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a: 11x+4=-3/2
=>\(11x=-\dfrac{3}{2}-4=-\dfrac{11}{2}\)
=>\(x=-\dfrac{1}{2}\)
b: \(x^2-9+2\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(x+3\right)+2\left(x-3\right)=0\)
=>\(\left(x-3\right)\left(x+3+2\right)=0\)
=>(x-3)(x+5)=0
=>\(\left[{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
c: \(\dfrac{x-3}{5}+\dfrac{1+2x}{3}=6\)
=>\(\dfrac{3\left(x-3\right)+5\left(2x+1\right)}{15}=6\)
=>\(3x-9+10x+5=90\)
=>13x-4=90
=>13x=94
=>\(x=\dfrac{94}{13}\)
d: \(\dfrac{2}{x+1}-\dfrac{1}{x-2}=\dfrac{3x-11}{\left(x+1\right)\left(x-2\right)}\)(ĐKXĐ: \(x\notin\left\{-1;2\right\}\))
=>\(\dfrac{2\left(x-2\right)-\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{3x-11}{\left(x-2\right)\left(x+1\right)}\)
=>3x-11=2x-4-x-1
=>3x-11=x-5
=>2x=6
=>x=3(nhận)
Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
\(x-5=\frac{1}{3\left(x+2\right)}\left(đkxđ:x\ne-2\right)\)
\(< =>3\left(x-5\right)\left(x+2\right)=1\)
\(< =>3\left(x^2-3x-10\right)=1\)
\(< =>x^2-3x-10=\frac{1}{3}\)
\(< =>x^2-3x-\frac{31}{3}=0\)
giải pt bậc 2 dễ r
\(\frac{x}{3}+\frac{x}{4}=\frac{x}{5}-\frac{x}{6}\)
\(< =>\frac{4x+3x}{12}=\frac{6x-5x}{30}\)
\(< =>\frac{7x}{12}=\frac{x}{30}< =>12x=210x\)
\(< =>x\left(210-12\right)=0< =>x=0\)
a. 3(x-2)-10=5(2x + 1)
<=> 3x - 6 - 10 = 10x + 5
<=> 3x - 10x = 5 + 6 + 10
<=> -7x = 21
<=> x = -3
b. 3x + 2=8 -2(x-7)
<=> 3x + 2 = 8 - 2x + 14
<=> 3x + 2x = 8 + 14 - 2
<=> 5x = 20
<=> x = 4
c. 2x-(2+5x)= 4(x + 3)
<=> 2x - 2 - 5x = 4x + 12
<=> 2x - 5x - 4x = 12 + 2
<=> -7x = 14
<=> x = -2
d. 5-(x +8)=3x + 3(x-9)
<=> 5 - x - 8 = 3x + 3x - 27
<=> -x - 3x - 3x = -27 + 8 - 5
<=> -7x = -24
<=> x = 24/7
e. 3x - 18 + x= 12-(5x + 3)
<=> 3x - 18 + x = 12 - 5x - 3
<=> 3x + x - 5x = 12 - 3 + 18
<=> -x = 27
<=> x = - 27
a. 3(x-2)-10=5(2x + 1)
<=> 3x - 6 - 10 = 10x + 5
<=> 3x - 10x = 5 + 6 + 10
<=> -7x = 21
<=> x = -3
b. 3x + 2=8 -2(x-7)
<=> 3x + 2 = 8 - 2x + 14
<=> 3x + 2x = 8 + 14 - 2
<=> 5x = 20
<=> x = 4
c. 2x-(2+5x)= 4(x + 3)
<=> 2x - 2 - 5x = 4x + 12
<=> 2x - 5x - 4x = 12 + 2
<=> -7x = 14
<=> x = -2
d. 5-(x +8)=3x + 3(x-9)
<=> 5 - x - 8 = 3x + 3x - 27
<=> -x - 3x - 3x = -27 + 8 - 5
<=> -7x = -24
<=> x = 24/7
e. 3x - 18 + x= 12-(5x + 3)
<=> 3x - 18 + x = 12 - 5x - 3
<=> 3x + x - 5x = 12 - 3 + 18
<=> -x = 27
<=> x = - 27
a: =>3,6-1,7x=2,3-1,4-4=0,9-4=-3,1
=>1,7x=6,7
hay x=67/17
b: \(\Leftrightarrow30\left(5x+4\right)-15\left(3x+5\right)=24\left(4x+9\right)-40\left(x-9\right)\)
=>150x+120-45x-75=96x+216-40x+360
=>105x+45=56x+576
=>49x=531
hay x=531/49
a,\(\left(x-4-5\right)\left(x-4+5\right)=0\Leftrightarrow\left(x-9\right)\left(x+1\right)=0\Leftrightarrow x=9;x=-1\)
b, \(\left(x-3-x-1\right)\left(x-3+x+1\right)=0\Leftrightarrow2x-2=0\Leftrightarrow x=1\)
c, \(\left(x^2-4\right)\left(2x-3\right)-\left(x^2-4\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(2x-3-x+1\right)=0\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(x-2\right)=0\Leftrightarrow x=-2;x=2\)
d, \(\left(3x-7\right)^2-\left(2x+2\right)^2=0\Leftrightarrow\left(3x-7-2x-2\right)\left(3x-7+2x+2\right)=0\)
\(\Leftrightarrow\left(x-9\right)\left(5x-5\right)=0\Leftrightarrow x=1;x=9\)
a) Ta có: \(2\left(3x+1\right)-4\left(5-2x\right)>2\left(4x-3\right)-6\)
\(\Leftrightarrow6x+2-20+8x>8x-6-6\)
\(\Leftrightarrow14x-18-8x+12>0\)
\(\Leftrightarrow6x-6>0\)
\(\Leftrightarrow6x>6\)
hay x>1
Vậy: S={x|x>1}
b) Ta có: \(9x^2-3\left(10x-1\right)< \left(3x-5\right)^2-21\)
\(\Leftrightarrow9x^2-30x+3< 9x^2-30x+25-21\)
\(\Leftrightarrow9x^2-30x+3-9x^2+30x-4< 0\)
\(\Leftrightarrow-1< 0\)(luôn đúng)
Vậy: S={x|\(x\in R\)}
b) 
d) 
f) 
h) 
k) 
n) 
bạn đăng tách cho mn cùng giúp nhé
Bài 1 :
a, \(\Leftrightarrow11-x=12-8x\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)
b, \(\Leftrightarrow2x\left(x^2+4x+4\right)-8x^2=2\left(x^3-8\right)\)
\(\Leftrightarrow2x^3+8x^2+8x-8x^2=2x^3-16\Leftrightarrow x=-2\)
c, \(\Leftrightarrow3-2x=-x-4\Leftrightarrow x=7\)
d, \(\Leftrightarrow x^3-6x^2+12x-8+9x^2-1=x^3+3x^2+3x+1\)
\(\Leftrightarrow3x^2+12x-9=3x^2+3x+1\Leftrightarrow x=\dfrac{10}{9}\)
e, \(\Leftrightarrow2x^2-x-3=2x^2+9x-5\Leftrightarrow x=5\)
f, \(\Leftrightarrow x^3-3x^2+3x-1-x^3-2x^2-x=10x-5x^2-11x-22\)
\(\Leftrightarrow-5x^2+2x-1=-5x^2-x-22\Leftrightarrow3x=-21\Leftrightarrow x=-7\)
a) x = 10 3 b) x = - 31 12
c) x = 7 5 d) x = 4