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19 tháng 6 2017

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19 tháng 6 2017

\(\left(3a-1\right)^2=9a^2-6a+1\)

\(\left(a-2\right)^2=a^2-4a+4\)

\(\left(1-5a\right)^2=1-10a+25a^2\)

\(\left(3a-2b\right)^2=9a^2-12ab+4a^2\)

\(\left(4-3a\right)^2=16-24a+9a^2\)

\(\left(5a-4b\right)^2=25a^2-40ab+16b^2\)

\(\left(5a-3b\right)\left(5a+3b\right)=25a^2-9b^2\)

\(\left(3x+1\right)\left(3x-1\right)=9x^2-1\)

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

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

\(\left(3x^2-y\right)\left(3x^2+y\right)=9x^4-y^2\)

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

\(\left(\dfrac{3}{4}x+2\right)\left(\dfrac{3}{4}x-2\right)=\dfrac{9}{16}x^2-4\)

\(\left(5x-\dfrac{3}{2}\right)\left(5x+\dfrac{3}{2}\right)=25x^2-\dfrac{9}{4}\)

\(\left(2a^2-7\right)\left(2a^2+7\right)=4a^2-49\)

14 tháng 7 2017

a, Vì \(a^2-b^2=4c^2\Rightarrow16a^2-16b^2=64c^2\) (1)

Ta có:\(\left(5a-3b+8c\right)\left(5a-3b-8c\right)=\left(5a-3b\right)^2-\left(8c\right)^2\)

\(=25a^2-30ab+9b^2-64c^2\) (2)

Thay (1) vào (2) ta được

\(\left(5a-3b+8c\right)\left(5a-3b-8c\right)=25a^2-30ab+9b^2-16a^2+16b^2\)

\(=9a^2-30ab+25b^2=\left(3a-5b\right)^2\)

=> đpcm

b, \(M=\left(2a+2b-c\right)^2+\left(2b+2c-a\right)^2+\left(2c+2b-b\right)^2\)

\(=4a^2+4b^2+c^2+4b^2+4c^2+a^2+4c^2+4a^2+b^2\)

\(+8ab-4ac-4bc+8bc-4ab-4ac+8ac-4bc-4ab\)

\(=9.\left(a^2+b^2+c^2\right)=9.2017=18153\)

Vậy M=18153

13 tháng 6 2021

Có \(ab+bc+ac=abc\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1\)

Áp dụng các bđt sau:Với x;y;z>0 có: \(\dfrac{1}{x+y+z}\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) và \(\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\) 

Có \(\dfrac{1}{a+3b+2c}=\dfrac{1}{\left(a+b\right)+\left(b+c\right)+\left(b+c\right)}\le\dfrac{1}{9}\left(\dfrac{1}{a+b}+\dfrac{2}{b+c}\right)\)\(\le\dfrac{1}{9}.\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{b}+\dfrac{2}{c}\right)=\dfrac{1}{36}\left(\dfrac{1}{a}+\dfrac{3}{b}+\dfrac{2}{c}\right)\)

CMTT: \(\dfrac{1}{b+3c+2a}\le\dfrac{1}{36}\left(\dfrac{1}{b}+\dfrac{3}{c}+\dfrac{2}{a}\right)\)

\(\dfrac{1}{c+3a+2b}\le\dfrac{1}{36}\left(\dfrac{1}{c}+\dfrac{3}{a}+\dfrac{2}{b}\right)\)

Cộng vế với vế => \(VT\le\dfrac{1}{36}\left(\dfrac{6}{a}+\dfrac{6}{b}+\dfrac{6}{c}\right)=\dfrac{1}{36}.6\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{6}\)

Dấu = xảy ra khi a=b=c=3

13 tháng 6 2021

Có \(a+b=2\Leftrightarrow2\ge2\sqrt{ab}\Leftrightarrow ab\le1\)

\(E=\left(3a^2+2b\right)\left(3b^2+2a\right)+5a^2b+5ab^2+2ab\)

\(=9a^2b^2+6\left(a^3+b^3\right)+4ab+5ab\left(a+b\right)+20ab\)

\(=9a^2b^2+6\left(a+b\right)^3-18ab\left(a+b\right)+4ab+5ab\left(a+b\right)+20ab\)

\(=9a^2b^2+48-18ab.2+4ab+5.2.ab+20ab\)

\(=9a^2b^2-2ab+48\)

Đặt \(f\left(ab\right)=9a^2b^2-2ab+48;ab\le1\), đỉnh \(I\left(\dfrac{1}{9};\dfrac{431}{9}\right)\)

Hàm đồng biến trên khoảng \(\left[\dfrac{1}{9};1\right]\backslash\left\{\dfrac{1}{9}\right\}\)

 \(\Rightarrow f\left(ab\right)_{max}=55\Leftrightarrow ab=1\)

\(\Rightarrow E_{max}=55\Leftrightarrow a=b=1\)

Vậy...

10 tháng 7 2019

\(1,\)\(\left(2x+3\right)^2=4x^2+12x+9\)

\(2,\)\(\left(3x+2y\right)^2=9x^2+12xy+4x^2\)

\(3,\)\(\left(3a-1\right)^2=9x^2-6x+1\)

\(4,\)\(\left(a-2\right)^2=a^2-4a+4\)

\(5,\)\(\left(1-5a\right)^2=1-10a+25a^2\)

\(6,\)\(\left(x-4\right)^3=x^3-12a^2+48a-64.\)

\(7,\)\(\left(x^2-2y\right)^2=x^4-4x^2y-4y^2\)

\(8,\)\(\left(5x^2-2\right)\left(5x^2+2\right)=25x^4-4\)

\(9,\)\(\left(2a^2-7\right)\left(2a^2+7\right)=4a^4-49\)

\(10,\)\(\left(x-1\right)\left(x^2+x+1\right)=x^3-1\)

\(11,\)\(\left(x^3-2\right)\left(x^6+2x^3+4\right)=x^9-8\)

\(12,\)\(\left(3x+2\right)\left(9x^2-6x+4\right)=27x^3+8\)

\(13,\)\(\left(x^2+3\right)\left(x^4-3x^2+9\right)=x^6+27\)

10 tháng 7 2019

1, ( 2x + 3 )2 = 4x2 + 12x + 9

2, ( 3x + 2y )2 = 9x2 +12xy + 4y2

3 ( 3a - 1 )2 = 9a2 - 6x + 1

4, ( a - 2 )2 = a2 - 4a + 4

5, ( 1 - 5a )2 = 1 - 10a + 25a2

6,  ( x- 4 )3 = x3 - 12x2 + 48x - 64

7, ( x2 - 2y )2 = x4 - 4x2y + 4y2

8, ( 5X2 - 2 ).( 5X2 + 2 ) = 25X2 - 4

9, ( 2a2 - 7 ).( 2a2 + 7 ) = 4a4 - 49

10, ( x - 1 ).( x2 + x + 1 ) = x3 - 1

16 tháng 9 2016

a/  (2a+3b)^2 = (2a)^2+2.2a.3b+(3b)^2 = 4a^2+12ab+9b^2

b/   ta nhân đa thức với đa thức thì kết quả sẽ = -9a^2+25

c/   (x^2-3y)^2= (x^2)^2-2.x^2.3y+(3y)^2= x^4-6x^2y+9y^2

                                                     

a: =(5a-a+b)(5a+a-b)

=(4a+b)(5a-b)

 

b: =(2a-a-b)(2a+a+b)

=(a-b)(3a+b)

c: =(7a-2a+b)(7a+2a-b)

=(5a+b)(9a-b)

d: =(6a-3a+2b)(6a+3a-2b)

=(3a+2b)(9a-2b)

e: =(9a-5a+3b)(9a+5a-3b)

=(4a+3b)(14a-3b)

AH
Akai Haruma
Giáo viên
10 tháng 9 2023

Lời giải:

$25a^2-(a-b)^2=(5a)^2-(a-b)^2=[5a-(a-b)][5a+(a-b)]=(4a+b)(6a-b)$

$4a^2-(a+b)^2=(2a)^2-(a+b)^2=[2a-(a+b)][2a+(a+b)]=(a-b)(3a+b)$

$49a^2-(2a-b)^2=(7a)^2-(2a-b)^2=[7a-(2a-b)][7a+(2a-b)]=(5a+b)(9a-b)$

$36a^2-(3a-2b)^2=(6a)^2-(3a-2b)^2=[6a-(3a-2b)][6a+(3a-2b)]$

$=(3a+2b)(9a-2b)$

$81a^2-(5a-3b)^2=(9a)^2-(5a-3b)^2=[9a-(5a-3b)][9a+(5a-3b)]$

$=(4a+3b)(14a-3b)$

8 tháng 11 2016

a/ Ta có \(a\left(2a-5c\right)=2a^2-5ac=2bc-5ac=c\left(2b-5a\right)\Rightarrow\frac{c}{2a-5c}=\frac{a}{2b-5a}\)

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