ĐKXĐ:...

\(\sqrt{3x^2-5x-1}-\sqrt{3x^2-7x+9}+\sqrt{x^2-2}-\sqrt{x^2-3x+13}=0\)

\(\Leftrightarrow\frac{2\left(x-5\right)}{\sqrt{3x^2-5x-1}+\sqrt{3x^2-7x+9}}+\frac{3\left(x-5\right)}{\sqrt{x^2-2}+\sqrt{x^2-3x+13}}=0\)

\(\Leftrightarrow\left(x-5\right)\left(\frac{2}{\sqrt{3x^2-5x-1}+\sqrt{3x^2-7x+9}}+\frac{3}{\sqrt{x^2-2}+\sqrt{x^2-3x+13}}\right)=0\)

\(\Leftrightarrow x-5=0\) (ngoặc to phía sau luôn dương)

\(\Rightarrow x=5\)

Akai Haruma @Nguyễn Việt Lâm

+) \(\left(2x-1\right)^9-\left(2x-1\right)^{11}=0\)

\(\Leftrightarrow\left(2x-1\right)^9=\left(2x-1\right)^{11}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\2x-1=1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}}\)

+) \(\left(2-3x\right)^{13}=\left(3x-2\right)^{12}\)

\(\Leftrightarrow\left(2-3x\right)^{13}=\left(2-3x\right)^{12}\)

\(\Leftrightarrow\orbr{\begin{cases}2-3x=0\\2-3x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{2}{3}\\x=\frac{1}{3}\end{cases}}\)

vô nghiện

theo mik thì vô no

chắc sai đề

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy: \(x=\dfrac{1}{2}\)

===========

b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

\(\Leftrightarrow x=\dfrac{13}{3}\)

Vậy: \(x=\dfrac{13}{3}\)

==========

c/  \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)

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

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

\(\Leftrightarrow x=\dfrac{2}{7}\)

Vậy: \(x=\dfrac{2}{7}\)

==========

f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow-x^3=8\)

\(\Leftrightarrow x=-2\)

Vậy: \(x=-2\)

Lời giải:
PT $\Leftrightarrow (3x+1)(3x+5)(x+1)^2=13$

$\Leftrightarrow (9x^2+18x+5)(x^2+2x+1)=13$

Đặt $x^2+2x+1=t\Rightarrow 9x^2+18x+5=9t-4$

Khi đó pt đã cho trở thành:

$(9t-4)t=13$

$\Leftrightarrow 9t^2-4t-13=0$

$\Leftrightarrow (t+1)(9t-13)=0\Rightarrow t=-1$ hoặc $9t=13$. Vì $t=x^2+2x+1=(x+1)^2\geq 0$ nên suy ra $9t=13$

$\Leftrightarrow 9x^2+18x+9=13$

$\Leftrightarrow 9x^2+18x-4=0$

$\Leftrightarrow x=\frac{-3\pm \sqrt{13}}{3}$

Vậy..........

b. `|x + 1| + |2x - 3| = |3x - 2|`

Ta có: \(\left|x+1\right|+\left|2x-3\right|\ge\left|x+1+2x-3\right|=\left|3x-2\right|\)

\(\Leftrightarrow\left|3x-2\right|=\left|3x-2\right|\) (luôn đúng với mọi x)

Vậy phương trình có vô số nghiệm.

3x - 2x = 13 - 19

   x      = -6

Vậy x = -6

3x + 7   = 4 - 3x

3x + 3x = 7 - 4

 6x        = 3

   x        = 3 : 6

   x        = \(\frac{1}{2}\)

Vậy x = \(\frac{1}{2}\)

5 ( x + 1 ) -3 x   = x + 9         

5 . x + 5 . 1 - 3x = x + 9

5x + 5 - 3x         = x + 9

5x - x - 3x + 5    = 9

4x - 3x               = 9 - 5

 x                      = 4

Vậy x = 4

A) 3x-2x=13-19

=>.      x=-6

   Vậy x=-6

b)3x+7=4-3x

=>3x+3x=4-7

=>.     6x=-3

=>.       x=-3/6

     Vậy x=-3/6

c)5(x+1)-3x=x+9

=>5x+5-3x=x+9

=>. 5x-3x-x=9-5

 =>.          x=4

     Vậy x=4

    Chúc bạn học tốt nhé!

    \(3x-2x=13-19\)

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

Vậy....

   \(3x+7=4-3x\)

\(\Leftrightarrow\)\(6x=-3\)

\(\Leftrightarrow\)\(x=-\frac{1}{2}\)

Vậy...

      \(5\left(x+1\right)-3x=x+9\)

\(\Leftrightarrow\)\(5x+5-3x=x+9\)

\(\Leftrightarrow\)\(x=4\)

Vậy...