3(x-4)-66=4x-80

3x−12−66=4x−80

3x−78=4x−80

−78=4x−80−3x

−78=x−80

−78+80=x

2=x

Dễ thế mà làm cũng ko đc.

Khỏi tích  thấy nói vậy giúp thôi . ok 

3(x-4)-66=4x-80

=>3x-12-66=4x-80

=>3x-78=4x-80

=>3x-4x=78-80

=>x(3-4)=-2

=>x*-1=-2

=>x=2

Vậy x=2

(x+2)+(4x+4)+(7x+6)+...+(25x+18)+(28x+20)=1560

(x+4x+7x+...+25x+28x)+(2+4+6+..+18+20)=1560

145x+110=1560

145x=1450

x=10

vậy........

\(3\cdot\left(x-4\right)=4x-80\)

\(3x-12=4x+80\)

\(3x-12-4x+80=0\)

\(-x+68=0\)

\(-x=-68\)

\(x=68\)

3*(x-4)=4x-80

3x-12=4x-80

4x-3x=80-12

x=68

(!) là giai thừa, 4!=1.2.3.4=24, 2!=1.2=2

a, (24-2).x=66 => 22.x=66 => x=3

KL.

b, (6+24).(2x-1)=25 => 30.(2x-1)=25 => 2x-1=5/6 =>2x=11/6 => x=11/12

KL.

14x6=

\(\left(x+2\right)\left(3-4x\right)=x^2+4x+4\\ \Leftrightarrow6-5x-4x^2=x^2+4x+4\\ \Leftrightarrow5x^2+9x-2=0\\ \Leftrightarrow5\left(x+\dfrac{9}{10}\right)^2=\dfrac{121}{20}\\ \Leftrightarrow\left(x+\dfrac{9}{10}\right)^2=\dfrac{\dfrac{121}{20}}{5}=\dfrac{121}{100}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{9}{10}=\dfrac{11}{10}\\x+\dfrac{9}{10}=-\dfrac{11}{10}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2\end{matrix}\right.\)

vậy x cần tìm là 0,2 và 2

Nguyễn Quốc Khánh xóa bớt tin nhắn đi rồi mà nhắn típ

a) Đặt x^2+2x+2=t

\(\frac{4}{t-1}+\frac{3}{t+1}=\frac{3}{2}\Leftrightarrow\frac{4t+4+3t-3}{t^2-1}=\frac{7t+1}{t^2-1}=\frac{3}{2}\)

\(\Leftrightarrow14t+2=3t^2-3\Leftrightarrow3t^2-14t-5=3t\left(t-5\right)+t-5=0\)\(\Leftrightarrow\left(t-5\right)\left(3t+1\right)=0\Rightarrow\left[\begin{matrix}t=5\\t=-\frac{1}{3}\left(loai\right)\end{matrix}\right.\)

Với t=5 ta có (x+1)^2=4\(\Rightarrow\left[\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=1\\x=-3\end{matrix}\right.\)