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28 tháng 11 2019

\(\frac{38}{10}:2x=\frac{1}{4}\times\frac{3}{8}\)

\(\frac{38}{10}:2x=\frac{3}{32}\)

\(2x=\frac{38}{10}:\frac{3}{32}\)

\(2x=\frac{38}{10}\times\frac{32}{3}\)

\(2x=\frac{608}{15}\)

\(x=\frac{608}{15}\times\frac{1}{2}\)

\(x=\frac{304}{15}\)

20 tháng 7 2015

a) (3x-2)-(5x+3)=(x+4)-(x-1)

<=>3x-2-5x-3=x+4-x+1

<=>3x-5x-x+x=4+1+2+3

<=>-2x=10

<=>x=-5

Vậy S={-5}

b)2x2+9x=5

<=>2x2+9x-5=0

<=>2x2-x+10x-5=0

<=>x(2x-1)+5(2x-1)=0

<=>(2x-1)(x+5)=0

<=>2x-1=0 hoặc x+5=0

<=>x=0,5 hoặc x=-5

Vậy S={0,5;-5}

17 tháng 12 2016

Ta có: \(\frac{1+x}{3}=\frac{3+x}{5}\)

=> 5.(1+x) = 3.(3+1)

=> 5 + 5x = 9 + 3x

=> 5x - 3x = 9 - 5

=> 2x = 4

=> x = 2

Thế x = 2 vào \(\frac{1+x}{3}=\frac{8+2x}{3y}\)

Ta được: \(\frac{1+2}{3}=\frac{8+2.2}{3.y}\)= 1 = \(\frac{12}{3y}\)

=> y = 4

Vậy x = 2; y = 4

Nguyễn Huy TúTrương Hồng Hạnhsoyeon_Tiểubàng giảiHoàng Lê Bảo NgọcTrần Việt Linh

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6 tháng 9 2016

a.(-2/3+3/7) : 4/5 + (-1/3+4/7) : 4/5

= [(-2/3 + 3/7) + (-1/3 + 4/7)] : 4/5

= [(-2/3 + (-1/3) + (3/7 + 4/7)] : 4/5

= [-1 + 1] : 4/5

= 0 : 4/5

= 0   

6 tháng 9 2016

a) \(\left(\frac{-2}{3}+\frac{3}{7}\right).\frac{5}{4}+\left(\frac{-1}{3}+\frac{4}{7}\right).\frac{5}{4}\)

=\(\left(\frac{-2}{3}+\frac{-1}{3}+\frac{3}{7}+\frac{4}{7}\right).\frac{5}{4}\)

\(0.\frac{5}{4}=0\)

b) \(\frac{5}{9}:\left(\frac{1}{11}-\frac{5}{22}+\frac{1}{15}-\frac{2}{3}\right)\)

=\(\frac{5}{9}:\frac{-81}{110}=\frac{-550}{729}\)

22 tháng 11 2022

\(B=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^{10}}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}=\dfrac{2^{10}\cdot3^8\left(1-3^2\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{1-9}{6}=\dfrac{-8}{6}=-\dfrac{4}{3}\)

31 tháng 5 2020

*\(M+\left(5x^2-2xy\right)=6x^2+9xy-y^2\)

\(M=6x^2+9xy-y^2-\left(5x^2-2xy\right)\)

\(M=6x^2+9xy-y^2-5x^2+2xy\)

\(M=\left(6-5\right)x^2+\left(9+2\right)xy-y^2\)

\(M=x^2+11xy-y^2\)

\(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\)

Ta có : \(\hept{\begin{cases}\left(2x-5\right)^{2018}\ge0\forall x\\\left(3y+4\right)^{2020}\ge0\forall y\end{cases}\Rightarrow}\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\ge0\forall x,y\)

Mà đề cho \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}\le0\)

=> \(\left(2x-5\right)^{2018}+\left(3y+4\right)^{2020}=0\)

=> \(\hept{\begin{cases}2x-5=0\\3y+4=0\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{5}{2}\\y=-\frac{4}{3}\end{cases}}\)

Thay x = 5/2 ; y = -4/3 vào M ta được :

\(M=\left(\frac{5}{2}\right)^2+11\cdot\frac{5}{2}\cdot\left(-\frac{4}{3}\right)-\left(-\frac{4}{3}\right)^2\)

\(M=\frac{25}{4}+\frac{-110}{3}-\frac{16}{9}\)

\(M=\frac{-1159}{36}\)

Vậy giá trị của M = -1159/36 khi x = 5/2 ; y = -4/3

Không chắc nha 

31 tháng 3 2018

1) A(x) = 3x - 2x+ x3 + 5 = x3 - 2x2 + 3x + 5

    B(x) = x3 - x + 3x4 + 5 - x = 3x+ 3x- x - x + 5 = 3x+ 3x- 2x + 5

2) A(x) + B(x) = ( x3 - 2x2 + 3x + 5 ) + ( 3x+ 3x- 2x + 5 )

                       =  x3 - 2x2 + 3x + 5   +   3x+ 3x- 2x + 5

                       =  3x4 + x3 + 3x- 2x+ 3x - 2x + 5 + 5 = 3x4 + 4x- 2x+ x + 10

3) A(x) - B(x) = ( x3 - 2x2 + 3x + 5 ) - ( 3x+ 3x- 2x + 5 )

                     = x- 2x2 + 3x + 5 - 3x4 - 3x+ 2x - 5

                     = -3x+ x- 3x3 - 2x2 + 3x + 2x + 5 - 5

                     =  -3x4 - 2x3 - 2x2 + 5x