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

\(a,x+13=5\\ \Rightarrow x=5-13\\ \Rightarrow x=-8\\ b,x-11=-18\\ \Rightarrow x=-18+11\\ \Rightarrow x=-7\)

a) x+13=5

x = 5 -13

x = -8

b) x-11=-18

x = -18 + 11

x = -7

\(\dfrac{3}{x+2}=\dfrac{5}{x-3}\left(x\ne-2;x\ne3\right)\)

suy ra: \(3\left(x-3\right)=5\left(x+2\right)\\ < =>3x-9=5x+10\\ < =>3x-5x=10+9\\ < =>-2x=19\\ < =>x=-\dfrac{19}{2}\left(tm\right)\)

\(\dfrac{3}{x+2}=\dfrac{5}{x-3}\)ĐKXĐ \(\left\{{}\begin{matrix}x+2\ne0\\x-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-2\\x\ne3\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{3\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{5\left(x+2\right)}{\left(x+2\right)\left(x-3\right)}\)

`<=> 3(x-3) =5 (x+2)`

`<=> 3x-9 = 5x+10`

`<=>3x -5x=10+9`

`<=> -2x=19`

`<=>x=-19/2`

\(\dfrac{3}{x+2}=\dfrac{5}{x-3}\text{ĐKXĐ}:x\ne-2;3\)

\(\Leftrightarrow\dfrac{3\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{5\left(x+2\right)}{\left(x+2\right)\left(x-3\right)}MTC:\left(x+2\right)\left(x-3\right)\)

\(\Rightarrow3x-9=5x+10\)

\(\Leftrightarrow3x-9-5x-10=0\)

\(\Leftrightarrow-2x-19=0\)

\(\Leftrightarrow-2x=19\)

\(\Leftrightarrow x=\dfrac{-19}{2}\left(\text{nhận}\right)\)

\(\text{Vậy phương trình có tập nghiệm là }S=\left\{\dfrac{-19}{2}\right\}\)

a, Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x+y}{3+5}=\dfrac{16}{8}=2\\ \Rightarrow\left\{{}\begin{matrix}x=6\\y=10\end{matrix}\right.\\ b,x:2=y:\left(-5\right)\Rightarrow\dfrac{x}{2}=\dfrac{y}{-5}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{2}=\dfrac{y}{-5}=\dfrac{x-y}{2-\left(-5\right)}=\dfrac{-7}{7}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-2\\y=5\end{matrix}\right.\)

e: Ta có: \(\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow2x=-7\)

hay \(x=-\dfrac{7}{2}\)

f: Ta có: \(x^3-6x^2+12x-19=0\)

\(\Leftrightarrow x^3-6x^2+12x-8-11=0\)

\(\Leftrightarrow\left(x-2\right)^3=11\)

hay \(x=\sqrt[3]{11}+2\)

a, ĐKXĐ : \(\left\{{}\begin{matrix}x\ge-1\\x>1\end{matrix}\right.\)\(\Rightarrow x>1\)

Ta có : \(PT\Leftrightarrow\sqrt{x+1}=2\sqrt{x-1}\)

\(\Leftrightarrow x+1=4x-4\)

\(\Leftrightarrow3x=5\)

\(\Leftrightarrow x=\dfrac{5}{3}\left(TM\right)\)

Vậy ...

b, ĐKXĐ : \(\left\{{}\begin{matrix}x\ge1\\x>-1\end{matrix}\right.\)\(\Rightarrow x\ge1\)

Ta có : \(PT\Leftrightarrow\sqrt{x-1}=2\sqrt{x+1}\)

\(\Leftrightarrow x-1=4x+4\)

\(\Leftrightarrow3x=-5\)

\(\Leftrightarrow x=-\dfrac{5}{3}\left(L\right)\)

Vậy phương trình vô nghiệm .

a) ĐKXĐ: \(x>1\)

Ta có: \(\dfrac{\sqrt{x+1}}{\sqrt{x-1}}=2\)

\(\Leftrightarrow\sqrt{x+1}=2\sqrt{x-1}\)

\(\Leftrightarrow x+1=4x-4\)

\(\Leftrightarrow x-4x=-4-1\)

\(\Leftrightarrow-3x=-5\)

hay \(x=\dfrac{5}{3}\left(nhận\right)\)

Vậy: \(S=\left\{\dfrac{5}{3}\right\}\)

b) ĐKXĐ: \(\left\{{}\begin{matrix}x>-1\\x\ne1\end{matrix}\right.\)

Ta có: \(\dfrac{\sqrt{x-1}}{\sqrt{x+1}}=2\)

\(\Leftrightarrow\sqrt{x-1}=2\sqrt{x+1}\)

\(\Leftrightarrow x-1=4x+4\)

\(\Leftrightarrow x-4x=4+1\)

\(\Leftrightarrow-3x=5\)

hay \(x=-\dfrac{5}{3}\)(loại)

Vậy: \(S=\varnothing\)

a) 

\(13x+3=16\\ 13x=16-3\\ 13x=13\\ x=13:13\\ x=1\)

b) 

\(2x-138=2^4:2^3\\ 2x-138=2\\ 2x=2+138\\ 2x=140\\ x=140:2\\ x=70\)

a) 

b) 

a) 13x + 3 =16

=> 13x = 13 => x =1

b) 2x - 138 = 2⁴ : 2³

=> 2x  = 2 + 138

=> x = 140 :2 => x = 70

Đúng thì cho 1 like

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

b) \(x^3-x^2-x+1=0\Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\)

\(\Leftrightarrow x-1=0\)   hoặc   \(x+1=0\)

\(\Leftrightarrow x=1\)         hoặc    \(x=-1\)

c) \(x^2-6x+8=0\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\)

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

a) \(x^3+x^2+x+1=0\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

(do \(x^2+1\ge1>0\))

a: Ta có: \(x^3+x^2+x+1=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow x+1=0\)

hay x=-1

b: Ta có: \(x^3-x^2-x+1=0\)

\(\Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\)

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

c: Ta có: \(x^2-6x+8=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=0\)

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