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

\(\frac{1}{3}:\left(2x-1\right)=-5-\frac{1}{4}\)

\(\frac{1}{3}:\left(2x-1\right)=-\frac{20}{4}-\frac{1}{4}\)

\(\frac{1}{3}:\left(2x-1\right)=-\frac{21}{4}\)

\(\left(2x-1\right)=\frac{1}{3}:-\frac{21}{4}\)

\(\left(2x-1\right)=\frac{1}{3}.-\frac{4}{21}\)

\(\left(2x-1\right)=-\frac{4}{63}\)

2x= -4/63 + 1

2x = 59/63

x = 59/63 : 2

x = 59/126

1/3:(2.x-1)=-5-1/4

1/3:(2.x-1)=-21/4

2.x-1=1/3:-21/4

2.x-1=-4/63

2.x=-4/63+1

2.x=\(3\frac{59}{63}\)

x=\(3\frac{59}{63}\):2

x=\(1\frac{61}{63}\)

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

\(\frac{1}{3}\div\left(2x-1\right)=-5-\frac{1}{4}\)

\(\frac{1}{3}\div\left(2x-1\right)=\frac{-20}{4}+\frac{-1}{4}\)

\(\frac{1}{3}\div\left(2x-1\right)=\frac{-21}{4}\)

\(2x-1=\frac{1}{3}\div\frac{-21}{4}\)

\(2x-1=\frac{1}{3}.\frac{-4}{21}\)

\(2x-1=\frac{-4}{63}\)

\(2x=\frac{-4}{63}+1\)

\(2x=\frac{-4}{63}+\frac{63}{63}\)

\(2x=\frac{59}{63}\)

\(x=\frac{59}{63}\div2\)

\(x=\frac{59}{63}.\frac{1}{2}\)

\(x=\frac{59}{126}\)

x=6x7:4

x=

\(\frac{-4}{7}=\frac{6}{x}\Rightarrow\left(-4\right).x=6.7\)

                  \(\Rightarrow\left(-4\right).x=42\) 

                  \(\Rightarrow x=42:\left(-4\right)\) 

                   \(\Rightarrow x=-10,5\)

k mik nha

x=6.7:(-4)=?

(1 - 1/2) x (1 - 1/3) x (1 - 1/4) x (1 - 1/5) x (1 - 1/6)

= 1/2 x 2/3 x 3/4 x 4/5 x 5/6

= 1/6 (tử và mẫu triệt tiêu cho nhau)

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

\(\frac{x+1}{2}=\frac{y+3}{4}\)\(=\frac{z+5}{6}\)\(=\frac{2.\left(x+1\right)+3.\left(y+3\right)+4.\left(z+5\right)}{2.2+3.4+4.6}\)

\(=\frac{2x+2+3y+9+4z+20}{4+12+24}\)\(=\frac{\left(2x+3y+4z\right)+\left(2+9+20\right)}{40}\)

\(=\frac{9+31}{40}=\frac{40}{40}=1\)

Cứ thế là tìm x+1 rồi tìm x

                    y+3           y

                    x+5           z

Bài làm:

a) Ta có: \(\left(-\frac{3}{8}x^2z\right).\left(\frac{2}{3}xy^2z^2\right).\left(\frac{4}{5}x^3y\right)\)

\(=-\frac{1}{5}x^6y^3z^3\)

b) Tại x=-1 ; y=-2 ; z=3 thì giá trị đơn thức là:

\(-\frac{1}{5}.\left(-1\right)^6.\left(-2\right)^3.3^3=\frac{216}{5}\)

a) Ta có : \(\left(\frac{-3}{8}x^2z\right)\cdot\frac{2}{3}xy^2z^2\cdot\frac{4}{5}x^3y=\left(-\frac{3}{8}\cdot\frac{2}{3}\cdot\frac{4}{5}\right)\cdot x^2xx^3\cdot y^2y\cdot zz^2=-\frac{1}{5}x^6y^3z^3\)

b) Với x = -1 ; y = -2 , z = 3

Thế vào ba đơn thức trên và đơn thức tích ta được :

\(\frac{-3}{8}x^2z=\frac{-3}{8}\left(-1\right)^2\cdot3=\frac{-3}{8}\cdot1\cdot3=\frac{-9}{8}\)

\(\frac{2}{3}xy^2z^2=\frac{2}{3}\cdot\left(-1\right)\cdot\left(-2\right)^2\cdot3^2=\frac{2}{3}\left(-1\right)\cdot4\cdot9=-24\)

\(\frac{4}{5}x^3y=\frac{4}{5}\left(-1\right)^3\cdot\left(-2\right)=\frac{4}{5}\left(-1\right)\left(-2\right)=\frac{8}{5}\)

\(-\frac{1}{5}x^6y^3z^3=-\frac{1}{5}\left(-1\right)^6\left(-2\right)^3\cdot3^3=-\frac{1}{5}\cdot1\cdot\left(-8\right)\cdot27=\frac{216}{5}\)

đề học sinh giỏi hồi chiều ak!!!!!!!!! khó v:

a, => |5/3.x| = 1/6

=> 5/3.x = -1/6 hoặc 5/3.x = 1/6

=> x = -1/10 hoặc x = 1/10

Tk mk nha

Lời giải :

Theo đề bài ta có \(\frac{x}{\frac{5}{2}}=\frac{y}{\frac{4}{3}}=\frac{z}{\frac{6}{5}}\Leftrightarrow\frac{2x}{5}=\frac{3y}{4}=\frac{5z}{6}\)

Đặt \(\frac{2x}{5}=\frac{3y}{4}=\frac{5z}{6}=k\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{5k}{2}\\z=\frac{6k}{5}\end{cases}}\)

Mặt khác : \(\frac{x}{2}=\frac{z-28}{3}\)

\(\Leftrightarrow3x-2z=-56\)

\(\Leftrightarrow3\cdot\frac{5k}{2}-2\cdot\frac{6k}{5}=-56\)

\(\Leftrightarrow k=\frac{-560}{51}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{-1400}{51}\\y=\frac{-2240}{153}\\z=\frac{-224}{17}\end{cases}}\)

\(B=x+y-z=\frac{-1400}{51}+\frac{-2240}{153}-\frac{-224}{17}=\frac{-4424}{153}\)