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LH
0
23 tháng 5 2022
a: (2x-3)(3x+6)>0
=>(2x-3)(x+2)>0
=>x<-2 hoặc x>3/2
b: (3x+4)(2x-6)<0
=>(3x+4)(x-3)<0
=>-4/3<x<3
c: (3x+5)(2x+4)>4
\(\Leftrightarrow6x^2+12x+10x+20-4>0\)
\(\Leftrightarrow6x^2+22x+16>0\)
=>\(6x^2+6x+16x+16>0\)
=>(x+1)(3x+8)>0
=>x>-1 hoặc x<-8/3
f: (4x-8)(2x+5)<0
=>(x-2)(2x+5)<0
=>-5/2<x<2
h: (3x-7)(x+1)<=0
=>x+1>=0 và 3x-7<=0
=>-1<=x<=7/3
24 tháng 9 2022
a: x+1>0
=>x>-1
b: -2x-3<0
=>-2x<3
=>x>-3/2
c: 4x+5>0
=>4x>-5
=>x>-5/4
d: -7x-3<0
=>-7x<3
=>x>-3/7
k: 3x+7>0
=>3x>-7
=>x>-7/3
l: -4x-1<0
=>-4x<1
=>x>-1/4
NT
1
31 tháng 5 2017
a) \(\left|4x+3\right|-x=15\)
\(\Rightarrow\left|4x+3\right|=15+x\)
\(\Rightarrow\left[{}\begin{matrix}4x+3=15+x\\4x+3=-15+x\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x-x=15-3\\4x-x=-15-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=12\\3x=-18\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-6\end{matrix}\right.\)
b) \(A=\left|3x-2\right|\)
Dấu = xảy ra khi \(3x-2=0\)
\(\Rightarrow3x=2\Rightarrow x=\dfrac{2}{3}\)
Amin = 1 <=> x = 2/3
c) \(B=\left|2x+3\right|\le5\)
Dấu = xảy ra <=> 2x + 3 = 0 => 2x = -3 => x = -3/2
Bmax = 5 <=> x = -3/2
3 tháng 6 2017
Bạn ơi hai câu b, c họ bảo tìm x chứ có phải tìm GTLN, NN đâu
FA
2
25 tháng 1 2020
a) \(\left|4x+3\right|-x=15\)\\
\(\Rightarrow\left|4x+3\right|=15+x.\)
\(\Rightarrow\left[{}\begin{matrix}4x+3=15+x\\4x+3=-15-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}4x-x=15-3\\4x+x=-15-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=12\\5x=-18\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{18}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{4;-\dfrac{18}{5}\right\}.\)
b) \(\left|3x-2\right|-x>1\)
\(\Rightarrow\left|3x-2\right|>1+x.\)
\(\Rightarrow\left[{}\begin{matrix}3x-2>1+x\\3x-2< -1-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-x>1+2\\3x+x< -1+2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x>3\\4x< 1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< \dfrac{1}{4}\end{matrix}\right.\Rightarrow\dfrac{1}{4}< x< \dfrac{3}{2}.\)
Vậy \(\dfrac{1}{4}< x< \dfrac{3}{2}\)
c) \(\left|2x+3\right|\le5\)
\(\Rightarrow\left[{}\begin{matrix}2x+3\le5\\2x+3\ge-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x\le2\\2x\ge-8\end{matrix}\right.
\(\Rightarrow\left[{}\begin{matrix}x\le1\\x\ge-4\end{matrix}\right.\Rightarrow-4\le x\le1.\)
Vậy \(-4\le x\le1\)
25 tháng 1 2020
a) \(\left|4x+3\right|-x=15\)
\(\Rightarrow\left|4x+3\right|=15+x.\)
\(\Rightarrow\left[{}\begin{matrix}4x+3=15+x\\4x+3=-15-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}4x-x=15-3\\4x+x=-15-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=12\\5x=-18\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{18}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{4;-\dfrac{18}{5}\right\}.\)
b) \(\left|3x-2\right|-x>1\)
\(\Rightarrow\left|3x-2\right|>1+x.\)
\(\Rightarrow\left[{}\begin{matrix}3x-2>1+x\\3x-2< -1-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-x>1+2\\3x+x< -1+2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x>3\\4x< 1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>\dfrac{3}{2}\\x< \dfrac{1}{4}\end{matrix}\right.\Rightarrow\dfrac{1}{4}< x< \dfrac{3}{2}.\)
Vậy \(\dfrac{1}{4}< x< \dfrac{3}{2}\)
c) \(\left|2x+3\right|\le5\)
\(\Rightarrow\left[{}\begin{matrix}2x+3\le5\\2x+3\ge-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x\le2\\2x\ge-8\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\le1\\x\ge-4\end{matrix}\right.\Rightarrow-4\le x\le1.\)
Vậy \(-4\le x\le1\)
MH
A) 5/4+x=2/3
B) -x-2=5/4
C)4x+1/3=3/2
Đ) 1/3-2/5+3x=3/4
E) 3x+7+2x=4x-3
G) 3x(2x-3)-2x(3x-4)=15
H) x^2-x=0
1
7 tháng 7 2017
a) \(x=-\frac{7}{12}\)
b) \(x=-\frac{13}{4}\)
c) \(x=\frac{7}{24}\)
d) \(x=\frac{49}{180}\)
e) \(x=-10\)
g) \(x=15\)
h) \(\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
25 tháng 1 2020
a) Ta có: |4x+3|-x=15
⇒|4x+3|=15+x
\(\Rightarrow\left\{{}\begin{matrix}\left(4x+3\right)^2=\left(15+x\right)^2\\15+x\ge0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}16x^2+24x+9=225+30x+x^2\\x\ge-15\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}15x^2-6x-216=0\\x\ge-15\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=-3,6\end{matrix}\right.\)
Vậy: x∈{-3,6;4}
b) Ta có: |3x-2|-x>1
⇒|3x-2|>1+x
\(\Leftrightarrow\left\{{}\begin{matrix}3x-2>1+x\\3x-2< -1-x\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x>3\\4x< 1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x>\frac{3}{2}\\x< \frac{1}{4}\end{matrix}\right.\)
Vậy: \(\frac{1}{4}< x< \frac{3}{2}\)
c) Ta có: \(\left|2x+3\right|\le5\)
\(\Leftrightarrow\left\{{}\begin{matrix}2x+3\le5\\2x+3\ge-5\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}2x\le2\\2x\ge-8\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x\ge-4\end{matrix}\right.\)
Vậy: \(-4\le x\le1\)
VM
25 tháng 1 2020
a) \(\left|4x+3\right|-x=15\)
\(\Rightarrow\left|4x+3\right|=15+x.\)
\(\Rightarrow\left[{}\begin{matrix}4x+3=15+x\\4x+3=-15-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}4x-x=15-3\\4x+x=-15-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}3x=12\\5x=-18\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=12:3\\x=\left(-18\right):5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4\\x=-\frac{18}{5}\end{matrix}\right.\)
Vậy \(x\in\left\{4;-\frac{18}{5}\right\}.\)
b) \(\left|3x-2\right|-x>1\)
\(\Rightarrow\left|3x-2\right|>1+x.\)
\(\Rightarrow\left[{}\begin{matrix}3x-2>1+x\\3x-2< -1-x\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}3x-x>1+2\\3x+x< -1+2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x>3\\4x< 1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x>\frac{3}{2}\\x< \frac{1}{4}\end{matrix}\right.\Rightarrow\frac{1}{4}< x< \frac{3}{2}.\)
Vậy \(\frac{1}{4}< x< \frac{3}{2}\) thì \(\left|3x-2\right|-x>1.\)
c) \(\left|2x+3\right|\le5\)
\(\Rightarrow\left[{}\begin{matrix}2x+3\le5\\2x+3\ge-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x\le2\\2x\ge-8\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x\le2:2\\x\ge\left(-8\right):2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x\le1\\x\ge-4\end{matrix}\right.\Rightarrow-4\le x\le1.\)
Vậy \(-4\le x\le1\) thì \(\left|2x+3\right|\le5.\)
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0