ai giúp nhanh mik vs mik tick cho
tìm x thỏa mãn
65x+2=363x-4
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a: \(x^2\cdot x^3=32\)
\(\Leftrightarrow x^5=32\)
hay x=2
a, \(^{\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(y-z\right)^2\ge0\\\left(z-x\right)^2\ge0\end{cases}\Rightarrow}x^2-2xy+y^2+y^2-2yz+z^2+z^2-2xz+z^2\ge0}\)
\(\Rightarrow x^2+y^2+z^2\ge xy+yz+zx\Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\Rightarrow xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}\)
\(\Rightarrow A\le\frac{a^2}{3}\). dấu = xảy ra khi và chỉ khi x=y=z=a/3
b,Ap dụng bđt bunhia ta đc \(\left(1^2+1^2+1^2\right)\left(x^2+y^2+z^2\right)\ge\left(x+y+z\right)^2=a^2\Rightarrow B\ge\frac{a^2}{3}\)
dấu = xảy ra khi x=y=z=a/3
\(\left(x^4-2x^2+1\right)+\left(y^4-2y^2+1\right)+\left(z^4-2z^2+1\right)=0\)
\(\Leftrightarrow\)\(\left(x^2-1\right)^2+\left(y^2-1\right)^2+\left(z^2-1\right)^2=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}\left(x-1\right)\left(x+1\right)=0\\\left(y-1\right)\left(y+1\right)=0\\\left(z-1\right)\left(z+1\right)=0\end{cases}}\)\(\Rightarrow\)\(x,y,z\in\left\{1;-1\right\}\)
Mà \(\hept{\begin{cases}x^{2022}\ge0\forall x\\y^{2020}\ge0\forall y\\z^{2018}\ge0\forall z\end{cases}}\) nên P nhận giá trị không đổi khi \(x,y,z\in\left\{1;-1\right\}\)
\(\Rightarrow\)\(P=1+1+1=3\)
\(\left\{{}\begin{matrix}x^2+x+3=\dfrac{1}{4}\left(2x+1\right)^2+\dfrac{11}{4}\\\left(2y\right)^2-\left(2x+1\right)^2=11\end{matrix}\right.\)
\(\left\{{}\begin{matrix}y=\pm3\\\left\{{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\end{matrix}\right.\) thử lại ok !
a) Ta có: |2x-3|=x-6
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=x-6\\2x-3=6-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-3-x+6=0\\2x-3-6+x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+3=0\\3x-9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\)
Vậy: \(x\in\left\{-3;3\right\}\)
b) \(\frac{4}{9}x-\frac{1}{2}=\frac{-5}{9}\)
\(\Rightarrow\frac{4}{9}x=\frac{-5}{9}+\frac{1}{2}\)
\(\Rightarrow\frac{4}{9}x=\frac{-1}{18}\)
\(\Rightarrow x=\frac{-1}{18}:\frac{4}{9}\)
\(\Rightarrow x=\frac{-1}{8}\)
\(6^{5x+2}=36^{3x-4}\)
\(\Rightarrow6^{5x+2}=6^{2.\left(3x-4\right)}\)
\(\Rightarrow5x+2=2\left(3x-4\right)\)
\(\Rightarrow5x+2=6x-8\)
\(\Rightarrow-x=-10\)
\(\Rightarrow x=10\)
Vậy x = 10
\(6^{5x+2}=36^{3x-4}\)
\(6^{5x+2}=6^{2.\left(3x-4\right)}\)
\(\Rightarrow5x+2=6x-8\)
\(2+8=6x-5x\Leftrightarrow x=10\)
Chúc bạn học tốt!!!