a: =>2x-x=-5/2-1/3

=>x=-17/6

b: =>4(x-2)²=36

=>(x-2)²=9

=>x-2=3 hoặc x-2=-3

hay x=5 hoặc x=-1

c: =>2x+1/2=5/6

=>2x=1/3

hay x=1/6

a: =>2x-x=-5/2-1/3

=>x=-17/6

b: =>4(x-2)²=36

=>(x-2)²=9

=>x-2=3 hoặc x-2=-3

hay x=5 hoặc x=-1

c: =>2x+1/2=5/6

=>2x=1/3

hay x=1/6

a: =>3^x=3^4*3=3^5

=>x=5

b: =>\(2^{x+1}=2^5\)

=>x+1=5

=>x=4

c: \(\Leftrightarrow3^{x+2-3}=3\)

=>x-1=1

=>x=2

d: \(\Leftrightarrow x^2=\dfrac{32}{2}=16\)

=>x=4 hoặc x=-4

e: (2x-1)^4=81

=>2x-1=3 hoặc 2x-1=-3

=>2x=4 hoặc 2x=-2

=>x=-1 hoặc x=2

f: (2x-6)^4=0

=>2x-6=0

=>x-3=0

=>x=3

a) \(3^x=81\cdot3\)

\(\Rightarrow3^x=3^4\cdot3\)

\(\Rightarrow3^x=3^5\)

\(\Rightarrow x=5\)

b) \(2^{x+1}=32\)

\(\Rightarrow2^{x+1}=2^5\)

\(\Rightarrow x+1=5\)

\(\Rightarrow x=4\)

c) \(3^{x+2}:27=3\)

\(\Rightarrow3^{x+2}:3^3=3\)

\(\Rightarrow3^{x+2-3}=3\)

\(\Rightarrow3^{x-1}=3\)

\(\Rightarrow x-1=1\)

\(\Rightarrow x=2\)

d) \(2x^2=32\)

\(\Rightarrow x^2=16\)

\(\Rightarrow x^2=4^2\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

e) \(\left(2x-1\right)^4=81\)

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

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

\(\Rightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)

f)  \(\left(2x-6\right)^4=0\)

\(\Rightarrow2x-6=0\)

\(\Rightarrow2x=6\)

\(\Rightarrow x=6:2\)

\(\Rightarrow x=3\)

\(a,3^x=81\cdot3\\ \Leftrightarrow3^x=3^4\cdot3\\ \Leftrightarrow3^x=3^5\\ \Leftrightarrow x=5\\ d,2^{x+1}=32\\ \Leftrightarrow x+1=5\\ \Leftrightarrow x=4\\ c,3^{x+2}:27=3\\ \Leftrightarrow3^{x+2}:3^3=3\\ \Leftrightarrow3^{x-1}=3\\ \Leftrightarrow x-1=1\\ \Leftrightarrow x=2\\ d,2x^2=32\\ \Leftrightarrow x^2=16\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\\ e,\left(2x-1\right)^4=81\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ f,\left(2x-6\right)^4=0\\ \Leftrightarrow2x-6=0\\ \Leftrightarrow x=3\)

2:

a: \(126⋮x;144⋮x\)

=>x thuộc ƯC(126;144)

mà x lớn nhất

nên x=UCLN(126;144)=18

b: 121 chia x dư 1

=>121-1 chia hết cho x

=>120 chia hết cho x(1)

183 chia x dư 3

=>183-3 chia hết cho 3

=>180 chia hết cho x(2)

Từ (1), (2) suy ra \(x\inƯC\left(120;180\right)\)

mà x lớn nhất

nên x=ƯCLN(120;180)=60

c: 240 và 384 đều chia hết cho x

=>\(x\inƯC\left(240;384\right)\)

=>\(x\inƯ\left(48\right)\)

mà x>6

nên \(x\in\left\{8;12;16;24;48\right\}\)

\(\dfrac{x}{3}=\dfrac{y}{7}\Rightarrow\dfrac{x}{y}=\dfrac{3}{7}\)

\(\dfrac{x}{y}-1=\dfrac{-5}{19}\Rightarrow\dfrac{x}{y}=\dfrac{14}{19}\)

Vô lí => không có x,y thỏa mãn

a) Ta có: \(\dfrac{x}{3}=\dfrac{y}{7}\)

nên \(\dfrac{x}{y}=\dfrac{3}{7}\)

b) Ta có: \(\dfrac{x}{y-1}=\dfrac{5}{-19}\)

\(\Leftrightarrow\dfrac{x}{5}=\dfrac{y-1}{-19}\)

hay \(\dfrac{x}{5}=\dfrac{1-y}{19}\)

a) `4\sqrt(2x-1)>8`

`<=>\sqrt(2x-1)>2`

`<=>2x-1>4`

`<=>x>5/2`

b) `2\sqrtx-1>3`

`<=>2\sqrtx>4`

`<=>\sqrtx>2`

`<=>x>4`

a) Ta có: \(4\sqrt{2x-1}>8\)

\(\Leftrightarrow2x-1>4\)

\(\Leftrightarrow2x>5\)

hay \(x>\dfrac{5}{2}\)

b) Ta có: \(2\sqrt{x}-1>3\)

\(\Leftrightarrow\sqrt{x}>2\)

hay x>4

a

\(x+x^2-x^3-x^4=0\\ \Leftrightarrow x\left(1+x\right)-x^3\left(1+x\right)=0\\ \Leftrightarrow\left(1+x\right)\left(x-x^3\right)=0\\ \Leftrightarrow\left(1+x\right).x.\left(1-x^2\right)=0\\ \Leftrightarrow\left(1+x\right).x.\left(1-x\right)\left(1+x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

b

x^3 chứ: )

\(x^3+27+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow x^3+3^3+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\\ \Leftrightarrow\left(x+3\right).x.\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)

a: Ta có: \(\left(2x+1\right)^2-4\left(x+2\right)^2=9\)

\(\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\)

\(\Leftrightarrow-12x=24\)

hay x=-2

b: Ta có: \(\left(x+3\right)^2-\left(x-4\right)\left(x+8\right)=1\)

\(\Leftrightarrow x^2+6x+9-x^2-4x+32=1\)

\(\Leftrightarrow2x=-40\)

hay x=-20

\(a,-\dfrac{13}{20}+x=\dfrac{-11}{15}\\ \Rightarrow x=\dfrac{-11}{15}+\dfrac{13}{20}\\ \Rightarrow x=-\dfrac{1}{12}\\ b,\left(x-3,5\right):3\dfrac{1}{2}-2,5=-1\dfrac{3}{4}\\ \Rightarrow\left(x-\dfrac{7}{2}\right):\dfrac{7}{2}-\dfrac{5}{2}=\dfrac{-7}{4}\\ \Rightarrow\left(x-\dfrac{7}{2}\right):\dfrac{7}{2}=\dfrac{3}{4}\\ \Rightarrow x-\dfrac{7}{2}=\dfrac{21}{8}\\ \Rightarrow x=\dfrac{49}{8}\)

\(a,\dfrac{5}{8}=\dfrac{x}{14}\)

\(\Rightarrow x=\dfrac{5.14}{8}=8,75\)

Vậy \(x=8,75\)

\(b,\dfrac{x}{6}=-\dfrac{1}{3}\)

\(\Rightarrow x=-\dfrac{1.6}{3}=-2\)

Vậy \(x=-2\)

\(c,-\dfrac{3}{5}=\dfrac{x}{10}\)

\(\Rightarrow x=-\dfrac{3.10}{5}=-6\)

Vậy \(x=-6\)

câu d đã có đáp án

mik đang cần gấp ak

\(+\text{)}\dfrac{5}{8}=14x\)⇒\(x=\dfrac{5}{112}\)

\(+\text{)}\dfrac{x}{6}=-\dfrac{1}{3}\)⇒\(x=-2\)

\(+\text{)}-\dfrac{3}{5}=\dfrac{x}{10}\)⇒\(x=-6\)

\(+\text{)}\dfrac{3}{5}=-\dfrac{9}{11}\)⇒\(\dfrac{33}{55}=\dfrac{45}{55}\)

\(+\text{)}\dfrac{x}{2}=\dfrac{2}{x}\)⇒\(x=+-2\)

\(+\text{)}\dfrac{x}{-5}=\dfrac{-5}{x}\)⇒\(x=+-5\)