tìm x : 1 phần x*(x+1) + 1 phần (x+1)*(x+2) + 1 phần (x+2)*(x+3) = 1 phần x + 1 phần 2017
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1)
a) 21/23 x 5/11 x 4/35
= \(\frac{21.5.4}{23.11.35}\)
= \(\frac{7.3.5.4}{23.11.5.7}\)
= \(\frac{3.4}{23.11}\) ( ta đã rút gọc 7 và 5 )
= \(\frac{12}{253}\)
b) 26 x 13/42 : 13
= \(\frac{26.13}{42.13}\)
= \(\frac{13.2.13}{21.2.13}\)
= \(\frac{13}{21}\)
c) 5/9 : 2/9 + 4/9 : 2/3
= \(\frac{5.9}{9.2}+\frac{4.3}{9.2}\)
= \(\frac{5}{2}+\frac{2.2.3}{3.3.2}\)
= \(\frac{5}{2}+\frac{2}{3}\)
= \(\frac{19}{6}\)
2 ) tìm x
a ) 2 - x = 1/3 + 1/6
2 - x = 1/2
x = 2 - 1/2
x = 3/2
b ) x + 5/8 = 1 + 3/4
x + 5/8 = 7/4
x = 7/4 - 5/8
x = 9/8
3)
a) x < 3/4 + 4/5
=> x < 31/20
=> x = 1
b) 1/2 + 3/5 < x < 5/4
=> 11/10 < x < 5/4
ta quy đồng phân số 11/10 và 5/4
\(\frac{11}{10}=\frac{11.4}{10.4}=\frac{44}{40}\)
\(\frac{5}{4}=\frac{5.10}{4.10}=\frac{50}{40}\)
=> 44/40 < x < 50/40
=> x không thỏa mãn
A=1/1.2+1/2.3+...1/x =49/50
A=1-1/2+1/2-1/3+...+1/x-1-1/x=49/50
A=1-1/x=49/50
A=50/50-1=x=49/50
x=1/50
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x}=\frac{49}{50}\)
\(\Rightarrow\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{x\left(x-1\right)}=\frac{49}{50}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x-1}-\frac{1}{x}=\frac{49}{50}\)
\(\Rightarrow1-\frac{1}{x}=\frac{49}{50}\)
\(\Rightarrow\frac{1}{x}=\frac{1}{50}\)
\(\Rightarrow x=50\)
a: \(A=\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{2\left(x-3\right)}{2-x}\)
\(=\dfrac{4+4x+x^2+4x^2-\left(2-x\right)^2}{\left(2-x\right)\left(2+x\right)}\cdot\dfrac{2-x}{2\left(x-3\right)}\)
\(=\dfrac{5x^2+4x+4-4+4x-x^2}{\left(2+x\right)}\cdot\dfrac{1}{2\left(x-3\right)}\)
\(=\dfrac{4x^2+8x}{x+2}\cdot\dfrac{1}{2\left(x-3\right)}=\dfrac{4x\left(x+2\right)}{2\left(x+2\right)}\cdot\dfrac{1}{x-3}=\dfrac{2x}{x-3}\)
b: |x-2|=2
=>x-2=2 hoặc x-2=-2
=>x=0(nhận) hoặc x=4(nhận)
Khi x=0 thì \(A=\dfrac{2\cdot0}{0-3}=\dfrac{-2}{3}\)
Khi x=4 thì \(A=\dfrac{2\cdot4}{4-3}=8\)
c: A>0
=>x/x-3>0
=>x>3 hoặc x<0
=>x>3
\(\frac{x}{3\frac{1}{2}.2\frac{2}{3}}=\frac{9}{56}\Rightarrow\frac{x}{\frac{7}{2}.\frac{8}{3}}=\frac{9}{56}\Rightarrow x=\frac{9}{56}.\frac{28}{3}=\frac{3}{2}\)
\(x:\left(3\frac{1}{2}.2\frac{2}{3}\right)=\frac{9}{56}\)
\(x:\left(\frac{7}{2}.\frac{8}{3}\right)=\frac{9}{56}\)
\(x:\left(\frac{7.4}{3}\right)=\frac{9}{56}\)
\(x.\frac{3}{28}=\frac{9}{56}\)
\(x=\frac{9}{56}.\frac{28}{3}=\frac{3}{2}\)
Vậy \(x=\frac{3}{2}\)
a)\(\frac{6}{x-5}=\frac{y+2}{-5}\)
=>(x-5)(y+2)=6*(-5)=-30
Ta xét bảng sau:
| x-5 | 1 | 2 | 3 | 5 | 6 | 10 | 15 | 30 | -1 | -2 | -3 | -5 | -6 | -10 | -15 | -30 |
| x | 6 | 7 | 8 | 10 | 11 | 15 | 20 | 35 | 4 | 3 | 2 | 0 | -1 | -5 | -10 | -25 |
| y+2 | -30 | -15 | -10 | -6 | -5 | -3 | -2 | -1 | 30 | 15 | 10 | 6 | 5 | 3 | 2 | 1 |
| y | -32 | -17 | -12 | -8 | -7 | -5 | -4 | -3 | 28 | 13 | 8 | 4 | 3 | 1 | 0 | -1 |
b)Để (x+1)/(x-5) có giá trị nguyên
thì (x+1) phải chia hết cho (x-5)
(x-5)+6 chia hết cho (x-5)
=>6 chia hết cho (x-5) hay (x-5)EƯ(6)={1;-1;2;-2;3;-3;6;-6}
=>xE{6;4;7;3;8;2;11;-1}
Vậy xE{6;4;7;3;8;2;11;-1}
ĐKXĐ: x∉{0;-1;-2;-3}
Ta có: \(\frac{1}{x\left(x+1\right)}+\frac{1}{\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}=\frac{1}{x}+\frac{1}{2017}\)
=>\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}=\frac{1}{x}+\frac{1}{2017}\)
=>\(\frac{1}{x}-\frac{1}{x+3}=\frac{1}{x}+\frac{1}{2017}\)
=>\(-\frac{1}{x+3}=\frac{1}{2017}\)
=>x+3=-2017
=>x=-2017-3=-2020(nhận)