a, \(\dfrac{x}{3}=\dfrac{8}{12};12:3=4\)\(;8:4=2\)

⇒\(x=\dfrac{2}{3}\)

a: \(\dfrac{x}{3}=\dfrac{8}{12}\)

nên x/3=2/3

hay x=2

b: \(x+\dfrac{1}{15}=\dfrac{1}{3}\)

\(\Leftrightarrow x=\dfrac{1}{3}-\dfrac{1}{15}=\dfrac{5}{15}-\dfrac{1}{15}=\dfrac{4}{15}\)

c: \(x\cdot\dfrac{4}{7}-\dfrac{2}{7}=\dfrac{5}{7}\)

\(\Leftrightarrow x\cdot\dfrac{4}{7}=\dfrac{5}{7}+\dfrac{2}{7}=1\)

hay x=7/4

d: \(\Leftrightarrow x\cdot\dfrac{3}{4}-\dfrac{5}{4}=-\dfrac{1}{2}\)

\(\Leftrightarrow x\cdot\dfrac{3}{4}=-\dfrac{2}{4}+\dfrac{5}{4}=\dfrac{3}{4}\)

hay x=1

a) Ta có: \(\left(x-3\right)=\left(3-x\right)^2\)

\(\Leftrightarrow\left(x-3\right)^2-\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-4\right)=0\)

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

b) Ta có: \(x^3+\dfrac{3}{2}x^2+\dfrac{3}{4}x+\dfrac{1}{8}=\dfrac{1}{64}\)

\(\Leftrightarrow x^3+3\cdot x^2\cdot\dfrac{1}{2}+3\cdot x\cdot\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3=\dfrac{1}{64}\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{1}{4}\right)^3\)

\(\Leftrightarrow x+\dfrac{1}{2}=\dfrac{1}{4}\)

hay \(x=-\dfrac{1}{4}\)

c) Ta có: \(8x^3-50x=0\)

\(\Leftrightarrow2x\left(4x^2-25\right)=0\)

\(\Leftrightarrow x\left(2x-5\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)

e) Ta có: \(x\left(x+3\right)-x^2-3x=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)

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

f) Ta có: \(x^3+27+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)

\(\Leftrightarrow x\left(x-2\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-3\end{matrix}\right.\)

(2x+3).(10x+2)=(5x+2).(4x+5)

\(a,2^{x+1}=64\\ \Rightarrow a,2^{x+1}=2^6\\ \Rightarrow x+1=6\\ \Rightarrow x=5\)

\(b,x=18\)

\(c,\left(4x-9\right)-\left(x+111\right)=0\\ \Rightarrow4x-9-x-111=0\\ \Rightarrow3x-120=0\\ \Rightarrow3x=120\\ \Rightarrow x=40\)

a
,
2
x
+
1
=
64
⇒
a
,
2
x
+
1
=
2
6
⇒
x
+
1
=
6
⇒
x
=
5

b
,
x
=
18

c
,
(
4
x
−
9
)
−
(
x
+
111
)
=
0
⇒
4
x
−
9
−
x
−
111
=
0
⇒
3
x
−
120
=
0
⇒
3
x
=
120
⇒
x
=
40

VD: 1/2 là 1 phần 2 đó nha.

a: x/3-1/6=1/5

=>x/3=11/30

hay x=11/90

b: =>1/2x=2

hay x=4

c: =>2/3:x=-7-1/3=-22/3

=>x=-1/11

a: x/3-1/6=1/5

=>x/3=11/30

hay x=11/90

b: =>1/2x=2

hay x=4

c: =>2/3:x=-7-1/3=-22/3

=>x=-1/11

a) 3/8 . x = 9/8 - 1

3/8 . x = 1/8

x = 1/8 : 3/8

x = 1/3

b) 4/5 . x = 7/5 - 1/5

4/5 . x = 6/5

x = 6/5 : 4/5

x = 3/2

c) 12/7 : x + 2/3 = 7/5

12/7 : x = 7/5 - 2/3

12/7 : x = 11/15

x = 12/7 : 11/15

x = 180/77

d) 3.(x + 7) - 15 = 27

3.(x + 7) = 27 + 15

3.(x + 7) = 42

x + 7 = 42 : 3

x + 7 = 14

x = 14 - 7

x = 7

a) \(\dfrac{3}{8}x=\dfrac{9}{8}-1\)

\(\Rightarrow\dfrac{3}{8}x=\dfrac{1}{8}\)

\(\Rightarrow x=\dfrac{1}{8}:\dfrac{3}{8}=\dfrac{1}{3}\)

b) \(\dfrac{4}{5}x=\dfrac{7}{5}-\dfrac{1}{5}\)

\(\Rightarrow\dfrac{4}{5}x=\dfrac{6}{5}\)

\(\Rightarrow x=\dfrac{6}{5}:\dfrac{4}{5}=\dfrac{3}{2}\)

c) \(\dfrac{12}{7}:x+\dfrac{2}{3}=\dfrac{7}{5}\)

\(\Rightarrow\dfrac{12}{7}:x=\dfrac{7}{5}-\dfrac{2}{3}\)

\(\Rightarrow\dfrac{12}{7}:x=\dfrac{11}{15}\)

\(\Rightarrow x=\dfrac{12}{7}:\dfrac{11}{15}=\dfrac{180}{77}\)

d) \(3\left(x+7\right)-15=27\)

\(\Leftrightarrow3\left(x+7\right)=42\)

\(\Leftrightarrow x+7=14\Leftrightarrow x=7\)

\(a,1-\left(\dfrac{\dfrac{5}{3}}{8}+x-\dfrac{\dfrac{7}{5}}{24}\right)-\dfrac{\dfrac{16}{2}}{3}=0\\ \Leftrightarrow1-\left(\dfrac{5}{24}+x-\dfrac{7}{120}\right)=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{3}{20}+x=1-\dfrac{8}{3}=-\dfrac{5}{3}\\ \Leftrightarrow x=-\dfrac{5}{3}-\dfrac{3}{20}=-\dfrac{109}{60}\)

Ta có: \(1-\left(\dfrac{5}{\dfrac{3}{8}}+\dfrac{x-7}{\dfrac{5}{24}}\right)-\dfrac{16}{\dfrac{2}{3}}=0\)

\(\Leftrightarrow\dfrac{40}{3}+\dfrac{24\left(x-7\right)}{5}=23\)

\(\Leftrightarrow\left(x-7\right)\cdot\dfrac{24}{5}=\dfrac{29}{3}\)

\(\Leftrightarrow x-7=\dfrac{145}{72}\)

hay \(x=\dfrac{649}{72}\)

bạn đã kiểm tra kĩ chưa vậy?mình đọc đề câu B mà loạn não luôn á;-;

mik kiểm tra rùi

a)

\(\left(x-\dfrac{2}{3}\right):\dfrac{1}{2}=\dfrac{5}{7}\)

\(\Rightarrow x-\dfrac{2}{3}=\dfrac{5}{7}\) x \(\dfrac{1}{2}\)

\(\Rightarrow x-\dfrac{2}{3}=\dfrac{5}{14}\)

\(\Rightarrow x=\dfrac{5}{14}+\dfrac{2}{3}\)

\(\Rightarrow x=\dfrac{15}{42}+\dfrac{2}{42}\)

\(\Rightarrow x=\dfrac{17}{42}\)

b)

\(x\) x \(\dfrac{1}{2}=1-\dfrac{1}{3}\)

\(x\) x \(\dfrac{1}{2}=\dfrac{3}{3}-\dfrac{1}{3}\)

\(x\) x \(\dfrac{1}{2}=\dfrac{2}{3}\)

\(\Rightarrow x=\dfrac{2}{3}:\dfrac{1}{2}\)

\(\Rightarrow x=\dfrac{2}{3}\) x \(2=\dfrac{4}{3}\)

c)

\(\dfrac{26}{5}-x=\dfrac{9}{15}\) x \(\dfrac{25}{3}\)

\(\dfrac{26}{5}-x=5\)

\(\Rightarrow x=\dfrac{26}{5}-5\)

\(\Rightarrow x=\dfrac{26}{5}-\dfrac{25}{5}\)

\(\Rightarrow x=\dfrac{1}{5}\)