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a) 7x - 35 = 0
<=> 7x = 0 + 35
<=> 7x = 35
<=> x = 5
b) 4x - x - 18 = 0
<=> 3x - 18 = 0
<=> 3x = 0 + 18
<=> 3x = 18
<=> x = 5
c) x - 6 = 8 - x
<=> x - 6 + x = 8
<=> 2x - 6 = 8
<=> 2x = 8 + 6
<=> 2x = 14
<=> x = 7
d) 48 - 5x = 39 - 2x
<=> 48 - 5x + 2x = 39
<=> 48 - 3x = 39
<=> -3x = 39 - 48
<=> -3x = -9
<=> x = 3
a) ĐKXĐ: \(x\notin\left\{-1;0\right\}\)
Ta có: \(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2\)
\(\Leftrightarrow\dfrac{x\left(x+3\right)}{x\left(x+1\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{x\left(x+1\right)}=\dfrac{2x\left(x+1\right)}{x\left(x+1\right)}\)
Suy ra: \(x^2+3x+x^2-3x+2=2x^2+2x\)
\(\Leftrightarrow2x^2+2-2x^2-2x=0\)
\(\Leftrightarrow-2x+2=0\)
\(\Leftrightarrow-2x=-2\)
hay x=1(nhận)
Vậy: S={1}
b) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)
Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)
\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)
\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)
\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)
\(\Leftrightarrow-56x-1=0\)
\(\Leftrightarrow-56x=1\)
hay \(x=-\dfrac{1}{56}\)(nhận)
Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)
c) ĐKXĐ: \(x\ne-\dfrac{2}{3}\)
Ta có: \(\dfrac{5}{3x+2}=2x-1\)
\(\Leftrightarrow5=\left(3x+2\right)\left(2x-1\right)\)
\(\Leftrightarrow6x^2-3x+4x-2-5=0\)
\(\Leftrightarrow6x^2+x-7=0\)
\(\Leftrightarrow6x^2-6x+7x-7=0\)
\(\Leftrightarrow6x\left(x-1\right)+7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\6x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\6x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{7}{6}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{1;-\dfrac{7}{6}\right\}\)
d) ĐKXĐ: \(x\ne\dfrac{2}{7}\)
Ta có: \(\left(2x+3\right)\cdot\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)
\(\Leftrightarrow\left(2x+3\right)\cdot\left(\dfrac{3x+8+2-7x}{2-7x}\right)-\left(x-5\right)\left(\dfrac{3x+8+2-7x}{2-7x}\right)=0\)
\(\Leftrightarrow\left(2x+3-x+5\right)\cdot\dfrac{-4x+6}{2-7x}=0\)
\(\Leftrightarrow\left(x+8\right)\cdot\left(-4x+6\right)=0\)(Vì \(2-7x\ne0\forall x\) thỏa mãn ĐKXĐ)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\-4x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\-4x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\left(nhận\right)\\x=\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{-8;\dfrac{3}{2}\right\}\)
a, -2x>15 x>-15/2 c, th1 x+2>0 vs x+3 <0 suy ra x>-2 vs x<-3 . th2 x+2<0,x+3>0 suy ra x<-2 ,x>-3
b, 112-x2>0
x2<112 x<11
a) \(3x-8>5x+7\)
\(\Leftrightarrow-8>5x+7-3x\)
\(\Leftrightarrow-8>2x+7\)
\(\Leftrightarrow-8-7>2x\)
\(\Leftrightarrow-15>2x\)
\(\Leftrightarrow-\frac{15}{2}>x\)
\(\Rightarrow x< -\frac{15}{2}\)
b) \(\left(11-x\right)\left(11+x\right)>0\)
\(\Leftrightarrow x=\pm11\)
\(\Rightarrow-11< x< 11\)
c) \(\left(x+2\right)\left(x+3\right)< 0\)
\(\Leftrightarrow x=-2;-3\)
\(\Rightarrow-3< x< -2\)
\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)
\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)
\(\Leftrightarrow x^2-9-x^2+3x=0\)
\(\Leftrightarrow3x-9=0\)
\(\Leftrightarrow3x=9\)
\(\Leftrightarrow x=3\left(n\right)\)
Vậy \(S=\left\{3\right\}\)
\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)
\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)
\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)
\(\Leftrightarrow12x-9-12x+20+2x-7>0\)
\(\Leftrightarrow2x+4>0\)
\(\Leftrightarrow2x>-4\)
\(\Leftrightarrow x>-2\)
1.a)|−7x|=3x+16
Vì |-7x| ≥ 0 nên 3x+16 ≥ 0 ⇔ x ≥ \(\dfrac{-16}{3}\) (*)
Với đk (*), ta có: |-7x|=3x+16
\(\left[\begin{array}{} -7x=3x+16\\ -7x=-3x-16 \end{array} \right.\) ⇔ \(\left[\begin{array}{} -7x-3x=16\\ -7x+3x=-16 \end{array} \right.\)
⇔ \(\left[\begin{array}{} x=-1,6 (t/m)\\ x= 4 (t/m) \end{array} \right.\)
b) \(\dfrac{x-1}{x+2}\) - \(\dfrac{x}{x-2}\) = \(\dfrac{5x-8}{x^2-4}\)
⇔ \(\dfrac{(x-1)(x-2)}{x^2-4}\) - \(\dfrac{x(x+2)}{x^2-4}\) = \(\dfrac{5x-8}{x^2-4}\)
⇒ x2 - 2x - x + 2 - x2 - 2x = 5x - 8
⇔ -5x - 5x = -8 - 2
⇔ -10x = -10
⇔ x=1
2.7x+5 < 3x−11
⇔ 7x - 3x < -11 - 5
⇔ 4x < -16
⇔ x < -4
bạn tự biểu diễn trên trục số nha !
Cách giải
a, 2x - x (3x + 1 ) < 15 - 3x(x + 2)
<=> 2x - 3x2 - x < 15 - 3x2 - 6x
<=> 7x < 15
<=> x < 15/7 Vậy Tập nghiệm của BPT là : { x / x < 15/7 }
b , BPT <=> 2(1 - 2x ) - 16 < 1 - 5x + 8x
<=> -7x < 15
<=> x > -15/7 Vậy tập nghiệm của BPT là : { x / x > -15/7 }
a) 2x-x(3x+1) < 15-3x(x+2)
<=> 2x-3x2-x < 15-3x2-6x
<=> 2x-3x2-x+3x2+6x < 15
<=> 7x < 15
<=> x < 15/7
Vậy tập nghiệm của bất phương trình là x < 15/7
b) \(\frac{1-2x}{4}-2\le\frac{1-5x}{8}+x\)
Quy đồng mẫu ta được :
\(\frac{2-4x}{8}-\frac{16}{8}\le\frac{1-5x}{8}+\frac{8x}{8}\)
Khử mẫu
=> \(2-4x-16\le1-5x+8x\)
<=> \(-4x+5x-8x\le1-2+16\)
<=> \(-7x\le15\)
<=> \(x\ge-\frac{15}{7}\)
Vậy tập nghiệm của bất phương trình là \(x\ge-\frac{15}{7}\)
a. (x−1)(5x+3)=(3x−8)(x−1)(x−1)(5x+3)=(3x−8)(x−1)
⇔(x−1)(5x+3)−(3x−8)(x−1)=0⇔(x−1)[(5x+3)−(3x−8)]=0⇔(x−1)(5x+3−3x+8)=0⇔(x−1)(2x+11)=0⇔(x−1)(5x+3)−(3x−8)(x−1)=0⇔(x−1)[(5x+3)−(3x−8)]=0⇔(x−1)(5x+3−3x+8)=0⇔(x−1)(2x+11)=0
⇔x−1=0⇔x−1=0hoặc 2x+11=02x+11=0
+ x−1=0⇔x=1x−1=0⇔x=1
+ 2x+11=0⇔x=−5,52x+11=0⇔x=−5,5
Phương trình có nghiệm x = 1 hoặc x = -5,5
b. 3x(25x+15)−35(5x+3)=03x(25x+15)−35(5x+3)=0
⇔15x(5x+3)−35(5x+3)=0⇔(15x−35)(5x+3)=0⇔15x(5x+3)−35(5x+3)=0⇔(15x−35)(5x+3)=0
⇔15x−35=0⇔15x−35=0 hoặc 5x+3=05x+3=0
+ 15x−35=0⇔x=3515=7315x−35=0⇔x=3515=\(\frac{7}{3}\)
+ 5x+3=0⇔x=−355x+3=0⇔x=−\(\frac{3}{5}\)
Phương trình có nghiệm x=\(\frac{7}{3}\)x=\(\frac{7}{3}\) hoặc x=−\(\frac{3}{5}\)
a: 7x+35=0
=>7x=-35
=>x=-5
b: \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)
=>8-x-8(x-7)=1
=>8-x-8x+56=1
=>-9x+64=1
=>-9x=-63
hay x=7(loại)
a, \(7x=-35\Leftrightarrow x=-5\)
b, đk : x khác 7
\(8-x-8x+56=1\Leftrightarrow-9x=-63\Leftrightarrow x=7\left(ktm\right)\)
vậy pt vô nghiệm
2, thiếu đề