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10 tháng 1 2017

1/ 3(x - 4) - (8 - x) = 12 => 3x - 12 - 8 + x = 12 => 4x = 32 => x = 8

2/ 4(x - 5) - 3(x - 7) = -9 => 4x - 20 - 3x + 21 = -9 => x = -10

3/ 7(x - 3) -5(3 - x) = 11x - 5 => 7x - 21 - 15 + 5x = 11x - 5 => x = 31

4/ -5(2 - x) + 4(x - 3) = 10x => -10 + 5x + 4x - 12 = 10x => x = -22

10 tháng 1 2017

1) 3.(x-4)-(8-x) = 12

=> 3x-12-8+x = 12

=> 4x - 20 = 12

=> 4x = 12 +20 = 32

=> x = 32: 4 = 8

2) 4.(x-5)-3(x-7) = -9

=> 4x-20-3x+21 = -9

=>x +1 = -9

=>x = -9-1= -10

3) 7.(x-3)-5(3-x) = 11x-5

=> 7x-21-15+5x = 11x -5

=> 12x - 36 = 11x -5

=> 12x-11x = -5+36

hay x = 31

4) -5.(2-x)+4.(x-3) = 10x

=> -10+5x+4x-12 = 10x

=> -22 + 9x = 10x

=> -22 = 10x -9x

hay x = -22

XONG ok

6 tháng 7 2017

1) Ta có : 2x+ 3x - 5

= 2x2 - 2x + 5x - 5

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

= (x - 1) (2x + 5) 

3) x2 + x - 6

= x2 + 2x - 3x - 6

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

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

= (x - 3)(x + 2) 

9 tháng 1 2021

Câu 1 : 

a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)

\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)

\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)

Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)

tương tự 

16 tháng 5 2021

\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)

\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)

\(< =>95-24x+40=6-4x-15x+5\)

\(< =>-24x+135=-19x+11\)

\(< =>5x=135-11=124\)

\(< =>x=\frac{124}{5}\)

17 tháng 8 2020

a, \(12-2\left(1-x\right)^2=\left(3x-2\right)\left(2x-3\right)\)

\(< =>12-2\left(1-2x+x^2\right)=6x^2-9x-4x+6\)

\(< =>12-2+4x-2x^2=6x^2-13x+6\)

\(< =>10+4x-2x^2-6x^2+13x-6=0\)

\(< =>-8x^2+17x+4=0< =>\orbr{\begin{cases}x=\frac{17-\sqrt{417}}{16}\\x=\frac{17+\sqrt{417}}{16}\end{cases}}\)

b, \(10x+3-5x=4x+12< =>5x+3-4x-12=0\)

\(< =>x-9=0< =>x=9\)

c, \(11x+42-2x=100-9x-22< =>9x+42-100+9x+22=0\)

\(< =>18x+64-100=0< =>18x-36=0< =>x=\frac{36}{18}=2\)

d, \(2x-\left(3-5x\right)=4\left(x+3\right)< =>2x-3+5x=4x+12\)

\(< =>7x-3-4x-12=0< =>3x-15=0< =>x=\frac{15}{3}=5\)

e, \(2\left(x-3\right)+5x\left(x-1\right)=5x^2< =>2x-6+5x^2-5=5x^2\)

\(< =>2x-11+5x^2-5x^2=0< =>2x-11=0< =>x=\frac{11}{2}\)

f, \(-6\left(1,5-2x\right)=3\left(-15+2x\right)< =>-6\left(\frac{3}{2}-2x\right)=3\left(2x-15\right)\)

\(< =>-9+12x-6x+45=0< =>6x+36=0< =>x=-6\)

g, \(14x-\left(2x+7\right)=3x+12x-13< =>14x-2x-7=15x-13\)

\(< =>12x-7-15x+13=0< =>-3x+6=0< =>x=-2\)

h, \(\left(x-4\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)

\(< =>x^2-16-6x+4=x^2-8x+16\)

\(< =>x^2-6x-12-x^2+8x-16=0\)

\(< =>2x-28=0< =>x=\frac{28}{2}=14\)

q, \(4\left(x-2\right)-\left(x-3\right)\left(2x-5\right)=?\)thiếu đề

7 tháng 8 2017

mọi người giải giúp mình với ạ

12 tháng 12 2017

1.3x^2-11x+6=3x^2-9x-2x+6=3x(x-3)-2(x-3)=(3x-2)(x-3)

2.8x^2+10x-3=8x^2-2x+12x-3=2x(4x-1)+3(4x-1)=(2x+3)(4x-1)

20 tháng 12 2020

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})`

11 tháng 3 2023

bạn tách một câu vài câu hỏi chứ đừng gộp như thế này ko ai trả lời đâu

a: =>\(4x-5=2x-2+x=3x-2\)

=>x=3

b: \(\Leftrightarrow7x-35=3x+6\)

=>4x=41

=>x=41/4

c: =>(2x+5)(x+5)-2x^2=0

=>2x^2+10x+5x+25-2x^2=0

=>15x=-25

=>x=-5/3

e: \(\Leftrightarrow\dfrac{11}{x}=\dfrac{9x-36+2x+2}{\left(x+1\right)\left(x-4\right)}\)

=>11(x^2-3x-4)=x(11x-34)

=>11x^2-33x-44=11x^2-34x

=>x=44

17 tháng 11 2021

\(1,\Leftrightarrow x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=9\\x=0\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\Leftrightarrow-4x=7\Leftrightarrow x=-\dfrac{7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\Leftrightarrow5x=15\Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(x-7\right)\left(3x+4\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-\dfrac{4}{3}\end{matrix}\right.\)

\(7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ 8,\Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{5}\\x=4\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\\ 11,\Leftrightarrow\left(4x-3\right)\left(3-2x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=\dfrac{3}{2}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\Leftrightarrow-10x=3\Leftrightarrow x=-\dfrac{3}{10}\)

17 tháng 11 2021

\(1,\Leftrightarrow x\left(x-9\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ 2,\Leftrightarrow x^2-4x-x^2=7\\ \Leftrightarrow-4x=7\\ \Leftrightarrow x=\dfrac{-7}{4}\\ 3,\Leftrightarrow3x+2x-10=5\\ \Leftrightarrow5x=15\\ \Leftrightarrow x=3\\ 4,\Leftrightarrow\left(5x-1\right)\left(5x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=-\dfrac{1}{5}\end{matrix}\right.\)

\(5,\Leftrightarrow\left(x-2\right)\left(3x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{5}{3}\end{matrix}\right.\\ 6,\Leftrightarrow\left(3x+4\right)\left(x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=7\end{matrix}\right.\\ 7,\Leftrightarrow\left(2x-3\right)\left(2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

\(8,\Leftrightarrow10x\left(x-4\right)+2\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(10x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{1}{5}\end{matrix}\right.\\ 9,\Leftrightarrow2x^2-5x-2x^2=0\\ \Leftrightarrow-5x=0\\ \Leftrightarrow x=0\\ 10,\Leftrightarrow2x\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

\(11,\Leftrightarrow\left(2x-3\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{3}{4}\end{matrix}\right.\\ 12,\Leftrightarrow2x^2-10x-2x^2=3\\ \Leftrightarrow-10x=3\\ \Leftrightarrow x=-\dfrac{3}{10}\)

a: \(\Leftrightarrow\left(x+12-3x\right)\left(x+12+3x\right)=0\)

=>(-2x+12)(4x+12)=0

=>x=-3 hoặc x=6

b: \(\Leftrightarrow20x^3-15x^2+45x-45=0\)

=>\(x\simeq0.93\)

d: =>-4x+28+11x=-x+3x+15

=>7x+28=2x+15

=>5x=-13

=>x=-13/5

e: \(\Leftrightarrow4x^3-12x+x=4x^3-3x+5\)

=>-9x=-3x+5

=>-6x=5

=>x=-5/6