a)

\(\left(3x+\dfrac{1}{3}\right)\left(x-\dfrac{1}{2}\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x+\dfrac{1}{3}=0\\x-\dfrac{1}{2}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{9}\\x=\dfrac{1}{2}\end{matrix}\right.\)

b)

\(\left(x-\dfrac{3}{2}\right)\left(2x+1\right)>0\\ \Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-\dfrac{3}{2}>0\\2x+1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-\dfrac{3}{2}< 0\\2x+1< 0\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>\dfrac{3}{2}\\x>-\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x< \dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< -\dfrac{1}{2}\end{matrix}\right.\)

tiếp đi bạn

mình làm lại câu b) nha

b) |x-3|=-4

th1: x-3=-4

x=3+(-4)

x=-1

th2: x-3=4

x=3+4

x=7

b) \(\left|x-3\right|=-4\)

t/h1:\(x-3=-4\)

\(x=3-\left(-4\right)\)

\(x=7\)

t/h2:\(x-3=4\)

\(x=3-4\)

\(x=-1\)

h) \(5^x+5^{x+2}=650\)

\(\Leftrightarrow5^x+5^x.5^2=650\)

\(\Leftrightarrow5^x\left(1+25\right)=650\)

\(\Leftrightarrow5^x.26=650\)

\(\Leftrightarrow5^x=25\)

\(\Leftrightarrow x=2\)

haizzz,đăng ít thôi,chứ nhìn hoa mắt quá =.=

bây định làm j ở chỗ này vậy??? có j ib ns vs nhao chớ sao ns ở đây

1) a) \(\left|7x-5y\right|+\left|2z-3y\right|+\left|xy+yz+xz-2000\right|\ge0\)

Dấu "=" xảy ra khi: \(\left\{{}\begin{matrix}7x=5y\\2z=3y\\xy+yz+xz=2000\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{7}y\\z=\dfrac{3}{2}y\\xy+yz+xz=2000\end{matrix}\right.\)

Ta có: \(xy+yz+xz=2000\)

\(\Rightarrow\dfrac{5}{7}y^2+\dfrac{3}{2}y^2+\dfrac{15}{14}y^2=2000\)

\(\Rightarrow y^2\left(\dfrac{5}{7}+\dfrac{3}{2}+\dfrac{15}{14}\right)=2000\Leftrightarrow\dfrac{23}{7}y^2=2000\)

Tìm \(y\) và suy ra \(x;z\) là được,Bài này nghiệm khá xấu

b) \(\left|3x-7\right|+\left|3x+2\right|+8=\left|7-3x\right|+\left|3x+2\right|+8\ge\left|7-3x+3x+2\right|+8\ge9+8=17\)Dấu "=" xảy ra khi: \(-\dfrac{3}{2}\le x\le\dfrac{7}{3}\)

2) a)Ta có: \(\left\{{}\begin{matrix}\left|x-5\right|+\left|1-x\right|\ge\left|x-5+1-x\right|=4\\\dfrac{12}{\left|y+1\right|+3}\le\dfrac{12}{3}=4\end{matrix}\right.\)

Mà theo đề bài: \(\left|x-5\right|+\left|1-x\right|=\dfrac{12}{\left|y+1\right|+3}\)

\(\Rightarrow\left|x-5\right|+\left|1-x\right|=\dfrac{12}{\left|y+1\right|+3}=4\)

\(\Rightarrow\left\{{}\begin{matrix}1\le x\le5\\y=-1\end{matrix}\right.\)

b) Ta có: \(\left\{{}\begin{matrix}\left|y+3\right|+5\ge5\\\dfrac{10}{\left(2x-6\right)^2+2}\le\dfrac{10}{2}=5\end{matrix}\right.\)

Mà theo đề bài: \(\left|y+3\right|+5=\dfrac{10}{\left(2x-6\right)^2+2}\)

\(\Rightarrow\left|y+3\right|+5=\dfrac{10}{\left(2x-6\right)^2+2}=5\)

\(\Rightarrow\left\{{}\begin{matrix}y=-3\\x=3\end{matrix}\right.\)

c) Ta có: \(\left\{{}\begin{matrix}\left|x-1\right|+\left|3-x\right|\ge\left|x-1+3-x\right|=2\\\dfrac{6}{\left|y+3\right|+3}\le\dfrac{6}{3}=2\end{matrix}\right.\)

Mà theo đề bài: \(\left|x-1\right|+\left|3-x\right|=\dfrac{6}{\left|y+3\right|+3}\)

\(\Rightarrow\left|x-1\right|+\left|3-x\right|=\dfrac{6}{\left|y+3\right|+3}=2\)

\(\Rightarrow\left\{{}\begin{matrix}1\le x\le3\\y=-3\end{matrix}\right.\)

1. Tìm x:

a) \(\left(x+36\right)^2=1936\Leftrightarrow x+36=\pm44.\) Vậy x = 8 hoặc x = -80

b) \(\left(\dfrac{3}{5}\right)^{x+2}=\dfrac{81}{625}\Leftrightarrow\left(\dfrac{3}{5}\right)^{x+2}=\left(\dfrac{3}{5}\right)^4\Leftrightarrow x+2=4\Leftrightarrow x=2\)

c) Xem lại đề

d) \(\left(\dfrac{9}{16}\right)^{x-5}=\left(\dfrac{4}{3}\right)^4\Leftrightarrow\left(\dfrac{3}{4}\right)^{2\left(x-5\right)}=\left(\dfrac{3}{4}\right)^{-4}\Leftrightarrow2\left(x-5\right)=-4\Leftrightarrow x=3\)

e) \(\left(\dfrac{3}{5}\right)^x.\left(\dfrac{125}{27}\right)^x=\dfrac{81}{625}\Leftrightarrow\left(\dfrac{3}{5}.\dfrac{125}{27}\right)^x=\left(\dfrac{3}{5}\right)^4\Leftrightarrow\left(\dfrac{5}{3}\right)^{2x}=\left(\dfrac{5}{3}\right)^{-4}\Leftrightarrow2x=-4\) Vậy x = -2

3. Tính giá trị của biểu thức:

\(A=\left\{-\left[\left(\dfrac{1}{x}\right)^2\right]^3\right\}^5.\left\{-\left[\left(-x\right)^5\right]^2\right\}^3\) \(\left(x\notin0\right)\)

\(=\left\{-\left[-\dfrac{1}{x^2}\right]^3\right\}^5.\left\{-\left[-\left(-x\right)^5\right]^2\right\}^3=\left\{-\left[-\dfrac{1}{x^6}\right]\right\}^5.\left\{-\left[x^5\right]^2\right\}^3\)

\(=\left\{\dfrac{1}{x^6}\right\}^5.\left\{-x^{10}\right\}^3=\dfrac{1}{x^{30}}.\left(-x^{30}\right)=-1\)

a. \(\left(x+5\right)^3=-64\)

\(\left(x+5\right)^3=\left(-4\right)^3\)

\(\Rightarrow x+5=-4\)

=> x = -9

b. \(|2x-5|=8\)

\(\left[{}\begin{matrix}2x-5=8\\2x-5=-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=13\\2x=-3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{2}\\x=\dfrac{-3}{2}\end{matrix}\right.\)

c. \(\left|\dfrac{3}{4}x-\dfrac{1}{5}\right|=2\)

\(\left[{}\begin{matrix}\dfrac{3}{4}x-\dfrac{1}{5}=2\\\dfrac{3}{4}x-\dfrac{1}{5}=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}\dfrac{3}{4}x=\dfrac{11}{5}\\\dfrac{3}{4}x=\dfrac{-9}{5}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{44}{15}\\x=\dfrac{-12}{5}\end{matrix}\right.\)

d. \(\left|3x-6\right|=x+4\)

\(\left[{}\begin{matrix}3x-6=x+4\\3x-6=-x-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-x=4+6\\3x+x=-4+6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=10\\4x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{2}\end{matrix}\right.\)

e. \(\left|x-3\right|=2x+1\)

\(\left[{}\begin{matrix}x-3=2x+1\\x-3=-2x-1\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x-2x=1+3\\x+2x=-1+3\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}-x=4\\3x=2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)

thank bn nha