\(ĐKXĐ:x\ne-1;x\ne\dfrac{1}{2}\)

Ta có : \(\dfrac{6x+5}{3x+3}=\dfrac{5-4x}{1-2x}\)

\(\Leftrightarrow\left(6x+5\right)\left(1-2x\right)=\left(5-4x\right)\left(3x+3\right)\)

\(\Leftrightarrow6x-12x^2+5-10x=15x+15-12x^2-12x\)

\(\Leftrightarrow5-4x-12x^2-15-3x+12x^2=0\)

\(\Leftrightarrow-10-7x=0\)

\(\Rightarrow x=\dfrac{-10}{7}\)

GIUP MINH BAI NAY DUOC KO

\(a)5\left(3-2x\right)+5\left(x-4\right)=6-4x\)

\(\Leftrightarrow5\left(3-2x+x-4\right)=6-4x\)

\(\Leftrightarrow5\left(-x-1\right)=6-4x\)

\(\Leftrightarrow-5x-5=6-4x\)

\(\Leftrightarrow-x=11\Leftrightarrow x=-11\)

Vậy \(x=-11\)

\(b)-5\left(2-x\right)+4\left(x-3\right)=10x-15\)

\(\Leftrightarrow-10+5x+4x-12=10x-15\)

\(\Leftrightarrow-22+9x=10x-15\)

\(\Leftrightarrow-x=7\Leftrightarrow x=-7\)

Vậy \(x=-7\)

nhiều quá :((

\(a,2\left(x-5\right)-3\left(x+7\right)=14\)

\(2x-10-3x-21=14\)

\(-x-31=14\)

\(-x=45\)

\(x=45\)

\(b,5\left(x-6\right)-2\left(x+3\right)=12\)

\(5x-30-2x-6=12\)

\(3x-36==12\)

\(3x=48\)

\(x=16\)

\(c,3\left(x-4\right)-\left(8-x\right)=12\)

\(3x-12-8+x=0\)

\(4x-20=0\)

\(4x=20\)

\(x=5\)

Cố nốt nha bn ! 

cảm ơn, bn nha:)))

mà hình như bạn TOP 3 trả lời câu hỏi pải ko nhỉ???

d, \(-7\left(3x-5\right)+2\left(7x-14\right)=28\)

\(\Leftrightarrow-21x+35+14x-28=28\Leftrightarrow-7x=21\Leftrightarrow x=-3\)

e, \(5\left(3-2x\right)+5\left(x-4\right)=6-4x\)

\(\Leftrightarrow15-6x+5x-20=6-4x\Leftrightarrow-5-x=6-4x\)

\(\Leftrightarrow-11+3x=0\Leftrightarrow x=\frac{11}{3}\)

f, \(-5\left(2-x\right)+4\left(x-3\right)=10x-15\)

\(\Leftrightarrow-10+5x+4x-12=10x-15\Leftrightarrow-22+9x=10x-15\)

\(\Leftrightarrow-7-x=0\Leftrightarrow x=-7\)

a, \(-15\left(x+2\right)+7\left(2x-3\right)=-50\\ -15x-30+14x-21=-50\\ -x-51=-50\\ -x=-50+51\\ -x=1\\ \Rightarrow x=-1\)Vậy \(x=-1\)

b, \(3\left(x+2\right)-6\left(x-5\right)=2\left(5-2x\right)\\ 3x+6-6x+30=10-4x\\ -3x+36=10-4x\\ -3x+4x=10-36\\ x=-26\)Vậy \(x=-26\)

a, $-15\left(x+2\right)+7\left(2x-3\right)=-50-15x-30+14x-21=-50-x-51=-50-x=-50+51-x=1\Rightarrow x=-1$Vậy $x=-1$

b, $3\left(x+2\right)-6\left(x-5\right)=2\left(5-2x\right)3x+6-6x+30=10-4x-3x+36=10-4x-3x+4x=10-36x=-26$Vậy $x=-26$

1) |x + 2| = 4

\(\Leftrightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-6\end{cases}}\)

2) 3 – |2x + 1| = (-5)

\(\Leftrightarrow\left|2x+1\right|=8\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}\)

3) 12 + |3 – x| = 9

\(\Leftrightarrow\left|3-x\right|=-3\)(vô lí)

=>\(x=\varnothing\) 

1) I x+2 I=4

\(\Rightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-6\end{cases}}}\)

2) \(3-|2x+1|=-5\)

\(\Leftrightarrow|2x+1|=8\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}}\)

3) \(12+|3-x|=9\)

\(\Leftrightarrow|3-x|=-3\)(vô lí vì I 3-x I \(\ge\)0)

1; |\(x\) + 2| = 4

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

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

     \(\left[{}\begin{matrix}x=-6\\x=2\end{matrix}\right.\)

Vậy \(x\) \(\in\) {- 6; 2}

ĐKXĐ: \(x\ne\pm5\)

\(pt\Leftrightarrow\dfrac{3}{4\left(x-5\right)}-\dfrac{15}{2\left(x-5\right)\left(x+5\right)}+\dfrac{7}{6\left(x+5\right)}=0\)

\(\Leftrightarrow\dfrac{9\left(x+5\right)}{12\left(x-5\right)\left(x+5\right)}-\dfrac{90}{12\left(x-5\right)\left(x+5\right)}+\dfrac{14\left(x-5\right)}{12\left(x-5\right)\left(x+5\right)}=0\)

\(\Leftrightarrow9\left(x+5\right)-90+14\left(x-5\right)=0\)

\(\Leftrightarrow23x=-115\Leftrightarrow x=-5\) (không t/m)

Vậy pt vô nghiệm

a, thiếu đề 

b, -5 ( 2 - x ) + 4 ( x - 3 ) = 10x - 15 

<=> -10 + 5x + 4x - 12 = 10x - 15 

<=> -x = 7 <=> x = 7 

c,-7(5-x)-2(x-10)=15

<=> -35 + 7x - 2x + 20 = 15 <=> 5x = 30 <=> x = 6 

d, -4(x+1) + 89x - 3 = 24 

<=> 85x = 31 <=> x = 31/85 

e, 5(x-30-2(x+6)) = 9 

<=> 5 (-x-49) = 9 <=> -5x = 254 <=> x = -254/5 

b: =>-10+5x+4x-12=10x-15

=>9x-10x=-15+22

=>-x=7

hay x=-7

c: =>-35+7x-2x+20=15

=>5x-15=15

=>x=6

d: =>-4x-4+72x-24=24

=>68x-32=24

hay x=14/17

1. \(6x^3-8=40\\ 6x^3=48\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2

2. \(4x^5+15=47\\ 4x^5=32\\ x^5=8\\ \Rightarrow x\in\varnothing\left(\text{vì }x\in N\right)\)Vậy x ∈ ∅

3. \(2x^3-4=12\\ 2x^3=16\\ x^3=8\\ \Rightarrow x=2\)Vậy x = 2

4. \(5x^3-5=0\\ 5x^3=5\\ x^3=1\\ \Rightarrow x=1\)Vậy x = 1

5. \(\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Rightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=6\end{matrix}\right.\)Vậy \(x\in\left\{5;6\right\}\)

6. \(\left(3x-2\right)^{20}=\left(3x-1\right)^{20}\\ \Rightarrow3x-2=3x-1\\ 3x-3x=2-1\\ 0=1\left(\text{vô lí}\right)\)Vậy x ∈ ∅

7. \(\left(3x-1\right)^{10}=\left(3x-1\right)^{20}\\ \left(3x-1\right)^{10}=\left[\left(3x-1\right)^2\right]^{10}\\ \Rightarrow\left(3x-1\right)^2=3x-1\\ \left(3x-1\right)^2-\left(3x-1\right)=0\\ \left(3x-1\right)\left[\left(3x-1\right)-1\right]=0\\ \left(3x-1\right)\left(3x-2\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-1=0\\3x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x=1\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\frac{1}{3}\left(\text{loại vì }x\in N\right)\\x=\frac{2}{3}\left(\text{loại vì }x\in N\right)\end{matrix}\right.\)Vậy x ∈ ∅

8. \(\left(2x-1\right)^{50}=2x-1\\ \left(2x-1\right)^{50}-\left(2x-1\right)=0\\ \left(2x-1\right)\left[\left(2x-1\right)^{49}-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\\left(2x-1\right)^{49}=1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=1\\2x-1=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\frac{1}{2}\left(\text{loại vì }x\in N\right)\\x=1\left(t/m\right)\end{matrix}\right.\)Vậy x = 1

9. \(\left(\frac{x}{3}-5\right)^{2000}=\left(\frac{x}{3}-5\right)^{2008}\\ \left(\frac{x}{3}-5\right)^{2008}-\left(\frac{x}{3}-5\right)^{2000}=0\\ \left(\frac{x}{3}-5\right)^{2000}\left[\left(\frac{x}{3}-5\right)^8-1\right]=0\\ \Rightarrow\left[{}\begin{matrix}\left(\frac{x}{3}-5\right)^{2000}=0\\\left(\frac{x}{3}-5\right)^8=1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}-5=0\\\frac{x}{3}-5=1\\\frac{x}{3}-5=-1\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\frac{x}{3}=5\\\frac{x}{3}=6\\\frac{x}{3}=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\cdot3=15\\x=6\cdot3=18\\x=4\cdot3=12\end{matrix}\right.\)Vậy \(x\in\left\{15;18;12\right\}\)

\(1.6x^3-8=40\\ \Leftrightarrow6x^3=48\\ \Leftrightarrow x^3=8\Leftrightarrow x^3=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{2;-2\right\}\)

\(2.4x^3+15=47\) (T nghĩ đề là mũ 3)

\(\Leftrightarrow4x^3=32\Leftrightarrow x^3=8=2^3=\left(-2\right)^3\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{2;-2\right\}\)

Câu 3, 4 tương tự nhé.

\(5.\left(x-5\right)^{2016}=\left(x-5\right)^{2018}\\ \Leftrightarrow\left(x-5\right)^{2018}-\left(x-5\right)^{2016}=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left[\left(x-5\right)^2-1\right]=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left(x-5-1\right)\left(x-5+1\right)=0\\ \Leftrightarrow\left(x-5\right)^{2016}\left(x-6\right)\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x-5\right)^{2016}=0\\x-6=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\x=6\\x=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)

Vậy \(x\in\left\{4;5;6\right\}\)