\(3x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}\Rightarrow\dfrac{2x}{4}=\dfrac{3y}{9}\) \(\left(1\right)\)
\(y=2z\Rightarrow\dfrac{3y}{3}=2z\Rightarrow\dfrac{3y}{9}=\dfrac{2z}{3}\) \(\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)
\(\Rightarrow\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{2z}{3}\)
Áp dụng tính chất dãy tỉ số bằng nhau
ta có: \(\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{2z}{3}=\dfrac{2x+3y-2z}{4+9-3}=\dfrac{40}{10}=4\)
+\(\dfrac{2x}{4}=4\Rightarrow2x=16\Rightarrow x=8\)
+\(\dfrac{3y}{9}=4\Rightarrow3y=36\Rightarrow y=12\)
+\(\dfrac{2z}{3}=4\Rightarrow2z=12\Rightarrow z=6\)
Vậy \(x=8;y=12;z=6\)

Cứu tui vs:((((

Điểm \(E\) ở đâu vậy bạn?

\(\dfrac{x+y}{x-y}=\dfrac{2}{5}\Rightarrow5\left(x+y\right)=2\left(x-y\right)\)

\(\Rightarrow5x+5y=2x-2y\)

\(\Rightarrow5x-2x=-2y-5y\)

\(\Rightarrow3x=-7y\)

\(\Rightarrow\dfrac{x}{y}=-\dfrac{7}{3}\)

Lời giải:
a. Xét tam giác $BAD$ và $BHD$ có:

$\widehat{BAD}=\widehat{BHD}=90^0$

$BD$ chung

$\widehat{ABD}=\widehat{HBD}$ (do $BD$ là phân giác $\widehat{B}$)

$\Rightarrow \triangle BAD=\triangle BHD$ (ch-gn)

$\Rightarrow AB=BH$

b. Từ tam giác bằng nhau phần a suy ra $AD=DH$ (1)

Xét tam giác vuông $DHC$ vuông tại $H$ nên $DC> DH$ (do $DC$ là cạnh huyền) (2)

Từ $(1); (2)\Rightarrow DC> AD$

c.

Xét tam giác $BIH$ và $BCA$ có:

$\widehat{B}$ chung

$BH=BA$ (cmt)

$\widehat{BHI}=\widehat{BAC}=90^0$

$\Rightarrow \triangle BIH=\triangle BCA$ (g.c.g)

$\Rightarrow BI=BC$
$\Rightarrow BIC$ cân tại $I$

Hình vẽ:

Số gạo xay được từ 15kg thóc là:

\(8:10\times15=12\left(kg\right)\)

15 kg thóc xay được số ki-lô-gam gạo là:

   15 : 10 x 8 = 12 (kg)

Kết luận:..

Em nên đặt câu hỏi vào đúng lớp sẽ dễ được hỗ trợ hơn. Việc đặt 1 bài toán BĐT vào khu vực lớp 7 rất không ổn.

Ta có:

\(P=\dfrac{a}{2a+1}+\dfrac{b}{2b+1}+\dfrac{c}{2c+1}=\dfrac{a}{a+a+1}+\dfrac{b}{b+b+1}+\dfrac{c}{c+c+1}\)

\(P\le\dfrac{a}{9}\left(\dfrac{1}{a}+\dfrac{1}{a}+1\right)+\dfrac{b}{9}\left(\dfrac{1}{b}+\dfrac{1}{b}+1\right)+\dfrac{c}{9}\left(\dfrac{1}{c}+\dfrac{1}{c}+1\right)\)

\(P\le\dfrac{2a}{9a}+\dfrac{2b}{9b}+\dfrac{2c}{9c}+\dfrac{a+b+c}{9}=1\)

Dấu "=" xảy ra khi \(a=b=c=1\)

\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)

\(\dfrac{1}{a}+\dfrac{1}{c}\ge\dfrac{4}{a+c}\)

\(\dfrac{4}{a+b}+\dfrac{4}{a+c}\ge4\left(\dfrac{4}{a+b+a+c}\right)=\dfrac{16}{2a+b+c}\)

\(\Rightarrow\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{16}{2a+b+c}\)

Tương tự ta có:

\(\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{c}\ge\dfrac{16}{a+2b+c}\) ; \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{c}\ge\dfrac{16}{a+b+2c}\)

Cộng vế:

\(4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{16}{2a+b+c}+\dfrac{16}{a+2b+c}+\dfrac{16}{a+b+2c}\)

\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{2a+b+c}+\dfrac{4}{a+2b+c}+\dfrac{4}{a+b+2c}\)

Dấu "=" xảy ra khi \(a=b=c\)

Ta có:

\(VP=\dfrac{4}{2a+b+c}+\dfrac{4}{2b+a+c}+\dfrac{4}{2c+a+b}\)

\(\le\dfrac{1}{2a}+\dfrac{1}{b+c}+\dfrac{1}{2b}+\dfrac{1}{c+a}+\dfrac{1}{2c}+\dfrac{1}{a+b}\)

\(=\dfrac{1}{2a}+\dfrac{1}{4}\left(\dfrac{4}{b+c}\right)+\dfrac{1}{2b}+\dfrac{1}{4}\left(\dfrac{4}{c+a}\right)+\dfrac{1}{2c}+\dfrac{1}{4}\left(\dfrac{4}{a+b}\right)\)

\(\le\dfrac{1}{2a}+\dfrac{1}{4}\left(\dfrac{1}{b}+\dfrac{1}{c}\right)+\dfrac{1}{2b}+\dfrac{1}{4}\left(\dfrac{1}{c}+\dfrac{1}{a}\right)+\dfrac{1}{2c}+\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\)

\(=\dfrac{1}{2a}+\dfrac{1}{4b}+\dfrac{1}{4c}+\dfrac{1}{2b}+\dfrac{1}{4c}+\dfrac{1}{4a}+\dfrac{1}{2c}+\dfrac{1}{4a}+\dfrac{1}{4b}\)

\(=\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\)

\(=VT\)

 Ta có đpcm. Dấu "=" xảy ra \(\Leftrightarrow a=b=c\)

 Chú ý: Trong bài ta đã sử dụng bất đẳng thức \(\dfrac{4}{x+y}\le\dfrac{1}{x}+\dfrac{1}{y}\) với \(x,y>0\) hai lần

Ta có \(VT=\dfrac{1}{a}+\dfrac{1}{4b}\)

\(=\dfrac{1}{a}+\dfrac{\dfrac{1}{4}}{b}\)

\(=\dfrac{1^2}{a}+\dfrac{\left(\dfrac{1}{2}\right)^2}{b}\)

\(\ge\dfrac{\left(1+\dfrac{1}{2}\right)^2}{a+b}\) (áp dụng BĐT \(\dfrac{x^2}{m}+\dfrac{y^2}{n}\ge\dfrac{\left(x+y\right)^2}{m+n}\))

\(=\dfrac{\left(\dfrac{3}{2}\right)^2}{1}\) (vì \(a+b=1\))

\(=\dfrac{9}{4}\)

Ta có đpcm. Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}a+b=1\\\dfrac{1}{a}=\dfrac{1}{2b}\end{matrix}\right.\) \(\Leftrightarrow\left(a,b\right)=\left(\dfrac{2}{3},\dfrac{1}{3}\right)\)

Ta có:

\(\dfrac{1}{a}+\dfrac{1}{4b}=\dfrac{1}{a}+\dfrac{\left(\dfrac{1}{2}\right)^2}{b}\ge\dfrac{\left(1+\dfrac{1}{2}\right)^2}{a+b}=\dfrac{9}{4}\)

Dấu "=" xảy ra khi \(\left(a;b\right)=\left(\dfrac{2}{3};\dfrac{1}{3}\right)\)