfrac{94}{97}.\dfrac{97}{100}\)
\(=\dfrac{1.4.7.10.....94.97}{4.7.10.13.....97.100}\)
\(=\dfrac{1}{100}\)
Chúc bạn học tốt!
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(\left(x+1\right)^3-\left(x+2\right)\left(x-1\right)^2-3\left(x-3\right)\left(x+3\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1-\left(x+2\right)\left(x^2-2x+1\right)-3\left(x^2-9\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1-\left(x^3-2x^2+x+2x^2-4x+2\right)-3\left(x^2-9\right)=5\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x-2-3x^2+9=5\)
\(\Leftrightarrow6x=-3\)
hay \(x=-\dfrac{1}{2}\)
b: Ta có: \(\left(x+1\right)^3+\left(x-1\right)^3=\left(x+2\right)^3+\left(x-2\right)^3\)
\(\Leftrightarrow x^3+3x^2+3x+1+x^3-3x^2+3x-1=x^3+6x^2+12x+8+x^3-6x^2+12x-8\)
\(\Leftrightarrow2x^3+6x=2x^3+24x\)
\(\Leftrightarrow x=0\)
c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)^2=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-1=-10\)
\(\Leftrightarrow12x=-11\)
hay \(x=-\dfrac{11}{12}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)\left(1-\frac{1}{1+2+3+4}\right)...\left(1-\frac{1}{1+2+3+4+...+2018}\right)\)
\(A=\frac{2}{1+2}\cdot\frac{2+3}{1+2+3}\cdot\frac{2+3+4}{1+2+3+4}\cdot...\cdot\frac{2+3+4+5+...+2018}{1+2+3+4+5+...+2018}\)
Đến chỗ này đố ai tính được ?!!?!
gạch các số của tử số và các số của mẫu số giống nhau
ví dụ như bạn nói:
\(\dfrac{2+3+4+5+...+2018}{1+2+3+4+5+...+2018} =1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
Ta có: \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)
\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)
Suy ra: \(x^2+2x+1-\left(x^2-2x+1\right)=4\)
\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4\)
\(\Leftrightarrow4x=4\)
hay x=1(loại)
Vậy: \(S=\varnothing\)
2) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+2}{x-2}+\dfrac{x}{x+2}=2\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+4x+4+x^2-2x=2x^2-8\)
\(\Leftrightarrow2x^2+2x+4-2x^2-8=0\)
\(\Leftrightarrow2x-4=0\)
\(\Leftrightarrow2x=4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1-x}{1+x}+3=\frac{2x+3}{x+1}\left(ĐKXĐ:x\ne-1\right)\)
\(\Leftrightarrow\frac{1-x}{x+1}+\frac{3\left(x+1\right)}{x+1}=\frac{2x+3}{x+1}\)
\(\Leftrightarrow\frac{1-x+3\left(x+1\right)}{x+1}=\frac{2x+3}{x+1}\)
\(\Rightarrow1-x+3\left(x+1\right)=2x+3\)
\(\Leftrightarrow1-x+3x+3=2x+3\)
\(\Leftrightarrow2x+4=2x+3\)
\(\Leftrightarrow0x=-1\)(vô nghiệm)
Vậy phương trình vô nghiệm.
\(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2-10}{2x-3}\left(ĐKXĐ:x\ne\frac{3}{2}\right)\)
\(\Leftrightarrow\frac{x^2+4x+4}{2x-3}-\frac{2x-3}{2x-3}=\frac{x^2-10}{2x-3}\)
\(\Leftrightarrow\frac{x^2+4x+4-2x+3}{2x-3}=\frac{x^2-10}{2x-3}\)
\(\Rightarrow x^2+4x+4-2x+3=x^2-10\)
\(\Leftrightarrow2x+7=-10\)
\(\Leftrightarrow2x=-17\)
\(\Leftrightarrow x=\frac{-17}{2}\)(thỏa mãn ĐKXĐ)
Vậy phương trình có nghiệm duy nhất : \(x=\frac{-17}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 3 3/4 . x = 1 1/2
<=> 15/4 . x = 3/2
<=> x = 3/4 . 4/15
<=> x = 1/5
Vậy x = 1/5
b) 1 1/4 x + 1 1/2 = 1 1/4
<=> 5/4 . x + 3/2 = 5/4
<=> 5/4 . x = 5/4 - 3/2
<=> 5/4 . x = -1/4
<=> x = -1/4 . 4/5
<=> x = -1/5
Vậy x = -1/5
c) ( 3 1/3 - 1 1/2 x ) : 5/6 = 1 1/2
<=> ( 10/3 - 3/2 x ) : 5/6 = 3/2
<=> 10/3 - 3/2 x = 3/2 . 5/6
<=> 10/3 - 3/2 x = 5/4
<=> 3/2 . x = 10/3 - 5/4
<=> 3/2 . x = 25/12
<=> x = 25/12 . 2/3
<=> x = 25/18
Vậy x = 25/18
d) ( 3/7 x - 1 ) : 4 = -1/28
<=> 3/7 . x - 1 = -1/28 . 1/4
<=> 3/7 . x - 1 = -1/112
<=> 3/7 . x = -1/112 + 1
<=> 3/7 . x = 111/112
<=> x = 111/112 . 7/3
<=> x = 37/16
Vậy x = 37/16
e) | x - 3/4 | = 1
<=> x - 3/4 = 1
hoặc x - 3/4 = -1
<=> x = 1 + 3/4
hoặc x = -1 + 3/4
<=> x = 7/4
hoặc x = -1/4
Vậy x = 7/4 ; x = -1/4
f) | 2/3 . x + 1/3 | = 5/6
<=> 2/3 . x + 1/3 = 5/6
hoặc 2/3 . x + 1/3 = -5/6
<=> 2/3 . x = 5/6 - 1/3
hoặc 2/3 . x = -5/6 - 1/3
<=> 2/3 . x = 1/2
hoặc 2/3 . x = -7/6
<=> x = 1/2 . 3/2
hoặc x = -7/6 . 3/2
<=> x = 3/4
hoặc x = -7/4
Vậy x = 3/4 ; x = -7/4
kết quả là 5 nhân chia trước cộng trừ sau
1+1x3+1=5