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a) \(2\frac{1}{3}+\left(x-\frac{3}{2}\right)=\left(3-\frac{3}{2}\right)x\)
\(2\frac{1}{3}+x-\frac{3}{2}=3x-\frac{3}{2}x\)
\(2\frac{1}{3}-\frac{3}{2}=3x-\frac{3}{2}x-x\)
\(\frac{5}{6}=3x-\frac{3}{2}x-x\)
\(\frac{5}{6}=\left(3-\frac{3}{2}-1\right)x\)
\(\frac{5}{6}=\frac{1}{2}x\)
\(x=\frac{5}{6}:\frac{1}{2}\)
\(x=\frac{5}{3}\)
b) |3x-4|+|3y+5|=0
ĐK : \(\hept{\begin{cases}\left|3x-4\right|\ge0\\\left|3y+5\right|\ge0\end{cases}}\Leftrightarrow\left|3x-4\right|+\left|3y+5\right|\ge0\)
Mà |3x-4|+|3y+5|=0 nên :
\(\Rightarrow\hept{\begin{cases}3x-4=0\\3y+5=0\end{cases}}\Rightarrow\hept{\begin{cases}3x=4\\3y=-5\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{4}{3}\\y=\frac{-5}{3}\end{cases}}\)
Vậy x=4/3 ; y=-5/3
c) \(\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2004\right|=0\)
ĐK : \(\hept{\begin{cases}\left|x+\frac{19}{5}\right|\ge0\\\left|y+\frac{1890}{1975}\right|\ge0\\\left|z-2004\right|\ge0\end{cases}}\Leftrightarrow\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2004\right|\ge0\)
Mà \(\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2004\right|=0\) nên :
\(\Rightarrow\hept{\begin{cases}x+\frac{19}{5}=0\\y+\frac{1890}{1975}=0\\z-2004=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{19}{5}\\y=-\frac{1890}{1975}\\z=2004\end{cases}}\)
Vậy ...
\(\frac{1}{4}+\frac{1}{3}:2x=-5\)
\(\frac{1}{3}:2x=-5-\frac{1}{4}\)
\(\frac{1}{3}:2x=\frac{-21}{4}\)
\(2x=\frac{1}{3}:\frac{-21}{4}\)
\(2x=\frac{-4}{63}\)
\(x=\frac{-4}{63}:2\)
\(x=\frac{-2}{63}\)
\(\)
\(\frac{1}{4}+\frac{1}{3}:2x=-5\)
\(\Rightarrow\frac{1}{3}:2x=-\frac{21}{4}\)
\(\Rightarrow2x=\frac{-4}{63}\)
\(\Rightarrow x=\frac{-2}{63}\)
\(\left(3x-\frac{1}{4}\right)\left(x+\frac{1}{2}\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-\frac{1}{4}=0\\x+\frac{1}{2}=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-1}{2}\end{cases}}}\)
\(\left(2x-5\right)\left(\frac{3}{2}x+9\right)\left(0,3x-12\right)=0\)
Th1 : \(2x-5=0\Rightarrow x=\frac{5}{2}\)
Th2 : \(\frac{3}{2}x+9=0\Rightarrow x=-6\)
Th3 : \(0,3x-12=0\Rightarrow x=\frac{12}{0,3}\)
a) 3x-1(1+5)=162
3x-1.6=162
3x-1=162:6=27=33
=>x-1=3
x=4
b) x(x+3)=0
=>x=0 hoặc x+3=0
=>x=0 hoặc x=-3
c) Vì tích nhỏ hơn 0 nên có 1 thừa số dương và 1 thừa số âm
Có x-1>x-3
=>x-1>0 và x-3<0
=>x>1 và x<3
Vậy x=2
a) 3x-1 + 5. 3x-1 = 162
1. 3x-1 + 5. 3x-1 = 162
( 1 + 5 ) . 3x-1 = 162
6. 3x-1 = 162
3x-1 = 162 : 6
3x-1 = 27
3x-1 = 33
x - 1 =3
x = 3 + 1
x = 4
a)\(\left(2x-3\right)\left(x+1\right)< 0\)
\(\Leftrightarrow\begin{cases}2x-3>0\\x+1< 0\end{cases}\) hoặc \(\begin{cases}2x-3< 0\\x+1>0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>\frac{3}{2}\\x< -1\end{cases}\) (loại) hoặc \(\begin{cases}x< \frac{3}{2}\\x>-1\end{cases}\)
\(\Leftrightarrow-1< x< \frac{3}{2}\)
b) \(\left(x-\frac{1}{2}\right)\left(x+3\right)>0\)
\(\Leftrightarrow\begin{cases}x-\frac{1}{2}>0\\x+3>0\end{cases}\) hoặc \(\begin{cases}x-\frac{1}{2}< 0\\x+3< 0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>\frac{1}{2}\\x>-3\end{cases}\) hoặc \(\begin{cases}x< \frac{1}{2}\\x< -3\end{cases}\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x>\frac{1}{2}\\x< -3\end{array}\right.\)
c) Sai đề phải là \(\frac{x}{\left(x+3\right)\left(x+7\right)}\)
Có: \(\frac{3}{\left(x+3\right)\left(x+5\right)}+\frac{5}{\left(x+5\right)\left(x+10\right)}+\frac{7}{\left(x+10\right)\left(x+17\right)}=\frac{x}{\left(x+3\right)\left(x+17\right)}\)
\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+10}+\frac{1}{x+10}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow\)\(\frac{1}{x+3}-\frac{1}{x+7}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow\)\(\frac{4}{\left(x+3\right)\left(x+7\right)}=\frac{x}{\left(x+3\right)\left(x+7\right)}\)
\(\Leftrightarrow x=4\)
mấy cái này đơn dãng vô cùng nhưng có đều bn ra đề dài quá nha
a) \(3x+4\ge7\Leftrightarrow3x\ge7-4\Leftrightarrow3x\ge3\Leftrightarrow x\ge1\) vậy \(x\ge1\)
b) \(-5x+1< 11\Leftrightarrow-5x< 11-1\Leftrightarrow-5x< 10\Leftrightarrow x>\dfrac{10}{-5}\)
\(\Leftrightarrow x>-2\) vậy \(x>-2\)
c) \(\dfrac{5}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow x< 3\) vậy \(x< 3\)
d) \(\dfrac{-7}{2-x}\ge0\Leftrightarrow2-x\le0\Leftrightarrow x\ge2\) vậy \(x\ge2\)
e) \(x^2+4x>0\Leftrightarrow x\left(x+4\right)>0\) \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>0\\x+4>0\end{matrix}\right.\\\left[{}\begin{matrix}x< 0\\x+4< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>0\\x>-4\end{matrix}\right.\\\left[{}\begin{matrix}x< 0\\x< -4\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x>0\\x< -4\end{matrix}\right.\) vậy \(x>0\) hoặc \(x< -4\)
f) \(\dfrac{x-2}{x-6}< 0\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x-2>0\\x-6>0\end{matrix}\right.\\\left[{}\begin{matrix}x-2< 0\\x-6< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>2\\x>6\end{matrix}\right.\\\left[{}\begin{matrix}x< 2\\x< 6\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>6\\x< 2\end{matrix}\right.\)
vậy \(x>6\) hoặc \(x< 2\)
g) \(\left(x-1\right)\left(x+2\right)\left(3-x\right)< 0\Leftrightarrow-\left[\left(x-1\right)\left(x+2\right)\left(x-3\right)\right]< 0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x-3\right)>0\)
th1: 3 số hạng đều dương : \(\Leftrightarrow\left[{}\begin{matrix}x-1>0\\x+2>0\\x-3>0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x>1\\x>-2\\x>3\end{matrix}\right.\) \(\Rightarrow x>3\)
th2: 2 âm 1 dương : (vì trong 3 số hạng ta có : \(\left(x+2\right)\) lớn nhất \(\Rightarrow\left(x+2\right)\) dương)
\(\Leftrightarrow\left[{}\begin{matrix}x-1< 0\\x+2>0\\x-3< 0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x< 1\\x>-2\\x< 3\end{matrix}\right.\) \(\Rightarrow-2< x< 1\)
vậy \(x>3\) hoặc \(-2< x< 1\)
h) \(\dfrac{x^2-1}{x}>0\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x^2-1>0\\x>0\end{matrix}\right.\\\left[{}\begin{matrix}x^2-1< 0\\x< 0\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x^2>1\\x>0\end{matrix}\right.\\\left[{}\begin{matrix}x^2< 1\\x< 0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\\x>0\end{matrix}\right.\\\left[{}\begin{matrix}-1< x< 1\\x< 0\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>1\\-1< x< 0\end{matrix}\right.\) vậy \(x>1\) hoặc \(-1< x< 0\)
i) \(x^2+x-2< 0\Leftrightarrow x^2+x+\dfrac{1}{4}-\dfrac{9}{4}< 0\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2-\dfrac{9}{4}< 0\)
\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2< \dfrac{9}{4}\Leftrightarrow\dfrac{-3}{2}< \left(x+\dfrac{1}{2}\right)< \dfrac{3}{2}\Leftrightarrow-2< x< 1\)
vậy \(-2< x< 1\)
Mysterious Person, Đoàn Đức Hiếu, Nguyễn Đình Dũng , ... giúp mình!
1. Tìm x, biết :
a. ( x - \(\frac{3}{4}\)) \(^2\)= 0
=> x - \(\frac{3}{4}\)= 0
=> x = 0 + \(\frac{3}{4}\)
=> x = \(\frac{3}{4}\)
b. ( x + \(\frac{1}{2}\)) \(^2\)= \(\frac{9}{64}\)
=> ( x + \(\frac{1}{2}\)) \(^2\)= ( \(\frac{3}{8}\)) \(^2\)
=> x + \(\frac{1}{2}\)= \(\frac{3}{8}\)
=> x = \(\frac{3}{8}\)- \(\frac{1}{2}\)
=> x = \(\frac{-1}{8}\)
c. \(\frac{\left(-2\right)^x}{16}=-8\)
=> \(\frac{\left(-2\right)^x}{16}=\frac{-8}{1}=\frac{-128}{16}\)
=> ( -2)\(^x\)= -128
=> ( -2 ) \(^x\)= ( -2) \(^7\)
=> x = 7