a/ \(\frac{1}{3}.3^x+3^{x+2}=3^{16}+3^{13}\)

\(\Leftrightarrow3^{x-1}+3^{x+2}=3^{13}+3^{16}\)

\(\Leftrightarrow3^{x-1}\left(1+3^3\right)=3^{13}\left(1+3^3\right)\)

\(\Leftrightarrow3^{x-1}=3^{13}\Rightarrow x-1=13\Rightarrow x=14\)

b/ \(\frac{1}{6}6^x+6^{x+2}=6^{15}+6^{18}\)

\(\Leftrightarrow6^{x-1}+6^{x+2}=6^{15}+6^{18}\)

\(\Leftrightarrow6^{x-1}\left(1+6^3\right)=6^{15}\left(1+6^3\right)\)

\(\Rightarrow x=16\)

c/ \(\frac{1}{2}2^{x+3}-2^x=2^{22}-2^{20}\)

\(\Leftrightarrow2^x\left(2^2-1\right)=2^{20}\left(2^2-1\right)\)

\(\Rightarrow x=20\)

a/ Đề?

b/ \(\frac{1}{6}6^x+6^{x+2}=6^{10}+6^7\)

\(\Leftrightarrow6^{x-1}+6^{x+2}=6^{10}+6^7\)

\(\Leftrightarrow6^{x-1}\left(1+6^3\right)=6^7\left(1+6^3\right)\)

\(\Leftrightarrow6^{x-1}=6^7\Rightarrow x-1=7\Rightarrow x=8\)

c/ Hoàn toàn tương tự câu trên:

\(2^{x-1}+2^{x+1}=2^{12}+2^{10}\)

\(\Leftrightarrow2^{x-1}\left(1+2^2\right)=2^{10}\left(1+2^2\right)\)

\(\Leftrightarrow x=11\)

Nguyễn Việt Lâm

a, \(3.5^{x+2}+4.5^{x-3}=19.5^{10}\)

\(\dfrac{5}{x}+\dfrac{y}{4}=\dfrac{1}{8}\)

\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{y}{4}\)

\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{8}-\dfrac{2y}{8}\)

\(\Rightarrow\dfrac{5}{x}=\dfrac{1-2y}{8}\)

\(\Rightarrow x\left(1-2y\right)=40\)

\(\Rightarrow x;1-2y\in U\left(40\right)\)

\(U\left(40\right)=\left\{\pm1;\pm2;\pm4;\pm5;\pm8;\pm10;\pm20;\pm40\right\}\)

Mà 1-2y lẻ nên:

\(\left\{{}\begin{matrix}1-2y=1\Rightarrow2y=0\Rightarrow y=0\\x=40\\1-2y=-1\Rightarrow2y=2\Rightarrow y=1\\x=-40\end{matrix}\right.\)

\(\left\{{}\begin{matrix}1-2y=5\Rightarrow2y=-4\Rightarrow y=-2\\x=8\\1-2y=-5\Rightarrow2y=6\Rightarrow y=3\\x=-8\end{matrix}\right.\)

b tương tự.

c) \(\left(x+1\right)\left(x-2\right)< 0\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+1< 0\Rightarrow x< -1\\x-2>0\Rightarrow x>2\end{matrix}\right.\\\left\{{}\begin{matrix}x+1>0\Rightarrow x>-1\\x-2< 0\Rightarrow x< 2\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow-1< x< 2\Rightarrow x\in\left\{0;1\right\}\)

d tương tự

Mấy bài dễ tự làm nhé:D

1)

Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\\\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\end{matrix}\right.\)

Ta có điều phải chứng minh

\(\left\{{}\begin{matrix}\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\\\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\end{matrix}\right.\)

Ta có điều phải chứng minh

a,|$\frac{x}{2}-\frac{1}{3}$| = $\frac{3}{2}$

b, $\frac{3}{2}-\frac{1}{2}$ ( 2x-1)=$\frac{3}{4}$

c, |x-1|+2x=2

a)\(\left|\dfrac{x}{2}-\dfrac{1}{3}\right|=\dfrac{3}{2}\)

TH1

\(\dfrac{x}{2}-\dfrac{1}{3}=\dfrac{3}{2}\)

=>\(\dfrac{x}{2}=\dfrac{11}{6}\)

=>x=\(\dfrac{11.2}{6}\)

=>x=\(\dfrac{11}{3}\)

TH2

\(\dfrac{x}{2}-\dfrac{1}{2}=-\dfrac{3}{2}\)

=>\(\dfrac{x}{2}=-\dfrac{3}{2}+\dfrac{1}{2}\)

=>\(\dfrac{x}{2}=-1\)

=>x=-2

1.Tính

a.\(\dfrac{7}{23}\left[(-\dfrac{8}{6})-\dfrac{45}{18}\right]=\dfrac{7}{23}.-\dfrac{12}{6}=-\dfrac{7}{6}\)

b.\(\dfrac{1}{5}\div\dfrac{1}{10}-\dfrac{1}{3}(\dfrac{6}{5}-\dfrac{9}{4})=2-(-\dfrac{7}{20})=\dfrac{47}{20}\)

c.\(\dfrac{3}{5}.(-\dfrac{8}{3})-\dfrac{3}{5}\div(-6)=-\dfrac{3}{2}\)

d.\(\dfrac{1}{2}.(\dfrac{4}{3}+\dfrac{2}{5})-\dfrac{3}{4}.(\dfrac{8}{9}+\dfrac{16}{3})=-\dfrac{19}{5}\)

e.\(\dfrac{6}{7}\div(\dfrac{3}{26}-\dfrac{3}{13})+\dfrac{6}{7}.(\dfrac{1}{10}-\dfrac{8}{5})=-\dfrac{61}{7}\)

Bài 2

a.\(1^2_5x+\dfrac{3}{7}=\dfrac{4}{5}\)

\(x=\dfrac{13}{49}\)

b.\(\left|x-1,5\right|=2\)

Xảy ra 2 trường hợp

TH1

\(x-1,5=2\)

\(x=3,5\)

TH2

\(x-1,5=-2\)

\(x=-0,5\)

Vậy \(x=3,5\) hoặc \(x=-0,5\) .

Ngại làm quá trời ơi,lần sau bn tách ra nhá làm vậy mỏi tay quá.

Ths bn nhé

a. \(x:3\dfrac{1}{15}=1\dfrac{1}{2}\)

\(x:\dfrac{46}{15}=\dfrac{3}{2}\)

\(x=\dfrac{3}{2}.\dfrac{46}{15}=\dfrac{23}{5}\)

b. \(x.\dfrac{3}{2}=-\dfrac{7}{6}\)

\(x=-\dfrac{7}{6}:\dfrac{3}{2}=-\dfrac{7}{9}\)

c. \(\dfrac{5}{6}+\dfrac{1}{4}:x=-\dfrac{2}{3}\)

\(\dfrac{13}{12}:x=-\dfrac{2}{3}\)

\(x=\dfrac{13}{12}:\left(-\dfrac{2}{3}\right)=-\dfrac{13}{8}\)

Còn lại tương tự thôi

\(\)

lm mấy câu còn lại đi bn

b, \(-x-2=\dfrac{5}{4}\Rightarrow-x=\dfrac{13}{4}\Rightarrow x=-\dfrac{13}{4}\)

c, \(\dfrac{4}{3}-\left(x-\dfrac{1}{5}\right)=\left|-\dfrac{3}{10}+\dfrac{1}{2}\right|-\dfrac{1}{6}\)

\(\Rightarrow\dfrac{4}{3}-x+\dfrac{1}{5}=\left|\dfrac{1}{5}\right|-\dfrac{1}{6}\)

\(\Rightarrow-x=\dfrac{1}{5}-\dfrac{1}{6}-\dfrac{4}{3}-\dfrac{1}{5}\)

\(\Rightarrow-x=-\dfrac{3}{2}\Rightarrow x=\dfrac{3}{2}\)

d, \(\dfrac{1}{3}-\left(\dfrac{2}{3}-x+\dfrac{5}{4}\right)=\dfrac{7}{12}-\left(\dfrac{5}{2}-\dfrac{13}{6}\right)\)

\(\Rightarrow\dfrac{1}{3}-\dfrac{2}{3}+x-\dfrac{5}{4}=\dfrac{7}{12}-\dfrac{5}{2}+\dfrac{13}{6}\)

\(\Rightarrow x=\dfrac{7}{12}-\dfrac{5}{2}+\dfrac{13}{6}-\dfrac{1}{3}+\dfrac{2}{3}+\dfrac{5}{4}\)

\(\Rightarrow x=\dfrac{11}{6}\)

Chúc bạn học tốt!!!