1) 6x\(^2\) + 5x - 11 = 0

<=> 6x\(^2\) - 6x + 11x - 11 = 0

<=> 6x . (x - 1) + 11 . (x - 1) = 0

<=> (x - 1)(6x + 11) = 0

<=> \(\orbr{\begin{cases}x-1=0\\6x+11=0\end{cases}}\) <=> \(\orbr{\begin{cases}x=1\\6x=-11\end{cases}}\) <=> \(\orbr{\begin{cases}x=1\\x=-\frac{11}{6}\end{cases}}\)

2) 7x\(^2\) - 4x - 3 = 0

<=> 7x\(^2\) - 7x + 3x - 3 = 0

.<=> 7x . (x - 1) + 3 . (x - 1) = 0

<=> (x - 1)(7x + 3) = 0

<=> \(\orbr{\begin{cases}x-1=0\\7x+3=0\end{cases}}\) <=> \(\orbr{\begin{cases}x=1\\7x=-3\end{cases}}\) <=> \(\orbr{\begin{cases}x=1\\x=-\frac{3}{7}\end{cases}}\)

3) 5x\(^2\) - 2x - 3 = 0

<=> 5x\(^2\) - 5x + 3x - 3 = 0

<=> 5x . (x - 1) + 3 . (x - 1) = 0

<=> (x - 1)(5x + 3) = 0

<=> \(\orbr{\begin{cases}x-1=0\\5x+3=0\end{cases}}\) <=> \(\orbr{\begin{cases}x=1\\5x=-3\end{cases}}\) <=> \(\orbr{\begin{cases}x=1\\x=-\frac{3}{5}\end{cases}}\)

a) Ta có |2x + 3x| - 3x + 2 = 0

=> |2x + 3x| = 3x - 2

ĐK : 3x - 2 \(\ge0\Rightarrow x\ge\frac{2}{3}\)

Khi đó |2x + 3x| = 3x - 2

<=> \(\orbr{\begin{cases}2x+3x=3x-2\\2x+3x=-3x+2\end{cases}}\Rightarrow\orbr{\begin{cases}2x=-2\\8x=2\end{cases}}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{1}{4}\end{cases}}\)(loại)

Vậy không tìm được giá trị của x thỏa mãn

b) ĐK 4x - 3 \(\ge0\Rightarrow x\ge\frac{3}{4}\)

Khi đó |2 + 3x| = 4x - 3

<=> \(\orbr{\begin{cases}2+3x=4x-3\\2+3x=-4x+3\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\left(tm\right)\\x=\frac{1}{7}\left(\text{loại}\right)\end{cases}}\)

Vậy x = 5 là giá trị cần tìm

c) |7x + 1| - |5x + 6| = 0

=> |7x + 1| = |5x + 6|

=> \(\orbr{\begin{cases}7x+1=5x+6\\7x+1=-5x-6\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{7}{12}\end{cases}}\)

Vậy \(x\in\left\{\frac{5}{2};-\frac{7}{12}\right\}\)là giá trị cần tìm

a) \(\left|2x+3x\right|-3x+2=0\)

<=> \(\left|5x\right|-3x+2=0\)

<=> \(\orbr{\begin{cases}5x-3x+2=0\left(x\ge0\right)\\-5x-3x+2=0\left(x< 0\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}x=-1\\x=\frac{1}{4}\end{cases}\left(ktm\right)}\)

b) \(\left|2+3x\right|=4x-3\)

<=> \(\orbr{\begin{cases}2+3x=4x-3\left(x\ge-\frac{2}{3}\right)\\-2-3x=4x-3\left(x< -\frac{2}{3}\right)\end{cases}}\)

<=> \(\orbr{\begin{cases}3x-4x=-3-2\\-3x-4x=-3+2\end{cases}}\)

<=> \(\orbr{\begin{cases}-x=-5\\-7x=-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=5\left(tm\right)\\x=\frac{1}{7}\left(ktm\right)\end{cases}}\)

c) \(\left|7x+1\right|-\left|5x+6\right|=0\)

<=> \(\left|7x+1\right|=\left|5x+6\right|\)

<=> \(\orbr{\begin{cases}7x+1=5x+6\\7x+1=-5x-6\end{cases}}\)

<=> \(\orbr{\begin{cases}x=\frac{5}{2}\\x=-\frac{7}{12}\end{cases}}\)

a: \(=2x^3:\dfrac{-3}{2}x+4x:\dfrac{3}{2}x-5:\dfrac{3}{2}\)

=-4/3x^2+8/3-10/3

=-4/3x^2-2/3

d: \(\dfrac{3x^3-5x+2}{x-3}=\dfrac{3x^3-9x^2+9x^2-27x+22x-66+68}{x-3}\)

\(=3x^2+9x+22+\dfrac{68}{x-3}\)

a)\(\dfrac{1}{6}x+\dfrac{1}{10}x-\dfrac{4}{15}x+1=0\)
\(\left(\dfrac{1}{6}+\dfrac{1}{10}-\dfrac{4}{15}\right).x+1=0\)
\(\left(\dfrac{5}{30}+\dfrac{3}{30}-\dfrac{8}{30}\right).x+1=0\)
\(0.x+1=0\)
\(0.x=-1\)
=> Không có giá trị nào của x.
Vậy...
b)\(\left(\dfrac{1}{7}x-\dfrac{2}{7}\right).\left(-\dfrac{1}{5}x+\dfrac{3}{5}\right).\left(\dfrac{1}{3}x+\dfrac{4}{3}\right)=0\)
=> \(\dfrac{1}{7}x-\dfrac{2}{7}=0hoặc-\dfrac{1}{5}x+\dfrac{3}{5}=hoăc\dfrac{1}{3}x+\dfrac{4}{3}=0\)
+)\(~\dfrac{1}{7}x-\dfrac{2}{7}=0\) +) \(-\dfrac{1}{5}x+\dfrac{3}{5}=0\) +) \(\dfrac{1}{3}x+\dfrac{4}{3}=0\)
\(\dfrac{1}{7}x=-\dfrac{2}{7}\) \(-\dfrac{1}{5}x=-\dfrac{3}{5}\) \(\dfrac{1}{3}x=-\dfrac{4}{3}\)
\(x=2\) \(x=3\) \(x=-4\)
Vậy...

a 1/6x+1/10x-4/15x+1=0

(1/6+1/10-4/15)x+1=0

0x+1=0

0x=-1

x=-1/0

Vậy không có x (vì không có số nào chia cho 0)

Chịu rhif ki cần trả lời à

a)  |2x+3x|=|4x-3|

\(\Rightarrow\orbr{\begin{cases}2x+3x=4x-3\\2x+3x=-4x+3\end{cases}\Rightarrow\orbr{\begin{cases}5x-4x=-3\\5x+4x=3\end{cases}\Rightarrow}\orbr{\begin{cases}x=-3\\9x=3\end{cases}\Rightarrow}\orbr{\begin{cases}x=-3\\x=\frac{1}{3}\end{cases}}}\)

b) |7x|-|5x+6|=0

=>|7x|=|5x+6|

\(\Rightarrow\orbr{\begin{cases}7x=5x+6\\7x=-5x-6\end{cases}\Rightarrow\orbr{\begin{cases}7x-5x=6\\7x+5x=-6\end{cases}\Rightarrow}\orbr{\begin{cases}2x=6\\12x=-6\end{cases}\Rightarrow}\orbr{\begin{cases}x=3\\x=\frac{-1}{2}\end{cases}}}\)

c) |3/2+1/2|=|4x-1|

=>|4x-1|=2

\(\Rightarrow\orbr{\begin{cases}4x-1=2\\4x-1=-2\end{cases}\Rightarrow\orbr{\begin{cases}4x=3\\4x=-1\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{3}{4}\\x=\frac{-1}{4}\end{cases}}}\)