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31 tháng 3 2019

CÂU 1:

a) \(2x+4+x^2=-2x+x-3x+2x\)

\(\Leftrightarrow2x+4+x^2=-2x\)

\(\Leftrightarrow x^2+4x+4=0\)

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

\(\Leftrightarrow x+2=0\)

\(\Leftrightarrow x=-2\)

b) \(2x^2-5x-x=x^2+6x\)

\(\Leftrightarrow2x^2-5x-x-x^2-6x=0\)

\(\Leftrightarrow3x^2-12x=0\)

\(\Leftrightarrow3x\left(x-4\right)=0\)

Hoặc \(3x=0\Leftrightarrow x=0\)

Hoặc \(x-4=0\Leftrightarrow x=4\)

nhiều quá :((

\(a,2\left(x-5\right)-3\left(x+7\right)=14\)

\(2x-10-3x-21=14\)

\(-x-31=14\)

\(-x=45\)

\(x=45\)

\(b,5\left(x-6\right)-2\left(x+3\right)=12\)

\(5x-30-2x-6=12\)

\(3x-36==12\)

\(3x=48\)

\(x=16\)

\(c,3\left(x-4\right)-\left(8-x\right)=12\)

\(3x-12-8+x=0\)

\(4x-20=0\)

\(4x=20\)

\(x=5\)

Cố nốt nha bn ! 

19 tháng 3 2020

cảm ơn, bn nha:)))

mà hình như bạn TOP 3 trả lời câu hỏi pải ko nhỉ???

20 tháng 6 2015

Tính nhanh mỗi biểu thức sau:

a, 0 + 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20

= (0 + 20) + (1 + 19) + (2 + 18) + (3 + 17) + (4 + 16) + (5 + 15) + (6 + 14) + (7 + 13) + (8 + 12) + (9 + 11) + 10

= 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 20 + 10

= 20 x 10 + 10

= 200 + 10

= 210

b, 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x (4 x 9 - 36)

= 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x (36 - 36)

= 1 x 2 x 3 x 4 x 5 x 6 x 7 x 8 x 9 x 0

= A x 0

= 0

c, (81 - 7 x 9 - 18) : (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)

= (81 - 63 - 18) : (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)

= (18 - 18) : (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)

= 0 :(1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9)

= 0 : A

= 0

d, (6 x 5 + 7 - 37) x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)

= (30 + 7 - 37)  x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)

= (37 - 37)  x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)

= 0  x (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10)

= 0 x A

= 0

e, (11 x 9 - 100 + 1) : (1 x 2 x 3 x 4 x ... x 10)

= (99 - 100 + 1) : (1 x 2 x 3 x 4 x ... x 10)

= (99 + 1 - 100) : (1 x 2 x 3 x 4 x ... x 10)

= (100 - 100) : (1 x 2 x 3 x 4 x ... x 10)

= 0 : (1 x 2 x 3 x 4 x ... x 10)

= 0 : A

= 0

g, (m : 1 - m x 1) : (m x 2008 + m x 2008)

= (m - m) : (m x 2008 + m x 2008)

= 0 : (m x 2008 + m x 2008)

= 0 : A

= 0

h, (2 + 4 + 6 + 8 + m x n) x (324 x 3 - 972)

= (2 + 4 + 6 + 8 + m x n) x (972 - 972)

= (2 + 4 + 6 + 8 + m x n) x 0

= A x 0

= 0

l, (1 + 2 + 3 + ... + 99)  x (13 x 15 - 12 x 15 - 15)

= (1 + 2 + 3 + ... + 99) x (15 x (13 - 12 - 1))

= (1 + 2 + 3 + ... + 99) x (15 x 0)

= (1 + 2 + 3 + ... + 99) x 0

= A x 0

= 0

i, (0 x 1 x 2 x...x 99 x 100) : (2 + 4 + 6 +...+ 98)

= 0 x : (2 + 4 + 6 +...+ 98)

= 0 x A

= 0

k, (0 + 1 + 2 +...+ 97 + 99) x (45 x 3 - 45 x 2 - 45)

= (0 + 1 + 2 +...+ 97 + 99) x (45 x (3 - 2 - 4))

= (0 + 1 + 2 +...+ 97 + 99) x (45 x 0)

= (0 + 1 + 2 +...+ 97 + 99) x 0

= A x 0

= 0

11 tháng 1 2022

GIÚP MÌNH VỚI  

(-7) NHÂN (-24) + (-36) : (-3)^2 - (-5)^3

a: =7/8:(2/9-18+1/36)-5/12

=-7/142-5/12=-397/852

b: =3/7(4/9+5/9:6/12)=2/3

c: =5^8(16/31-47/31)+1/3=-5^8+1/3

d: =7/2(3/8+5/8:4/15)=609/64

14 tháng 6 2023

Cho nên kết quả cau c đi

1) |x + 2| = 4

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

2) 3 – |2x + 1| = (-5)

\(\Leftrightarrow\left|2x+1\right|=8\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}\)

3) 12 + |3 – x| = 9

\(\Leftrightarrow\left|3-x\right|=-3\)(vô lí)

=>\(x=\varnothing\) 

25 tháng 2 2020

1) I x+2 I=4

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

2) \(3-|2x+1|=-5\)

\(\Leftrightarrow|2x+1|=8\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}}\)

3) \(12+|3-x|=9\)

\(\Leftrightarrow|3-x|=-3\)(vô lí vì I 3-x I \(\ge\)0)

28 tháng 4 2022

a, 4x+1=13-2x <-->6x=12 <-->x=2

b, (2x-5)(x-4)=0 <-->x=5/2  hoặc x=4

c,Đề bài -->x(x-2)+6(x+2)=2x+12 -->x^2+2x=0 -->x=0  hoặc x=-2

d,|x-3|=9-2x -->TH1: x-3=9-2x -->x=x=4           TH2:3-x=9-2x -->x=6

 
AH
Akai Haruma
Giáo viên
3 tháng 8 2021

a. ĐKXĐ: $x\geq 0$

PT $\Leftrightarrow -5x-5\sqrt{x}+12\sqrt{x}+12=0$

$\Leftrightarrow -5\sqrt{x}(\sqrt{x}+1)+12(\sqrt{x}+1)=0$

$\Leftrightarrow (\sqrt{x}+1)(12-5\sqrt{x})=0$

Dễ thấy $\sqrt{x}+1>1$ với mọi $x\geq 0$ nên $12-5\sqrt{x}=0$

$\Leftrightarrow \sqrt{x}=\frac{12}{5}$

$\Leftrightarrow x=5,76$ (thỏa mãn)

 

AH
Akai Haruma
Giáo viên
3 tháng 8 2021

b. ĐKXĐ: $x^2\geq 5$

PT $\Leftrightarrow \frac{1}{3}\sqrt{4}.\sqrt{x^2-5}+2\sqrt{\frac{1}{9}}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$

$\Leftrightarrow \frac{2}{3}\sqrt{x^2-5}+\frac{2}{3}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$

$\Leftrightarrow -\frac{5}{3}\sqrt{x^2-5}=0$

$\Leftrightarrow \sqrt{x^2-5}=0$

$\Leftrightarrow x=\pm \sqrt{5}$

a: x*3/4=1/5

=>x=1/5:3/4=1/5*4/3=4/15

b: x*3/7=2/5

=>x=2/5:3/7=2/5*7/3=14/15

c: 1/3+2/9=2/12x

=>1/6x=3/9+2/9=5/9

=>x=5/9*6=30/9=10/3

d: 4/15*x-2/3=1/5

=>4/15*x=2/3+1/5=10/15+3/15=13/15

=>4x=13

=>x=13/4

e: x:1/7=2/3

=>x=2/3*1/7=2/21

f: 1/9:x=7/3

=>x=1/9:7/3=1/9*3/7=3/63=1/21

j: 1/4+5/12=8/3:x

=>8/3:x=3/12+5/12=8/12=2/3

=>x=4

h: =>7/4:x=1/5+1/2=7/10

=>x=7/4:7/10=10/4=5/2

30 tháng 8 2023

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10 tháng 1 2018

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.

10 tháng 1 2018

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.

a: \(\dfrac{-8}{3}\cdot\dfrac{15}{24}=\dfrac{-8}{24}\cdot\dfrac{15}{3}=\dfrac{-1}{3}\cdot5=-\dfrac{5}{3}\)

b: \(=\dfrac{3}{4}\cdot\dfrac{1}{-9}=\dfrac{-1}{12}\)

c: \(x=-\dfrac{7}{6}+\dfrac{5}{8}=-\dfrac{13}{24}\)

d: \(x=-\dfrac{14}{25}-\dfrac{3}{4}=\dfrac{-56-75}{100}=\dfrac{-131}{100}\)

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 : 

\(t^2-5t+4=t^2-t-4t+4=t\left(t-1\right)-4\left(t-1\right)\)

\(=\left(t-4\right)\left(t-1\right)=\left(x^2-4\right)\left(x^2-1\right)=\left(x-2\right)\left(x+2\right)\left(x-1\right)\left(x+1\right)\)

29 tháng 3 2022

`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})`