a) A = x( 5 - 3x ) = -3x² + 5x = -3( x² - 5/3x + 25/36 ) + 25/12

= -3( x - 5/6 )² + 25/12 ≤ +25/12 ∀ x

Dấu "=" xảy ra khi x = 5/6

Vậy MaxA = 25/12 <=> x = 5/6

b) Từ x + y = 7 => x = 7 - y

Ta có : xy = ( 7 - y ).y = 7y - y² = -( y² - 7y + 49/4 ) + 49/4 = -( y - 7/2 )² + 49/4 ≤ 49/4 ∀ y

Dấu "=" xảy ra <=> y = 7/2 => x = 7/2

Vậy Max(xy) = 49/4 <=> x = y = 7/2

( nếu cho x,y dương thì Cauchy nhanh gọn luôn :)) )

c) F = x( x - 3 )( x - 4 )( x - 7 )

= [ x( x - 7 ) ][ ( x - 3 )( x - 4 ) ]

= ( x² - 7x )( x² - 7x + 12 )

Đặt t = x² - 7x

F = t( t + 12 ) = t² + 12t = ( t² + 12t + 36 ) - 36 = ( t + 6 )² - 36

= ( x² - 7x + 6 )² - 36 ≥ -36 ∀ x

Dấu "=" xảy ra khi x² - 7x + 6 = 0 <=> x = 1 hoặc x = 6

Vậy MinF = -36 <=> x = 1 hoặc x = 6

a: \(\Leftrightarrow\dfrac{x}{-4}=\dfrac{21}{y}=\dfrac{z}{-80}=\dfrac{3}{4}\)

=>x=-3; y=28; z=-60

b: 5/12=x/-72

=>x=-72*5/12=-6*5=-30

c: =>x+3=-5

=>x=-8

A=10 

B=-7

C=-5

D=-3

E=15

F=3

bạn giải chi tiết ra giúp mình đc ko?

a) A = |x - 3| + 10

Vì |x - 3| >= 0

=> A = |x - 3| + 10 >= 10

A = 10 <=> |x - 3| = 0=> x - 3 = 0 => x = 3

Vậy: A_min = 10 <=> x = 3

b) B = -7 + (x - 1)²

Vì (x - 1)² >= 0

=> B = -7 + (x - 1)² >= -7

B = -7 <=> (x - 1)² = 0 => x - 1 = 0 => x = 1 

Vậy: B_min = -7 <=> x = 1

a) GTNN cua A=10 khi x=3

b) GTNN cua B=-7 khi x=1

c) GTLN cua C=-3 khi x=-2

d) GTLN cua D=15 khi x=2

e) GTLN cua E=3 khi x=-5   y=7

1) \(M=\frac{x^2+y^2+7}{x^2+y^2+5}=1+\frac{2}{x^2+y^2+5}\)

Ta có: \(x^2+y^2\ge0,\forall x;y\)

=> \(x^2+y^2+5\ge5\) với mọi x; y 

=> \(\frac{2}{x^2+y^2+5}\le\frac{2}{5}\)

=> \(M\le1+\frac{2}{5}=\frac{7}{5}\)

Dấu "=" xảy ra <=> x = y = 0 

Vậy max M = 7/5 đạt tại x = y = 0 

2) \(f\left(x-1\right)=x^2-3x+5=x^2-x-2x+2+3\)

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

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

=> \(f\left(x\right)=x.x-x+3=x^2-x+3\)

\(y\times\dfrac{2}{3}=\dfrac{1}{4}+\dfrac{1}{1}\)

\(y\times\dfrac{2}{3}=\dfrac{1}{4}+\dfrac{4}{4}\)

\(y\times\dfrac{2}{3}=\dfrac{5}{4}\)

\(y=\dfrac{5}{4}:\dfrac{2}{3}\)

\(y=\dfrac{5}{4}\times\dfrac{3}{2}\)

\(y=\dfrac{15}{8}\)

e: \(\left\{{}\begin{matrix}\dfrac{1}{x}-\dfrac{1}{y}=1\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{3}{x}-\dfrac{3}{y}=3\\\dfrac{3}{x}+\dfrac{4}{y}=5\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-7}{y}=-2\\\dfrac{1}{x}-\dfrac{1}{y}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{7}{2}\\\dfrac{1}{x}=1+\dfrac{2}{7}=\dfrac{9}{7}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=\dfrac{7}{2}\\x=\dfrac{7}{9}\end{matrix}\right.\)