\(\left(x+1\right)^2-3\left(x+1\right)=\left(x+1\right)\left(x+1-3\right)=\left(x+1\right)\left(x-2\right)\)

\(2x\left(x-2\right)-\left(x-2\right)^2=\left(x-2\right)\left[2x-\left(x-2\right)\right]=\left(x-2\right)\left(2x-x+2\right)=\left(x-2\right)\left(x+2\right)\)

\(4x^2-20xy+25y^2=\left(2x\right)^2-2.2x.5y+\left(5y\right)^2=\left(2x-5y\right)^2\)

\(x^2+3x-x-3=x\left(x+3\right)-\left(x+3\right)=\left(x-1\right)\left(x+3\right)\)

\(x^2-xy+x-y=x\left(x-y\right)+\left(x-y\right)=\left(x-y\right)\left(x+1\right)\)

\(2y\left(x+2\right)-3x-6=2y\left(x+2\right)-3\left(x+2\right)=\left(x+2\right)\left(2y-3\right)\)

Yêu cầu đề bài là gi bạn?

tim X nha

giup

a)    bạn nhóm 2 cái cuối thành 1 nhóm,  2 cái ở giữa thành 1 nhóm,  rồi đặt ẩn phụ là ra

         \(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)

\(\Leftrightarrow\)\(\left(x^2+3x\right)\left(x^2+3x+2\right)-24=0\)

Đặt  \(x^2+3x=t\)   ta có:

                  \(t\left(t+2\right)-24=0\)

\(\Leftrightarrow\)\(t^2+2t-24=0\)

\(\Leftrightarrow\)\(\left(t-4\right)\left(t+6\right)=0\)

đến đây bn thay trở lại rồi tìm nghiệm nhé

a, x(x+3)(x+1)(x+2)-24=0

=> (x^2+3x)(x^2+3x+2)-24=0

đặt x^2+3x=a

ta có : a(a+2)-24=0

=> a^2+2a-24=0 => \(\orbr{\begin{cases}a=4\\a=-6\end{cases}}\) giải ra ta được x^2+3x=4 hay \(\orbr{\begin{cases}x=-4\\x=1\end{cases}}\)

và x^2+3x=-6 => vô nghiệm vậy x=-4 hoặc x=1

b, 5x(2-3x)-(4-6x)=0

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

=>(5x-2)(2-3x)=0 

=>\(\orbr{\begin{cases}x=\frac{3}{2}\\x=\frac{2}{5}\end{cases}}\)

\(a,\left(x-2\right)^2-\left(x-3\right)\left(x^2+3x+9\right)+6\left(x+1\right)^2=15\)\(\Leftrightarrow x^3-6x^2+12x-8-x^3+27+6\left(x^2+2x+1\right)=15\)\(\Leftrightarrow-6x^2+12x+19+6x^2+12x+6=15\)

\(\Leftrightarrow24x=-10\)

\(\Leftrightarrow x=-\dfrac{5}{12}\)

Vậy:....

\(b,\left(5x+1\right)^2-\left(5x+3\right)\left(5x-3\right)=30\)

\(\Leftrightarrow25x^2+10x+1-25^2+9=30\)

\(\Leftrightarrow10x=20\)

\(\Rightarrow x=2\)

Vậy :....

\(c,\left(x+3\right)\left(x^2-3x+9\right)-x\left(x-2\right)\left(x+2\right)=15\)\(\Leftrightarrow x^3+27-x\left(x^2-4\right)=15\)

\(\Leftrightarrow x^3+27-x^3+4x=15\)

\(\Leftrightarrow4x=15-27=-12\)

\(\Leftrightarrow x=-3\)

vậy : .....

1/3x+2/5(x+1)=0

1/3x+2/5x+2/5=0

11/15x            =-2/5

 x                   =-2/5:11/15

x                     =-6/11

\(\left|3x-2018\right|+\left|x-2017\right|=\left|2x-1\right|\)

\(\Rightarrow\orbr{\begin{cases}3x-2018+x-2017=2x-1\\-\left(3x-2018\right)+\left[-\left(x-2017\right)\right]=2x-1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}4x-4035=2x-1\\\left(-3x-x\right)+\left(2018+2017\right)=2x-1\end{cases}}\)

Làm tiếp

TH2:

\(\left|3x-2018\right|+\left|x-2017\right|=\left|2x-1\right|\)

\(\Rightarrow\orbr{\begin{cases}3x-2018+x-2017=-2x+1\\-\left(3x-2018\right)+\left[-\left(x-2017\right)\right]=-2x+1\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}4x-4035=-2x+1\\\left(-3x-x\right)+\left(2018+2017\right)=-2x+1\end{cases}}\)

Tự tiếp tiếp nha bạn

Bài sau cũng tg tự vậy mà làm