thiếu đề

Theo đề bài, ta có:

\(\dfrac{3x}{4}=\dfrac{y}{2}=\dfrac{3z}{5}\) và x - z = 15

\(\Rightarrow\dfrac{3x}{4}=\dfrac{y}{2}\Rightarrow6x=4y\Rightarrow\dfrac{x}{4}=\dfrac{y}{6}\) (1)

\(\Rightarrow\dfrac{y}{2}=\dfrac{3z}{5}\Rightarrow5y=6z\Rightarrow\dfrac{y}{6}=\dfrac{z}{5}\) (2)

(1)(2) \(\Rightarrow\dfrac{x}{4}=\dfrac{y}{6}=\dfrac{z}{5}\)

Áp dụng t/c của dãy tỉ số bằng nhau, ta có:

\(\dfrac{x}{4}=\dfrac{y}{6}=\dfrac{z}{5}=\dfrac{x-z}{4-5}=-\dfrac{15}{1}=-15\)

\(\Rightarrow x=-60;y=-90;z=-75\)

\(\Rightarrow x+y+z=-225\)

-25

\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{81}\)

<=> \(\left\{{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{9}\\x+\dfrac{1}{2}=-\dfrac{1}{9}\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}x=-\dfrac{7}{18}\\x=-\dfrac{11}{18}\end{matrix}\right.\)

\(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{81}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{9}\\x+\dfrac{1}{2}=-\dfrac{1}{9}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{18}\\x=-\dfrac{11}{18}\end{matrix}\right.\)

Vậy \(x_1=-\dfrac{7}{18};x_2=-\dfrac{11}{18}\).

bn rút gọn đi r tính thui ???

6.(\(\dfrac{-2}{3}\))+12.\(\dfrac{-2^2}{3}\)+18.\(\dfrac{-2^3}{3}\)

= -4+(-16)+(-48)

=-68

Đáp án là 1

1

a, Ta có: \(\dfrac{a}{a+b+c}< \dfrac{a}{a+b}< \dfrac{a+c}{a+b+c}\) (1)

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

\(\dfrac{c}{a+b+c}< \dfrac{c}{c+a}< \dfrac{c+b}{a+b+c}\) (3)

Từ (1), (2), (3) \(\Rightarrow\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< \dfrac{a+c}{a+b+c}+\dfrac{b+a}{a+b+c}+\dfrac{c+b}{a+b+c}\Rightarrow1< \dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}< 2\)

Thầy mk hướng dẫn phần a như thế còn phần b mk ko bt lm, chúc p hk tốt

thks bn <3

mik chưa học đến bài này bn ag

hì hì

a)Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

Suy ra \(\begin{cases}a=bk\\c=dk\end{cases}\)\(\Rightarrow\frac{a-b}{b}=\frac{c-d}{d}\)\(\Leftrightarrow\frac{bk-b}{b}=\frac{dk-d}{d}\)

Xét VT \(\frac{bk-b}{b}=\frac{b\left(k-1\right)}{b}=k-1\left(1\right)\)

Xét VP \(\frac{dk-d}{d}=\frac{d\left(k-1\right)}{d}=k-1\left(2\right)\)

Từ (1) và (2) =>Đpcm

b)Đặt tương tự ta xét VT:

\(\frac{11bk+3b}{11dk+3d}=\frac{b\left(11k+3\right)}{d\left(11k+3\right)}=\frac{b}{d}\left(1\right)\)

Xét VP \(\frac{3bk-11b}{3dk-11d}=\frac{b\left(3k-11\right)}{d\left(3k-11\right)}=\frac{b}{d}\left(2\right)\)

Từ (1) và (2) =>Đpcm

c)Cũng đặt tương tự

Xét VT \(\frac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\frac{b^2k^2+d^2k^2}{b^2+d^2}=\frac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(1\right)\)

Xét VP \(\frac{bk\cdot dk}{b\cdot d}=\frac{b\cdot d\cdot k^2}{b\cdot d}=k^2\left(2\right)\)

Từ (1) và (2) =>Đpcm

d)Đặt cũng như vậy 

Xét VT \(\frac{4\left(bk\right)^4+5b^4}{4\left(dk\right)^4+5d^4}=\frac{4b^4k^4+5b^4}{4d^4k^4+5d^4}=\frac{b^4\left(4k^4+5\right)}{d^4\left(4k+5\right)}=\frac{b^4}{d^4}\left(1\right)\)

Xét VP \(\frac{\left(bk\right)^2b^2}{\left(dk\right)^2d^2}=\frac{b^2k^2b^2}{d^2k^2d^2}=\frac{k^2b^4}{k^2d^4}=\frac{b^4}{d^4}\left(2\right)\)

Từ (1) và (2) =>Đpcm

a) \(\frac{a-b}{b}=\frac{c-d}{d}\)

Xét d. ( a - b ) = a . d - b . d

      b. ( c - d ) = b . c - b . d

Vì \(\frac{a}{b}=\frac{c}{d}\) => a . d = b . c

hay d. ( a - b ) = b. ( c - d )

=> \(\frac{a-b}{b}=\frac{c-d}{d}\)

Vậy \(\frac{a-b}{b}=\frac{c-d}{d}\)