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2 tháng 4 2018

Trước hết ta hãy so sánh :

\(\dfrac{10^{100}+1}{10^{101}+1}\)với \(\dfrac{10^{100}+1}{10^{102}+1}\)

Ta có: Cả hai phân số trên cùng tử.

\(\Rightarrow\dfrac{10^{100}+1}{10^{101}+1}>\dfrac{10^{100}+1}{10^{102}+1}\)

Tiếp đó so sánh : \(\dfrac{10^{101}+1}{10^{102}+1}\)với \(1\)

Ta được: \(\dfrac{10^{101}+1}{10^{102}+1}< 1\)

Ta lại so sánh được:\(\dfrac{10^{100}+1}{10^{102}+1}< 1\) (*)

Từ (*) suy ra \(\dfrac{10^{100}+1}{10^{101}+1}< \dfrac{10^{101}+1}{10^{102}+2}< \dfrac{10^{101}+1}{10^{102}+1}< 1\Rightarrow\dfrac{10^{100}+1}{10^{101}+1}< \dfrac{10^{101}+1}{10^{102}+1}\)

Ngoài ra còn một cách như sau:

\(\dfrac{10^{101}+1}{10^{102}+1}=\dfrac{10^{\left(100+1\right)}+1}{10^{\left(101+1\right)}+1}=\dfrac{10}{10}.\dfrac{10^{100}+1}{10^{101}+1}>\dfrac{10^{100}+1}{10^{101}+1}\) hay B > A hay A < B

3 tháng 4 2018

Bài 1:

d)

\(\dfrac{x+5}{95}+\dfrac{x+10}{90}+\dfrac{x+15}{85}+\dfrac{x+20}{80}=-4\)

\(\Leftrightarrow\dfrac{x+5}{95}+1+\dfrac{x+10}{90}+1+\dfrac{x+15}{85}+1+\dfrac{x+20}{80}+1=-4+1+1+1+1\)

\(\Leftrightarrow\dfrac{x+100}{95}+\dfrac{x+100}{90}+\dfrac{x+100}{85}+\dfrac{x+100}{80}=0\)

\(\Leftrightarrow\left(x+100\right)\left(\dfrac{1}{95}+\dfrac{1}{90}+\dfrac{1}{85}+\dfrac{1}{80}\right)=0\)

\(\Leftrightarrow x+100=0\) ( vì: \(\dfrac{1}{95}+\dfrac{1}{90}+\dfrac{1}{85}+\dfrac{1}{80}\ne0\))

\(\Leftrightarrow x=-100\)

6 tháng 3 2022

\(a,\dfrac{x}{7}=\dfrac{6}{12}\\ x\cdot12=7\cdot6=42\\ x=42:12\\ x=\dfrac{7}{2}\\ b,\dfrac{-5}{x}=\dfrac{20}{28}\\ x\cdot20=\left(-5\right)\cdot28=-140\\ x=\left(-140\right):20\\ x=-7\\ c,\dfrac{x-2}{8}=\dfrac{3}{4}\\ \left(x-2\right)4=8\cdot3=24\\ x-2=24:4\\ x-2=6\\ x=6+2\\ x=8\\ d,\dfrac{x}{-5}=\dfrac{-5}{x}\\ x^2=\left(-5\right)\cdot\left(-5\right)=25\\ x=5\)

6 tháng 3 2022

cảm ơn bạn

a:=>0,75x-x+1,25x=0,2

=>x=0,2

b: =>1/3-x=-3/6+4/6=1/6

=>x=1/3-1/6=1/6

c: =>(x-1)/45=-6/30=-1/5

=>x-1=-9

=>x=-8

d: =>(2/5x-1)=-5/7

=>2/5x=2/7

=>x=5/7

18 tháng 8 2017

a, \(\frac{2}{3}x=\frac{3}{4}y=\frac{4}{5}z\)

\(\Rightarrow\frac{2x}{3.12}=\frac{3y}{4.12}=\frac{4z}{5.12}\)

\(\Rightarrow\frac{x}{18}=\frac{y}{16}=\frac{z}{15}=\frac{x+y+z}{18+16+15}=\frac{45}{49}\)

Đến đây tự làm tiếp nhé

b, \(2x=3y=5z\Rightarrow\frac{2x}{30}=\frac{3y}{30}=\frac{5z}{30}\Rightarrow\frac{x}{15}=\frac{y}{10}=\frac{z}{6}=\frac{x+y-z}{15+10-6}=\frac{95}{19}=5\)

=> x = 75, y = 50, z = 30

c, \(\frac{3}{4}x=\frac{5}{7}y=\frac{10}{11}z\)

\(\Rightarrow\frac{3x}{4.30}=\frac{5y}{7.30}=\frac{10z}{11.30}\)

\(\Rightarrow\frac{x}{40}=\frac{y}{42}=\frac{z}{33}\)

\(\Rightarrow\frac{2x}{80}=\frac{3y}{126}=\frac{4z}{132}=\frac{2x-3y+4z}{80-126+132}=\frac{8,6}{86}=\frac{1}{10}\)

=> x=... , y=... , z=...

d, Đặt \(\frac{x}{2}=\frac{y}{5}=k\Rightarrow x=2k,y=5k\)

Ta có: xy = 90 => 2k.5k = 90 => 10k2 = 90 => k2 = 9 => k = 3 hoặc -3

Với k = 3 => x = 6, y = 15

Với k = -3 => x = -6, y = -15

Vậy...

e, Tương tự câu d

18 tháng 8 2017

b) Ta có :\(\text{ 2x = 3y = 5z }=\frac{x}{\frac{1}{2}}=\frac{y}{\frac{1}{3}}=\frac{z}{\frac{1}{5}}=\frac{x+y-z}{\frac{1}{2}+\frac{1}{3}-\frac{1}{5}}=\frac{95}{\frac{19}{30}}=\frac{1}{6}\)

=> \(2x=\frac{1}{6}\Rightarrow x=\frac{1}{12}\)

     \(3y=\frac{1}{6}\Rightarrow y=\frac{1}{18}\)

      \(5z=\frac{1}{6}\Rightarrow z=\frac{1}{30}\)

25 tháng 3 2023

a, \(x\cdot\dfrac{-5}{8}=\dfrac{15}{32}\)

\(x=\dfrac{15}{32}:\dfrac{-5}{8}\)

\(x=\dfrac{-3}{4}\)

 

b, \(\dfrac{3}{10}:x=\dfrac{-9}{20}\)

\(x=\dfrac{3}{10}:\dfrac{-9}{20}\)

\(x=-\dfrac{2}{3}\)

 

c, \(\dfrac{-1}{4}x+\dfrac{4}{5}=\dfrac{3}{4}\)

\(\dfrac{-1}{4}x=\dfrac{3}{4}-\dfrac{4}{5}\)

\(\dfrac{-1}{4}x=-\dfrac{1}{20}\)

\(x=-\dfrac{1}{20}:\dfrac{-1}{4}\)

\(x=\dfrac{1}{5}\)

 

d, \(\dfrac{-7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}\cdot\dfrac{-5}{12}\)

\(\dfrac{-7}{8}+\dfrac{2}{3}:x=-\dfrac{1}{4}\)

\(\dfrac{2}{3}:x=-\dfrac{1}{4}+\dfrac{-7}{8}\)

\(\dfrac{2}{3}:x=\dfrac{5}{8}\)

\(x=\dfrac{2}{3}:\dfrac{5}{8}\)

\(x=\dfrac{16}{15}\)

 

 

 

#YVA6

25 tháng 3 2023

\(a,x.\dfrac{-5}{8}=\dfrac{15}{32}\)

\(\Leftrightarrow x=\dfrac{15}{32}:\dfrac{-5}{8}\)

\(\Leftrightarrow x=\dfrac{15}{32}.\dfrac{-8}{5}\)

\(\Leftrightarrow x=-\dfrac{3}{4}\)

\(b,\dfrac{3}{10}:x=-\dfrac{9}{20}\)

\(\Leftrightarrow x=\dfrac{3}{10}:\dfrac{-9}{20}\)

\(\Leftrightarrow x=\dfrac{3}{10}.\dfrac{-20}{9}\)

\(\Leftrightarrow x=-\dfrac{2}{3}\)

\(c,-\dfrac{1}{4}x+\dfrac{4}{5}=\dfrac{3}{4}\)

\(\Leftrightarrow-\dfrac{1}{4}x=\dfrac{3}{4}-\dfrac{4}{5}\)

\(\Leftrightarrow-\dfrac{1}{4}x=-\dfrac{1}{20}\)

\(\Leftrightarrow x=-\dfrac{1}{20}\times\left(-4\right)\)

\(\Leftrightarrow x=\dfrac{1}{5}\)

\(d,-\dfrac{7}{8}+\dfrac{2}{3}:x=\dfrac{3}{5}.\dfrac{-5}{12}\)

\(\Leftrightarrow-\dfrac{7}{8}+\dfrac{2}{3}:x=-\dfrac{1}{4}\)

\(\Leftrightarrow\dfrac{2}{3}:x=-\dfrac{1}{4}+\dfrac{7}{8}\)

\(\Leftrightarrow\dfrac{2}{3}:x=\dfrac{5}{8}\)

\(\Leftrightarrow x=\dfrac{2}{3}:\dfrac{5}{8}\)

\(\Leftrightarrow x=\dfrac{16}{15}\)

AH
Akai Haruma
Giáo viên
17 tháng 4 2023

Lời giải:
a.

$x=\frac{-5}{6}-\frac{2}{3}=\frac{-3}{2}$

b.

$\frac{2}{3}x=\frac{1}{10}-\frac{1}{2}=\frac{-2}{5}$

$x=\frac{-2}{5}: \frac{2}{3}=\frac{-3}{5}$

c.

$\frac{7}{8}x=\frac{2}{9}-\frac{1}{3}=\frac{-1}{9}$
$x=\frac{-1}{9}: \frac{7}{8}=\frac{-8}{63}$

 

 

AH
Akai Haruma
Giáo viên
17 tháng 4 2023

d.

$\frac{5}{7}: x=\frac{1}{6}-\frac{4}{5}=\frac{-19}{30}$

$x=\frac{5}{7}: \frac{-19}{30}=\frac{-150}{133}$

e.

$(\frac{2}{5}-1\frac{2}{3}):x=\frac{2}{5}+\frac{3}{5}=1$

$\frac{-19}{15}: x=1$

$x=\frac{-19}{15}:1 =\frac{-19}{15}$

f.

$(-\frac{3}{4}+x).2\frac{2}{3}=1$

$\frac{-3}{4}+x=1: 2\frac{2}{3}=\frac{3}{8}$

$x=\frac{3}{8}+\frac{3}{4}=\frac{9}{8}$

20 tháng 9 2023

a, - \(\dfrac{1}{10}\) + \(\dfrac{2}{5}\)\(x\) + \(\dfrac{7}{20}\) = \(\dfrac{1}{10}\)

              \(\dfrac{2}{5}\)\(x\)            = \(\dfrac{1}{10}\) - \(\dfrac{7}{20}\) + \(\dfrac{1}{10}\)

              \(\dfrac{2}{5}\) \(x\)           = - \(\dfrac{3}{20}\)

                  \(x\)          =  - \(\dfrac{3}{20}\)\(\dfrac{2}{5}\)

                  \(x\)          = - \(\dfrac{3}{8}\)

20 tháng 9 2023

b, \(\dfrac{1}{3}\) +   \(\dfrac{1}{2}\)\(x\) = - \(\dfrac{1}{5}\)

            \(\dfrac{1}{2}\)\(x\) =  - \(\dfrac{1}{5}\) - \(\dfrac{1}{3}\)

             \(\dfrac{1}{2}\)\(x\) = - \(\dfrac{8}{15}\)

                   \(x\) =  \(\dfrac{1}{2}\): (- \(\dfrac{8}{15}\))

                    \(x\) =  - \(\dfrac{15}{16}\)

30 tháng 12 2021

\(\dfrac{-3}{5}-x=\dfrac{21}{10}\)

\(x=\dfrac{-3}{5}-\dfrac{21}{10}\)

\(x=\)-\(\dfrac{27}{10}\)

 

\(x:\dfrac{2}{9}=\dfrac{9}{2}\)

\(x.\dfrac{9}{2}=\dfrac{9}{2}\)

\(x=\dfrac{9}{2}:\dfrac{9}{2}\)

\(x=1\)

 

\(\dfrac{x}{9}=\dfrac{5}{3}\)

\(x.3=5.9\)

\(x.3=45\)

\(x=45:3=15\)

 

\(x:\left(\dfrac{2}{5}\right)^3=\left(\dfrac{5}{2}\right)^3\)

\(x:\dfrac{8}{125}=\dfrac{125}{8}\)

\(x.\dfrac{125}{8}=\dfrac{125}{8}\)

\(x=\dfrac{125}{8}:\dfrac{125}{8}=1\)

 

AH
Akai Haruma
Giáo viên
13 tháng 12 2021

Lời giải:

a.

 \(\frac{10}{x+2}=\frac{60}{6(x+2)}=\frac{60(x-2)}{6(x+2)(x-2)}=\frac{60(x-2)}{6(x^2-4)}\)

\(\frac{5}{2x-4}=\frac{15(x+2)}{6(x-2)(x+2)}=\frac{15(x+2)}{6(x^2-4)}\)

\(\frac{1}{6-3x}=\frac{x+2}{3(2-x)}=\frac{2(x+2)^2}{6(2-x)(2+x)}=\frac{-2(x+2)^2}{6(x^2-4)}\)

b.

\(\frac{1}{x+2}=\frac{x(2-x)}{x(x+2)(2-x)}=\frac{x(2-x)}{x(4-x^2)}\)

\(\frac{8}{2x-x^2}=\frac{8(x+2)}{(x+2)x(2-x)}=\frac{8(x+2)}{x(4-x^2)}\)

c.

\(\frac{4x^2-3x+5}{x^3-1}\)

\(\frac{1-2x}{x^2+x+1}=\frac{(1-2x)(x-1)}{(x-1)(x^2+x+1)}=\frac{-2x^2+3x-1}{x^3-1}\)

\(-2=\frac{-2(x^3-1)}{x^3-1}\)

 

5 tháng 10 2021

a) \(\dfrac{x}{y}=\dfrac{9}{7}\)\(\dfrac{x}{9}=\dfrac{y}{7}\)

\(\dfrac{y}{z}=\dfrac{7}{3}\)\(\dfrac{y}{7}=\dfrac{z}{3}\)

\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}\)

Áp dụng tính chất dãy tỉ số bằng nhau,ta có:

\(\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=-\dfrac{15}{5}=-3\)

\(\left\{{}\begin{matrix}x=-3.9=-27\\y=-3.7=-21\\z=-3.3=-9\end{matrix}\right.\)

c: Ta có: 5x=8y=20z

nên \(\dfrac{x}{\dfrac{1}{5}}=\dfrac{y}{\dfrac{1}{8}}=\dfrac{z}{\dfrac{1}{20}}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{\dfrac{1}{5}}=\dfrac{y}{\dfrac{1}{8}}=\dfrac{z}{\dfrac{1}{20}}=\dfrac{x-y-z}{\dfrac{1}{5}-\dfrac{1}{8}-\dfrac{1}{20}}=\dfrac{3}{\dfrac{1}{40}}=120\)

Do đó: x=24; y=15; z=6