a)\(x=\left(\dfrac{3}{56}\cdot\dfrac{28}{9}\right):\dfrac{-3}{7}=\dfrac{1}{6}:\dfrac{-3}{7}=-\dfrac{7}{18}\)

b)\(x=\left(\dfrac{7}{15}\cdot\dfrac{5}{3}\right)+\dfrac{3}{16}=\dfrac{7}{9}+\dfrac{3}{16}=\dfrac{139}{144}\)

c)\(x=\left(\dfrac{5}{6}-\dfrac{2}{5}\right).5=\dfrac{13}{6}\)

d)\(=>x\left(\dfrac{3}{4}-\dfrac{2}{5}\right)=\dfrac{1}{6}\cdot\left(\dfrac{3}{7}+\dfrac{5}{7}\right)\)

\(x\cdot\dfrac{7}{20}=\dfrac{4}{21}=>x=\dfrac{4}{21}\cdot\dfrac{20}{7}=\dfrac{80}{147}\)

TK

https://lazi.vn/edu/exercise/giai-phuong-trinh-4x-5-x-1-2-x-x-1-7-x-2-3-x-5

a: \(\Leftrightarrow4x-5=2x-2+x\)

=>4x-5=3x-2

=>x=3(nhận)

b: =>7x-35=3x+6

=>4x=41

hay x=41/4(nhận)

c: \(\Leftrightarrow\dfrac{14}{3\left(x-4\right)}-\dfrac{x+2}{x-4}=\dfrac{-3}{2\left(x-4\right)}-\dfrac{5}{6}\)

\(\Leftrightarrow\dfrac{28}{6\left(x-4\right)}-\dfrac{6\left(x+2\right)}{6\left(x-4\right)}=\dfrac{-9}{6\left(x-4\right)}-\dfrac{5\left(x-4\right)}{6\left(x-4\right)}\)

\(\Leftrightarrow28-6x-12=-9-5x+20\)

=>-6x+16=-5x+11

=>-x=-5

hay x=5(nhận)

d: \(\Leftrightarrow x^2+2x+1-\left(x^2-2x+1\right)=16\)

\(\Leftrightarrow4x=16\)

hay x=4(nhận)

a: =>x-3/4=1/6-1/2=1/6-3/6=-2/6=-1/3

=>x=-1/3+3/4=-4/12+9/12=5/12

b: =>x(1/2-5/6)=7/2

=>-1/3x=7/2

hay x=-21/2

c: (4-x)(3x+5)=0

=>4-x=0 hoặc 3x+5=0

=>x=4 hoặc x=-5/3

d: x/16=50/32

=>x/16=25/16

hay x=25

e: =>2x-3=-1/4-3/2=-1/4-6/4=-7/4

=>2x=-7/4+3=5/4

hay x=5/8

Hinh câu 1

a) Ta có: \(\dfrac{2}{3}x-1=\dfrac{3}{2}\)

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

hay \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)

b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)

\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)

c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)

\(\Leftrightarrow\left(x-3\right)^2=64\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)

d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)

\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)

\(\Leftrightarrow x\in\left\{1;2;3\right\}\)

ko biết =))

a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Leftrightarrow x+5=4\)

hay x=-1

b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

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

b) \(\Leftrightarrow x=\dfrac{7}{9}-\dfrac{2}{3}\\ \Leftrightarrow x=\dfrac{1}{9}\)

c) \(\Leftrightarrow x=\dfrac{8}{7}:\dfrac{6}{7}\\ \Leftrightarrow x=\dfrac{4}{3}\)

d) \(\Leftrightarrow x=\dfrac{9}{5}-\dfrac{3}{7}\\ \Leftrightarrow x=\dfrac{48}{35}\)

a) x = 3

b) x = \(\dfrac{1}{9}\)

c) x = \(\dfrac{4}{3}\)

d)\(\dfrac{48}{35}\)

a, \(x^2=\dfrac{1}{16}\Rightarrow x=\pm\dfrac{1}{4}\)

b, \(x^5:x^2=-\dfrac{1}{64}\Rightarrow x^3=\left(-\dfrac{1}{4}\right)^3\Rightarrow x=-\dfrac{1}{4}\)

c, \(x^3:x^2=\dfrac{32}{243}\Rightarrow x=\dfrac{32}{243}\)

d, \(\left(x^2\right)^2=\dfrac{81}{16}\Rightarrow x^4=\left(\dfrac{3}{2}\right)^4\Rightarrow x=\pm\dfrac{3}{2}\)

Chúc bạn học tốt!!!

3) Tìm x

a) \(^{x^2}\)=\(\dfrac{1}{16}\)

<=> x = \(\sqrt{-\dfrac{1}{16}}\)

\(\sqrt{\dfrac{1}{16}}\)

<=> x = -14

+14

b) \(x^{5^{ }}\): \(x^2\) = \(-\dfrac{1}{64}\)

<=> \(^{x^{5-2}}\) =\(-\dfrac{1}{64}\)

<=> \(x^3\) = \(-\dfrac{1}{64}\)

<=> x = \(-\dfrac{1}{4}\)

c)\(x^3:x^2\) = \(\dfrac{32}{243}\)

<=> \(^{x^{3-2}}\) = \(\dfrac{32}{243}\)

<=> x = \(\dfrac{32}{243}\)

d) \((x^2)^2\) = \(\dfrac{81}{16}\)

<=>\(^{x^{2.2}}\) = \(\dfrac{81}{16}\)

<=> \(x^4\) = \(\dfrac{81}{16}\)

<=> x = \(\dfrac{3}{2}\)

\(-\dfrac{3}{2}\)

a) \(x^2=\dfrac{1}{16}\)

=> \(x^2=\pm\left(\dfrac{1}{4}\right)^2\)

=> \(x=\pm\dfrac{1}{4}\)

b) \(x^5:x^2=\dfrac{-1}{64}\)

=> \(x^{5-2}=\dfrac{-1}{64}\)

=> \(x^3=\dfrac{-1}{64}\)

=> \(x^3=\left(\dfrac{-1}{4}\right)^3\)

=> \(x=\dfrac{-1}{4}\)

c) \(x^3:x^2=\dfrac{32}{243}\)

=> \(x^1=\dfrac{32}{243}\)

=> \(x=\dfrac{32}{243}\)

d) \(\left(x^2\right)^2=\dfrac{81}{16}\)

=> \(x^4=\dfrac{81}{16}\)

=> \(x^4=\pm\left(\dfrac{3}{2}\right)^4\)

=> \(x=\pm\dfrac{3}{2}\)

a)4/5+x=2/3

x=2/3-4/5

x=-2/15

b)-5/6-x=2/3

x=-5/6-2/3

x=-3/2

c)1/2x+3/4=-3/10

1/2x=-3/10-3/4

1/2x=-21/20

x=-21/20:1/2

x=-21/10

d)x/3-1/2=1/5

x/3=1/5+1/2

x/3=7/10

10x/30=21/30

10x=21

x=21:10

x=21/10