b, \(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)

\(x\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)+1=0\)

\(x.0=-1\)

\(\Rightarrow x\in rỗng\)

loz mẹ làm như cái loz

Bài 3:

1. \(\left(x-1\right)\left(x+2\right)+5x-5=0\)

\(\Rightarrow\left(x-1\right)\left(x+2\right)+5\left(x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(x+2+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)

Vậy.......................

2. \(\left(3x+5\right)\left(x-3\right)-6x-10=0\)

\(\Rightarrow\left(3x+5\right)\left(x-3\right)-2\left(3x+5\right)=0\)

\(\Rightarrow\left(3x+5\right)\left(x-3-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)

Vậy........................

3. \(\left(x-2\right)\left(2x+3\right)-7x^2+14x=0\)

\(\Rightarrow\left(x-2\right)\left(2x+3\right)-7x\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(2x+3-7x\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\-5x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{3}{5}\end{matrix}\right.\)

Vậy............................

4, 5 tương tự nhé bn!

bài 3

1 (x-1)(x+2)+5x-5=0

=>(x-1)(x+2)+(5x-5)=o

=>(x-1)(x+2)+5(x-1)=0

=>(x-1)(x+2+5)=0

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

=>\(\left[{}\begin{matrix}x-1=0\\x+7=0\end{matrix}\right.\) =>\(\left[{}\begin{matrix}x=1\\x=-7\end{matrix}\right.\)

vậy x=1 hoặc x=-7

2. (3x+5)(x-3)-6x-10=0

=>(3x+5)(x-3)-(6x+10)=0

=>(3x+5)(x-3)-2(3x+5)=0

=>(3x+5)(x-3-2)=0

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

=>\(\left[{}\begin{matrix}3x+5=0\\x-5=0\end{matrix}\right.\)=>\(\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=5\end{matrix}\right.\)

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, \(\left(x+2\right)^3-x\left(x^2+6x-3\right)=0\Leftrightarrow x^3+4x^2+4x+2x^2+8x+8-x^3-6x^2+3x=0\)

\(\Leftrightarrow15x+8=0\Leftrightarrow x=-\frac{8}{15}\)

b, \(\left(x+4\right)^3-x\left(x+6\right)^2=7\Leftrightarrow12x+64=0\Leftrightarrow x=-\frac{19}{4}\)làm tắt:P 

Tự làm nốt nhé 

b)x^3 - 6x^2 +11x-6=0

<=>x^3 - x^2 - 5x^2 +5x + 6x - 6=0

<=>x^2(x - 1) - 5x(x - 1) +6(x - 1)=0

<=>(x-1).(x^2 - 5x + 6)=0

<=>(x - 1).(x^2 - 2x - 3x + 6)=0

<=>(x - 1).[(x(x-2)-3(x-2)]=0

<=>(x-1)(x-2)(x-3)=0

<=>x-1=0hoac x-2=0 hoac x-3=0

<=>x=1hoac x=2 hoac x=3

bạn mua cái máy tính vinacal xog giải nghiệm ra hết thui

mình cần làm cụ thể =)))

\(a,\left(x+8\right)\left(x+6\right)-x^2=104\)

\(\Rightarrow x^2+14x+48-x^2=104\)

\(\Rightarrow14x=56\)

\(\Rightarrow x=4\)

Vậy x=4