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x - y = 7

<=> ( x - y )² = 49

<=> x² - 2xy + y² = 49

<=> x² + y² - 2.60 = 49

<=> x² + y² - 120 = 49

<=> x² + y² = 169

<=> x² + 2xy + y² - 2xy = 169

<=> ( x + y )² - 2.60 = 169

<=> ( x + y )² - 120 = 169

<=> ( x + y )² = 289

<=> x + y = ±17

x² - y² = ( x - y )( x + y ) = 7( x + y )

Với x + y = 17 => x² - y² = 7.17 = 119

Với x + y = -17 => x² - y² = 7.(-17) = -119

( a + 2 )³ - a( a - 3 )²

= a³ + 6a² + 12a + 8 - a( a² - 6a + 9 )

= a³ + 6a² + 12a + 8 - a³ + 6a² - 9a

= 12a² + 3a + 8

cách của symbolab:

\(\left(a+2\right)^3-a\left(a-3\right)^2\)

\(=a^3+6a^2+12a+8-a\left(a-3\right)^2\)

\(=a^3+6a^2+12a+8-a\left(a^2-6a+9\right)\)

\(=a^3+6a^2+12a+8-a^3+6a^2-9a\)

\(=12a^2+3a+8\)

\(\Leftrightarrow x^2+2x+3-3x^2-3x-3=\left(x^2+x+1\right)\left(x^4+x^2+1\right)\)

\(\Leftrightarrow-2x^2-x=\left(x^2+x+1\right)\left(x^4+x^2+1\right)\)

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

\(\Leftrightarrow-2\left(x^2+\frac{x}{2}+\frac{1}{16}\right)+\frac{1}{8}=\left(x^2+x+1\right)\left(x^4+x^2+1\right)\)

\(\Leftrightarrow\frac{1}{8}-2\left(x+\frac{1}{4}\right)^2=\left[\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\right]\left[\left(x^2+\frac{1}{2}\right)^2+\frac{3}{4}\right]\)

Vế trái của pt luôn luôn nhỏ hơn 1/8, còn vế phải luôn luôn lớn hơn 9/16=> pt vô nghiệm

\(A=2x^2-20x+7=2\left(x^2-10x+25\right)-43=2\left(x-5\right)^2-43\ge-43\left(\forall x\right)\)

=> Chưa thể khẳng định A dương

\(B=9x^2-6xy+2y^2+1\)

\(B=\left(9x^2-6xy+y^2\right)+y^2+1\)

\(B=\left(3x-y\right)^2+y^2+1\ge1>0\left(\forall x\right)\)

=> đpcm

\(C=x^2-2x+y^2+4y+6\)

\(C=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+1\)

\(C=\left(x-1\right)^2+\left(y+2\right)^2+1\ge1>0\left(\forall x\right)\)

=> đpcm

\(D=x^2-2x+2=\left(x^2-2x+1\right)+1=\left(x-1\right)^2+1\ge1>0\left(\forall x\right)\)

=> đpcm

a) 2x - x² - 4 = -( x² - 2x + 1 ) - 3 = -( x - 1 )² - 3 ≤ -3 ∀ x

Dấu "=" xảy ra khi x = 1

=> GTLN của bthuc = -3 <=> x = 1

b) -x² - 4x = -( x² + 4x + 4 ) + 4 = -( x + 2 )² + 4 ≤ 4 ∀ x

Dấu "=" xảy ra khi x = -2

=> GTLN của bthuc = 4 <=> x = -2

c) -9x² + 24x - 18 = -9( x² - 8/3x + 16/9 ) - 17 = -9( x - 4/3 )² - 17 ≤ -17 ∀ x

Dấu "=" xảy ra khi x = 4/3

=> GTLN của bthuc = -17 <=> x = 4/3

d) 4x - x² - 1 = -( x² - 4x + 4 ) + 3 = -( x - 2 )² + 3 ≤ 3 ∀ x

Dấu "=" xảy ra khi x = 2

=> GTLN của bthuc = 3 <=> x = 2

e) 5 - x² + 2x - 4y² - 4y

= -( x² - 2x + 1 ) - ( 4y² + 4y + 1 ) + 7

= -( x - 1 )² - ( 2y + 1 )² + 7 ≤ 7 ∀ x, y

Dấu "=" xảy ra khi x = 1 ; y = -1/2

=> GTLN của bthuc = 7 <=> x = 1 ; y = -1/2

\(a+b+c=1\Rightarrow\left(a+b+c\right)^3=1\Rightarrow a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)=1\Rightarrow3\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)(Do \(a^3+b^3+c^3=1\))

* Nếu a + b = 0 suy ra c = 1 và b = -a suy ra \(a^5+b^5+c^5=a^5+\left(-a\right)^5+1^5=1\)

Tương tự với b + c = 0 và c + a = 0 ta cũng được\(a^5+b^5+c^5=1\)

\(\frac{a^2}{x}+\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\Leftrightarrow\frac{a^2y+b^2x}{xy}\ge\frac{\left(a+b\right)^2}{x+y}\Leftrightarrow\left(a^2y+b^2x\right)\left(x+y\right)\ge xy\left(a+b\right)^2\Leftrightarrow a^2xy+b^2x^2+a^2y^2+b^2xy\ge a^2xy+b^2xy+2abxy\Leftrightarrow a^2y^2-2abxy+b^2x^2\ge0\Leftrightarrow\left(ay-bx\right)^2\ge0\)*đúng*

Đẳng thức xảy ra khi ^a/_b = ^x/_y

12x² - 4x( 3x - 5 ) = 10x - 18 < đã sửa >

<=> 12x² - 12x² + 20x = 10x - 18

<=> 20x = 10x - 18

<=> 20x - 10x = -18

<=> 10x = -18

<=> x = -18/10 = -9/5

\(A=\left(\frac{x}{25+5x}+\frac{5x+50}{x^2+5x}-\frac{10-2x}{x}\right)\div\frac{3x+15}{7}\)

ĐK : \(\hept{\begin{cases}x\ne0\\x\ne-5\end{cases}}\)

\(=\left(\frac{x}{5\left(x+5\right)}+\frac{5\left(x+10\right)}{x\left(x+5\right)}-\frac{2\left(5-x\right)}{x}\right)\div\frac{3\left(x+5\right)}{7}\)

\(=\left(\frac{x^2}{5x\left(x+5\right)}+\frac{5\cdot5\cdot\left(x+10\right)}{5x\left(x+5\right)}-\frac{2\left(5-x\right)\cdot5\left(x+5\right)}{5x\left(x+5\right)}\right)\div\frac{3\left(x+5\right)}{7}\)

\(=\left(\frac{x^2}{5x\left(x+5\right)}+\frac{25x+250}{5x\left(x+5\right)}-\frac{10\left(25-x^2\right)}{5x\left(x+5\right)}\right)\div\frac{3\left(x+5\right)}{7}\)

\(=\left(\frac{x^2+25x+250-250+10x^2}{5x\left(x+5\right)}\right)\div\frac{3\left(x+5\right)}{7}\)

\(=\frac{11x^2+25x}{5x\left(x+5\right)}\times\frac{7}{3\left(x+5\right)}\)

\(=\frac{77x^2+175x}{15x\left(x+5\right)^2}\)

\(=\frac{77x^2+175x}{15x\left(x^2+10x+25\right)}=\frac{77x^2+175x}{15x^3+150x^2+375x}\)

\(=\frac{77x+175}{15x^2+150x+375}\)