1) 54 : x - x : x = 3 x 2 - 1
2) 42 : x + x - x = 4 x 3 - 4
3) 24 : x - 2 x 4 = 24 - 2 x 3 x 4
X-7 = -5 (x+2) = 0 (3x - 24) . 73 =2.74 | x - 5 | = |-7| x- [42 + ( -28) = -8 ( 7x - x ) - ( 25+7 ) = -25 |x-3| = |5| + | -7 | ( x : 3-4 ) : 5 =15 | x -3 | = 7 - (-2) 128 -3 . (x+4) = 23 4 - ( 7-x ) = x - (13-4 ) BÀI 2 : TÍNH NHANH S1 = 1+2+3+.......999 S1= 10 +12 + 14 +.......2010 S3 + 21 + 23 + 25 + .......10001
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
giải pt
a 4(x+3)^2=(2x+6)^2
b (3x+4)^2=4(x+3)
c (6x+3)^2=(x-4)^2
d (x^2+3x+2)(x^2+3x+3)-2=0
e(x^2-5x)+10(x^2-5x)+24=0
f(x^2+x+1)(x^2+x+2)=12
gx(x+1)(x-1)(x+2)=24
h(x+1)(x+2)(x+3)(x+4)-24=0
lm giúp mik nha các bn
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 \(S=\left\{0;-5\right\}\)
bài 1:
1) 20 - 2(x+4)= 4
2) 30 - [ 4(x-2)+ 15]= 3
3) 14 ⋮(2.x + 3)
4) (3x - 24).7 = 2.73
5) (x + 1)3 = 27
6) (2x - 6).47=49
7) (10 + 2x) : 42011 = 42013
8) [(8x - 12) :4].33 = 36
9) 6⋮(x - 1)
10) (x + 5)⋮(x + 2)
mong mn giúp mình gấp với ạ ^^
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
Phân tích đa thức thành nhân tử:
1, x^3-x+y^3-4
2, 4x^2-y^2+4x+1
3, x^4+2x^3+x^2
4, x^2+5x-6
5, 7x-6x^2-2
6, 5x^2+5xy-x-y
7, 2x^2+3x-5
8,x^4-5x^2+4
9, x^3-5x^2+45-9x
10, x^4-2x^3-2x^2-2x-3
11, 81x^4+4
12,x^5+x+1
13, x^4+6x^3+7x^2-6x+1
14, x(x+4)(x+6)(x+10)+128
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)
Phân tích đa thức thành nhân tử a) A = x . y - x^2 - y^2 + 4 . x + 5
b) B = 19 . x^2 + 54 . y^2 + 16 . z^2 - 16 . x . z -24 .y . z +36 . x . y + 5
c) C = x . y . ( x - 2 ) . ( y + 6 ) + 12 . x^2 - 24 . x + 3 . y^2 + 18 . y + 2016
d) 12.x^4 +3.x^3+x^2-2
e) x^4+3x^3-x^2-4x+2
g) x^4+6x^3+7x^2-6x+1
h)x^9-x^7-x^6-x^5+x^4+x^3+x^2+1
tim a,b,c,d thoa man a^2+b^2+c^2+d^2-ab-bc-cd-d+2/5=0
1. x - 8 +32= 68
2. x +8+32=68
3. 98-x +34=43
4. 98+x-34=43
5. x :5:4=800
6. 18+x=384:8
7. (84,6-2 x x):3,02= 5,1
8. x x5= 120:6
9. (15x 24-x) :0,25= 100:0,25
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. \(\left(360-x\right):0,25=400\Rightarrow360-x=400.0,25=100\Rightarrow x=400-100=300\)
tìm x:
a,1-(43/8+x-173/24):50/3=0
b,1/2.x-2/3.x=3/4
c,11/3.(1/5-1/2)<:=x<;=3/11.(1/5+2/3-1/2)
phân tích đa thức thành nhân tử
a,(x+1).(x+2).(x+3).(x+4)+1
b,(x+1).(x+2).(x+3).(x+4)-24
c,(x+1).(x+3).(x+5).(x+7)+15
d,.(x+2).(x+3).(x+4).(x+5)-24
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)