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24 tháng 8 2021

b) (2x)^2 -1 - (3x)^2 + 4=-2

   -5x^2=1

   x^2=-1/5

x không tồn tại

11 tháng 7 2021

`a)|2x+1|=5`

`<=>` \(\left[ \begin{array}{l}2x+1=5\\2x+1=-5\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}2x=4\\2x=-6\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=2\\x=-3\end{array} \right.\) 

`b)|2x+1|=0`

`<=>2x+1=0`

`<=>2x=-1`

`<=>x=-1/2`

`c)|2x+1|=7`

`<=>` \(\left[ \begin{array}{l}2x+1=7\\2x+1=-7\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}2x=6\\2x=-8\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=4\\x=-4\end{array} \right.\) 

`d)|2x+5|=|3x-7|`

`<=>` \(\left[ \begin{array}{l}2x+5=3x-7\\2x+5=7-3x\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=12\\5x=2\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=12\\x=\dfrac25\end{array} \right.\) 

`e)|2x+7|=1`

`<=>` \(\left[ \begin{array}{l}2x+7=1\\2x+7=-1\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}2x=-6\\2x=-8\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=3\\x=-4\end{array} \right.\) 

`g)|x-2|+|2x-3|=2`

Nếu `x>=2=>|x-2|=x-2,|2x-3|=2x-3`

`pt<=>x-2+2x-3=2`

`<=>3x-5=2`

`<=>3x=7`

`<=>x=7/3(tm)`

Nếu `x<=3/2=>|x-2|=2-x,|2x-3|=3-2x`

`pt<=>2-x+3-2x=2`

`<=>5-3x=2`

`<=>3x=3`

`<=>x=1(tm)`

Nếu `3/2<=x<=2=>|x-2|=2-x,|2x-3|=2x-3`

`pt<=>2-x+2x-3=2`

`<=>x-1=2`

`<=>x=3(l)`

`h)|x+2|+|1-x|=3x+2`

Vì `VT>=0=>3x+2>=0=>x>=-2/3`

`=>|x+2|=x+2`

`pt<=>x+2+|1-x|=3x+2`

`<=>|1-x|=2x(x>=0)`

`<=>` \(\left[ \begin{array}{l}2x=1-x\\2x=x-1\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}3x=1\\x=-1\end{array} \right.\) 

`<=>` \(\left[ \begin{array}{l}x=\dfrac13(TM)\\x=-1(KTM)\end{array} \right.\) 

AH
Akai Haruma
Giáo viên
11 tháng 7 2021

a.

$|2x+1|=5$
\(\Leftrightarrow \left[\begin{matrix} 2x+1=5\\ 2x+1=-5\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=2\\ x=-3\end{matrix}\right.\)

b.

$|2x+1|=0$

$\Leftrightarrow 2x+1=0$

$\Leftrightarrow x=-\frac{1}{2}$
c.

$|2x+1|=7$

\(\Leftrightarrow \left[\begin{matrix} 2x+1=7\\ 2x+1=-7\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=3\\ x=-4\end{matrix}\right.\)

 

1: (3x+2)(x+2)(2x-1)

=(3x^2+6x+2x+4)(2x-1)

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

=6x^3-3x^2+16x^2-8x+8x-4

=6x^3+13x^2-4

2: (5x+1)(x-1)+3x(2x+2)

=5x^2-5x+x-1+6x^2+6x

=11x^2+10x-1

3: 4x(2x+1)(x-1)+(x+5)(x-3)

=4x(2x^2-2x+x-1)+x^2+2x-15

=8x^3-4x^2-4x+x^2+2x-15

=8x^3-3x^2-2x-15

4: (2x-1)(x+2)(x-2)+(3x-1)(x-1)

=(2x-1)(x^2-4)+3x^2-4x+1

=2x^3-8x-x^2+4+3x^2-4x+1

=2x^3+2x^2-12x+5

5 tháng 7 2017

a) \(A=\left(2x-1\right)\left(x+3\right)-\left(x-2\right)\left(3x-4\right)+5x\)

\(=\left(2x^2+6x-x-3\right)-\left(3x^2-4x-6x+8\right)+5x\)

\(=\left(2x^2+5x-3\right)-\left(3x^2-10x+8\right)+5x\)

\(=2x^2+5x-3-3x^2+10x-8+5x\)

\(=x^2+20x-11\)

b) \(5x\left(2x^2-3x+1\right)-2x\left(x+1\right)\left(x-2\right)\)

\(=10x^3-15x^2+5x-2x\left(x^2-2x+x-2\right)\)

\(=10x^3-15x^2+5x-2x^3+4x^2-2x^2+4x\)

\(=8x^3-13x^2+9x\)

c) \(\left(3x+2\right)\left(x+1\right)-2x\left(x+3\right)-2x+1\)

\(=3x^2+3x+2x+2-2x^2-6x-2x+1\)

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

c: Ta có: \(\left(2x-3\right)^2-\left(2x-3\right)\left(x-10\right)=7\)

\(\Leftrightarrow4x^2-12x+9-2x^2+20x+3x-30=7\)

\(\Leftrightarrow11x=28\)

hay \(x=\dfrac{28}{11}\)

d: Ta có: \(\left(3x-4\right)^2-9\left(x-3\right)\left(x+3\right)=8\)

\(\Leftrightarrow9x^2-24x+16-9x^2+81=8\)

\(\Leftrightarrow-24x=-89\)

hay \(x=\dfrac{89}{24}\)

f: Ta có: \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)

\(\Leftrightarrow x^2+8x+16-x^2+1=16\)

\(\Leftrightarrow8x=-1\)

hay \(x=-\dfrac{1}{8}\)

4 tháng 12 2021

hello

4 tháng 12 2021

ủa dễ mà.

2 tháng 10 2021

a) (x+2)2+x(x-4)

   =x2+4x+4+x2-4x

   =2x2+4

b)(x-3)2-(x+3)(x-4)

  =x2-6x+9-x2+4x-3x+12

  =-5x+12

c) (3x+1)2+3x(2-4x)

   =9x2+6x+1+6x-12x2

   =-3x2+12x+1

d) (2x-4y)2-(2x-3)(2x-3y)

  =4x2-16xy+16y2-4x2+6xy+6x-9y

  =16y2-10xy+6x-9y

Tôi học lớp 3 ok

4 tháng 12 2021

jmkhgfda????