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a) Có \(\left|x-3y\right|^5\ge0\);\(\left|y+4\right|\ge0\)
\(\rightarrow\left|x-3y\right|^5+\left|y+4\right|\ge0\)
mà \(\left|x-3y\right|^5+\left|y+4\right|=0\)
\(\rightarrow\left\{{}\begin{matrix}\left|x-3y\right|^5=0\\\left|y+4\right|=0\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}x=3y\\y=-4\end{matrix}\right.\)
\(\rightarrow\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\)
b) Tương tự câu a, ta có:
\(\left\{{}\begin{matrix}\left|x-y-5\right|=0\\\left(y-3\right)^4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=3\end{matrix}\right.\)
c. Tương tự, ta có:
\(\left\{{}\begin{matrix}\left|x+3y-1\right|=0\\\left|y+2\right|=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=1-3y\\y=-2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=7\\y=-2\end{matrix}\right.\)
a. \(\left|x-3y\right|^5\ge0,\left|y+4\right|\ge0\Rightarrow\left|x-3y\right|^5+\left|y+4\right|\ge0\) \(\Rightarrow VT\ge VP\)
Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x-3y\right|^5=0\\\left|y+4\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3y\\y=-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\) Vậy...
b. \(\left|x-y-5\right|\ge0,\left(y-3\right)^4\ge0\Rightarrow\left|x-y-5\right|+\left(y-3\right)^4\ge0\) \(\Rightarrow VT\ge VP\)
Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x-y-5\right|=0\\\left(y-3\right)^4=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=y+5\\y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=8\\y=3\end{matrix}\right.\) Vậy ...
c. \(\left|x+3y-1\right|\ge0,3\cdot\left|y+2\right|\ge0\Rightarrow\left|x+3y-1\right|+3\left|y+2\right|\ge0\) \(\Rightarrow VT\ge VP\) Dấu bằng xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}\left|x+3y-1\right|=0\\3\left|y+2\right|=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1-3y\\y=-2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1-\left(-2\right)\cdot3=7\\y=-2\end{matrix}\right.\) Vậy...
\(\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+100\right)=5750\)
\(100x+\left(1+2+3+4+...+100\right)=5750\)
Áp dụng công thức tính dãy số ta có
\(\left(100-1\right):1+1.\left(100+1\right):2=100.101:2=5050\)
\(\Rightarrow100x+5050=5750\)
\(\Rightarrow100x=700\)
\(\Rightarrow x=7\)
( x+x+x+....+x)+(1+2+3+4+.....+ 100)=5750
=(x.100)+(101.1):100:2=5750
=> (x.100)+5050=5750
=>x.100=700
=>x=7
( 3x + 1 ) ( 5 - 2x ) > 0
---> 3x + 1 và 5 - 2x cùng dấu
+, \(\hept{\begin{cases}3x+1>0\\5-2x>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>\frac{-1}{3}\\\frac{5}{2}>x\end{cases}}\Leftrightarrow\frac{5}{2}>x>\frac{-1}{3}\)
+, \(\hept{\begin{cases}3x+1< 0\\5-2x< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< \frac{-1}{3}\\\frac{5}{2}< x\end{cases}}\Leftrightarrow\frac{5}{2}< x< \frac{-1}{3}\)VÔ LÝ
xin tiick
\(a.\)
\(\left(x-8\right)\left(x^3+8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\x^3+8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x^3=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
\(S=\left\{8,-2\right\}\)
\(b.\)
\(\left(4x-3\right)-\left(x+5\right)=3\cdot\left(10-x\right)\)
\(\Leftrightarrow4x-3-x-5-30+3x=0\)
\(\Leftrightarrow6x-38=0\)
\(\Leftrightarrow x=\dfrac{38}{6}\)
\(S=\left\{\dfrac{38}{6}\right\}\)
a) (x - 8 )( x3 + 8) = 0
\(\Rightarrow\left[{}\begin{matrix}x-8=0\\x^3=-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
b)(4x - 3) – ( x + 5) = 3(10 - x)
\(\Leftrightarrow4x-3-x-5=30-3x\)
\(\Leftrightarrow3x-8=30-3x\)
\(\Leftrightarrow3x-8-30+3x=0\)
\(\Leftrightarrow6x-38=0\)
\(\Leftrightarrow x=\dfrac{19}{3}\)
a) \(\left|4x-1\right|-\left|3x-\dfrac{1}{2}\right|=0\\ \Leftrightarrow\left|4x-1\right|=\left|3x-\dfrac{1}{2}\right|\\ \Leftrightarrow\left[{}\begin{matrix}4x-1=3x-\dfrac{1}{2}\\4x-1=\dfrac{1}{2}-3x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}4x-3x=1-\dfrac{1}{2}\\4x+3x=\dfrac{1}{2}+1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\7x=\dfrac{3}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{3}{14}\end{matrix}\right.\)
Vậy \(x\in\left\{\dfrac{1}{2};\dfrac{3}{14}\right\}\) là nghiệm của pt.
b) \(\left|x-1\right|-2x=\dfrac{1}{2}\\ \Leftrightarrow\left|x-1\right|=2x+\dfrac{1}{2}\left(ĐK:x\ge\dfrac{-1}{4}\right)\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x+\dfrac{1}{2}\\x-1=-2x-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x-2x=1+\dfrac{1}{2}\\x+2x=1-\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-x=\dfrac{3}{2}\\3x=\dfrac{1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-3}{2}\left(ktmđk\right)\\x=\dfrac{1}{6}\left(tmđk\right)\end{matrix}\right.\)
Vậy \(x=\dfrac{1}{6}\) là nghiệm của pt.
Lời giải:
a.
$|4x-1|-|3x-\frac{1}{2}|=0$
$\Leftrightarrow |4x-1|=|3x-\frac{1}{2}$
\(\Leftrightarrow \left[\begin{matrix} 4x-1=3x-\frac{1}{2}\\ 4x-1=\frac{1}{2}-3x\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=\frac{1}{2}\\ x=\frac{3}{14}\end{matrix}\right.\)
b. Nếu $x\geq 1$ thì:
$|x-1|-2x=\frac{1}{2}$
$\Leftrightarrow x-1-2x=\frac{1}{2}$
$\Leftrightarrow -x-1=\frac{1}{2}$
$\Leftrightarrow x=\frac{-3}{2}$ (vô lý vì $x\geq 1$)
Nếu $x< 1$ thì:
$1-x-2x=\frac{1}{2}$
$\Leftrightarrow x=\frac{1}{6}$ (tm)
Bài 2
P(x) + Q(x) = x3 – 6x + 2 + 2x2 - 4x3 + x - 5 = - 3x3 + 2x2 – 5x - 3
P(x) - Q(x) = x3 – 6x + 2 - 2x2 + 4x3 - x + 5 = 5x3 − 2x2 − 7x+7
a) \(\left(x+5\right)\left(x-2\right)< 0\)
\(\Leftrightarrow\hept{\begin{cases}x+5>0\\x-2< 0\end{cases}}\) hoặc \(\hept{\begin{cases}x+5< 0\\x-2>0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x>-5\\x< 2\end{cases}}\) hoặc \(\hept{\begin{cases}x< -5\\x>2\end{cases}}\) (loại)
Vậy -5 < x < 2
b) \(\left(x+2\right)\left(x-\frac{3}{5}\right)>0\)
\(\Leftrightarrow\hept{\begin{cases}x+2>0\\x-\frac{3}{5}>0\end{cases}}\) hoặc \(\hept{\begin{cases}x+2< 0\\x-\frac{3}{5}< 0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}x>-2\\x>\frac{3}{5}\end{cases}}\) hoặc \(\hept{\begin{cases}x< -2\\x< \frac{3}{5}\end{cases}}\)
Vậy x > 3/5 hoặc x < -2
a ) ( x + 5 )( x - 2 ) < 0
=> x + 5 duong va x - 2 am hoac x + 5 am va x - 2 duong
Neu x + 5 duong va x - 2 am thi
-5 < x < 2
=> x \(\in\left\{1;0;-1;-2;-3;-4\right\}\)
Neu x + 5 am va x - 2 duong thi :
x < -5 va x > 2
Vi 2 dieu kien tren mau thuan vs nhau nen x\(\varnothing\)trong truong hop nay