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3/4+1/4x=1/2+1/2x

1/2x+1/4x=1/2+3/4

1/2x+1/4x=5/4

x×(1/2+1/4)=5/4

x×3/4=5/4

x=5/4:3/4

x=5/4×4/3

x=5/3

vậy x=5/3

27 tháng 7 2016

Hỏi đáp Toán

27 tháng 7 2016

ở hàng thứ 3 tính cả đề, ở phân số thứ 2 trên tử là số 3 ak bn???

19 tháng 2 2018

      \(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)

\(\Leftrightarrow\)\(\frac{x+2}{327}+1+\frac{x+3}{326}+1+\frac{x+4}{325}+1+\frac{x+5}{324}+1 +\frac{x+349}{5}-4=0\)

\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

\(\Leftrightarrow\)\(x+329=0\)   (vì  1/327 + 1/326 + 1/325 + 1/324 + 1/5  khác  0  )

\(\Leftrightarrow\)\(x=-329\)

19 tháng 2 2018

Bài 1 : 

\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)

\(\Leftrightarrow\)\(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\left(\frac{x+349}{5}-4\right)=0\)

\(\Leftrightarrow\)\(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}=0\)

\(\Leftrightarrow\)\(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)

Vì \(\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)\ne0\)

\(\Rightarrow\)\(x+329=0\)

\(\Rightarrow\)\(x=-329\)

Vậy \(x=-329\)

HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023

a)

\(\begin{array}{l}\frac{2}{9}:x + \frac{5}{6} = 0,5\\\frac{2}{9}:x = \frac{1}{2} - \frac{5}{6}\\\frac{2}{9}:x = \frac{3}{6} - \frac{5}{6}\\\frac{2}{9}:x = \frac{{ - 2}}{6}\\x = \frac{2}{9}:\frac{{ - 2}}{6}\\x = \frac{2}{9}.\frac{{ - 6}}{2}\\x = \frac{{ - 2}}{3}\end{array}\)                        

Vậy \(x = \frac{{ - 2}}{3}\).

b)

\(\begin{array}{l}\frac{3}{4} - \left( {x - \frac{2}{3}} \right) = 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - 1\frac{1}{3}\\x - \frac{2}{3} = \frac{3}{4} - \frac{4}{3}\\x - \frac{2}{3} = \frac{9}{{12}} - \frac{{16}}{{12}}\\x - \frac{2}{3} = \frac{{ - 7}}{{12}}\\x = \frac{{ - 7}}{{12}} + \frac{2}{3}\\x = \frac{{ - 7}}{{12}} + \frac{8}{{12}}\\x = \frac{1}{12}\end{array}\)

Vậy\(x = \frac{1}{12}\).

c)

\(\begin{array}{l}1\frac{1}{4}:\left( {x - \frac{2}{3}} \right) = 0,75\\\frac{5}{4}:\left( {x - \frac{2}{3}} \right) = \frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}:\frac{3}{4}\\x - \frac{2}{3} = \frac{5}{4}.\frac{4}{3}\\x - \frac{2}{3} = \frac{5}{3}\\x = \frac{5}{3} + \frac{2}{3}\\x = \frac{7}{3}\end{array}\)               

Vậy \(x = \frac{7}{3}\).

d)

\(\begin{array}{l}\left( { - \frac{5}{6}x + \frac{5}{4}} \right):\frac{3}{2} = \frac{4}{3}\\ - \frac{5}{6}x + \frac{5}{4} = \frac{4}{3}.\frac{3}{2}\\ - \frac{5}{6}x + \frac{5}{4} = 2\\ - \frac{5}{6}x = 2 - \frac{5}{4}\\ - \frac{5}{6}x = \frac{8}{4} - \frac{5}{4}\\ - \frac{5}{6}x = \frac{3}{4}\\x = \frac{3}{4}:\left( { - \frac{5}{6}} \right)\\x = \frac{3}{4}.\frac{{ - 6}}{5}\\x = \frac{{ - 9}}{{10}}\end{array}\)

Vậy \(x = \frac{{ - 9}}{{10}}\).

28 tháng 9 2016

d) \(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}\)

<=> \(\frac{x+1}{2009}+\frac{x+2}{2008}+\frac{x+3}{2007}-\frac{x+10}{2000}-\frac{x+11}{1999}-\frac{x+12}{1998}=0\)

<=> \(\left(\frac{x+1}{2009}+1\right)+\left(\frac{x+2}{2008}+1\right)+\left(\frac{x+3}{2007}+1\right)-\left(\frac{x+10}{2000}+1\right)-\left(\frac{x+11}{1999}+1\right)-\left(\frac{x+12}{1998}+1=0\right)\)

<=> \(\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0\)

<=>\(\left(x+2010\right).\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\right)=0\)

<=> x+2010 = 0 vì \(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}-\frac{1}{2000}-\frac{1}{1999}-\frac{1}{1998}\ne0\)

<=> x = -2010

28 tháng 9 2016

Làm câu khó nhất rồi, còn lại tự làm nha <(") /_\

Bài 1 : Thực hiện phép tính(1) D = \(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+...+16\right)\)(2) M =\(\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)Bài 2 : Tìm x biết(1) \(x-\left\{x-\left[x-\left(-x+1\right)\right]\right\}=1\)(2) \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right]\cdot...
Đọc tiếp

Bài 1 : Thực hiện phép tính

(1) D = \(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+...+16\right)\)

(2) M =\(\frac{\frac{1}{99}+\frac{2}{98}+\frac{3}{97}+...+\frac{99}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

Bài 2 : Tìm x biết

(1) \(x-\left\{x-\left[x-\left(-x+1\right)\right]\right\}=1\)

(2) \(\left[\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2016}\right]\cdot x=\frac{2015}{1}+\frac{2014}{2}+...+\frac{1}{2015}\)

(3) \(\frac{x}{\left(a+5\right)\left(4-a\right)}=\frac{1}{a+5}+\frac{1}{4-a}\)

(4) \(\frac{x+2}{11}+\frac{x+2}{12}+\frac{x+2}{13}=\frac{x+2}{14}+\frac{x+2}{15}\)

(5) \(\frac{x+1}{2015}+\frac{x+2}{2014}+\frac{x+3}{2013}+\frac{x+4}{2012}+4=0\)

Bài 3 : 

(1) Cho : A =\(\frac{9}{1}+\frac{8}{2}+\frac{7}{3}+...+\frac{1}{9}\); B =\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{10}\)

CMR : \(\frac{A}{B}\)Là 1 số nguyên

(2) Cho : D =\(\frac{1}{1001}+\frac{1}{1002}+\frac{1}{1003}+...+\frac{1}{2000}\)CMR : \(D< \frac{3}{4}\)

Bài 4 : Ký hiệu [x] là số nguyên lớn nhất không vượt quá x , gọi là phần nguyên của x.

VD : [1.5] =1 ; [3] =3 ; [-3.5] = -4

(1) Tính :\(\left[\frac{100}{3}\right]+\left[\frac{100}{3^2}\right]+\left[\frac{100}{3^3}\right]+\left[\frac{100}{3^4}\right]\)

(2) So sánh : A =\(\left[X\right]+\left[X+\frac{1}{5}\right]+\left[X+\frac{2}{5}\right]+\left[X+\frac{3}{5}\right]+\left[X+\frac{4}{5}\right]\)và B = [5x]. Biết x=3.7

0
Bài 1: Thu gọn a) \(\frac{1}{5}x^4y^3-3x^4y^3\) b) \(5x^2y^5-\frac{1}{4}x^2y^5\) c) \(\frac{1}{7}x^2y^3.\left(-\frac{14}{3}xy^2\right)-\frac{1}{2}xy.\left(x^2y^{\text{4}}\right)\) d) \(\left(3xy\right)^2.\left(-\frac{1}{2}x^3y^2\right)\) e) \(-\frac{1}{4}xy^2+\frac{2}{5}x^2y+\frac{1}{2}xy^2-x^2y\) f) \(\frac{1}{2}x^4y.\left(-\frac{2}{3}x^3y^2\right)-\frac{1}{3}x^7y^3\) g) \(\frac{1}{2}x^2y.\left(-10x^3yz^2\right).\frac{1}{4}x^5y^3z\) h)...
Đọc tiếp

Bài 1: Thu gọn

a) \(\frac{1}{5}x^4y^3-3x^4y^3\)

b) \(5x^2y^5-\frac{1}{4}x^2y^5\)

c) \(\frac{1}{7}x^2y^3.\left(-\frac{14}{3}xy^2\right)-\frac{1}{2}xy.\left(x^2y^{\text{4}}\right)\)

d) \(\left(3xy\right)^2.\left(-\frac{1}{2}x^3y^2\right)\)

e) \(-\frac{1}{4}xy^2+\frac{2}{5}x^2y+\frac{1}{2}xy^2-x^2y\)

f) \(\frac{1}{2}x^4y.\left(-\frac{2}{3}x^3y^2\right)-\frac{1}{3}x^7y^3\)

g) \(\frac{1}{2}x^2y.\left(-10x^3yz^2\right).\frac{1}{4}x^5y^3z\)

h) \(4.\left(-\frac{1}{2}x\right)^2-\frac{3}{2}x.\left(-x\right)+\frac{1}{3}x^2\)

i) \(1\frac{2}{3}x^3y.\left(\frac{-1}{2}xy^2\right)^2-\frac{5}{4}.\frac{8}{15}x^3y.\left(-\frac{1}{2}xy^2\right)^2\)

k) \(-\frac{3}{2}xy^2.\left(\frac{3}{4}x^2y\right)^2-\frac{3}{5}xy.\left(-\frac{1}{3}x^4y^3\right)+\left(-x^2y\right)^2.\left(xy\right)^2\)

n) \(-2\frac{1}{5}xy.\left(-5x\right)^2+\frac{3}{4}y.\frac{2}{3}\left(-x^3\right)-\frac{1}{9}.\left(-x\right)^3.\frac{1}{3}y\)

m) \(\left(-\frac{1}{3}xy^2\right)^2.\left(3x^2y\right)^3.\left(-\frac{5}{2}xy^2z^3\right)^{^2}\)

p) \(-2y.\left|2\right|x^4y^5.\left|-\frac{3}{4}\right|x^3y^2z\)

1
26 tháng 7 2019

Bài 1:

a) \(\frac{1}{5}x^4y^3-3x^4y^3\)

= \(\left(\frac{1}{5}-3\right)x^4y^3\)

= \(-\frac{14}{5}x^4y^3.\)

b) \(5x^2y^5-\frac{1}{4}x^2y^5\)

= \(\left(5-\frac{1}{4}\right)x^2y^5\)

= \(\frac{19}{4}x^2y^5.\)

Mình chỉ làm 2 câu thôi nhé, bạn đăng nhiều quá.

Chúc bạn học tốt!

29 tháng 7 2019

cảm ơn nha

chúc bạn học tốt