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21 tháng 10 2020

x+4/7=3/4                   x-3/4=2/5                 x x 2/3=3/4x1/2

x=3/4-4/7                     x=2/5+3/4               x x2/3=3/8

x=21/28-16/28             x=8/20+15/20          x=3/8:2/3

x=5/28                          x=23/20                   x=9/16

21 tháng 10 2020

1. Đề:...........

<=> x = 3/4 - 4/7

<=> x = 5/28

Vậy x = 5/28

2. Đề:............

<=> x = 2/5 + 3/4

<=> x = 23/20

Vậy x = 23/20

3. Đề:..........

<=> x. 2/3 = 3/8

<=> x = 3/8. 3/2

<=> x = 9/16

Vậy x = 9/16

2 tháng 3 2022

ai giúp mình với

2 tháng 3 2022

plz

4 tháng 7 2017

a, \(4\left(18-5x\right)-12\left(3x-7\right)=15\left(2x-16\right)-6\left(x+14\right)\)

\(\Rightarrow72-20x-36x+84=30x-240-6x-84\)

\(\Rightarrow-20x-36x-30x+6x=-240-84-72-84\)

\(\Rightarrow-80x=-480\Rightarrow x=6\)

b, \(5\left(3x+5\right)-4\left(2x-3\right)=5x+3\left(2x+12\right)+1\)

\(\Rightarrow15x+25-8x+12=5x+6x+36+1\)

\(\Rightarrow15x-8x-5x-6x=36+1-25-12\)

\(\Rightarrow-4x=0\Rightarrow x=0\)

c, \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Rightarrow10x-16-12x+15=12x-16+11\)

\(\Rightarrow10x-12x-12x=-16+11+16-15\)

\(\Rightarrow-14x=-4\Rightarrow x=\dfrac{2}{7}\)

d, \(5x-3\left\{4x-2\left[4x-3\left(5x-2\right)\right]\right\}=182\)

\(\Rightarrow5x-3\left[4x-2\left(4x-15x+6\right)\right]=182\)

\(\Rightarrow5x-3\left(4x-8x+30x-12\right)=182\)

\(\Rightarrow5x-12x+24x-90x+36=182\)

\(\Rightarrow-73x=182-36\)

\(\Rightarrow-73x=146\Rightarrow x=-2\)

Chúc bạn học tốt!!!

15 tháng 2 2020
https://i.imgur.com/zKeoHqB.jpg

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

\(=x^2+x+2x+2\)

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

=(x+1)(x+2)

2: \(x^2+4x+3\)

\(=x^2+x+3x+3\)

=x(x+1)+3(x+1)

=(x+1)(x+3)

3: \(x^2+5x+4\)

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

=x(x+1)+4(x+1)

=(x+1)(x+4)

4: \(x^2-4x+3\)

\(=x^2-x-3x+3\)

=x(x-1)-3(x-1)

=(x-1)(x-3)

5: \(x^2-4x+4=x^2-2\cdot x\cdot2+2^2=\left(x-2\right)^2\)

6: \(x^2-5x+4\)

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

=x(x-1)-4(x-1)

=(x-1)(x-4)

7: \(x^2-5x+6\)

\(=x^2-2x-3x+6\)

=x(x-2)-3(x-2)

=(x-2)(x-3)

8: \(x^2+6x+5\)

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

=x(x+1)+5(x+1)

=(x+1)(x+5)

9: \(x^2-7x+10\)

\(=x^2-2x-5x+10\)

=x(x-2)-5(x-2)

=(x-2)(x-5)

10: \(x^2+8x+12\)

\(=x^2+2x+6x+12\)

=x(x+2)+6(x+2)

=(x+2)(x+6)

11: \(x^2-8x+16=x^2-2\cdot x\cdot4+4^2=\left(x-4\right)^2\)

12: \(x^2+8x+15\)

\(=x^2+3x+5x+15\)

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

=(x+3)(x+5)

13: \(x^2-8x+7\)

\(=x^2-x-7x+7\)

=x(x-1)-7(x-1)

=(x-1)(x-7)

14: \(x^2+9x+8\)

\(=x^2+x+8x+8\)

=x(x+1)+8(x+1)

=(x+1)(x+8)

15: \(x^2-9x+14\)

\(=x^2-2x-7x+14\)

=x(x-2)-7(x-2)

=(x-2)(x-7)

16: \(x^2+9x+18\)

\(=x^2+3x+6x+18\)

=x(x+3)+6(x+3)

=(x+3)(x+6)

17: \(x^2-9x+20\)

\(=x^2-4x-5x+20\)

=x(x-4)-5(x-4)

=(x-4)(x-5)

18: \(2x^2-3x+1\)

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

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

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

S
19 tháng 8

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

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

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

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

5. \(x^2-4x+4=\left(x-2\right)^2\)

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

7. \(x^2-5x+6=\left(x-2\right)\left(x-3\right)\)

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

9. \(x^2-7x+10=\left(x-2\right)\left(x-5\right)\)

10. \(x^2+8x+12=\left(x+2\right)\left(x+6\right)\)

11. \(x^2-8x+16=\left(x-4\right)^2\)

12. \(x^2+8x+15=\left(x+3\right)\left(x+5\right)\)

13. \(x^2-8x+7=\left(x-1\right)\left(x-7\right)\)

14. \(x^2+9x+8=\left(x+1\right)\left(x+8\right)\)

15. \(x^2-9x+14=\left(x-2\right)\left(x-7\right)\)

16. \(x^2+9x+18=\left(x+3\right)\left(x+6\right)\)

17. \(x^2-9x+20=\left(x-4\right)\left(x-5\right)\)

\(18.2x^2-3x+1=2x^2-x-2x+1\)

\(=x\cdot\left(2x-1\right)-\left(2x-1\right)=\left(2x-1\right)\left(x-1\right)\)

d: =>4x+6=15x-12

=>4x-15x=-12-6=-18

=>-11x=-18

hay x=18/11

e: =>\(45x+27=12+24x\)

=>21x=-15

hay x=-5/7

f: =>35x-5=96-6x

=>41x=101

hay x=101/41

g: =>3(x-3)=90-5(1-2x)

=>3x-9=90-5+10x

=>3x-9=10x+85

=>-7x=94

hay x=-94/7

24 tháng 1 2022

làm rõ ra giúp với ạ, ghi v k hỉu j hết ;-;

Mình thu gọn 2 đa thức trước r mới cộng nhé

\(P\left(x\right)=3x^2+7+2x^4-3x^2-4-5x+2x^3\)

\(P\left(x\right)=\left(3x^2-3x^2\right)+\left(7-4\right)+2x^4-5x+2x^3\)

\(P\left(x\right)=2x^4+2x^3-5x+3\)

\(Q\left(x\right)=-3x^3+2x^2-x^4+x+x^3+4x-2+5x^4\)

\(Q\left(x\right)=\left(-3x^3+x^3\right)+2x^2+\left(-x^4+5x^4\right)+\left(x+4x\right)-2\)

\(Q\left(x\right)=-2x^3+4x^4+2x^2+5x-2\)

\(P\left(x\right)+Q\left(x\right)=2x^4+2x^3-5x+3-2x^3+4x^4+2x^2+5x-2\)

\(P\left(x\right)+Q\left(x\right)=\left(2x^4+4x^4\right)+\left(2x^3-2x^3\right)+\left(-5x+5x\right)+\left(3-2\right)+2x^2\)

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

a: =>x=3/7+3/5=15/35+21/35=36/35

b: =>x/35=4/5-5/7=28/35-25/35=3/35

=>x=3

c: =>x<3/4+8/4=11/4

=>\(x\in\left\{0;1;2;3\right\}\)

d: =>5/3<x<5/6+24/6=29/6

=>\(x\in\left\{2;3;4\right\}\)

e: =>x<10/12-9/12=1/12

=>x=0

f: =>2/3<x<12/6-5/6=7/6

=>x=1

1 tháng 8 2021

a) \(\left(x-4\right)^2-\left(x-4\right)=0\)

\(\left(x-4\right)\left(x-4-1\right)=0\)

\(\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

b) \(5x^2\left(x-7\right)+7\left(x-7\right)=0\)

\(\left(x-7\right)\left(5x^2+7\right)=0\)

\(\left[{}\begin{matrix}x-7=0\\5x^2+7=0\end{matrix}\right.\)

\(\left[{}\begin{matrix}x=7\\x^2=\dfrac{-7}{5}\end{matrix}\right.\)

\(x=7\)

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

\(\left(x-3\right)\left(x^2-1\right)=0\)

\(\left[{}\begin{matrix}x=3\\x=\pm1\end{matrix}\right.\)

1 tháng 8 2021

a) (x - 4)^2=(x - 4)

(x - 4) (x -4)=(x -4 )

(x - 4) (x - 4)-(x - 4)=0

(x-4) (x-4-1)=0

(x-4) (x-5)=0

TH1:x-4=0                          TH2:x-5=0

            x=4                                      x=5

 

1) \(\left(\dfrac{1}{2}x+3\right)\left(x^2-4x-6\right)\)

\(=\dfrac{1}{2}x^3-2x^2-3x+3x^2-12x-18\)

\(=\dfrac{1}{2}x^3+x^2-15x-18\)

2) \(\left(6x^2-9x+15\right)\left(\dfrac{2}{3}x+1\right)\)

\(=4x^3+6x^2-6x^2-9x+10x+15\)

\(=4x^3+x+15\)

3) Ta có: \(\left(3x^2-x+5\right)\left(x^3+5x-1\right)\)

\(=3x^5+15x^2-3x^2-x^4-5x^2+x+5x^3+25x-5\)

\(=3x^5-x^4+5x^3+10x^2+26x-5\)

4) Ta có: \(\left(x-1\right)\left(x+1\right)\left(x-2\right)\)

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

\(=x^3-2x^2-x+2\)