a)

\(\left\{{}\begin{matrix}\left(4x-1\right)^4\ge0\\\left|2x-3y\right|\ge0\end{matrix}\right.\) \(\Rightarrow A\ge25,6\) tự tìm cận

không có Max

b) giống vậy

c) \(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\Rightarrow-\left(x-3\right)^2\le0\\\left|4x-3y\right|\ge0\Rightarrow-\left|4x-3y\right|\le0\end{matrix}\right.\)

\(C\le40,5\) tự tìm cận

không có GTNN

a.\(\left(3x-2\right)^2=16\)

Ta có: \(\left(3x-2\right)^2=16\)

\(\Rightarrow\left(3x-2\right)^2=\left(4\right)^2\)

\(\Rightarrow3x-2=4\)

\(\Rightarrow3x=6\)

\(\Rightarrow x=2\)

b. \(\left(\dfrac{4}{5}x-\dfrac{3}{4}\right)^3=\dfrac{-8}{125}\)

\(\Rightarrow\left(\dfrac{4}{5}x-\dfrac{3}{4}\right)^3=\left(\dfrac{-2}{5}\right)^3\)

\(\Rightarrow\dfrac{4}{5}x-\dfrac{3}{4}=\dfrac{-2}{5}^{ }\)

\(\Rightarrow\dfrac{4}{5}x-=\dfrac{7}{20}\)

\(\Rightarrow x=\dfrac{7}{16}\)

c. \(5^{x+2}+5^x=3250\)

\(5^x.5^2+5^x=3250\)

\(5^x.\left(5^2+1\right)=3250\)

\(5^x.26=3250\)

\(5^x=125\)

\(5^x=5^3\)

\(x=5\)

\(\left\{{}\begin{matrix}x\left(x+y+z\right)=13\\y\left(x+y+z\right)=7\\z\left(x+y+z\right)=-4\end{matrix}\right.\) \(\Leftrightarrow x\left(x+y+z\right)+y\left(x+y+z\right)+z\left(x+y+z\right)=13+7-4\)

\(\Rightarrow\left(x+y+z\right)\left(x+y+z\right)=16\)

\(\Rightarrow\left(x+y+z\right)^2=16\)

\(\Rightarrow\left[{}\begin{matrix}x+y+z=4\\x+y+z=-4\end{matrix}\right.\)

Với \(x+y+z=4\)

\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{13}{4}\\y=\dfrac{7}{4}\\z=-1\end{matrix}\right.\)

Với \(x+y+z=-4\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{13}{4}\\y=-\dfrac{7}{4}\\z=1\end{matrix}\right.\)

câu hỏi là j

a) \(\left|x\left(x-7\right)\right|=x\)

\(\Rightarrow\orbr{\begin{cases}x\left(x-7\right)=x\\x\left(x-7\right)=-x\end{cases}\Leftrightarrow\orbr{\begin{cases}x-7=1\\x-7=-1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=8\\x=6\end{cases}}}\)

b) \(\left|x-1,1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|=5x\)

\(\Rightarrow x-1,1+x+1,2+x+1,3+x+1,4=5x\)

\(\Leftrightarrow4x+2,8=5x\)

\(\Leftrightarrow x=2,8\)

\(a.\)\(\left|x.\left(x-7\right)\right|=x\)( Đk: \(x\ge0\))

\(\Leftrightarrow\orbr{\begin{cases}x.\left(x-7\right)=x\\x.\left(x-7\right)=-x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=x:x\\x-7=-x:x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=1\\x-7=-1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1+7\\x=-1+7\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=8\\x=6\end{cases}}\)

\(b.\)\(\left|x-1,1\right|+\left|x+1,2\right|+\left|x+1,3\right|+\left|x+1,4\right|=5x\)( Đk: \(5x\ge0\Leftrightarrow x\ge0\))

\(\Rightarrow x-1,1+x+1,2+x+1,3+x+1,4=5x\)

\(\Leftrightarrow\left(x+x+x+x\right)+\left(-1,1+1,2+1,3+1,4\right)=5x\)

\(\Leftrightarrow4x+2,8=5x\)

\(\Leftrightarrow2,8=5x-4x\)

\(\Leftrightarrow x=2,8\)

\(c.\)\(7^{x+2}+2.7^{x-1}=345\)

\(\Leftrightarrow7^{x-1}.7^{x+3}+2.7^{x-1}=345\)

\(\Leftrightarrow7^{x-1}.\left(7^{x+3}+2\right)=345\)

            \(......................\)

Đến đây mk ko bt làm nữa, tự lm nhé !

a) \(\left|2,5-x\right|-1,3=0\)

th1: \(2,5-x\ge0\Leftrightarrow x\le2,5\)

\(\Rightarrow\left|2,5-x\right|-1,3=0\Leftrightarrow2,5-x-1,3=0\Leftrightarrow x=1,2\left(tmđk\right)\)

th2: \(2,5-x< 0\Leftrightarrow x>2,5\)

\(\Rightarrow\left|2,5-x\right|-1,3=0\Leftrightarrow x-2,5-1,3=0\Leftrightarrow x=3,8\left(tmđk\right)\)

vậy \(x=1,2;x=3,8\)

b) \(1,6.\left|x-0,2\right|=0\Leftrightarrow\left|x-0,2\right|=0\Leftrightarrow x-0,2=0\Leftrightarrow x=0,2\) vậy \(x=0,2\)

c) \(\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\)

th1: \(\dfrac{1}{3}-x\ge0\Leftrightarrow x\le\dfrac{1}{3}\)

\(\Rightarrow\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\Leftrightarrow\dfrac{1}{3}-x-\dfrac{3}{7}=0\Leftrightarrow x=\dfrac{-2}{21}\left(tmđk\right)\)

th2: \(\dfrac{1}{3}-x< 0\Leftrightarrow x>\dfrac{1}{3}\)

\(\Rightarrow\left|\dfrac{1}{3}-x\right|-\left|\dfrac{-3}{7}\right|=0\Leftrightarrow x-\dfrac{1}{3}-\dfrac{3}{7}=0\Leftrightarrow x=\dfrac{16}{21}\left(tmđk\right)\)

vậy \(x=\dfrac{-2}{21};x=\dfrac{16}{21}\)

d) \(\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\)

th1: \(x+\dfrac{4}{15}\ge0\Leftrightarrow x\ge\dfrac{-4}{15}\)

\(\Rightarrow\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\Leftrightarrow x+\dfrac{4}{15}-3,75=-2,15\)

\(\Leftrightarrow x=\dfrac{4}{3}\left(tmđk\right)\)

th2: \(x+\dfrac{4}{15}< 0\Leftrightarrow x< \dfrac{-4}{15}\)

\(\Rightarrow\left|x+\dfrac{4}{15}\right|-\left|-3,75\right|=-\left|-2,15\right|\Leftrightarrow-x-\dfrac{4}{15}-3,75=-2,15\)

\(\Leftrightarrow x=\dfrac{-28}{15}\left(tmđk\right)\)

vậy \(x=\dfrac{4}{3};x=\dfrac{-28}{15}\)

e) ta có : \(\left|x-1,5\right|\ge0\forall x\) và \(\left|2,5-x\right|\ge0\forall x\)

\(\Rightarrow\left|x-1,5\right|+\left|2,5-x\right|=0\Leftrightarrow\left\{{}\begin{matrix}x-1,5=0\\2,5-x=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=1,5\\x=2,5\end{matrix}\right.\) 2 giá trị này khác nhau \(\Rightarrow\) phương trình vô nghiệm