l can't help you :))))

câu hỏi 

\(\left(2x-5\right)^3=\left(2x-5\right)\)

trả lời 

\(x=3\)

\(\left(2x-5\right)^3=2x-5\)

\(\Leftrightarrow\orbr{\begin{cases}2x-5=1\\2x-5=-1\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=6\\2x=2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3\\x=1\end{cases}}}\)

Vậy x=3; x=1

5(3x+5)-4(2x-3)=5x+3(2x+12)+1

15x+25-8x+12=5x+6x+36+1

7x+37=11x+37

7x+37-11x-37=0

-4x=0

x=0

you are my angel~

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

\(\Rightarrow2x-\dfrac{3}{4}=3-x\)

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

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

*Gọi a=x-1, b=2x-3, c=3x-5.

-Phương trình trở thành:

a³+b³+c³-3abc=0 ⇔(a+b)³+c³-3ab(a+b)-3abc=0

⇔(a+b+c)[(a+b)²-c(a+b)+c²]-3ab(a+b+c)=0

⇔(a+b+c)(a²+2ab+b²-ac-bc+c²-3ab)=0

⇔(a+b+c)(a²+b²+c²-ab-ac-bc)=0

⇔a+b+c=0 hay a²+b²+c²-ab-ac-bc=0

*a+b+c=0 ⇔x-1+2x-3+3x-5=0 ⇔6x-9=0 ⇔x=\(\dfrac{3}{2}\)

*a²+b²+c²-ab-ac-bc=0

Vì a²+b²+c²-ab-ac-bc≥0 và dấu bằng xảy ra khi và chỉ khi a=b=c nên

=>x-1=2x-3 ⇔x=2

=>x-1=3x-5 ⇔x=2

=>2x-3=3x-5⇔ x=2

\(-\dfrac{4}{7}-x=\dfrac{3}{5}-2x\\ \Rightarrow-x+2x=\dfrac{3}{5}+\dfrac{4}{7}\\ \Rightarrow x=\dfrac{21}{35}+\dfrac{20}{35}\\ \Rightarrow x=\dfrac{41}{35}\)

Vậy `x=41/35`

__

\(\dfrac{3}{7}x-\dfrac{2}{3}x=\dfrac{10}{21}\\ \Rightarrow\left(\dfrac{3}{7}-\dfrac{2}{3}\right)x=\dfrac{10}{21}\\ \Rightarrow\left(\dfrac{9}{21}-\dfrac{14}{21}\right)x=\dfrac{10}{21}\\ \Rightarrow\dfrac{-5}{21}x=\dfrac{10}{21}\\ \Rightarrow x=\dfrac{10}{21}:\left(-\dfrac{5}{21}\right)\\ \Rightarrow x=-2\)

Vậy `x=-2`

a)

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

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

x = 41/35

Vậy x = 41/35

b)

3/7.x - 2/3.x = 10/21

x(3/7 - 2/3) = 10/21

x.(-5/21) = 10/21

x = 10/21 : (-5/21) = -2

Vậy x = -2

\(\frac{38}{10}:2x=\frac{1}{4}\times\frac{3}{8}\)

\(\frac{38}{10}:2x=\frac{3}{32}\)

\(2x=\frac{38}{10}:\frac{3}{32}\)

\(2x=\frac{38}{10}\times\frac{32}{3}\)

\(2x=\frac{608}{15}\)

\(x=\frac{608}{15}\times\frac{1}{2}\)

\(x=\frac{304}{15}\)

a) (2x - 1) x - x (2x + 3) = 7

<=> x (2x - 1 - 2x - 3) = 7

<=> -4x = 7

<=> x = \(-\dfrac{7}{4}\)

b) 3 (2x - 1) - 5 (x - 3) + 6 (3x - 4) = 24

<=> 6x - 3 - 5x + 15 + 18x - 24 = 24

<=> 19x - 12 = 24

<=> 19x = 36

<=> x = \(\dfrac{36}{19}\)