khó quá

a. Vì \(\left|x-y-5\right|\ge0\forall x;y;2019\left|y-3\right|^{2020}\ge0\forall y\)

\(\Rightarrow\left|x-y-5\right|+2019\left|y-3\right|^{2020}\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}\left|x-y-5\right|=0\\2019\left|y-3\right|^{2020}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-y-5=0\\y-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-y=5\\y=3\end{cases}}\)

b. \(2\left(x-5\right)^4\ge0\forall x;5\left|2y-7\right|^5\ge0\forall y\)

\(\Rightarrow2\left(x-5\right)^4+5\left|2y-7\right|^5\ge0\)

Dấu "=" xảy ra \(\Leftrightarrow\orbr{\begin{cases}2\left(x-5\right)^4=0\\5\left|2y-7\right|^5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-5=0\\2y-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\\y=\frac{7}{2}\end{cases}}\)

a ra ~ cop bài lộ liễu quá r đó bạn :v nam trần

a, $\frac{x}{5}=\frac{2}{5}$

⇔ x = 2

Vậy x = 2

b, $\frac{3}{8}=\frac{6}{x}$

⇔ 3x = 48

⇔ x = 16

Vậy x = 16

c, $\frac{1}{9}=\frac{x}{27}$

⇔ $\frac{3}{27}=\frac{x}{27}$

⇔ x = 3

Vậy ...

d, $\frac{4}{x}=\frac{2}{5}$

⇔ 2x = 4.5

⇔ x = 2 . 5 = 10

Vậy ....

e, K rõ đề :vv

f, $\frac{x}{-2}=-\frac{8}{x}$

⇔ x2 = - 2 . ( - 8 ) = 16

⇔ $x=\pm 4$

Vậy ..

a. 

| x | + \(\left|-\frac{2}{5}\right|=-\frac{5}{3}\)

| x | + \(\frac{2}{5}=-\frac{5}{3}\)

| x | = \(-\frac{5}{3}-\frac{2}{5}\)

| x | = \(-\frac{31}{15}\)

\(\Rightarrow x\in\varnothing\)vì trị đối \(\ge\)0

Vậy x \(\in\varnothing\)

b.

| x - 3 | = \(\frac{4}{5}\)

\(\Rightarrow\)x - 3 = \(\frac{4}{5}\)hoặc \(-\frac{4}{5}\)

\(\Rightarrow\)x = \(\frac{4}{5}+3\)hoặc \(-\frac{4}{5}+3\)

\(\Rightarrow\)x = \(\frac{19}{5}\)hoặc \(\frac{11}{5}\)

Vậy x \(\in\){ \(\frac{19}{5}\); \(\frac{11}{5}\)} 

c.

| x - 7 | = \(\frac{5}{3}\)

\(\Rightarrow\)x - 7 = \(\frac{5}{3}\)hoặc \(-\frac{5}{3}\)

\(\Rightarrow\)x = \(\frac{5}{3}+7\)hoặc \(-\frac{5}{3}+7\)

\(\Rightarrow\)x = \(\frac{26}{3}\)hoặc \(\frac{16}{3}\)

Vậy x \(\in\){ \(\frac{26}{3}\); \(\frac{16}{3}\)}

d.

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

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

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

\(\Rightarrow\)x = \(\frac{3}{4}\)hoặc \(\frac{1}{4}\)

Vậy x \(\in\){ \(\frac{3}{4}\); \(\frac{1}{4}\)}

e.

| x - 7 | = \(-\frac{5}{3}\)

\(\Rightarrow\)x - 7 = \(-\frac{5}{3}\)hoặc \(\frac{5}{3}\)

\(\Rightarrow\)x = \(-\frac{5}{3}+7\)hoặc \(\frac{5}{3}+7\)

\(\Rightarrow\)x = \(\frac{16}{3}\)hoặc \(\frac{26}{3}\)

Vậy x \(\in\){ \(\frac{16}{3}\); \(\frac{26}{3}\)}

a) Ta có:

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

\(\Rightarrow3\left(2x+1\right)=5\left(x+2\right)\)

\(\Rightarrow6x+3=5x+10\)

\(\Rightarrow6x-5x=10-3\)

\(\Rightarrow x=7\)

b)Ta có:

\(\frac{5}{8x-2}=\frac{-4}{7-x}\)

\(\Rightarrow5\left(7-x\right)=-4\left(8x-2\right)\)

\(\Rightarrow35-5x=-32x+8\)

\(\Rightarrow-5x+32x=8-35\)

\(\Rightarrow27x=-27\)

\(\Rightarrow x=-1\)

c) Ta có:

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

\(\Rightarrow4x=3\left(2x-1\right)\)

\(\Rightarrow4x=6x-3\)

\(\Rightarrow3=6x-4x=2x\)

\(\Rightarrow x=\frac{3}{2}\)

d)Ta có:

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

\(\Rightarrow4\left(2x-1\right)=3\left(3x+1\right)\)

\(\Rightarrow8x-4=9x+3\)

\(\Rightarrow8x-9x=3+4\)

\(\Rightarrow-x=7\Rightarrow x=-7\)

e)Ta có:

\(\frac{4}{x+2}=\frac{7}{3x+1}\)

\(\Rightarrow4\left(3x+1\right)=7\left(x+2\right)\)

\(\Rightarrow12x+4=7x+14\)

\(\Rightarrow12x-7x=14-4\)

\(\Rightarrow5x=10\)

\(\Rightarrow x=2\)

f)Ta có:

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

\(\Rightarrow-3\left(2-2x\right)=4\left(x+1\right)\)

\(\Rightarrow-6+6x=4x+4\)

\(\Rightarrow6x-4x=4+6\)

\(\Rightarrow2x=10\)

\(\Rightarrow x=5\)

a, -5/6 -x = 7/12 + -1/3
⇔-10/12 - 12x/12 = 7/12 + -4/12
⇒-10 - 12x = 7 - 4
⇔-12x = 7 - 4 +10
⇔-12x = 13
⇔x = -13/12
b, x+13/-15 = 1/3
⇔-(x+13)/15 = 5/15
⇒ -x - 13 = 5
⇔-x = 5 +13
⇔-x = 18
⇔x = -18
c,-15/x-1 = -3/5
⇔-75/(x-1).5 = -3.(x-1)/5.(x-1)
⇒-75 = -3x + 3
⇔3x = 3 + 75
⇔3x = 78
⇔x = 26
d, (1/2).x + -2/5 = 1/5
⇔5x/10 + -4/10 = 1/10
⇒5x - 4 = 1
⇔5x = 1 + 4
⇔5x = 5
⇔x = 1
e, (-2/3).x + 1/5 = 1/10
⇔-20x/30 + 6/30 = 3/30
⇒-20x + 6 = 3
⇔-20x = 3 - 6
⇔-20x = -3
⇔x = 3/20
f, 4/5 - (1/2).x = 1/10
⇔8/10 - 5x/10 = 1/10
⇒8 - 5x = 1
⇔-5x = 1 - 8
⇔-5x = -7
⇔x=7/5

a) \(\frac{5}{6}=\frac{x-1}{x}\)

\(5x=6x-6\)

\(6x-5x=6\)

\(x=6\)

các câu còn lại lm tương tự

hok tốt!!

b) \(\frac{1}{2}=\frac{x+1}{3x}\)

\(\Rightarrow1.3x=2.\left(x+1\right)\)

      \(3x=2x+2\)

      \(3x-2x=2\)

            \(x=2\)

Vậy x=2

các câu khác bạn làm tương tự

b)\(\frac{n+2}{n-5}=\frac{n-5+7}{n-5}=\frac{n-5}{n-5}+\frac{7}{n-5}=1+\frac{7}{n-5}\)

=> n-5 thuộc Ư(7)

a) Ta có: \(\frac{x+1}{3}=\frac{2}{6}\)

⇔\(x=\frac{2\cdot3}{6}-1=\frac{6}{6}-1=1-1=0\)

Vậy: x=0

b) Ta có: \(\frac{x-1}{4}=\frac{1}{-2}\)

⇔\(x=\frac{1\cdot4}{-2}+1=\frac{4}{-2}+1=-1\)

Vậy: x=-1

c) Ta có: \(\frac{-1}{6}=\frac{3}{2x}\)

⇔\(2x=\frac{3\cdot6}{-1}=-18\)

hay x=-9

Vậy: x=-9

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

⇔\(\left(x+1\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=3\\x+1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

Vậy: x∈{2;-4}

e) Ta có: \(\frac{4}{5}=\frac{-12}{9-x}\)

⇔\(9-x=\frac{-12\cdot5}{4}=-15\)

hay x=24

Vậy: x=24

f) Ta có: \(\frac{x-1}{-4}=\frac{-4}{x-1}\)

⇔\(\left(x-1\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-3\end{matrix}\right.\)

Vậy: x∈{5;-3}

g) Ta có: \(\frac{5-x}{2}=\frac{2}{5-x}\)

⇔\(\left(5-x\right)^2=4\)

⇔\(\left[{}\begin{matrix}5-x=2\\5-x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=7\end{matrix}\right.\)

Vậy: x∈{3;7}

h) Ta có: \(\frac{4-x}{-5}=\frac{-5}{4-x}\)

⇔\(\left(4-x\right)^2=25\)

⇔\(\left[{}\begin{matrix}4-x=5\\4-x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=9\end{matrix}\right.\)

Vậy: x∈{-1;9}

Cảm ơn bạn