\(\Rightarrow\frac{a}{\frac{1}{2}}=\frac{b}{\frac{1}{3}}=\frac{c}{\frac{1}{4}}\)

+ Áp dụng tính chất bằng nhau ta có : 

\(\frac{a}{\frac{1}{2}}=\frac{b}{\frac{1}{3}}=\frac{c}{\frac{1}{4}}=\frac{a+b+c}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}=\frac{117}{\frac{13}{12}}=108\)

Suy ra \(\frac{a}{\frac{1}{2}}=108\Rightarrow a=54\)

              \(\frac{b}{\frac{1}{3}}=108\Rightarrow b=36\)

          \(\frac{c}{\frac{1}{4}}=108\Rightarrow c=27\)

Vậy \(a=54;b=36;c=27\)

Chúc bạn học tốt !!!

\(\Rightarrow a=2c;b=\frac{4}{3}c\Rightarrow a+b+c=\left(3+\frac{4}{3}\right)c=\frac{13}{3}=117\Rightarrow c=27\Rightarrow b=36;a=54\)

\(\frac{x}{3}=\frac{y}{5}=\frac{-4x}{-12}=\frac{3y}{15}=\frac{-4x+3y}{-12+15}=\frac{12}{3}=4\Rightarrow x=12;y=20\)

\(\Rightarrow\frac{-4x}{-12}=\frac{3y}{15}\)

+ Áp dụng tính chất bằng nhau ta có : 

\(\frac{-4x}{-12}=\frac{3y}{15}=\frac{-4x+3y}{-12+15}=\frac{12}{3}=4\)

Suy ra : \(\frac{-4x}{-12}=4\Rightarrow x=132\)

               \(\frac{3y}{15}=4\Rightarrow y=20\)

Vậy \(x=132;y=20\)

Chúc bạn học tốt !!!

Ta có: |2x - 5| \(\ge\)0 \(\forall\)x

=> |2x - 5| + 1,(3) \(\ge\)1,(3)

hay |2x - 5| + 4/3 \(\ge\)4/3

Dấu "=" xảy ra <=> 2x - 5 = 0 <=>  x = 5/2

Vậy Min F = 4/3 <=> x = 5/2

Ta có: G = |x - 3| + |x + 3/2|

G = |3 - x| + |x + 3/2| \(\ge\)|3 - x + x + 3/2| = |3/2| = 3/2

Dấu "=" xảy ra <=> (3 - x)(x + 3/2) \(\ge\)0

<=> -3/2 \(\le\)x \(\le\)3

Vậy MinG = 3/2 <=> -3/2 \(\le\)x \(\le\)3

Làm lại cho Edogawa Conan

\(G=\left|x-3\right|+\left|x+\frac{3}{2}\right|\)

\(G=\left|3-x\right|+\left|x+\frac{3}{2}\right|\ge\left|\left(3-x\right)+\left(x+\frac{3}{2}\right)\right|\)

\(=\frac{9}{2}\)

Vậy \(G_{min}=\frac{9}{2}\Leftrightarrow\left(3-x\right)\left(x+\frac{3}{2}\right)\ge0\)

\(Th1:\hept{\begin{cases}3-x\ge0\\x+\frac{3}{2}\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le3\\x\ge\frac{3}{2}\end{cases}}\Leftrightarrow\frac{3}{2}\le x\le2\)

\(Th2:\hept{\begin{cases}3-x\le0\\x+\frac{3}{2}\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge3\\x\le\frac{3}{2}\end{cases}}\left(L\right)\)

8\(\sqrt{x}\)= x^2

bình phương 2 vế, ta được:

64x = x^4

64x - x^4 = 0

x(64 - x³) = 0

x = 0 hoặc x = 4

\(8\sqrt{x}=x^2\)

\(\Leftrightarrow8\sqrt{x}-x^2=0\)

\(\Leftrightarrow\sqrt{x}\left(8-\sqrt{x^3}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x}=0\\8-\sqrt{x^3}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4\end{cases}}\)

mình đang cần gấp có trước 4 rưỡi nha

Để \(\frac{a-5}{2}>0=>a-5>2=>\)A lớn hơn hoặc = 7

ĐỂ\(\frac{a-5}{2}< 0=>a-5< 2=>\)A bé hơn hoặc bằng 4

ĐẺ \(\frac{a-5}{2}=0=>a-5\)chia hết cho 2 => A bằng 5

Thấy đúng thì cho mik 1 tk nha mn

\(5^a+25\)

\(+,a=0\Rightarrow5^a+25=26\left(l\right)\)

\(+,a=1\Rightarrow5^a+25=30\left(l\right)\)

\(+,a=2\Rightarrow5^a+25=50\left(l\right)\)

\(+,a=3\Rightarrow5^a+25=150\left(l\right)\)

\(+,a\ge4\Rightarrow5^a=\left(....25\right)+25=\left(....50\right)\Rightarrow\hept{\begin{cases}5^a+25⋮2\\5^a+25⋮4̸\end{cases}}\left(l\right)\)

shitbo ơi, TH cuối 5^n không chia hết cho 4 đúng không

Nếu đề đúng.

\(a^2+b^3-\sqrt{5^2}c=a+b^3-\frac{5}{3}c\)

<=> \(a+\frac{10}{3}c=a^2\)

Mặt khác:

\(a=\frac{3}{2}c\)=> \(a=\frac{\frac{10}{3}c}{\frac{20}{9}.}=\frac{a+\frac{10}{3}c}{1+\frac{20}{9}}=\frac{a^2}{\frac{29}{9}}\)

=> \(\frac{29}{9}a=a^2\Leftrightarrow\orbr{\begin{cases}a=0\\a=\frac{29}{9}\end{cases}}\)

Với a=0 => b=c =0

Với \(a=\frac{29}{9}\Rightarrow b=\frac{29}{18};c=\frac{58}{27}\)

Bn tự vẽ hình nha

Ta có:\(\widehat{ABD}+\widehat{DBC}=\widehat{ABC}\)

mà\(\widehat{ABD}=\widehat{DBC}\)(BD là đg phân giác của\(\widehat{ABC}\))

\(\Rightarrow\widehat{ABD}=\widehat{DBC}=\frac{\widehat{ABC}}{2}=\frac{60^o}{2}=30^o\)

\(\widehat{ACE}+\widehat{ECB}=\widehat{ACB}\)

mà\(\widehat{ACE}=\widehat{ECB}\)(AC là đg phân giác của​\(\widehat{ACB}\))​

\(\Rightarrow\widehat{ACE}=\widehat{ECB}=\frac{\widehat{ACB}}{2}=\frac{40^o}{2}=20^o\)

Xét\(\Delta BIC\)có:\(\widehat{IBC}+\widehat{BIC}+\widehat{ICB}=180^o\)(ĐL tổng 3 góc của 1\(\Delta\))

hay\(30^o+\widehat{BIC}+20^o=180^o\)

\(\Rightarrow\widehat{BIC}=180^o-30^o-20^o=130^o\)

Ta lại có:\(\widehat{BIC}+\widehat{CID}=180^o\)(2 góc kề bù)

hay\(130^o+\widehat{CID}=180^o\)

\(\Rightarrow\widehat{CID}=180^o-130^o=50^o\)

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

\(\Rightarrow f\left(1\right)+f\left(2\right)+....+f\left(x\right)=1-\frac{1}{2^2}+\frac{1}{2^2}-....-\frac{1}{\left(x+1\right)^2}\)

\(\Rightarrow\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-19+x=\frac{x\left(x+2\right)}{\left(x+1\right)^2}\)

\(\Leftrightarrow\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-19+x=\frac{2y\left(x+1\right)^3-1}{\left(x+1\right)^2}-20+\left(x+1\right)=\frac{x\left(x+2\right)}{\left(x+1\right)^2}\)

Dat:\(x+1=a\Rightarrow\frac{\left(2y+1\right)a^3-20a^2-1}{a^2}=\frac{a^2-1}{a^2}\Leftrightarrow\left(2y+1\right)a^3-20a^2-1=a^2-1\)

\(\Leftrightarrow\left(2y+1\right)a^3-20a^2=a^2\Leftrightarrow\left(2ay+a\right)-20=1\left(coi:x=-1cophailanghiemko\right)\)

\(\Leftrightarrow2ay+a=21\Leftrightarrow a\left(2y+1\right)=21\Leftrightarrow\left(x+1\right)\left(2y+1\right)=21\)