a) 17 - 14( x + 1 ) = 13 - 4( x + 1 ) - 5( x - 3 )

<=> 17 - 14x - 14 = 13 - 4x - 4 - 5x + 15

<=> 17 - 14 - 13 + 4 - 15 = -4x - 5x + 14x 

<=> -21 = 5x

<=> x = -21/5

b) 7( 4x + 3 ) - 4( x - 1 ) = 15( x + 0, 75 ) + 7

<=> 28x + 21 - 4x + 4 = 15x + 45/4 + 7

<=> 28x - 4x - 15x = 45/4 + 7 - 21 - 4

<=> 9x = -27/4

<=> x = -3/4

c) 3x( x + 1 ) - 2x( x + 2 ) = x² - 1

<=> 3x² + 3x - 2x² - 4x = x² - 1

<=> 3x² + 3x - 2x² - 4x - x² = -1

<=> -x = -1

<=> x = 1 

a, \(17-14\left(x+1\right)=13-4\left(x+1\right)-5\left(x-3\right)\)

\(\Leftrightarrow17-14x-14=13-4x-4-5x+15\)

\(\Leftrightarrow3-14x=24-9x\Leftrightarrow3-14x-24+9x=0\)

\(\Leftrightarrow-21-5x=0\Leftrightarrow5x=-21\Leftrightarrow x=-\frac{21}{5}\)

b, \(7\left(4x+3\right)-4\left(x-1\right)=15\left(x+0,75\right)+7\)

\(\Leftrightarrow28x+21-4x+1=15x+\frac{45}{4}+7\)

\(\Leftrightarrow9x=-\frac{27}{4}\Leftrightarrow x=-\frac{3}{4}\)

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

\(\Leftrightarrow3x^2+3x-2x^2-4x=x^2-1\)

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

a) ( x² - 5 )( x + 3 ) = x³ + 3x² - 5x - 15

b) ( x + 4 )( x - x² ) = x² - x³ + 4x - 4x² = -x³ - 3x² + 4x 

c) ( x² - 6 )( x + 2 ) + ( x + 3 )( x - x² ) = x³ + 2x² - 6x - 12 + x² - x³ + 3x - 3x² = -3x - 12 = -3( x + 4 )

d) x( x - y ) - y( x - y ) = ( x - y )( x - y ) = ( x - y )²

e) x²( x + y ) - x( x² - y ) = x³ + x²y - x³ + xy = x²y + xy = xy( x + 1 ) 

f) 3x( 12x - 4 ) - 9x( 4x - 3 ) = 36x² - 12x - 36x² + 27x = 15x 

Bài làm

a) ( x² - 5 )( x + 3 ) 

= x³ + 3x² - 5x - 15

b) ( x + 4 )( x - x² )

= ( x + 4 ) . x( 1 - x )

= x( x + 4 )( 1 - x )

= x( x - x² + 4 - 4x )

= x( 4 - x² - 3x )

= 4x - x³ - 3x² 

c) ( x² - 6 )( x + 2 ) + ( x + 3 )( x - x² )

= ( x - 3 )( x + 3 )( x + 2 ) + ( x + 3 )( x - x² )

= ( x + 3 )[ ( x - 3 )( x + 2 ) + ( x - x² )]

= ( x + 3 ) [ x² + 2x - 3x - 6 + x² - x² ]

= ( x + 3 ) ( x² - x - 6 )

= x³ - x² - 6x + 3x² - 3x - 18

= x³ + 2x² - 9x - 18

d) x( x - y ) - y( x - y )

= ( x - y )( x - y )

= ( x - y )² 

= x² - 2xy + y

e) x²( x + y ) - x( x² - y )

= x³ + x²y - x³ + xy

= x²y + xy

f) 3x( 12x - 4 ) - 9x( 4x - 3 )

= 3x . 3( 4x - 1 ) - 9x( 4x - 3 )

= 9x( 4x - 1 ) - 9x( 4x - 3 )

= 9x( 4x - 1 - 4x + 3 )

= 9x . 2

= 18x

Bạn vào thống kê hỏi đáp của mình để xem lời giải nhé !

\(\widehat{EIF}=\frac{\widehat{A}+\widehat{C}}{2}=\frac{180^o}{2}=90^o\)  (ĐPCM)

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. \(=\left(x-\frac{1}{2}y\right)^2\)

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

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