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giải chi tiết giúp mk vs nha 
Bài 2:
a) Ta có : Từ \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Rightarrow\dfrac{5a}{5c}=\dfrac{7b}{7d}\)
Theo tính chất dãy tỉ số bằng nhau, ta có :
\(\dfrac{5a}{5c}=\dfrac{7b}{7d}=\dfrac{5a+7b}{5c+7d}\left(1\right)\)
Và \(\dfrac{5a}{5c}=\dfrac{7b}{7d}=\dfrac{5a-7b}{5c-7d}\left(2\right)\)
Từ (1) và (2)=> \(\dfrac{5a+7b}{5c+7d}=\dfrac{5a-7b}{5c-7d}\Rightarrow\dfrac{5a+7b}{5a-7b}=\dfrac{5c+7d}{5c-7d}\)Vậy...
b) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
Thay các đẳng thức vừa tìm được , ta có :
\(\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=k^2\left(1\right)\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}\)
\(=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)
từ (1) và (2)=> đpcm
tik mik nha !!!
1. Bạn xem lại đề bài nhé! Mình nghĩ là \(2x=3y=5z\) thì đúng hơn!
2.
a) Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\)
\(\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
\(\Rightarrow\dfrac{5a}{5c}=\dfrac{7b}{7d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{5a}{5c}=\dfrac{7b}{7d}=\dfrac{5a+7b}{5c+7d}=\dfrac{5a-7b}{5c-7d}\)
Từ \(\dfrac{5a+7b}{5c+7d}=\dfrac{5a-7b}{5c-7d}\Rightarrow\dfrac{5a+7b}{5a-7b}=\dfrac{5c+7d}{5c-7d}\)(đpcm)
Vậy \(\dfrac{5a+7b}{5a-7b}=\dfrac{5c+7d}{5c-7d}\)
b) Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có:
\(VT=\dfrac{ac}{bd}=\dfrac{bk.dk}{bd}=\dfrac{bd.k^2}{bd}=k^2\left(1\right)\)
\(VP=\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{\left(bk\right)^2+\left(dk\right)^2}{b^2+d^2}=\dfrac{b^2.k^2+d^2.k^2}{b^2+d^2}=\dfrac{k^2.\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)
Từ \(\left(1\right)\) và \(\left(2\right)\)
\(\Rightarrow\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\left(đpcm\right)\)
Vậy \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
Để \(\dfrac{a}{b}< \dfrac{a+c}{b+d}\) thì a(b+d) < b(a+c)
<=> ab + ad < ba + cb
<=> ad < cb
<=> \(\dfrac{a}{b}< \dfrac{c}{d}\)
Để \(\dfrac{a+c}{b+d}< \dfrac{c}{d}\) thì (a+c)d < (b+d)c
<=> ad + cd < bc + dc
<=> ad < bc
<=> \(\dfrac{a}{b}< \dfrac{c}{d}\)
Chúc bạn học tốt!
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 
Xét 2 t.h là ra mà bn : a âm - b dương
a dương -b âm ( loại vì thế k thỏa mãn bài )
minhf cũng làm theo cach này nhưng cô bảo là chưa chắc đã dc điểm
ĐKXĐ: \(x\ne5\)
a) \(\dfrac{7-x}{x-5}=\dfrac{1}{2}\)
\(\Leftrightarrow2\left(7-x\right)=x-5\)
\(\Leftrightarrow14-2x=x-5\)
\(\Leftrightarrow-2x-x=-5-14\)
\(\Leftrightarrow-3x=-19\)
\(\Leftrightarrow x=\dfrac{19}{3}\)
b, c) cách duy nhất mình biết là dùng Table :v
\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\)
\(\Leftrightarrow\left(a+5\right)\left(b-6\right)=\left(a-5\right)\left(b+6\right)\)
\(\Leftrightarrow ab-6a+5b-30=ab+6a-5b-30\)
=>-6a+5b=6a-5b
=>-12a=-10b
=>6a=5b
hay a/b=5/6
a) \(\left(x+\dfrac{1}{2}\right)+\left(x+\dfrac{1}{6}\right)+\left(x+\dfrac{1}{12}\right)+....+\left(x+\dfrac{1}{9900}\right)\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)=1\)
\(\Leftrightarrow50x+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=1\)
\(\Leftrightarrow50x+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=1\)
\(\Leftrightarrow50x+\left(1-\dfrac{1}{100}\right)=1\)
\(\Leftrightarrow50x+\dfrac{99}{100}=1\)
\(\Leftrightarrow50x=\dfrac{1}{100}\Rightarrow x=\dfrac{1}{5000}\)
b) \(A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+...+\dfrac{3^2}{202.205}\)
\(A=\dfrac{3^2}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{202}-\dfrac{1}{205}\right)\)
\(A=\dfrac{9}{3}\cdot\left(1-\dfrac{1}{205}\right)\)
\(A=\dfrac{9}{3}\cdot\dfrac{204}{205}=\dfrac{615}{205}\)
a) \(\left(x+\dfrac{1}{2}\right)+\left(x+\dfrac{1}{6}\right)+\left(x+\dfrac{1}{12}\right)+....+\left(x+\dfrac{1}{9900}\right)=1\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)=1\)
\(\Leftrightarrow\left(x+x+x+...+x\right)+\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)=1\)
Có tất cả : (99 - 1) : 1 + 1 = 99 (số x)
\(\Rightarrow99x+\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)=1\)
\(\Rightarrow99x+\left(1-\dfrac{1}{100}\right)=1\)
\(\Rightarrow99x+\dfrac{99}{100}=1\Rightarrow99x=1-\dfrac{99}{100}\)
\(\Rightarrow99x=\dfrac{1}{100}\Rightarrow x=\dfrac{1}{100.99}=\dfrac{1}{9900}\)
b) \(A=\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+....+\dfrac{3^2}{202.205}\)
\(A=\dfrac{3^2}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{202}-\dfrac{1}{205}\right)\)
\(A=\dfrac{9}{3}\cdot\left(1-\dfrac{1}{205}\right)\)
\(A=3\cdot\dfrac{204}{205}=\dfrac{615}{205}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\) \(\begin{cases} a = bk \\ c = dk \end{cases}\)
Ta có: \(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(1\right)\)
\(\dfrac{a.c}{b.d}=\dfrac{bk.dk}{b.d}=\dfrac{k^2.b.d}{b.d}=k^2\left(2\right)\)
Từ (1) và (2) suy ra: \(\dfrac{a.c}{b.d}=\dfrac{a^2+c^2}{b^2+d^2}\) \(\rightarrow đpcm\).
Bài 2:
a: \(\left|x\right|=-x\)
nên x<=0
b: \(\left|x\right|>x\)
=>x<0