tìm x biết (x+1/6)+(x+1/12)+(x+1/20)+(x+1/30)=25/3
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1: =>x=8-35=-27
2: =>15-4+x=6
=>x+11=6
hay x=-5
3: =>-30+25-x=-1
=>x+5=1
hay x=-4
4: =>x-(-13)=-8
=>x+13=-8
hay x=-21
5: =>x-29-17+38=-9
=>x-8=-9
hay x=-1
1/2 + 1/6 + 1/12 + 1/20 + 1/30 1/x = 41/42
5/6 + 1/x = 41/42
1/x = 41/42 - 5/6
1/x = 1/7
vậy x = 7
Lời giải:
$\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{x(x+1)}=\frac{3}{8}$
$\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{x(x+1)}=\frac{3}{8}$
$1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-.....+\frac{1}{6}-\frac{1}{7}+\frac{1}{x(x+1)}=\frac{3}{8}$
$1-\frac{1}{7}+\frac{1}{x(x+1)}=\frac{3}{8}$
$\frac{6}{7}+\frac{1}{x(x+1)}=\frac{3}{8}$
$\frac{1}{x(x+1)}=\frac{3}{8}-\frac{6}{7}=\frac{-27}{56}$
Kết quả này không phù hợp lắm.
Bạn xem lại đề nhé.
1.8100
2. 34 x 18 + 18 x 66
= 18 x ( 34 + 66)
= 18 x 100 = 1800
3. X × ( 8 + 12) = 160 + 20 × 12
= X x 20 = 160 + 240
= X x 20 = 400
X = 400 : 20 = 20
4. X x 12 - X x 2 = 2020
(12 - 2) x X = 2020
10 x X = 2020
X = 2020 : 10 = 202
15-(4-x)=6
⇒15-4+x=6
⇒11+x=6
⇒x=-5
x-(12-25)=-8
⇒x+13=-8
⇒x=-21
(x-29)-(17-38)=-9
⇒x-29+21=-9
⇒x-8=-9
⇒x=-1
-30+(25-x)=-1
⇒-30+25-x=-1
⇒-5-x=-1
⇒x=-4
24:
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
\(\Leftrightarrow\left(x+2\right)\left(x+6\right)=8\left(x+6\right)-8\left(x+2\right)\)
\(\Leftrightarrow x^2+8x+12=8x+48-8x-16=32\)
=>(x+10)(x-2)=0
=>x=-10 hoặc x=2
25: \(\Leftrightarrow\dfrac{\left(x+1\right)^2+1}{x+1}+\dfrac{\left(x+4\right)^2+4}{x+4}=\dfrac{\left(x+2\right)^2+2}{x+2}+\dfrac{\left(x+3\right)^2+3}{x+3}\)
\(\Leftrightarrow x+1+\dfrac{1}{x+1}+x+4+\dfrac{4}{x+4}=x+2+\dfrac{2}{x+2}+x+3+\dfrac{3}{x+3}\)
\(\Leftrightarrow\dfrac{1}{x+1}+\dfrac{4}{x+4}=\dfrac{2}{x+2}+\dfrac{3}{x+3}\)
\(\Leftrightarrow x+5=0\)
hay x=-5
\(a,-12\left(x-5\right)+7\left(3-x\right)=5\)
\(-12x+60+21-7x=5\)
\(-12x-7x+81=5\)
\(-19x=5-81\)
\(-19x=-76\)
\(x=-76:\left(-19\right)\)
\(x=4\)
\(Vậyx=4\)
\(b,30\left(x+2\right)-6\left(x-5\right)-24x=100\)
\(30x+60-6x-30-24x=100\)
\(30x-6x-24x+60-30=100\)
\(0x+30=100\)
\(\Rightarrow Vôlý\)
Vậy không có giá trị nào của x thỏa mãn đề bài.
\(c,-5\left(x+\frac{1}{5}\right)-\frac{1}{2}\left(x-\frac{2}{3}\right)=\frac{3}{2}x-\frac{5}{6}\)
\(-5x-1-\frac{1}{2}x-\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(-5x-\frac{1}{2}x-1-\frac{1}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(-\frac{11}{2}x-\frac{2}{3}=\frac{3}{2}x-\frac{5}{6}\)
\(-\frac{2}{3}+\frac{5}{6}=\frac{3}{2}x+\frac{11}{2}x\)
\(-\frac{4}{6}+\frac{5}{6}=\frac{14}{2}x\)
\(\frac{1}{6}=7x\)
\(x=\frac{1}{6}:7\)
\(x=\frac{1}{6}.\frac{1}{7}\)
\(x=\frac{1}{42}\)
\(Vậyx=\frac{1}{42}\)
\(d,-3\left(x-\frac{1}{2}\right)-5\left(x+\frac{3}{5}\right)=-x+\frac{1}{5}\)
\(-3x+\frac{3}{2}-5x-3=-x+\frac{1}{5}\)
\(-3x-5x+\frac{3}{2}-3=-x+\frac{1}{5}\)
\(-8x+\frac{3}{2}-\frac{6}{2}=-x+\frac{1}{5}\)
\(-8x-\frac{3}{2}=-x+\frac{1}{5}\)
\(-\frac{3}{2}-\frac{1}{5}=-x+8x\)
\(\frac{15}{10}-\frac{2}{10}=7x\)
\(7x=\frac{13}{10}\)
\(x=\frac{13}{10}:7\)
\(x=\frac{13}{10}.\frac{1}{7}\)
\(x=\frac{13}{70}\)
\(Vậyx=\frac{13}{70}\)
\(\left(x+\dfrac{1}{6}\right)+\left(x+\dfrac{1}{12}\right)+\left(x+\dfrac{1}{20}\right)+\left(x+\dfrac{1}{30}\right)=\dfrac{25}{3}\)
=>\(4x+\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}\right)=\dfrac{25}{3}\)
=>\(4x+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}\right)=\dfrac{25}{3}\)
=>\(4x+\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=\dfrac{25}{3}\)
=>\(4x+\dfrac{1}{3}=\dfrac{25}{3}\)
=>\(4x=\dfrac{24}{3}=8\)
=>x=2