giúp toi với tôi đang cần gấp

a) \(A=1,25+\left|2,5-x\right|\)

Vì \(\left|2,5-x\right|\ge0\forall x\)

\(\Rightarrow1,25+\left|2,5-x\right|\ge1,25\forall x\)

hay \(A\ge1,25\)

Dấu " = " xảy ra \(\Leftrightarrow2,5-x=0\)\(\Leftrightarrow x=2,5\)

Vậy \(minA=1,25\)\(\Leftrightarrow x=2,5\)

b) \(B=-\left(2x+\frac{1}{3}\right)^4-1\)

Vì \(\left(2x+\frac{1}{3}\right)^4\ge0\forall x\)

\(\Rightarrow-\left(2x+\frac{1}{3}\right)^4\le0\forall x\)

\(\Rightarrow-\left(2x+\frac{1}{3}\right)^4-1\le-1\forall x\)

hay \(B\le-1\)

Dấu " = " xảy ra \(\Leftrightarrow2x+\frac{1}{3}=0\)\(\Leftrightarrow2x=-\frac{1}{3}\)

\(\Leftrightarrow x=-\frac{1}{6}\)

Vậy \(maxB=-1\)\(\Leftrightarrow x=-\frac{1}{6}\)

A = 1, 25 + | 2, 5 - x |

Ta có : | 2, 5 - x | ≥ 0 ∀ x => 1, 25 + | 2, 5 - x | ≥ 1, 25 ∀ x

Dấu "=" xảy ra khi x = 2, 5

=> MinA = 1, 25 <=> x = 2, 5

B = -( 2x + 1/3 )⁴ - 1

Ta có -( 2x + 1/3 )⁴ ≤ 0 ∀ x => -( 2x + 1/3 )⁴ - 1 ≤ -1 ∀ x

Dấu "=" xảy ra khi x = -1/6

=> MaxB = -1 <=> x = -1/6

B1:

Từ \(b=\frac{a+c}{2}\Rightarrow2b=a+c\left(1\right)\)

Từ \(c=\frac{2bd}{b+a}\)thay vào (1) ta được:

\(2b=a+\frac{2bd}{b+a}\)

\(\Leftrightarrow2b\left(b+a\right)=a\left(b+a\right)+2bd\)

\(\Leftrightarrow2b^2+2ab=ab+a^2+2bd\)

\(\Leftrightarrow2b^2+ab-a^2-2bd=0\)

\(\Leftrightarrow2b\left(b-d\right)+a\left(b-a\right)=0\)

\(\Leftrightarrow2b\left(b-d\right)=a\left(a-b\right)\Leftrightarrow\frac{2b}{a}=\frac{a-b}{b-d}\)

B2: Từ \(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\Rightarrow\frac{1}{c}=\frac{a+b}{2ab}hay2ab=c\left(a+b\right)\)

\(\Rightarrow ab+ab=ac+bc\Rightarrow ab-bc=ac-ab\Rightarrow b\left(a-c\right)=a\left(c-b\right)\)

Do đó: \(\frac{a-c}{c-b}=\frac{a}{b}\)(đpcm)

i donnot

\(\left(x-\frac{5}{9}\right)^2=\frac{16}{25}\)

=> \(\left(x-\frac{5}{9}\right)^2=\left(\frac{4}{5}\right)^2\)

=> \(\orbr{\begin{cases}x-\frac{5}{9}=\frac{4}{5}\\x-\frac{5}{9}=-\frac{4}{5}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{61}{45}\\x=-\frac{11}{45}\end{cases}}\)

Vậy \(x\in\left\{\frac{61}{45};-\frac{11}{45}\right\}\)là giá trị cần tìm

( x - 5/9 )² = 16/25

<=> ( x - 5/9 )² = ( ±4/5 )²

<=> x - 5/9 = 4/5 hoặc x - 5/9 = -4/5

<=> x = 61/45 hoặc x = -11/45

Ta có\(\sqrt{x+5}\ge0\Rightarrow\sqrt{x+5}-3\ge-3\Rightarrow Bmin=-3\)

\(\sqrt{x^2-9}\ge0\Rightarrow\sqrt{x^2-9}+5\ge5\Rightarrow Cmin=5\)