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25 tháng 8 2016

a)(3x+4)2-10x-(x-4)(x+4)

    9x2+24x+16-10x-x2+16

    8x2+14x+32

b)(x+1)(x-2)(x2+1)(x+2)(x-1)(x2+4)

   (x+1)(x-1)(x+2)(x-2)(x2+1)(x2+4)

    (x2-1)(x2-4)(7x2+4)

    (-3x2+4)(7x2+4)

    -21x2-12x2+28x2+16

    16-x2

22 tháng 7 2018

a)(3x+4)2-10x-(x-4)(x+4)

9x2+24x+16-10x-x2+16

8x2+14x+32

b)(x+1)(x-2)(x2+1)(x+2)(x-1)(x2+4)

(x+1)(x-1)(x+2)(x-2)(x2+1)(x2+4)

(x2-1)(x2-4)(7x2+4)

(-3x2+4)(7x2+4)

-21x2-12x2+28x2+16

16-x2

a) Ta có: \(A=\left(4-x\right)\left(16+4x+x^2\right)-\left(4-x\right)^3\)

\(=64-x^3+\left(x-4\right)^3\)

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

\(=-12x^2+48x\)

b) Ta có: \(B=\left(3x+2\right)\left(9x^2-6x+4\right)-\left(3x-2\right)\left(9x^2+6x+4\right)\)

\(=27x^3+8-27x^3+8\)

=16

c) Ta có: \(C=\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)^2\)

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

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

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

a: \(A=\left(2x-5\right)^2-4x\left(x-5\right)\)

\(=4x^2-20x+25-4x^2+20x\)

=25

b: \(B=\left(4-3x\right)\left(4+3x\right)+\left(3x+1\right)^2\)

\(=16-9x^2+9x^2+6x+1\)

=6x+17

c: \(C=\left(x+1\right)^3-x\left(x^2+3x+3\right)\)

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

=1

d: \(D=\left(2021x-2020\right)^2-2\left(2021x-2020\right)\left(2020x-2021\right)+\left(2020x-2021\right)^2\)

\(=\left(2021x-2020-2020x+2021\right)^2\)

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

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

6 tháng 11 2021

\(a,=6x^2-4x-x^2-4x-4=5x^2-8x-4\\ b,=x^3+8-2\left(1-x^2\right)=x^3+8-2+2x^2=x^3+2x^2+6\\ c,=\left(2x-1\right)^2-2\left(2x-1\right)\left(2x+1\right)+\left(2x+1\right)^2\\ =\left(2x+1-2x+1\right)^2=4\)

6 tháng 11 2021

Có thể giúp mình thực hiện cách chi tiết ko ạ ? Gv dạy mik ko hiểu mấy

 

23 tháng 12 2021

a: \(=x^2+2x-8-x^2-2x-1=-9\)

b: \(=\dfrac{x^2+6x+9+3x-9+2x^2-18x}{x\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{3x^2-9x}{x\left(x-3\right)\left(x+3\right)}=\dfrac{3x\left(x-3\right)}{x\left(x-3\right)\left(x+3\right)}=\dfrac{3}{x+3}\)

AH
Akai Haruma
Giáo viên
9 tháng 7 2023

Lời giải:
1. 

$M=(x^2+6x+9)+(x^2-9)-2(x^2-2x-8)$

$=x^2+6x+9+x^2-9-2x^2+4x+16=(x^2+x^2-2x^2)+(6x+4x)+(9-9+16)$
$=10x+16=5(2x+1)+11=5.0+11=11$

2.

$V=(9x^2+24x+16)-(x^2-16)-10x=9x^2+24x+16-x^2+16-10x$

$=(9x^2-x^2)+(24x-10x)+(16+16)=8x^2+14x+32$

$=8(\frac{-1}{10})^2+14.\frac{-1}{10}+32=\frac{767}{25}$

3.

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

$=x^2+2x+1-4x^2+4x-1+3x^2-12$
$=(x^2-4x^2+3x^2)+(2x+4x)+(1-1-12)$

$=6x-12=6.1-12=-6$

4.

$Q=(x^2-9)+(x^2-4x+4)-2x^2+8x$

$=x^2-9+x^2-4x+4-2x^2+8x$
$=(x^2+x^2-2x^2)+(-4x+8x)-9+4$

$=4x-5=4(-1)-5=-9$

11 tháng 1 2022

\(A=\dfrac{x}{x-2}-\dfrac{x^2+x-2}{x^2-4}=\dfrac{x^2+2x-x^2-x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{1}{x-2}\)

11 tháng 1 2022

\(A=\dfrac{x}{x-2}+\dfrac{x^2+x-2}{4-x^2}\left(x\ne\pm2\right).\)

\(A=\dfrac{x}{x-2}-\dfrac{\left(x-1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{x}{x-2}-\dfrac{x-1}{x-2}=\dfrac{x-x+1}{x-2}=\dfrac{1}{x-2.}\)

21 tháng 8 2021

a) \(A=\left(5-x\right)\left(5+x\right)-x\left(4-x\right)-25\\ =25-x^2-4x+x^2-25\\ =-4x\)

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

c) \(\left(x+y-2\right)^2-2\left(x+y-2\right)\left(y+x\right)+\left(x+y\right)^2\)

\(=x^2+y^2+4+2xy-4y-4x-2\left(xy+y^2-2y+x^2+xy-2x\right)+x^2+2xy+y^2\)

\(=x^2+y^2+4+2xy-4y-4x-2\left(2xy+y^2-2y+x^2-2x\right)+x^2+2xy+y^2\)

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

\(=4\)

21 tháng 8 2021

a) \(A=\left(5-x\right)\left(5+x\right)-x\left(4-x\right)-25=25-x^2-4x+x^2-25=-4x\)b) \(B=\left(x^2+1\right)\left(x+1\right)-\left(x+1\right)^3=\left(x+1\right)\left[x^2+1-\left(x+1\right)^2\right]=\left(x+1\right)\left(x^2+1-x^2-2x-1\right)=\left(x+1\right)\left(-2x\right)\)c) \(C=\left(x+y-2\right)^2-2\left(x+y-2\right)\left(y+x\right)+\left(x+y\right)^2=\left(x+y-1-x-y\right)^2=\left(-1\right)^2=1\)

 

 

5 tháng 10 2021

Bài 2:

a) \(=x^2-36y^2\)

b) \(=x^3-8\)

Bài 3:

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

b) \(=6\left(x-1\right)\left(x+1\right)=6x^2-6\)

\(A=\dfrac{3x}{x-2}\cdot\sqrt{x^2-4x+4}\)

\(=\dfrac{3x}{x-2}\cdot\left(x-2\right)\)

=3x

\(B=\dfrac{-5y}{x+3}\cdot\sqrt{x^2+6x+9}\)

\(=\dfrac{-5y}{x+3}\cdot\left|x+3\right|\)

\(=\pm5y\)