Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
A=\(\dfrac{5}{4}\).(5-\(\dfrac{4}{3}\)).(\(-\dfrac{1}{11}\))
= \(\dfrac{5}{4}\).\(\dfrac{11}{3}\).(\(-\dfrac{1}{11}\))
=\(\dfrac{5}{4}\).[\(\dfrac{11}{3}.\left(-\dfrac{1}{11}\right)\text{]}\)
=\(\dfrac{5}{4}.\dfrac{1}{3}\)
=\(\dfrac{5}{12}\) (1)
B=\(\dfrac{3}{4}:\left(-12\right).\left(-\dfrac{2}{3}\right)\) =\(\dfrac{3}{4}:\text{[}\left(-12\right).\left(-\dfrac{2}{3}\right)\text{]}\)
=\(\dfrac{3}{4}:8\) =\(\dfrac{3}{4}.\dfrac{1}{8}\)=\(\dfrac{3}{32}\)(2)
C=\(\dfrac{5}{4}:\left(-15\right).\left(-\dfrac{2}{5}\right)\) =\(\dfrac{5}{4}:\text{[}\left(-15\right).\left(-\dfrac{2}{5}\right)\text{]}\)
=\(\dfrac{5}{4}:6=\dfrac{5}{4}.\dfrac{1}{6}=\dfrac{5}{24}\left(3\right)\)
D=(-3).\(\left(\dfrac{2}{3}-\dfrac{5}{4}\right):\left(-7\right)\) =(-3).\(\left(-\dfrac{7}{12}\right)\):(-7)=\(\dfrac{7}{4}:\left(-7\right)\)=\(\dfrac{7}{4}\).\(\left(\dfrac{-1}{7}\right)\)=\(\dfrac{-1}{4}\) (4)
Từ (1),(2),(3)và(4)=>Ta có thể sắp xếp các kết quả trên theo thứ tự tăng dần là:
(Bạn tự làm nhé! mình bận đi học rồi
)
\(A=\dfrac{5}{4}\cdot\dfrac{15-4}{3}\cdot\dfrac{-1}{11}=\dfrac{5}{4}\cdot\dfrac{11}{3}\cdot\dfrac{-1}{11}=\dfrac{-5}{12}\)=-50/120
\(B=\dfrac{3}{4}\cdot\dfrac{-1}{12}\cdot\dfrac{-2}{3}=\dfrac{3\cdot2}{4\cdot12\cdot3}=\dfrac{2}{4\cdot12}=\dfrac{1}{24}\)=5/120
\(C=\dfrac{5}{4}\cdot\dfrac{-1}{15}\cdot\dfrac{-2}{5}=\dfrac{2}{4\cdot15}=\dfrac{1}{30}\)=4/120
\(D=3\cdot\dfrac{8-15}{12}\cdot\dfrac{-1}{7}=\dfrac{1}{4}\)=30/120
Vì -50<4<5<30
nên A<C<B<D
\(A=\dfrac{5}{4}\left(5-\dfrac{4}{3}\right)\left(-\dfrac{1}{11}\right)\)
\(A=\dfrac{5}{4}.\dfrac{11}{3}.\left(-\dfrac{1}{11}\right)\)
\(A=-\dfrac{5}{12}\)
\(B=\dfrac{3}{4}:\left(-12\right).\left(-\dfrac{2}{3}\right)\)
\(B=\dfrac{3}{4}.\left(-\dfrac{1}{12}\right).\left(-\dfrac{2}{3}\right)\)
\(B=\dfrac{1}{24}\)
\(C=\dfrac{5}{4}:\left(-15\right).\left(-\dfrac{2}{5}\right)\)
\(C=\dfrac{5}{4}.\left(-\dfrac{1}{15}\right).\left(-\dfrac{2}{5}\right)\)
\(C=\dfrac{1}{30}\)
\(D=\left(-3\right)\left(\dfrac{2}{3}-\dfrac{5}{4}\right):\left(-7\right)\)
\(D=\left(-3\right)\left(-\dfrac{7}{12}\right)\left(-\dfrac{1}{7}\right)\)
\(D=-\dfrac{1}{4}\)
Sắp xếp theo thứ tự tăng dần:
\(A,D,C,B\)
Bài 1:
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
a, Ta có: \(\dfrac{a+c}{c}=\dfrac{bk+dk}{dk}=\dfrac{\left(b+d\right)k}{dk}=\dfrac{b+d}{d}\)
\(\Rightarrowđpcm\)
b, Ta có: \(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=\dfrac{k\left(b+d\right)}{b+d}=k\) (1)
\(\dfrac{a-c}{b-d}=\dfrac{bk-dk}{b-d}=\dfrac{k\left(b-d\right)}{b-d}=k\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
c, Ta có: \(\dfrac{a-c}{a}=\dfrac{bk-dk}{bk}=\dfrac{k\left(b-d\right)}{bk}=\dfrac{b-d}{b}\)
\(\Rightarrowđpcm\)
d, Ta có: \(\dfrac{3a+5b}{2a-7b}=\dfrac{3bk+5b}{2bk-7b}=\dfrac{b\left(3k+5\right)}{b\left(2k-7\right)}=\dfrac{3k+5}{2k-7}\)(1)
\(\dfrac{3c+5d}{2c-7d}=\dfrac{3dk+5d}{2dk-7d}=\dfrac{d\left(3k+5\right)}{d\left(2k-7\right)}=\dfrac{3k+5}{2k-7}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
e, Sai đề
f, \(\left(\dfrac{a-b}{c-d}\right)^{2012}=\left(\dfrac{bk-b}{dk-d}\right)^{2012}=\left[\dfrac{b\left(k-1\right)}{d\left(k-1\right)}\right]^{2012}=\dfrac{b^{2012}}{d^{2012}}\)(1)
\(\dfrac{a^{2012}+b^{2012}}{c^{2012}+d^{2012}}=\dfrac{b^{2012}k^{2012}+b^{2012}}{d^{2012}k^{2012}+d^{2012}}=\dfrac{b^{2012}\left(k^{2012}+1\right)}{d^{2012}\left(k^{2012}+1\right)}=\dfrac{b^{2012}}{d^{2012}}\) (2)
Từ (1), (2) \(\Rightarrowđpcm\)
A)\(\dfrac{a}{b}=\dfrac{c}{d}=\dfrac{a}{c}=\dfrac{b}{d}\)
áp dụng tính chất dãy tỉ số bằng nhau ta có
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}\)=\(\dfrac{a}{a-b}=\dfrac{c}{c-d}\) (đpcm)
a) \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2a-3c}{2b-3d}\)
Từ \(\dfrac{a}{b}\) = \(\dfrac{c}{d}\) = k ( k \(\in\) Q, k \(\ne\) 0 )
=> \(\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
VP = \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2.b.k+3.d.k}{2b+3d}\) = \(\dfrac{k.\left(2b+3d\right)}{2b+3d}\) = k (1)
VT = \(\dfrac{2a-3c}{2b-3d}\) = \(\dfrac{2.b.k-3.d.k}{2b-3d}\) = \(\dfrac{k.\left(2b-3d\right)}{2b-3d}\) = k (2)
Từ (1) và (2) ta có: \(\dfrac{2a+3c}{2b+3d}\) = \(\dfrac{2a-3c}{2b-3d}\)
hay: (2a+3c).(3b-3d) = (2a-3c).(2b+3d)
Bài 1:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\left\{\begin{matrix} \frac{2a+5b}{3a-4b}=\frac{2bk+5b}{3bk-4b}=\frac{b(2k+5)}{b(3k-4)}=\frac{2k+5}{3k-4}\\ \frac{2c+5d}{3c-4d}=\frac{2dk+5d}{3dk-4d}=\frac{d(2k+5)}{d(3k-4)}=\frac{2k+5}{3k-4}\end{matrix}\right.\)
\(\Rightarrow \frac{2a+5b}{3a-4b}=\frac{2c+5d}{3c-4d}\)
Ta có đpcm.
Bài 2:
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk; c=dk\)
Khi đó: \(\frac{ab}{cd}=\frac{bk.b}{dk.d}=\frac{b^2}{d^2}\)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{(bk)^2+b^2}{(dk)^2+d^2}=\frac{b^2(k^2+1)}{d^2(k^2+1)}=\frac{b^2}{d^2}\)
Do đó: \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}(=\frac{b^2}{d^2})\) . Ta có đpcm.
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\) \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Từ đó, ta được:\(\dfrac{\left(a+c\right)^3}{\left(b+d\right)^3}=\dfrac{\left(bk+dk\right)^3}{\left(b+d\right)^3}=\dfrac{\left[k\left(b+d\right)\right]^3}{\left(b+d\right)^3}=\dfrac{k^3.\left(b+d\right)^3}{\left(b+d\right)^3}=k^3\left(1\right)\) \(\dfrac{\left(a-c\right)^3}{\left(b-d\right)^3}=\dfrac{\left(bk-dk\right)^3}{\left(b-d\right)^3}=\dfrac{\left[k\left(b-d\right)\right]^3}{\left(b-d\right)^3}=\dfrac{k^3.\left(b-d\right)^3}{\left(b-d\right)^3}=k^3\left(2\right)\)
Từ (1) và (2) suy ra: \(\dfrac{\left(a+c\right)^3}{\left(b+d\right)^3}=\dfrac{\left(a-c\right)^3}{\left(b-d\right)^3}\)
\(a,C=\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\)
Ta có \(\left|\dfrac{1}{3}x+4\right|\ge0\)
\(\Rightarrow\left|\dfrac{1}{3}x+4\right|+1\dfrac{2}{3}\ge1\dfrac{2}{3}\)
Dấu "=" xảy ra khi \(\left|\dfrac{1}{3}x+4\right|=0\)
\(\Leftrightarrow\dfrac{1}{3}x+4=0\)
\(\Leftrightarrow\dfrac{1}{3}x=0-4=-4\)
\(\Leftrightarrow x=-4:\dfrac{1}{3}\)
\(\Leftrightarrow x=-12\)
Vậy \(\min\limits_C=1\dfrac{2}{3}\Leftrightarrow x=-12\)
\(b,D=\left|x-6\right|+\left|x+\dfrac{5}{4}\right|\)
Ta có : \(\left\{{}\begin{matrix}\left|x-6\right|\ge-x+6\\\left|x+\dfrac{5}{4}\right|\ge x+\dfrac{5}{4}\end{matrix}\right.\)
\(\Rightarrow\left|x-6\right|+\left|x+\dfrac{5}{4}\right|\ge-x+6+x+\dfrac{5}{4}=\dfrac{29}{4}\)
Dấu "=" xảy ra khi
\(\left\{{}\begin{matrix}-x+6\ge0\\x+\dfrac{5}{4}\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le6\\x\ge-\dfrac{5}{4}\end{matrix}\right.\)
Vậy \(\min\limits_D=\dfrac{29}{4}\Leftrightarrow-\dfrac{5}{4}\le x\le6\)
b) \(D=\left|x-6\right|+\left|x+\dfrac{5}{4}\right|\)
\(D=\left|6-x\right|+\left|x+\dfrac{5}{4}\right|\ge\left|6-x+x+\dfrac{5}{4}\right|=\dfrac{29}{4}\)
Dấu = xảy ra khi \(\left(6-x\right)\left(x+\dfrac{5}{4}\right)\ge0\Leftrightarrow-\dfrac{5}{4}\le x\le6\)
vậy \(D_{min}=\dfrac{29}{4}\) khi \(-\dfrac{5}{4}\le x\le6\)
Ta có :
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhay ta có :
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
\(\Rightarrow\dfrac{a^3}{c^3}=\dfrac{b^3}{d^3}=\dfrac{a^3+b^3}{c^3+d^3}=\dfrac{\left(a+b\right)^3}{\left(c+d\right)^3}\)
\(\Rightarrow\dfrac{a^3+b^3}{c^3+d^3}=\left(\dfrac{a+b}{c+d}\right)^3\)
\(\Rightarrowđpcm\)
thank you nha