\(\left(1+\frac{1}{x}\right).\left(1+\frac{1}{y}\right).\left(1+\frac{1}{z}\right)=2\)

Giả sử \(x\ge y\ge z>0\)

\(\Rightarrow\frac{1}{x}\le\frac{1}{y}\le\frac{1}{z}\)

\(\Rightarrow1+\frac{1}{x}\le1+\frac{1}{y}\le1+\frac{1}{z}\)

\(\Rightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{z}\right)\le \left(1+\frac{1}{z}\right)^3\)

\(\Rightarrow2\le\left(1+\frac{1}{z}\right)^3\)

\(\Rightarrow1+\frac{1}{z}\ge\sqrt[3]{2}\)
\(\Rightarrow\frac{1}{z}\ge\sqrt[3]{2}-1\)

\(\Rightarrow z\le\frac{1}{\sqrt[3]{2}-1}< 4\)

Mà z thuộc N^* \(\Rightarrow z\in\left\{1;2;3\right\}\)

TH1 : \(z=1\)

\(\Rightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{1}\right)=2\)

\(\Rightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=1\)

Ta có : \(1+\frac{1}{x}>1;1+\frac{1}{y}>1\)\(\Rightarrow\left(\frac{1}{x}+1\right)\left(1+\frac{1}{y}\right)>1\left(lọai\right)\)

TH2 : \(z=2\)

\(\Rightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)\left(1+\frac{1}{2}\right)=2\)

\(\Rightarrow\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=\frac{4}{3}\)

Ta có : \(\left(1+\frac{1}{y}\right)^2\ge\left(1+\frac{1}{x}\right)\left(1+\frac{1}{y}\right)=\frac{4}{3}\)

\(\Rightarrow1+\frac{1}{y}\ge\sqrt{\frac{4}{3}}\)

\(\Rightarrow\frac{1}{y}\ge\frac{2\sqrt{3}}{3}-1\)

\(\Rightarrow y\le\frac{1}{\frac{2\sqrt{3}}{3}-1}< 7\)

\(\Rightarrow y\in\left\{1;2;3;4;5;6\right\}\)

Nếu y = 1 \(\Rightarrow\left(1+1\right)\left(1+\frac{1}{x}\right)=\frac{4}{3}\)

= > x = -3 ( loại )

Nếu y = 2 \(\Rightarrow\left(1+\frac{1}{2}\right)\left(1+\frac{1}{x}\right)=\frac{4}{3}\)

= > x = -9 ( loại )

Nếu y = 3 \(\Rightarrow\left(1+\frac{1}{3}\right)\left(1+\frac{1}{x}\right)=\frac{4}{3}\)

= > \(x\in\varnothing\)

Nếu y = 4 \(\Rightarrow\left(1+\frac{1}{4}\right)\left(1+\frac{1}{x}\right)=\frac{4}{3}\)

= > x = 15 ( tm )

Nếu y = 5 \(\Rightarrow\left(1+\frac{1}{5}\right)\left(1+\frac{1}{x}\right)=\frac{4}{3}\)

= > x = 9 ( tm )

Nếu y = 6 \(\Rightarrow\left(1+\frac{1}{6}\right)\left(1+\frac{1}{x}\right)=\frac{4}{3}\)

= > x = 7 ( tm )

TH3 : z =3 thì bạn làm tương tự nhé

hiiiiiiiiiiiiiiiiiiiiiiiiiiii

ta có x= 2 - 2 =0 

vậy x=3=x=3

X x 3

x=3

mọi x dương

a) Xét \(\Delta ABC\)và \(\Delta HBA\), ta có:

\(\widehat{B}\)chung, \(\widehat{BAC}=\widehat{BHA}\left(=90^o\right)\)

\(\Rightarrow\Delta ABC~\Delta HBA\left(g.g\right)\)(đpcm)

b) \(\Delta ABC\)vuông tại A \(\Rightarrow BC^2=AB^2+AC^2\)\(\Rightarrow BC=\sqrt{AB^2+AC^2}=\sqrt{30^2+40^2}=\sqrt{900+1600}=\sqrt{2500}=50\left(cm\right)\)

Ta có \(S_{ABC}=\frac{1}{2}AB.AC=\frac{1}{2}AH.BC\)\(\Rightarrow AB.AC=AH.BC\)\(\Rightarrow AH=\frac{AB.AC}{BC}=\frac{30.40}{50}=24\left(cm\right)\)

Vậy \(AH=24cm\)

Bạn vào thống kê hỏi đáp của mình xem nhé.

\(x^2+3x+3+x^2-x-1-2x^2+2x+1=1\)

\(\Leftrightarrow4x+2=0\Leftrightarrow x=-\dfrac{1}{2}\)

`Answer:`

\(3x^2+17x+10=0\)

\(\Leftrightarrow3x^2+15x+2x+10=0\)

\(\Leftrightarrow3x.\left(x+5\right)+2.\left(x+5\right)=0\)

\(\Leftrightarrow\left(3x+2\right).\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x+2=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{2}{3}\\x=-5\end{cases}}}\)

Vậy......

Một mảnh đất hình chữ nhật có chu vi là 48m chiều dài gấp 3 lần chiều rộng .Tính chiều dài mảnh đất đó

1 giờ 30 phút = \(1,5giờ\)

Gọi độ dài quãng đường AB là a\(\left(km\right)\), ta có:

\(\frac{a}{40}-\frac{a}{50}=1,5\left(giờ\right)\)

\(\Rightarrow\frac{5a}{200}+\frac{4\left(-a\right)}{200}=1,5\left(giờ\right)\)

\(\Rightarrow\frac{5a+4\left(-a\right)}{200}=1,5\left(giờ\right)\)

\(\Rightarrow\frac{5a-4a}{200}=1,5\left(giờ\right)\)

\(\Rightarrow\frac{a}{200}=1,5\left(giờ\right)\)

\(\Rightarrow a=1,5\text{x}200\)

\(\Rightarrow a=300\)

Vậy quãng đường AB dài \(300km\).

Gọi quãng đường AB là x ( x > 0 ) 

Theo bài ra ta có pt 

\(\dfrac{x}{40}-\dfrac{x}{50}=1+\dfrac{1}{2}=\dfrac{3}{2}\Rightarrow x=300\left(tm\right)\)

(x - 2) (x - 1) (x + 1) (x + 2) = 4

x (-2 - 2 + 1 + 2) = 4

x . 0 = 4

0 = 4

\(\rightarrow\)vô nghiệm

ủa nhầm

x ( -2 - 1 + 1 + 2)

\(\dfrac{x+3}{x}-1< 0\Leftrightarrow\dfrac{x+3-x}{x}< 0\Leftrightarrow\dfrac{3}{x}< 0\Rightarrow x< 0\)