Ta có : \(\frac{x}{4y^2+1}=x-\frac{4xy^2}{4y^2+1};\frac{y}{4x^2+1}=y-\frac{4x^2y}{4x^2+1}\)

Áp dụng BĐT Cauchy ta có : 

\(4y^2+1\ge4y;4x^2+1\ge4x\)

\(\Rightarrow x-\frac{4xy^2}{4y^2+1}+y-\frac{4x^2y}{4x^2+1}\ge x-\frac{4xy^2}{4y}+y-\frac{4x^2y}{4x}\)

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

Đến đây ta áp dụng BĐT phụ \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)

\(x+y=4xy\Leftrightarrow\frac{1}{xy}=\frac{4}{x+y}\le\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=4\)

\(\Leftrightarrow\frac{1}{xy}\le4\Leftrightarrow2xy\ge\frac{1}{2}\)

\(\Leftrightarrow\frac{x}{4y^2+1}+\frac{y}{4x^2+1}\ge\frac{1}{2}\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x=y\\4y^2=1\\4x^2=1\end{cases}\Leftrightarrow x=y=\frac{1}{2}}\)

Bạn trên đã chứng minh \(xy\ge\frac{1}{4}\) rồi nên mình xin phép không trình bày

Áp dụng BĐT Cauchy Schwarz ta dễ có:

\(LHS=\frac{x^2}{4xy^2+x}+\frac{y^2}{4x^2y+y}\)

\(\ge\frac{\left(x+y\right)^2}{4xy\left(x+y\right)+\left(x+y\right)}=\frac{\left(x+y\right)^2}{\left(x+y\right)^2+\left(x+y\right)}\)

Ta cần đi chứng minh:

\(\frac{\left(x+y\right)^2}{\left(x+y\right)^2+\left(x+y\right)}\ge\frac{1}{2}\)

\(\Leftrightarrow\left(x+y\right)^2\ge x+y\Leftrightarrow x+y\ge1\)

Điều này là hiển nhiên vì theo AM - GM ta có:\(x+y\ge2\sqrt{xy}=1\)

Vậy ta có đpcm

do a>0, b>0 nên 1=a+b+3ab\(\ge3\sqrt[3]{3\left(ab\right)^2}\Leftrightarrow\frac{1}{3}\ge\sqrt[3]{3\left(ab\right)^2}\)

\(\Leftrightarrow\frac{1}{27}\ge3\left(ab\right)^2\Leftrightarrow\frac{1}{81}\ge\left(ab\right)^2\Leftrightarrow\frac{1}{9}\ge ab\Leftrightarrow\frac{1}{3}\ge\sqrt{ab}\)do đó

P=\(\frac{6ab}{a+b}-a^2-b^2=\frac{6ab}{a+b}-\left(a^2+b^2\right)\le\frac{6ab}{2\sqrt{ab}}-2ab=-2ab+3\sqrt{ab}=-2\left(ab-\frac{3}{2}\sqrt{ab}\right)\)

\(=-2\left[ab-2\sqrt{ab}\cdot\frac{1}{3}+\left(\frac{1}{3}\right)^2-\left(\frac{1}{3}\right)^2-\frac{5}{6}\sqrt{ab}\right]\)

\(=-2\left(\sqrt{ab}-\frac{1}{3}\right)^2+\frac{2}{9}+\frac{5}{3}\sqrt{ab}\le\frac{2}{9}+\frac{5}{3}\cdot\frac{1}{3}=\frac{7}{9}\)

vậy maxP=\(\frac{7}{9}\Leftrightarrow\hept{\begin{cases}a=b>0\\a+b+3ab=1\end{cases}\Leftrightarrow a=b=\frac{1}{3}}\)

Bài làm:

Ta có: \(\left(x-y\right)\left(x^2+xy+y^2\right)\)

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

\(=x^3-y^3\)(hằng đẳng thức)

( x - y )( x² + xy + y² ) = x( x² + xy + y² ) - y( x² + xy + y² )

                                   = x³ + x²y + xy² - x²y - xy² - y³

                                   = x³ - y³ ( HĐT số 7 ) 

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

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

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

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

Bài làm:

Ta có: \(x\left(x^2-y\right)+x^2\left(x+y\right)\)

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

\(=x^2y-xy\)

\(=xy\left(x-1\right)\)(nếu PTĐTTNT)

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

Bài làm:

Ta có: \(x\left(x-y\right)+y\left(x+y\right)\)

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

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

Bài làm

\(\left(x-\frac{1}{2}y\right)\left(x-\frac{1}{2}y\right)\)

\(=\left(x-\frac{1}{2}y\right)^2\)

\(=x^2-2\cdot x\cdot\frac{1}{2}y+\frac{1}{4}y^2\)

\(=x^2-xy+\frac{1}{4}y^2\)

\(\left(x-\frac{1}{2}y\right)\left(x-\frac{1}{2}y\right)\)

C1. \(=x\left(x-\frac{1}{2}y\right)-\frac{1}{2}y\left(x-\frac{1}{2}y\right)\)

\(=x^2-\frac{1}{2}xy-\frac{1}{2}xy+\frac{1}{4}y^2\)

\(=x^2-xy+\frac{1}{4}y^2\)

C2. \(=\left(x-\frac{1}{2}y\right)^2\)

\(=x^2-2\cdot x\cdot\frac{1}{2}y+\left(\frac{1}{2}y\right)^2\)

\(=x^2-xy+\frac{1}{4}y^2\)

ok bạn

\(\left(\frac{1}{2}xy\right)\cdot\left(\frac{1}{2}xy\right)=\frac{1}{2}\cdot\frac{1}{2}\cdot xx\cdot yy=\frac{1}{4}x^2y^2\)

\(a,\left(\frac{1}{2}x.y\right)\times\left(\frac{1}{2}x.y\right)\)

\(=\frac{1}{2}.\frac{1}{2}.x.x.y.y\)

\(=\frac{1}{4}x^2y^2\)

Học tốt

Sửa đề 2a) một chút

a) 1. ( x + y )² + ( x - y )²

= x² + 2xy + y + x² - 2xy + y²

= 2x² + 2y² = 2( x² + y² )

2. ( x + y )² - ( x - y )²

= x² + 2xy + y² - ( x² - 2xy + y² )

= x² + 2xy + y² - x² + 2xy - y²

= 4xy 

b) x² + 9x² + 27 = 10x² + 27 

c) 9x² - 6x + 1 = 9x² - 3x - 3x + 1 

                       = ( 9x² - 3x ) - ( 3x - 1 )

                       = 3x( 3x - 1 ) - 1( 3x - 1 )

                       = ( 3x - 1 )( 3x - 1 )

                       = ( 3x - 1 )²

d) 4x²y²( 2xy + 9 ) = 4x²y² . 2xy + 4x²y² . 9

                              = 8x³y³ + 36x²y²

A D B C 36 117

xét hình thang \(ACBD\)

CÓ \(AB//DC\)

\(\Rightarrow\widehat{ABC}+\widehat{BCD}=180^o\left(tcp\right)\)

thay\(\widehat{ABC}+117^o=180^o\)

\(\Rightarrow\widehat{ABC}=180^o-117^o=63^o\)

xét hình thang \(ACBD\)

có \(\widehat{ABC}+\widehat{BCD}+\widehat{ADC}+\widehat{BAD}=360^o\left(ĐL\right)\)

THAY \(63^o+117^o+\widehat{ADC}+36^o=360^o\)

\(\Rightarrow\widehat{ADC}=360^o-63^o-36^o-117^o=144^o\)