\(\in\)

\(\in\)

\(\notin\)

\(\in\)

\(\notin\)

\(\in\)

\(\notin\)

a,A=0,(000001).48571=1/999999.428571=3/7=B

b,C=0,(01).17+0,(01).82=0,(01).(17+82)=99/99=1=D

c,D=2³³²<2³³³=8¹¹¹<9¹¹¹=3²²²<3²²³

d,4F=1+1/4...+1/4⁹⁹⁹

=>4F-F=1-1/4¹⁰⁰⁰

=>3F=1-1/4¹⁰⁰⁰<1

=>F<1/3=>F<G

bmi=13,6

thể trạng hơi gầy cần ăn uống đầy đủ chất hơn

19,9 nha.cho mk biet tuoi cua ban de minh con danh gia

nhung theo mk thi ban can doi

\(\frac{a^3+b^3+c^3}{b^3+c^3+d^3}=\frac{a^3}{b^3}=\frac{b^3}{c^3}=\frac{c^3}{d^3}\)

\(Ư\left(48\right)\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;8;-8;12;-12;24;-24;48;-48\right\}\)

thiếu 16 bạn ơi

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

\(\Leftrightarrow\orbr{\begin{cases}2x-1=2x+3\\2x-1=-2x-3\end{cases}}\Leftrightarrow\orbr{\begin{cases}0=4\left(L\right)\\4x=-2\end{cases}}\Leftrightarrow x=\frac{-1}{2}\)

B1: Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{d}{a+b+c}=\frac{a+b+c+d}{3\left(a+b+c+d\right)}=\frac{1}{3}\)

Ta có: \(\frac{a}{b+c+d}=\frac{b}{a+c+d}=\frac{a+b}{a+b+2c+2d}=\frac{1}{3}\)

\(\Rightarrow\frac{a+b+2c+2d}{a+b}=3\)\(\Rightarrow1+\frac{2\left(c+d\right)}{a+b}=3\)\(\Rightarrow\frac{2\left(c+d\right)}{a+b}=2\)\(\Rightarrow\frac{c+d}{a+b}=1\)(1)

Lại có: \(\frac{b}{a+c+d}=\frac{c}{a+b+d}=\frac{b+c}{b+c+2\left(a+d\right)}=\frac{1}{3}\)

\(\Rightarrow\frac{b+c+2\left(a+d\right)}{b+c}=3\)\(\Rightarrow1+\frac{2\left(a+d\right)}{b+c}=3\)\(\Rightarrow\frac{2\left(a+d\right)}{b+c}=2\)\(\Rightarrow\frac{a+d}{b+c}=1\)(2)

Ta có: \(\frac{c}{a+b+d}=\frac{d}{a+b+c}=\frac{c+d}{c+d+2\left(a+b\right)}=\frac{1}{3}\)

\(\Rightarrow\frac{2\left(a+b\right)+c+d}{c+d}=3\)\(\Rightarrow\frac{2\left(a+b\right)}{c+d}+1=3\)\(\Rightarrow\frac{2\left(a+b\right)}{c+d}=2\)\(\Rightarrow\frac{a+b}{c+d}=1\)(3)

Lại có: \(\frac{a}{b+c+d}=\frac{d}{a+b+c}=\frac{a+d}{a+d+2\left(b+c\right)}=\frac{1}{3}\)

\(\Rightarrow\frac{2\left(c+b\right)+a+d}{a+d}=3\)\(\Rightarrow\frac{2\left(c+b\right)}{a+d}+1=3\)\(\Rightarrow\frac{2\left(b+c\right)}{a+d}=2\)\(\Rightarrow\frac{b+c}{a+d}=1\)(4)

Từ (1) , (2) , (3) , (4)

\(\Rightarrow P=\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=1+1+1+1=4\)

B2: a, Vì (x⁴ + 3)² ≥ 0

Dấu " = " xảy ra <=> x⁴ + 3 = 0

                          <=> x⁴ = 3

                          <=> x = ⁴√3

Vậy GTNN A = 0 khi x = ⁴√3

b, Vì |0,5 + x| ≥ 0 ; (y - 1,3)⁴ ≥ 0 

=> |0,5 + x| + (y - 1,3)⁴ ≥ 0

=> |0,5 + x| + (y - 1,3)⁴ + 20 ≥ 20

Dấu " = " xảy ra <=> \(\hept{\begin{cases}0,5+x=0\\y-1,3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=-0,5\\y=1,3\end{cases}}\)

Vậy GTNN V = 20 khi x = -0,5 và y = 1,3

c, Ta có: \(C=\frac{5x-19}{x-4}=\frac{5\left(x-4\right)+1}{x-4}=5+\frac{1}{x-4}\)

C đạt GTNN <=> \(\frac{1}{x-4}\)đạt GTNN <=> x - 4 đạt GTLN

<=> x > 4 , x nguyên dương

Vậy C có GTNN <=> x > 4 , x nguyên dương

(Ko chắc)

( t tham khảo 1 số bài khác thì ng` ta giải x = 3 thì C có GTNN = 4 )

Bài 3:

a, Để N có GTLN <=> 2(x - 2014)² + 3 có GTNN

Vì (x - 2014)² ≥ 0 => 2(x - 2014)² ≥ 0

=> 2(x - 2014)² + 3 ≥ 3

\(\Rightarrow\frac{1}{2\left(x-2014\right)^2+3}\le\frac{1}{3}\)

Dấu " = " xảy ra <=> x - 2014 = 0

                          <=> x = 2014

Vậy GTLN N = 1/3 khi x = 2014

b, Ta có: \(P=\frac{27-2x}{12-x}=\frac{2\left(12-x\right)+3}{12-x}=2+\frac{3}{12-x}\)

Để P có GTLN <=> \(\frac{3}{12-x}\)có GTLN <=> 12 - x có GTNN ( (12 - x) ∈ N ; 12 - x ≠ 0)

                                                                   <=> 12 - x = 1

                                                                   <=> x = 11

\(\Rightarrow P=2+\frac{3}{12-x}=2+3=5\)