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b: =x-2

d: \(=-x^3+\dfrac{3}{2}-2x\)

19 tháng 12 2021

\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)

8 tháng 9 2021

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

21 tháng 9 2021

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

= 2x2 - 10x + x2 + 3x - 2x - 6

= 3x2 - 9x - 6

2;3 tương tự 1

9 tháng 10 2019

a) ( 6x + 1 )2 + ( 6x - 1 )2 - 2( 1 + 6x )( 6x - 1 )

=  ( 6x + 1 )- 2( 1 + 6x )( 6x - 1 ) + ( 6x - 1 )2

= ( 6x + 1 - 6x + 1 )2 = 2= 4

AH
Akai Haruma
Giáo viên
9 tháng 9 2021

Lời giải:

a.

$(2x-3)^2+(2x+3)(5-2x)=(4x^2-12x+9)-(-4x^2+4x+15)$

$=4x^2-12x+9+4x^2-4x-15$

$=24-8x$
b.

$3(2x-3)+5(x+2)=6x-9+5x+10=11x+1$

c.

$3x(2x-8)+(6x-2)(5-x)=(6x^2-24x)+(-6x^2+32x-10)$

$=6x^2-24x-6x^2-32x+10$

$=8x-10$

d.

$(x-3)(x+3)-(x-5)^2=(x^2-9)-(x^2-10x+25)$

$=x^2-9-x^2+10x-25=10x-34$

e.

$(x-y)^3-(x-y)(x^2+xy+y^2)=(x^3-3x^2y+3xy^2-y^3)-(x^3-y^3)$

$=-3x^2y+3xy^2=3xy(y-x)$

a: ta có: \(\left(2x-3\right)^2+\left(2x+3\right)\left(5-2x\right)\)

\(=4x^2-12x+9+2x-4x^2+15-6x\)

\(=-16x+24\)

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

\(=6x-9+5x+10\)

\(=11x+1\)

c: ta có: \(3x\left(2x-8\right)+\left(6x-2\right)\left(5-x\right)\)

\(=6x^2-24x+30x-6x^2-10+2x\)

\(=8x-10\)

2 tháng 9 2021

a) \(3\left(2x-3\right)+5\left(x+2\right)=6x-9+5x+10=11x+1\)

b) \(3x\left(2x-8\right)+\left(6x+2\right)\left(5-x\right)=6x^2-24x+30x-6x^2+10-2x=4x+10\)

c) \(\left(x-3\right)\left(x+3\right)-\left(x-5\right)^2=x^2-9-x^2+10x-25=10x-34\)

d) \(\left(x-y\right)^3-\left(x-y\right)\left(x^2+xy+y^2\right)=x^3-3x^2y+3xy^2-y^3-x^3+y^3=3xy^2-3x^2y\)

19 tháng 4 2021

1, \(4x-10=0\\ \Leftrightarrow x=\dfrac{5}{2}\)

vậy tập no S=\(\left\{\dfrac{5}{2}\right\}\)

2, \(2x^3+6x^2=x^2+3x\\ \Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow x\left(2x-1\right)\left(x+3\right)=0\)

\(\Leftrightarrow\) \(x=0\) hoặc \(2x-1=0\) hoặc \(x+3=0\)

\(\Leftrightarrow\) \(x=0\) hoặc \(x=\dfrac{1}{2}\) hoặc \(x=-3\)

vậy tập no S=\(\left\{0,\dfrac{1}{2},-3\right\}\)

3, \(x-5=3-x\\ \Leftrightarrow2x=8\\ \Leftrightarrow x=4\)

vậy tập no S=\(\left\{4\right\}\)

4,\(\left(-10x+5\right)\left(2x-8\right)=0\)

\(\Leftrightarrow\) \(-10x+5=0\) hoặc \(2x-8=0\)

\(\Leftrightarrow\) \(x=\dfrac{1}{2}\) hoặc \(x=4\)

vậy tập no S=\(\left\{\dfrac{1}{2},4\right\}\)

 

5 tháng 9 2017

(6x+1)(2x-5)=12x2-30x+2x-5=12x2-28x-5

(2x+5)2-2x(2x+8)=4x2+20x+25-4x2-16x=4x+25

(3x-5)(2x-1)-(2x+3)(3x+7)+30x=6x2-3x-10x+5=6x2-13x+5

(X-1)2-(x+1)(x-1)=x2-2x+1-x2+1=-2x+2

(3x+2)(9x2-6x+4)-(3+x)(x-3)=27x3+8+9-x2=27x3-x2+17