1: 

\(\Leftrightarrow\left(x^2+5x+6\right)\left(x^2+5x+4\right)=24\)

\(\Leftrightarrow\left(x^2+5x\right)^2+10\left(x^2+5x\right)=0\)

\(\Leftrightarrow x^2+5x=0\)

=>x=0 hoặc x=-5

3: \(\Leftrightarrow\left(x^2+x+6\right)\left(x^2+x-2\right)=0\)

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

=>x=-2 hoặc x=1

a) \(4\left(x+3\right)^2=\left(2x+6\right)^2\)

\(\Leftrightarrow2^2\left(x+3\right)^2=\left(2x+6\right)^2\)

\(\Leftrightarrow\left(2x+6\right)^2=\left(2x+6\right)^2\)

Vậy tập nghiệm của phương trình là \(S=ℝ\)

b) \(\left(3x+4\right)^2=4\left(x+3\right)\)

\(\Leftrightarrow9x^2+24x+16=4x+12\)

\(\Leftrightarrow9x^2+20x+4=0\)

\(\Leftrightarrow\left(9x+2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}9x+2=0\\x+2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{2}{9}\\x=-2\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{-\frac{2}{9};-2\right\}\)

c) \(\left(6x+3\right)^2=\left(x-4\right)^2\)

\(\Leftrightarrow\orbr{\begin{cases}6x+3=x-4\\6x+3=4-x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}5x+7=0\\7x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{7}{5}\\x=\frac{1}{7}\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{-\frac{7}{5};\frac{1}{7}\right\}\)

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

Đặt \(t=x^2+3x+2\), ta có :

     \(t\left(t+1\right)-2=0\)

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

\(\Leftrightarrow\left(t+2\right)\left(t-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t+2=0\\t-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+3x+4=0\\x^2+3x+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{3}{2}\right)^2+\frac{7}{4}=0\left(ktm\right)\\\left(x+\frac{3}{2}\right)^2-1,25=0\left(tm\right)\end{cases}}\)

\(\Leftrightarrow x=\pm\sqrt{1,25}-\frac{3}{2}=-\frac{3\pm\sqrt{5}}{2}\)(tm)

Vậy tập nghiệm của phương trình là \(S=\left\{-\frac{3\pm\sqrt{5}}{2}\right\}\)

e)Đề bài sai ! Mik sửa :

 \(\left(x^2-5x\right)^2+10\left(x^2-5x\right)+24=0\)

Đặt \(t=x^2-5x\), ta có :

       \(t^2+10t-24=0\)

\(\Leftrightarrow\left(t+12\right)\left(t-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t+12=0\\t-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-5x+12=0\\x^2-5x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x-\frac{5}{2}\right)^2+\frac{23}{4}=0\left(ktm\right)\\\left(x-\frac{5}{2}\right)^2-\frac{33}{4}=0\left(tm\right)\end{cases}}\)

\(\Leftrightarrow x=\pm\frac{\sqrt{33}}{2}+\frac{5}{2}\)

Vậy tập nghiệm của phương trình là \(S=\left\{\frac{\sqrt{33}}{2}+\frac{5}{2};-\frac{\sqrt{33}}{2}+\frac{5}{2}\right\}\)

f) \(\left(x^2+x+1\right)\left(x^2+x+2\right)=12\)

\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2+x+2\right)-12=0\)

Đặt \(t=x^2+x+1\), ta có :

    \(t\left(t+1\right)-12=0\)

\(\Leftrightarrow t^2+t-12=0\)

\(\Leftrightarrow\left(t+4\right)\left(t-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}t+4=0\\t-3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+x+5=0\\x^2+x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{1}{2}\right)^2+\frac{19}{4}=0\left(ktm\right)\\\left(x+\frac{1}{2}\right)^2-\frac{9}{4}=0\left(tm\right)\end{cases}}\)

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

Vậy tập nghiệm của phương trình là \(S=\left\{1;-2\right\}\)

g) \(x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)

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

Đặt \(t=x^2+x\), ta có :

     \(t\left(t-2\right)-24=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}t+4=0\\t-6=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+x+4=0\\x^2+x-6=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{1}{2}\right)^2+\frac{15}{4}=0\left(ktm\right)\\\left(x+\frac{1}{2}\right)^2-\frac{25}{4}=0\left(tm\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{2}-\frac{1}{2}=2\left(tm\right)\\x=-\frac{5}{2}-\frac{1}{2}=-3\left(tm\right)\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{2;-3\right\}\)

h) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24=0\)

\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)

Đặt \(t=x^2+5x+4\), ta có :

     \(t\left(t+2\right)-24=0\)

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

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

\(\Leftrightarrow\orbr{\begin{cases}t+6=0\\t-4=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+5x+10=0\\x^2+5x=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\left(x+\frac{5}{2}\right)^2+\frac{15}{4}=0\left(ktm\right)\\x\left(x+5\right)=0\left(tm\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=-5\left(tm\right)\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{0;-5\right\}\)

cảm ơn bn nha

1: Ta có: \(20-2\left(x+4\right)=4\)

\(\Leftrightarrow2\left(x+4\right)=16\)

\(\Leftrightarrow x+4=8\)

hay x=4

5: Ta có: \(\left(x+1\right)^3=27\)

\(\Leftrightarrow x+1=3\)

hay x=2

2: =(2x+1)^2-y^2

=(2x+1+y)(2x+1-y)

3: =x^2(x^2+2x+1)

=x^2(x+1)^2

4: =x^2+6x-x-6

=(x+6)(x-1)

5: =-6x^2+3x+4x-2

=-3x(2x-1)+2(2x-1)

=(2x-1)(-3x+2)

6: =5x(x+y)-(x+y)

=(x+y)(5x-1)

7: =2x^2+5x-2x-5

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

8: =(x^2-1)*(x^2-4)

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

9: =x^2(x-5)-9(x-5)

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

1. \(x-8+32=68\)

\(x-8=68-32\)

\(x-8=36\)

\(x=36+8\)

\(x=44\)

_____

2. \(x+8+32=68\)

\(x+8=68-32\)

\(x+8=36\)

\(x=36-8\)

\(x=28\)

_____

3. \(98-x+34=43\)

\(98-x=43-34\)

\(98-x=9\)

\(x=98-9\)

\(x=89\)

_____

4. \(98+x-34=43\)

\(98+x=43+34\)

\(98+x=77\)

\(x=77-98\)

\(x=-21\)

_____

5. \(x:5:4=800\)

\(x:5=800.4\)

\(x:5=3200\)

\(x=3200.5\)

\(x=16000\)

_____

6. \(18+x=384:8\)

\(18+x=48\)

\(x=48-18\)

\(x=30\)

_____

7. \(\left(84,6-2\times x\right):3,02=5,1\)

\(84,6-2.x=5,1.3,02\)

\(84,6-2.x=15,402\)

\(2.x=84,6-15,402\)

\(2.x=69,198\)

\(x=69,198:2\)

\(x=34,599\)

_____

8. \(x\times5=120:6\)

\(x.5=20\)

\(x=20:5\)

\(x=4\)

_____

9. \(\left(15\times24-x\right):0,25=100:0,25\)

\(\left(15.24-x\right):0,25=400\)

\(15.24-x=400.0,25\)

\(15.24-x=100\)

\(360-x=100\)

\(x=360-100\)

\(x=260\)

1,x-8+32=68

=>x-8=68-32=36

=>x=36+8=42

2,x+8+32=68

=>x+8=68-32=36

=>x=36-8=28

3,98-x+34=43

=>98-x=43-34=9

=>x=98-9=89

4,98+x-34=43

=>98+x=43+34=77

=>x=77-98=-21

5,x:5:4=800

=>x:20=800

=>x=800x20=16000

6,18+x=384:8=48

=>x=48-18=30

7,(84,6-2x):3,02=5,1

=>84,6-2x=5,1x3,02=15,402

=>2x=84,6-15,402=69,198

=>x=69,198:2=34,599

8,5x=120:6=20

=>x=20:5=4

9,(15x24-x):0,25=100:0,25=400

=>360-x=400x0,25=100

=>x=360-100=260.

1. \(...\Rightarrow x+24=68\Rightarrow x=68-24=44\)

2. \(...\Rightarrow x+40=68\Rightarrow x=68-40=28\)

3. \(...\Rightarrow3^{2\left(8+x\right)}=9=3^2\Rightarrow2\left(8+x\right)=2\Rightarrow8+x=1\Rightarrow x=-7\)

4. \(...\Rightarrow3^{2\left(8+x\right)}=77\) (bạn xem lại đề)

5. \(...\Rightarrow x:5=800.4\Rightarrow x:5=3200\Rightarrow x=3200.5=16000\)

6. \(...\Rightarrow18+x=48\Rightarrow x=48-18=30\)

7. \(...\Rightarrow84,6-2x=5,1.3,02=15,402\Rightarrow2x=84,6-15,402\Rightarrow2x=69,198\Rightarrow x=34,599\)

8. \(...\Rightarrow5.x=20\Rightarrow x=20:5=4\)

9. \(\left(360-x\right):0,25=400\Rightarrow360-x=400.0,25=100\Rightarrow x=400-100=300\)

Bài làm:

a) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)

\(=\left[\left(x+1\right)\left(x+4\right)\right]\left[\left(x+2\right)\left(x+3\right)\right]+1\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)

Đặt \(x^2+5x+5=t\)\(\Rightarrow\left(t-1\right)\left(t+1\right)+1=t^2-1+1=t^2\)

\(=\left(x^2+5x+5\right)^2\)

b) Tương tự như a phân tích và đặt ra được: \(t^2-1-24=t^2-25=\left(t-5\right)\left(t+5\right)\)

\(=\left(x^2+5x\right)\left(x^2+5x+10\right)=x\left(x+5\right)\left(x^2+5x+10\right)\)

c) \(\left(x+1\right)\left(x+3\right)\left(x+5\right)\left(x+7\right)+15\)

\(=\left[\left(x+1\right)\left(x+7\right)\right]\left[\left(x+3\right)\left(x+5\right)\right]+15\)

\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)

Đặt \(x^2+8x+11=t\)\(\Rightarrow\left(t-4\right)\left(t+4\right)+15=t^2-16+15=t^2-1\)

\(=\left(t-1\right)\left(t+1\right)=\left(x^2+8x+10\right)\left(x^2+8x+12\right)\)

\(=\left(x^2+8x+10\right)\left(x+2\right)\left(x+6\right)\)

d) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

Đặt \(x^2+7x+11=t\)\(\Rightarrow\left(t-1\right)\left(t+1\right)-24=t^2-1-24=t^2-25\)

\(=\left(t-5\right)\left(t+5\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

Làm mẫu cho 1 vd:

a, (x+1)(x+2)(x+3)(x+4)+1

\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)+1\)

\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1\)(1)

Đặt \(y=x^2+5x+5\)

Khi đó ::

(1) = \(\left(y-1\right)\left(y+1\right)+1\)

\(=y^2-1+1=y^2\)

Thay vào ta được: \(\left(x^2+5x+5\right)^2\)

a) (x+1)(x+2)(x+3)(x+4)+1=[(x+1)(x+4)].[(x+2)(x+3)]+1=(x²+5x+4)(x²+5x+6)+1

đặt t=x²+5x+5 ta có đa thức (t-1)(t+1)+1=t²-1+1=t². mà t=x²+5x+5

=> (x+1)(x+2)(x+3)(x+4)+1=(x²+5x+5)²

b) (x+1)(x+2)(x+3)(x+4)-24. theo kết quả câu (a) ta được (x+1)(x+2)(x+3)(x+4)=(x²+5x+4)(x²+5x+6)

đặt t=x²+5x+5 ta có đa thức (t-1)(t+1)-24=t²-1-24=t²-25=(t-5)(t+5)

mà t=x²+5x+5 => (t-5)(t+5)=(x²+5x)(x²+5x+10)

c) (x+1)(x+3)(x+5)(x+7)+15=[(x+1)(x+7)].[(x+3)(x+5)]+15=(x²+8x+7)(x²+8x+15)+15

đặt x²+8x+11=t ta có đa thức (t-4)(t+4)+15=t²-16+15=t²-1=(t-1)(t+1)

mà t=x²+8x+11 => (t-1)(t+1)=(x²+8x-10)(x²+8x+12)

d) (x+2)(x+3)(x+4)(x+5)-24=[(x+2)(x+5)][(x+3)(x+4)]-24=(x²+7x+12)(x²+7x+10)-24

đặt t=x²+7x+11 ta có đa thức (t-1)(t+1)-24=t²-1-24=t²-25=(t+5)(t-5)

mà t=x²+7x+11 => (t-5)(t+5)=(x²+7x+6)(x²+7x+16)