Tìm x biết:
|3x - 2| - x > 1
|2x + 3| \(\le\) 5
Tìm x, biết:
a.|3x-2|-x>1
b.|2x+3|\(\le\)5
Bài giải
\(a,\text{ }\left|3x-2\right|-x>1\)
\(\left|3x-2\right|>x+1\)
TH1 : 3x - 2 < 0 => 3x < 3 => x < 1 thì :
\(3x-2>-x-1\)
\(3x+x>2-1\)
\(4x>1\)
\(x>\frac{1}{4}\)
=> \(\frac{1}{4}< x< 1\)
TH2 : 3x - 2 \(\ge\)0 => 3x \(\ge\)2 => x \(\ge\) \(\frac{2}{3}\) thì :
\(3x-2>x+1\)
\(3x-x>1+2\)
\(2x>3\)
\(x>\frac{3}{2}\)
Vậy \(\frac{1}{4}< x< 1\) hoặc \(x>\frac{3}{2}\)
Giai các bpt
a, 2x+2>4
b, 3x+2>-5
c,10-2x>2
d,1-2x<3
e,10x+3-5≤14x+12
f/ (3x-1)< 2x+4
g 4x-8 ≥3(2x-1)-2x+1
h/ x^x -x(x+2)> 3x-1
i/ x+8 >3x-1
j/ 3x- (2x+5) ≤(2x-3)
k/ (x-3) (x+3)<x(x+2)+3
l/ 2(3x-1) -2x<2x+1
m, (3-2x/5)> (2-x/3)
n, (x-2/6)-(x-1/3)≤x/2
o, (x+1/3)>(2x-1/6) -2
p, 1+ (2x+1)/3) >(2x-1/6) -2
q, (x+5/6)-(2x+1/3)≤ (x+3/2)
r, (5x+4/6) -(2x-1/12)≥4
a \(2x+2>4\\ \Leftrightarrow2\left(x+1\right)>4\\ \Leftrightarrow x+1>2\\ \Leftrightarrow x>1\)
b \(3x+2>-5\\ \Leftrightarrow3x>-7\\ \Leftrightarrow x>\dfrac{-7}{3}\)
c \(10-2x>2\\ \Leftrightarrow2\left(5-x\right)>2\\ \Leftrightarrow5-x>1\\ \Leftrightarrow-x>-4\\ \Leftrightarrow x< 4\)
d \(1-2x< 3\\ \Leftrightarrow-2x< 2\\ \Leftrightarrow2x>2\\ \Leftrightarrow x>1\)
a)2x+2>4
<=> 2x>4-2
<=>2x>2
<=>x>1
Vậy...
b)3x+2>-5
<=>3x>-5-2
<=>3x>-7
<=>x>\(\dfrac{-7}{3}\)
Vậy...
c)10-2x>2
<=>-2x>-10+2
<=>-2x>-8
<=>x<4
Vậy...
d)1-2x<3
<=>-2x<3-1
<=>-2x<2
<=>x>-1
Vậy...
e)10x+3-5\(\le\)14x+12
<=>10x-2\(\le\)14x+12
<=>10x-14x\(\le\)2+12
<=>-4x\(\le\)14
<=>x\(\ge\)\(\dfrac{-7}{2}\)
Vậy...
f)(3x-1)<2x+4
<=> 3x-2x<1+4
<=>x<5
Vậy...
Tìm x, biết:
a)/ 3x - 2 / - x = 7
b)/ 2x - 3 / > 5
c/ / 3x - 1 / ≤ 7
d) / 3x - 5 / + / 2x + 3/ = 7
a) / 3x - 2 / - x = 7
=> 3x-2-x= 7
=> 2x=9
=> x= 4,5
hoặc / 3x - 2 / - x = 7
=> 2x- 3x -x= 7
=> 2-4x=7
=> -5= 4x
=> x= -1,25
b)/ 2x - 3 / > 5
=> 2x> 8
=> x> 4
hoặc / 2x - 3 / > 5
=> 3-2x >5
=> 2x> -2
=> x> -1
nên ta => x> 4
Biết rằng bất phương trình \(\left\{{}\begin{matrix}x-1< 2x-3\\\frac{5-3x}{2}\le x-3\\3x\le x+5\end{matrix}\right.\)có tập nghiệm là 1 đoạn [a;b]. Hỏi a+b bằng?
\(\left\{{}\begin{matrix}-x< -2\\5-3x\le\\2x\le5\end{matrix}\right.2x-6}\)
<=>\(\left\{{}\begin{matrix}x>2\\-5x\le-11\\x\le\frac{5}{2}\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x>2\\x\ge\frac{11}{5}\\x\le\frac{5}{2}\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x\ge\frac{11}{5}\\x\le\frac{5}{2}\end{matrix}\right.\)
nên tập nghiệm của hệ là S=\(\left[\frac{11}{5};\frac{5}{2}\right]\)
a+b=\(\frac{11}{5}+\frac{5}{2}=\frac{47}{10}\)
tìm x biết :
a, |4x+3|-x=15
b, |3x-2|-x > 1
c, |2x+3 \(\le\) 5
tìm x biết
a, (2x-1)(3-x) \(\ge\)0
b, (5-x)(3x+2)\(\le\)0
1.|2x-1|≤x+2
2.(m+2)x² -3x+2m-3
3.5x-1>2x/5+3
4.(2x+1)² -3(x-3)>4x²+10
5.1<1/1-x
6.(x-5)²(x-3)/x+1≤0
Tìm x biết:
a) 3(2x-1)(3x-3)-(2x-1)(3x-3)=-3
b) (3x-1)(2x+7)-(x+1)(6x-5)=x+2-(x+5)
Bài 2 Tìm x biết 1, (2x-2).(3x+1)-(3x-2).(2x-3)=5 2,(1-3x).(3x-5)-(2x-4)(2-3x)=x-6 3,(2x-1).(4x^2+2x+1)-(2x+1)(4x^2-2x+1)=5x+6 Giúp tớ với
Tìm x biết:
a) 2x(3x+1) – (2x+3)(3x-2) = 12
b) (x+2)2 – (x-3)(x+3) = 5
b: \(\Leftrightarrow4x+13=5\)
hay x=-2
a) 2x(3x+1) – (2x+3)(3x-2) = 12
\(\Leftrightarrow6x^2+2x-\left(6x^2-4x+9x-6\right)=12\)
\(\Leftrightarrow6x^2+2x-6x^2+4x-9x+6=12\)
\(\Leftrightarrow-3x+6=12\)
\(\Leftrightarrow-3x=6\)
\(\Leftrightarrow x=-2\)
vậy x = -2
b) (x+2)2 – (x-3)(x+3) = 5
\(\Leftrightarrow\left(x+2\right)^2-\left(x^2-9\right)=5\)
\(\Leftrightarrow x^2+4x+4-x^2+9-5=0\)
\(\Leftrightarrow4x+8=0\)
\(\Leftrightarrow4x=-8\)
\(\Leftrightarrow x=-2\)
Vậy x = -2