`2x-2/3=1/2`

`2x=1/2+2/3`

`2x=7/6`

`x=7/6:2=7/12`

\(2x-\dfrac{2}{3}=\dfrac{1}{2}\Leftrightarrow2x=\dfrac{2}{3}+\dfrac{1}{2}=\dfrac{7}{6}\Leftrightarrow x=\dfrac{7}{6}:2=\dfrac{7}{12}\)

2x - 2/3 = 1/2

2x = 1/2 + 2/3 = 7/6

x = 7/6 : 2 = 7/12

vậy x = 7/12

a) |2x-3|+x=21

|2x-3|=21-x

\(\Rightarrow\)\(\orbr{\begin{cases}2x-3=21-x\\2x-3=-\left(21-x\right)\end{cases}}\)

TH1: 2x-3=21-x

2x-x=21+3

x=24

TH2: 2x-3=-(21-x)

2x-3 = -21+x

2x-x=-21+3

x=-18

Vậy x \(\varepsilon\){-18;24}

3/8 x 1/6 + 1/6 x 5/8 = 1/6 

HT

\(\frac{3}{8}\times\frac{1}{6}+\frac{1}{6}\times\frac{5}{8}\)

\(=\frac{1}{6}\times\left(\frac{3}{8}+\frac{5}{8}\right)\)

\(=\frac{1}{6}\times1=\frac{1}{6}\)

Học kĩ cộng trừ nhân chia phân số nhiều vào

tức bà tên Nguyễn Thị Thu Hương r nha. Giúp thì giúp cũng đừng chửi người khác, ko  thích giúp thì cút

Bài làm:

a) \(\left|\frac{1}{2}x-\frac{5}{2}\right|-1=-\frac{1}{2}\)

\(\Leftrightarrow\left|\frac{1}{2}x-\frac{5}{2}\right|=\frac{1}{2}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x-\frac{5}{2}=\frac{1}{2}\\\frac{1}{2}x-\frac{5}{2}=-\frac{1}{2}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{2}x=3\\\frac{1}{2}x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=6\\x=4\end{cases}}\)

+ Nếu x = 6

\(\left|12-\frac{1}{3}y\right|=\frac{5}{6}\)

\(\Leftrightarrow\orbr{\begin{cases}12-\frac{1}{3}y=\frac{5}{6}\\12-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{67}{6}\\\frac{1}{3}y=\frac{77}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{67}{2}\\y=\frac{77}{2}\end{cases}}\)

+ Nếu x = 4

\(\left|8-\frac{1}{3}y\right|=\frac{5}{6}\)

\(\Leftrightarrow\orbr{\begin{cases}8-\frac{1}{3}y=\frac{5}{6}\\8-\frac{1}{3}y=-\frac{5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}\frac{1}{3}y=\frac{43}{6}\\\frac{1}{3}y=\frac{53}{6}\end{cases}}\Rightarrow\orbr{\begin{cases}y=\frac{43}{2}\\y=\frac{53}{2}\end{cases}}\)

Vậy ta có 4 cặp số (x;y) thỏa mãn: \(\left(6;\frac{67}{2}\right);\left(6;\frac{77}{2}\right);\left(4;\frac{43}{2}\right);\left(4;\frac{53}{2}\right)\)

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

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

\(\Leftrightarrow x=\frac{4}{3}\)

Thay vào ta được:

\(\frac{2.\frac{4}{3}+y}{\frac{4}{3}-2y}=\frac{5}{4}\)

\(\Leftrightarrow\frac{32}{3}+4y=\frac{20}{3}-10y\)

\(\Leftrightarrow14y=-4\)

\(\Rightarrow y=-\frac{2}{7}\)

Vậy ta có 1 cặp số (x;y) thỏa mãn: \(\left(\frac{4}{3};-\frac{2}{7}\right)\)

lên google mà kham khảo

x=4

Với x\(\ge1\)\(x-1-\sqrt{x-1}=0< =>x-1=\sqrt{x-1}< =>\left(x-1\right)^2=x-1< =>\left(x-1\right)^2-\left(x-1\right)=0< =>\left(x-1\right)\left(x-1-1\right)=0< =>\left(x-1\right)\left(x-2\right)=0\)\(< =>\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=1\left(TM\right)\\x=2\left(TM\right)\end{matrix}\right.\)

\(\Leftrightarrow x-\sqrt{x}-2=0\left(1\right)\\Đặt:\sqrt{x}=t\left(x\ge0\right)\\ \left(1\right)\Leftrightarrow t^2-t-2=0\\ \Leftrightarrow t^2-2t+\left(t-2\right)=0\\ \Leftrightarrow t.\left(t-2\right)+\left(t-2\right)=0\\ \Leftrightarrow\left(t+1\right).\left(t-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t+1=0\\t-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}t=-1\left(Loại\right)\\t=2\left(Nhận\right)\end{matrix}\right.\\ Với:t=2\Leftrightarrow\sqrt{x}=2\\ \Leftrightarrow x=4\)

yx3/4=7/10-1/5

yx3/4=1/2

y      =1/2:3/4

y      =2/3

7/10 - y x 3/4 = 1/5

7/10 - y          = 1/5 : 3/4

7/10 - y          = 4/15

         y          = 7/10 - 4/15

         y          = 13/30

7/10 - y x 3/4 =1/5

=>y=1/5 :3/4

=>y =4/15

=>y =7/10 - 4/15

=>y =13/30

em khong biet lam vi em chi moi lop4

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