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a, \(5\left|2x-1\right|-3=7\Leftrightarrow5\left|2x-1\right|=10\Leftrightarrow\left|2x-1\right|=2\)
TH1 : \(2x-1=2\Leftrightarrow x=\frac{3}{2}\)
TH2 : \(2x-1=-2\Leftrightarrow x=-\frac{1}{2}\)
b, \(\left(2x+3\right)\left(x-2\right)-x^2+4=0\Leftrightarrow\left(2x+3\right)\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x+3-x-2\right)=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow x=-1;x=2\)
c, \(\frac{2x-3}{2}< \frac{1-3x}{-5}\Leftrightarrow\frac{2x-3}{2}+\frac{1-3x}{5}< 0\)
\(\Leftrightarrow\frac{10x-15+2-6x}{10}< 0\Rightarrow4x-13< 0\Leftrightarrow x< \frac{13}{4}\)
c: =>2x+4>=2x+2-3
=>4>=-1(luôn đúng)
a: 5x+10>3x+3
=>2x>-7
=>x>-7/2
a: 3x-5>15-x
=>4x>20
hay x>5
b: \(3\left(x-2\right)\left(x+2\right)< 3x^2+x\)
=>3x2+x>3x2-12
=>x>-12
3:
a: =>x=0 hoặc x+5=0
=>x=0 hoặc x=-5
b: =>x^2=4
=>x=2 hoặc x=-2
c: =>(x-5)(2x+1+x+6)=0
=>(x-5)(3x+7)=0
=>x=5 hoặc x=-7/3
1.
a. 2x - 6 > 0
\(\Leftrightarrow\) 2x > 6
\(\Leftrightarrow\) x > 3
S = \(\left\{x\uparrow x>3\right\}\)
b. -3x + 9 > 0
\(\Leftrightarrow\) - 3x > - 9
\(\Leftrightarrow\) x < 3
S = \(\left\{x\uparrow x< 3\right\}\)
c. 3(x - 1) + 5 > (x - 1) + 3
\(\Leftrightarrow\) 3x - 3 + 5 > x - 1 + 3
\(\Leftrightarrow\) 3x - 3 + 5 - x + 1 - 3 > 0
\(\Leftrightarrow\) 2x > 0
\(\Leftrightarrow\) x > 0
S = \(\left\{x\uparrow x>0\right\}\)
d. \(\dfrac{x}{3}-\dfrac{1}{2}>\dfrac{x}{6}\)
\(\Leftrightarrow\dfrac{2x}{6}-\dfrac{3}{6}>\dfrac{x}{6}\)
\(\Leftrightarrow2x-3>x\)
\(\Leftrightarrow2x-3-x>0\)
\(\Leftrightarrow x-3>0\)
\(\Leftrightarrow x>3\)
\(S=\left\{x\uparrow x>3\right\}\)
2.
a.
Ta có: a > b
3a > 3b (nhân cả 2 vế cho 3)
3a + 7 > 3b + 7 (cộng cả 2 vế cho 7)
b. Ta có: a > b
a > b (nhân cả 2 vế cho 1)
a + 3 > b + 3 (cộng cả 2 vế cho 3) (1)
Ta có; 3 > 1
b + 3 > b + 1 (nhân cả 2 vế cho 1b) (2)
Từ (1) và (2) \(\Rightarrow\) a + 3 > b + 1
c.
5a - 1 + 1 > 5b - 1 + 1 (cộng cả 2 vế cho 1)
5a . \(\dfrac{1}{5}\) > 5b . \(\dfrac{1}{5}\) (nhân cả 2 vế cho \(\dfrac{1}{5}\) )
a > b
3.
a. 2x(x + 5) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
\(S=\left\{0,-5\right\}\)
b. x2 - 4 = 0
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(S=\left\{0,4\right\}\)
d. (x - 5)(2x + 1) + (x - 5)(x + 6) = 0
\(\Leftrightarrow\left(x-5\right)\left(2x+1+x+6\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(3x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\3x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-7}{3}\end{matrix}\right.\)
\(S=\left\{5,\dfrac{-7}{3}\right\}\)
a) 5 - (x - 6) = 4(3 - 2x)
<=> 5 - x + 6 = 12 - 8x
<=> -x + 8x = 12 - 11
<=> 7x = 1
<=> x = 1/7
Vậy S = {1/7}
b) 2x(x - 3) + 5(x - 3) = 0
<=> (2x + 5)(x - 3) = 0
<=> \(\orbr{\begin{cases}2x+5=0\\x-3=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-\frac{5}{2}\\x=3\end{cases}}\)
Vậy S = {-5/2; 3}
c)ĐK: x \(\ne\)1; x \(\ne\)2
\(\frac{3x-5}{x-2}-\frac{2x-5}{x-1}=1\)
<=> \(\frac{\left(3x-5\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}-\frac{\left(2x-5\right)\left(x-2\right)}{\left(x-1\right)\left(x-2\right)}=\frac{\left(x-1\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)}\)
<=> 3x2 - 8x + 5 - 2x2 + 9x - 10 = x2 - 3x + 2
<=> x2 + x - 5 = x2 - 3x + 2
<=> x2 + x - x2 + 3x = 2 + 5
<=> 4x = 7
<=> x = 7/4
Vậy S = {7/4}
1:
a: =>3x=6
=>x=2
b: =>4x=16
=>x=4
c: =>4x-6=9-x
=>5x=15
=>x=3
d: =>7x-12=x+6
=>6x=18
=>x=3
2:
a: =>2x<=-8
=>x<=-4
b: =>x+5<0
=>x<-5
c: =>2x>8
=>x>4
ta có : x^5+2x^4+3x^3+3x^2+2x+1=0
\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0
\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0
\(\Leftrightarrow\)x^4(x+1)+x^3(x+1)+2x^2(x+1)+x(x+1)+(x+1)=0
\(\Leftrightarrow\)(x+1)(x^4+x^3+2x^2+x+1)=0
\(\Leftrightarrow\)(x+1)(x^4+x^3+x^2+x^2+x+1)=0
\(\Leftrightarrow\)(x+1)[x^2(x^2+x+1)+(x^2+x+1)]=0
\(\Leftrightarrow\)(x+1)(x^2+x+1)(x^2+1)=0
VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)
\(\Rightarrow\)x+1=0
\(\Rightarrow\)x=-1
CÒN CÂU B TỰ LÀM (02042006)
b: x^4+3x^3-2x^2+x-3=0
=>x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0
=>(x-1)(x^3+4x^2+2x+3)=0
=>x-1=0
=>x=1
mk chỉ giải đc có bài 1 thui nha bn
\(\frac{4}{x-2}+\frac{1}{x+3}=0\)
ĐKXĐ: x ≠ 2 và x ≠ -3
QĐKM:
⇔(x+3)4 + (x-2)1 = 0
⇔4x + 12 + x - 2 = 0
⇔4x + x = -12 + 2
⇔5x = -10
⇔x= -2
S={-2}