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a) \(5^{3x+1}=25^{x+2}\)
\(\Leftrightarrow5^{3x+1}=\left(5^2\right)^{x+2}\)
\(\Leftrightarrow5^{3x+1}=5^{2x+4}\)
\(\Leftrightarrow3x+1=2x+4\)
\(\Leftrightarrow3x-2x=4-1\)
\(\Leftrightarrow x=3\)
Bài 1:
\(\left|x+\frac{1}{2}\right|+\left|x+\frac{1}{6}\right|+...+\left|x+\frac{1}{101}\right|=101x\)
Ta thấy:
\(VT\ge0\Rightarrow VP\ge0\Rightarrow101x\ge0\Rightarrow x\ge0\)
\(\Rightarrow\left(x+\frac{1}{2}\right)+\left(x+\frac{1}{6}\right)+...+\left(x+\frac{1}{101}\right)=101x\)
\(\Rightarrow\left(x+x+...+x\right)+\left(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{101}\right)=0\)
\(\Rightarrow10x+\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{10.11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10}-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\left(1-\frac{1}{11}\right)=0\)
\(\Rightarrow10x+\frac{10}{11}=0\)
\(\Rightarrow10x=-\frac{10}{11}\Rightarrow x=-\frac{1}{11}\)(loại,vì x\(\ge\)0)
Bài 2:
Ta thấy: \(\begin{cases}\left(2x+1\right)^{2008}\ge0\\\left(y-\frac{2}{5}\right)^{2008}\ge0\\\left|x+y+z\right|\ge0\end{cases}\)
\(\Rightarrow\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|\ge0\)
Mà \(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\left(2x+1\right)^{2008}+\left(y-\frac{2}{5}\right)^{2008}+\left|x+y+z\right|=0\)
\(\Rightarrow\begin{cases}\left(2x+1\right)^{2008}=0\\\left(y-\frac{2}{5}\right)^{2008}=0\\\left|x+y+z\right|=0\end{cases}\)\(\Rightarrow\begin{cases}2x+1=0\\y-\frac{2}{5}=0\\x+y+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\x+y+z=0\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{2}+\frac{2}{5}+z=0\end{cases}\)
\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\-\frac{1}{10}=-z\end{cases}\)\(\Rightarrow\begin{cases}x=-\frac{1}{2}\\y=\frac{2}{5}\\z=\frac{1}{10}\end{cases}\)
a) Ta có: \(-2xy^2\cdot\left(x^3y-2x^2y^2+5xy^3\right)\)
\(=-2x^4y^3+4x^3y^4-10x^2y^5\)
b) Ta có: \(\left(-2x\right)\cdot\left(x^3-3x^2-x+1\right)\)
\(=-2x^4+6x^3+2x^2-2x\)
c) Ta có: \(3x^2\left(2x^3-x+5\right)\)
\(=6x^5-3x^3+15x^2\)
d) Ta có: \(\left(-10x^3+\frac{2}{5}y-\frac{1}{3}z\right)\cdot\left(-\frac{1}{2}xy\right)\)
\(=5x^4y-\frac{1}{5}xy^2+\frac{1}{6}xyz\)
e) Ta có: \(\left(3x^2y-6xy+9x\right)\cdot\left(-\frac{4}{3}xy\right)\)
\(=-4x^3y^2+8x^2y^2-12x^2y\)
f) Ta có: \(\left(4xy+3y-5x\right)\cdot x^2y\)
\(=4x^3y^2+3x^2y^2-5x^3y\)
a/ \(\left(\frac{1}{5}\right)^x=\left(\frac{1}{5^3}\right)^3=\left(\frac{1}{5}\right)^9\Rightarrow x=9\)
b/ \(\left(\frac{3}{5}\right)^x=\left(\frac{3^2}{5^2}\right)^3=\left(\frac{3}{5}\right)^6\Rightarrow x=6\)
c\(2^{3-2x}=\left(2^3\right)^3=2^9\Rightarrow3-2x=9\Rightarrow x=-3\)
d/ \(2^{3x+1}=32^2=\left(2^5\right)^2=2^{10}\Rightarrow3x+1=10\Rightarrow x=3\)
e/ \(3^{6-3x}=81^3=\left(3^4\right)^3=3^{12}\Rightarrow6-3x=12\Rightarrow x=-2\)
\(\left(\frac{1}{5}\right)^x=\left(\frac{1}{125}\right)^3\Leftrightarrow\left(\frac{1}{5}\right)^x=\left[\left(\frac{1}{5}\right)^3\right]^3\Leftrightarrow\left(\frac{1}{5}\right)^x=\left(\frac{1}{5}\right)^9\Leftrightarrow x=9\)
\(\left(\frac{3}{5}\right)^x=\left(\frac{9}{25}\right)^3\Leftrightarrow\left(\frac{3}{5}\right)^x=\left[\left(\frac{3}{5}\right)^2\right]^3\Leftrightarrow\left(\frac{3}{5}\right)^x=\left(\frac{3}{5}\right)^6\Leftrightarrow x=6\)
\(2^{3-2x}=8^3\Leftrightarrow2^{3-2x}=\left(2^3\right)^3\Leftrightarrow2^{3-2x}=2^9\Leftrightarrow3-2x=9\)
\(\Leftrightarrow2x=3-9\Leftrightarrow2x=-6\Leftrightarrow x=\left(-6\right):2\Leftrightarrow x=-3\)
Các phép còn lại làm tương tự bn nha !
Bài 1:
Ta có: \(x+\left(-\frac{31}{12}\right)^2=\left(\frac{49}{12}\right)^2-x\)
\(\Leftrightarrow2x=\frac{1440}{144}=10\)
\(\Rightarrow x=5\)
Khi đó: \(y^2=\left(\frac{49}{12}\right)^2-5=\frac{1681}{144}\)
=> \(\hept{\begin{cases}y=\frac{41}{12}\\y=-\frac{41}{12}\end{cases}}\)
Lời giải:
Áp dụng BĐT $|a|+|b|\geq |a+b|$ ta có:
$|3x+1|+|3x-5|=|3x+1|+|5-3x|\geq |3x+1+5-3x|=6$
$(y+3)^2+2\geq 2, \forall y\Rightarrow \frac{12}{(y+3)^2+2}\leq \frac{12}{2}=6$
Vậy:
$|3x+1|+|3x-5|\geq 6\geq \frac{12}{(y+3)^2+2}$
Dấu "=" xảy ra (3x+1)(5-3x)\geq 0$ và $y+3=0$
$\Leftrightarrow \frac{-1}{3}\leq x\leq \frac{5}{3}$ và $y=-3$