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13 tháng 8 2020

b) ( 2x + 3)2 – (2x + 1)(2x – 1) = 22

<=> 4x2 + 12x + 9 - 4x2 + 1 = 22

<=> 12x + 10 = 22

<=> 12x = 12

<=> x = 1

Vậy x = 1

c) (4x + 3)(4x – 3) – (4x - 5)2 = 16

<=> 16x2 - 9 - 16x2 + 40x - 25 = 16

<=> 40x - 34 = 16

<=> 40x = 50

<=> \(\frac{5}{4}\)

Vậy \(x=\frac{5}{4}\)

e) (x + 1)3 – x2(x + 3) = 2

<=> x3 + 3x2 + 3x + 1 - x3 - 3x2 = 2

<=> 3x + 1 = 2

<=> 3x = 2

<=> x = \(\frac{2}{3}\)

Vậy \(x=\frac{2}{3}\)

f) (x – 2)3 – x(x – 1)(x + 1) + 6x2 = 5

<=> x3 - 6x2 + 12x - 8 - x3 + x + 6x2 = 5

<=> 13x - 8 = 5

<=> 13x = 13

<=> x = 1

Vậy x = 1

25 tháng 8 2019

a) \(\left(x-3\right)^2-4=0\)

\(\left(x-3\right)^2=0+4\)

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

\(\left(x-3\right)^2=\pm4\)

\(\left(x-3\right)^2=\pm2^2\)

\(\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)

\(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)

\(4x^2+12x+9-4x^2+1=22\)

\(12x+10=22\)

\(12x=22-10\)

\(12x=12\)

\(x=1\)

c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)

\(16x^2-9-16x^2+40x-25=16\)

\(-34+40x=16\)

\(40x=16+34\)

\(40x=50\)

\(x=\frac{50}{40}=\frac{5}{4}\)

d) \(x^3-9x^2+27x-27=-8\)

\(x^3-9x^2+27x-27+8=0\)

\(x^3-9x^2+27x-19=0\)

\(\left(x^2-8x+19\right)\left(x-1\right)=0\)

Vì \(\left(x^2-8x+19\right)>0\) nên:

\(x-1=0\)

\(x=1\)

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

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

\(3x+1=2\)

\(3x=2-1\)

\(3x=1\)

\(x=\frac{1}{3}\)

25 tháng 8 2019

b) ( 2x+3)^2 - (2x+1)(2x-1) =22

=> 4x2+12x+9-4x2+1=22

=> 12x=12

=>x=1

c) (4x+3)(4x-3) -(4x-5)^2 =16

=>16x2-9-16x2+40x-25=16

=>40x=50

=>x=4/5

25 tháng 8 2019

a)\(\left(x-13\right)^2-4=0\\\left(x-13\right)^2=4\\ \left(x-13\right)^2=2^2\\ \Rightarrow\left\{{}\begin{matrix}x-13=2\\x-13=-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}15\\-11\end{matrix}\right.\)

vậy...

21 tháng 9 2020

a) ( x - 3 )2 - 4 = 0

<=> ( x - 3 )2 - 22 = 0

<=> ( x - 3 - 2 )( x - 3 + 2 ) = 0

<=> ( x - 5 )( x - 1 ) = 0

<=> x = 5 hoặc x = 1

b( 2x + 3 )2 - ( 2x + 1 )( 2x - 1 ) = 22

<=> 4x2 + 12x + 9 - ( 4x2 - 1 ) = 22

<=> 4x2 + 12x + 9 - 4x2 + 1 = 22

<=> 12x + 10 = 22

<=> 12x = 12

<=> x = 1

c) ( 4x + 3 )( 4x - 3 ) - ( 4x - 5 )2 = 16

<=> 16x2 - 9 - ( 16x2 - 40x + 25 ) = 16

<=> 16x2 - 9 - 16x2 + 40x - 25 = 16

<=> 40x - 34 = 16

<=> 40x = 50

<=> x = 50/40 = 5/4

d) x3 - 9x2 + 27x - 27 = -8

<=> ( x - 3 )3 = -8

<=> ( x - 3 )3 = (-2)3

<=> x - 3 = -2

<=> x = 1 

e) ( x + 1 )3 - x2( x + 3 ) = 2

<=> x3 + 3x2 + 3x + 1 - x3 - 3x2 = 2

<=> 3x + 1 = 2

<=> 3x = 1

<=> x = 1/3

f) ( x - 2 )3 - x( x - 1 )( x + 1 ) + 6x2 = 5

<=> x3 - 6x2 + 12x - 8 - x( x2 - 1 ) + 6x2 = 5

<=> x3 + 12x - 8 - x3 + x = 5

<=> 13x - 8 = 5

<=> 13x = 13

<=> x = 1

21 tháng 9 2020

a) \(\left(x-3\right)^2-4=0\)

=> \(\left(x-3\right)^2-2^2=0\)

=> \(\left(x-3-2\right)\left(x-3+2\right)=0\)

=> \(\left(x-5\right)\left(x-1\right)=0\)

=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)

b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)

=> \(\left(2x+3\right)^2-\left[\left(2x\right)^2-1^2\right]=22\)

=> \(\left(2x+3\right)^2-\left(4x^2-1\right)=22\)

=> \(\left(2x\right)^2+2\cdot2x\cdot3+3^2-4x^2+1=22\)

=> \(4x^2+12x+9-4x^2+1=22\)

=> \(12x+9+1=22\)

=> \(12x+10=22\)

=> 12x = 12

=> x = 1

c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)

=> \(\left(4x\right)^2-3^2-\left[\left(4x\right)^2-2\cdot4x\cdot5+5^2\right]=16\)

=> \(16x^2-9-\left(16x^2-40x+25\right)=16\)

=> \(16x^2-9-16x^2+40x-25=16\)

=> \(-9+40x-25=16\)

=> \(40x=16+25-\left(-9\right)=16+25+9=50\)

=> x = 50/40 = 5/4

d) \(x^3-9x^2+27x-27=-8\)

=> \(x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3=8\)

=> \(\left(x-3\right)^3=-8\)

=> \(\left(x-3\right)^3=\left(-2\right)^3\)

=> x - 3  = -2 => x = 1

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

=> \(x^3+3x^2+3x+1-x^3-3x^2=2\)

=> \(3x+1=2\)

=> \(3x=1\)=> x = 1/3

f) \(\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x^2=5\)

=> \(x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3-x\left(x^2-1\right)+6x^2=5\)

=> \(x^3-6x^2+12x-8-x^3+x+6x^2=5\)

=> \(\left(12x+x\right)-8=5\)

=> 13x  = 13

=> x = 1

28 tháng 7 2019

a) x3 - 9x2 + 27x - 27 = -8

<=> x3 - 3x2.3 + 3x.32 - 33 = -8

<=> (x - 3)3 = -23

<=> x - 3 = -2

<=> x = 1 (T/m)

Vậy x = 1.

28 tháng 7 2019

b) 64x3 + 48x2 + 12x + 1 = 27

<=> (4x)3 + 3.(4x)2.1 + 3.4x.12 + 13 = 27

<=> (4x + 1)3 = 33

<=> 4x + 1 = 3

<=> 4x = 2

<=> x = \(\frac{1}{2}\)(T/m)

Vậy x = \(\frac{1}{2}\).

28 tháng 7 2019

\(x^3-9x^2+27x-27=-8\Leftrightarrow\left(x^3-27\right)-\left(9x^2-27x\right)=\left(x-3\right)\left(x^2+3x+9\right)-9x\left(x-3\right)=\left(x-3\right)\left(x^2-6x+9\right)=\left(x-3\right)^3=-8=\left(-2\right)^3\Rightarrow x=\left(-2\right)+3=1\)

\(64x^3+48x^2+12x+1=\left(64x^3+1\right)+\left(48x^2+12x\right)=\left(4x+1\right)\left(16x^2-4x+1\right)+12x\left(4x+1\right)=\left(4x+1\right)\left(16x^2+8x+1\right)=\left(4x+1\right)^3=27\Rightarrow4x=2\Leftrightarrow x=\frac{1}{2}\)

29 tháng 7 2019

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

\(\Leftrightarrow\left(8x^3-12x^2+6x-1\right)-\left(8x^3-12x^2\right)=5\)

\(\Leftrightarrow8x^3-12x^2+6x-1-8x^3+12x^2=5\)

\(\Leftrightarrow6x-1=5\)

\(\Leftrightarrow6x=6\)

\(\Leftrightarrow x=1\)

d) \(\left(x+4\right)^3-x^2.\left(x+12\right)=16\)

\(\Leftrightarrow\left(x^3+12x^2+48x+64\right)-\left(x^3+12x^2\right)=16\)

\(\Leftrightarrow x^3+12x^2+48x+64-x^3-12x^2=16\)

\(\Leftrightarrow48x+64=16\)

\(\Leftrightarrow48x=-48\)

\(\Leftrightarrow x=-1\)

#vì câu a,b có người làm rồi nên mình chỉ làm c,d thôi nhé ! :)

Học Tốt !!

1 tháng 5 2017

a, 3y-2y=2y-3

    3y-2y-2y=3

    -y=3

     y=-3

b, 3-4x+24+6x=x+27+3x

   -4x+6x-x-3x =27-3-24

   -2x              =0

      x             =0

  

1 tháng 5 2017

c, 5-(6-x)=4.(3-2x)

   5-6+x =12-8x

   x+8x  =12+6-5

  9x      =13

   x       =13/9

d, 4.(x+3)=-7x+17

   4x+12  =-7x+17

4x+7x     =17-12

11x         =5

  x          =5/11

22 tháng 10 2021

\(a,=x^2-4x+4-\dfrac{15}{4}=\left(x-2\right)^2-\dfrac{15}{4}=\left(x-2-\dfrac{\sqrt{15}}{2}\right)\left(x-2+\dfrac{\sqrt{15}}{2}\right)\\ b,=?\\ c,\Rightarrow x^2+7x-8=0\\ \Rightarrow\left(x+8\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=-8\\x=1\end{matrix}\right.\\ d,Sửa:x^3-3x^2=-27+9x\\ \Rightarrow x^3-3x^2+9x-27=0\\ \Rightarrow x^2\left(x-3\right)+9\left(x-3\right)=0\\ \Rightarrow\left(x^2+9\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-9\left(vô.lí\right)\\x=3\end{matrix}\right.\\ \Rightarrow x=3\\ e,\Rightarrow x\left(x-3\right)-7x+21=0\\ \Rightarrow x\left(x-3\right)-7\left(x-3\right)=0\\ \Rightarrow\left(x-7\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\\ f,\Rightarrow x^2\left(x-2\right)+\left(x-2\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=2\end{matrix}\right.\\ \Rightarrow x=2\)

\(g,\Rightarrow x^2-4x+4=0\\ \Rightarrow\left(x-2\right)^2=0\\ \Rightarrow x=2\\ h,Sửa:x^3-x^2+x=1\\ \Rightarrow x^2\left(x-1\right)+\left(x-1\right)=0\\ \Rightarrow\left(x^2+1\right)\left(x-1\right)=0\Rightarrow\left[{}\begin{matrix}x^2=-1\left(vô.lí\right)\\x=1\end{matrix}\right.\\ \Rightarrow x=1\)

22 tháng 10 2021

cảm ơn kou nhaa:3

mà cái ý b đầu bài là 8x\(^2-25\), kou giải giúp tớ uwu