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2. \(x\left(x+2\right)\left(x+3\right)\left(x+5\right)=280\)
\(\Leftrightarrow x\left(x+5\right)\left(x+2\right)\left(x+3\right)=280\)
\(\Leftrightarrow\left(x^2+5x\right)\left(x^2+5x+6\right)=280\)
Đặt \(x^2+5x+3=t\)
\(\Rightarrow\left(t-3\right)\left(t+3\right)=280\)
\(\Leftrightarrow t^2-9=280\)
\(\Leftrightarrow t^2=289\Leftrightarrow\left[{}\begin{matrix}t=17\\t=-17\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+3=17\\x^2+5x+3=-17\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x-14=0\\x^2+5x+20=0\end{matrix}\right.\)
\(\Leftrightarrow x^2+5x-14=0\text{(vì }x^2+5x+20=\left(x+\dfrac{5}{2}\right)^2+\dfrac{55}{4}>0\forall x\text{)}\)
\(\Leftrightarrow x^2-2x+7x-14=0\)
\(\Leftrightarrow x\left(x-2\right)+7\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+7\right)=0\)
\(\Leftrightarrow\) x - 2 = 0 hoặc x + 7 = 0
\(\Leftrightarrow\) x = 2 hoặc x = - 7
Vậy x = 2 hoặc x = -7.
3. \(\left(x+3\right)\left(x+4\right)\left(x+5\right)=x\)
\(\Leftrightarrow\left(x+3\right)\left(x+4\right)\left(x+5\right)-x=0\)
\(\Leftrightarrow x^3+12x^2+47x+60-x=0\)
\(\Leftrightarrow x^3+12x^2+46x+60=0\)
\(\Leftrightarrow x^3+6x^2+6x^2+36x+10x+60=0\)
\(\Leftrightarrow x^2\left(x+6\right)+6x\left(x+6\right)+10\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x^2+6x+10\right)=0\)
\(\Leftrightarrow x+6=0\text{(vì }x^2+6x+10=\left(x+3\right)^2+1>0\forall x\text{)}\)
\(\Leftrightarrow x=-6\)
Vậy x = -6.
1/(2.x-5)+17=6
=> 2x - 5 = -11
=> 2x = -6
=> x = 3
vậy_
2/10-2.(4-3x)=-4
=> 2(4 - 3x) = 14
=> 4 - 3x = 7
=> 3x = -3
=> x = -1
3/-12+3.(-x+7)=-18
=> 3(-x+7) = -6
=> -x+7 = -2
=> -x = -9
=> x = 9
4/24:(3.x-2)=-3
=> 3x - 2 = -8
=> 3x = -6
=> x = -2
5/-45:5.(-3-2.x)=3
=> 5(-3 - 2x) = -15
=> -3 - 2x = -3
=> - 2x = 0
=> x = 0
6/x.(x+7)=0
=> x = 0 hoặc x + 7 = 0
=> x = 0 hoặc x = -7
7/(x+12).(x-3)=0
=> x + 12 = 0 hoặc x - 3 = 0
=> x = -12 hoặc x = 3
8/(-x+5).(3-x)=0
=> -x + 5 = 0 hoặc 3 - x = 0
=> x = 5 hoặc x = 3
9/x.(2+x).(7-x)=0
=> x = 0 hoặc 2 + x = 0 hoặc 7 - x = 0
=> x = 0 hoặc x = -2 hoặc x = 7
10/(x-1).(x+2).(-x-3)=0
=> x - 1 = 0 hoặc x + 2 = 0 hoặc -x-3 = 0
=> x = 1 hoặc x = -2 hoặc x = -3
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 \(S=\left\{0;-5\right\}\)
1, \(45+x^3-5x^2-9x=9\left(5-x\right)+x^2\left(x-5\right)\)
\(=\left(9-x^2\right)\left(x-5\right)=\left(3-x\right)\left(x+3\right)\left(x-5\right)\)
3, \(x^4-5x^2+4\)
Đặt \(x^2=t\left(t\ge0\right)\)ta có :
\(t^2-5t+4=t^2-t-4t+4=t\left(t-1\right)-4\left(t-1\right)\)
\(=\left(t-4\right)\left(t-1\right)=\left(x^2-4\right)\left(x^2-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)
`Answer:`
1. `45+x^3-5x^2-9x`
`=x^3+3x^2-8x^2-24x+15x+45x`
`=x^2 .(x+3)-8x.(x+3)+15.(x+3)`
`=(x+3).(x^2-8x+15)`
`=(x+3).(x^2-5x-3x+15)`
`=(x-3).(x-5).(x-3)`
2. `x^4-2x^3-2x^2-2x-3`
`=x^4+x^3-3x^3+x^2+x-3x-3`
`=x^3 .(x+1)-3x^2 .(x+1)+x.(x+1)-3.(x+1)`
`=(x+1).(x^3-3x^2+x-3)`
`=(x+1).[x^3 .(x-3).(x-3)]`
`=(x+1).(x-3).(x^2+1)`
3. `x^4-5x^2+4`
`=x^4-x^2-4x^2+4`
`=x^2 .(x^2-1)-4.(x^2-1)`
`=(x^2-1).(x^2-4)`
`=(x-1).(x+1).(x-2).(x+2)`
4. `x^4+64`
`=x^4+16x^2+64-16x^2`
`=(x^2+8)^2-16x^2`
`=(x^2+8-4x).(x^2+8+4x)`
5. `x^5+x^4+1`
`=x^5+x^4+x^3-x^3+1`
`=x^3 .(x^2+x+1)-(x^3-1)`
`=x^3 .(x^2+x+1)-(x-1).(x^2+x+1)`
`=(x^2+x+1).(x^3-x+1)`
6. `(x^2+2x).(x^2+2x+4)+3`
`=(x^2+2x)^2+4.(x^2+2x)+3`
`=(x^2+2x)^2+x^2+2x+3.(x^2+2x)+3`
`=(x^2+2x+1).(x^2+2x)+3.(x^2+2x+1)`
`=(x^2+2x+1).(x^2+2x+3)`
`=(x+1)^2 .(x^2+2x+3)`
7. `(x^3+4x+8)^2+3x.(x^2+4x+8)+2x^2`
`=x^6+8x^4+16x^3+16x^2+64x+64+3x^3+12x^2+24x+2x^2`
`=x^6+8x^4+19x^3+30x^2+88x+64`
8. `x^3 .(x^2-7)^2-36x`
`=x[x^2.(x^2-7)^2-36]`
`=x[(x^3-7x)^2-6^2]`
`=x.(x^3-7x-6).(x^3-7x+6)`
`=x.(x^3-6x-x-6).(x^3-x-6x+6)`
`=x.[x.(x^2-1)-6.(x+1)].[x.(x^2-1)-6.(x-1)]`
`=x.(x+1).[x.(x-1)-6].(x-1).[x.(x+1)-6]`
`=x.(x+1).(x-1).(x^2-3x+2x-6).(x^2+3x-2x-6)`
`=x.(x+1).(x-1).[x.(x-3)+2.(x-3)].[x.(x+3)-2.(x+3)]`
`=x.(x+1)(x-1).(x-2).(x+2).(x-3).(x+3)`
9. `x^5+x+1`
`=x^5-x^2+x^2+x+1`
`=x^2 .(x^3-1)+(x^2+x+1)`
`=x^2 .(x-1).(x^2+x+1)+(x^2+x+1)`
`=(x^2+x+1).(x^3-x^2+1)`
10. `x^8+x^4+1`
`=[(x^4)^2+2x^4+1]-x^4`
`=(x^4+1)^2-(x^2)^2`
`=(x^4-x^2+1).(x^4+x^2+1)`
`=[(x^4+2x^2+1)-x^2].(x^4-x^2+1)`
`=[(x^2+1)^2-x^2].(x^4-x^2+1)`
`=(x^2-x+1).(x^2+x+1).(x^4-x^2+1)
11. ` x^5-x^4-x^3-x^2-x-2`
`=x^5-2x^4+x^4-2x^3+x^3-2x^2+x^2-2x+x-2`
`=x^4 .(x-2)+x^3 ,(x-2)+x^2 .(x-2)+x.(x-2)+(x-2)`
`=(x-2).(x^4+x^3+x^2+x+1)`
12. `x^9-x^7-x^6-x^5+x^4+x^3+x^2-1`
`=(x^9-x^7)-(x^6-x^4)-(x^5-x^3)+(x^2-1)`
`=x^7 .(x^2-1)-x^4 .(x^2-1)-x^3 .(x^2-1)+(x^2-1)`
`=(x^2-1).(x^7-x^4-x^3+1)`
`=(x-1)(x+1)(x^3-1)(x^4-1)`
`=(x-1)(x+1)(x^2+x+1)(x-1)(x^2-1)(x^2+1)`
`=(x-1)^2 .(x+1)(x^2+x+1)(x-1)(x+1)(x^2+1)`
`=(x-1)^3 .(x+1)^2 .(x^2+x+1)(x^2+1)`
13. `(x^2-x)^2-12(x^2-x)+24`
`=[ (x^2-x)^2-2.6(x^2-x)+6^2]-12`
`=(x^2-x+6)^2-12`
`=(x^2-x+6-\sqrt{12})(x^2-x+6+\sqrt{12})`
3 x 25 x 8 + 4 x 6 x 37 + 2 x 38 x 12
= (3 x 8) x 25 + (4 x 6) x 37 + (2 x 12) x 38
= 24 x 25 + 24 x 37 + 24 x 38
= 24 x (25 + 37 + 38)
= 24 x 100
= 2400
3 x 25 x 8 + 4 x 6 x 37 + 2 x 38 x 12
= (3 x 8) x 25 + (4 x 6) x 37 + (2 x 12) x 38
= 24 x 25 + 24 x 37 + 24 x 38
= 24 x (25 + 37 + 38)
= 24 x 100
= 2400
Câu 2 sai đề nhé
Phải là:(x-999)/99+(x-896)/101+(x-789/103)=6
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