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11 tháng 8 2021
1: Ta có: \(\dfrac{x+4}{4}+\dfrac{3x-7}{5}=\dfrac{7x+2}{20}\)
\(\Leftrightarrow5x+20+12x-28=7x+2\)
\(\Leftrightarrow17x-7x=2+8=10\)
hay x=1
2: Ta có: \(\dfrac{x}{6}+\dfrac{1-3x}{9}=\dfrac{-x+1}{12}\)
\(\Leftrightarrow\dfrac{6x}{36}+\dfrac{4\left(1-3x\right)}{36}=\dfrac{3\left(-x+1\right)}{36}\)
\(\Leftrightarrow6x+4-12x=-3x+3\)
\(\Leftrightarrow-6x+3x=3-4\)
hay \(x=\dfrac{1}{3}\)
3: Ta có: \(\dfrac{x-3}{3}-\dfrac{x+2}{12}=\dfrac{2x-1}{4}\)
\(\Leftrightarrow4x-12-x-2=6x-3\)
\(\Leftrightarrow3x-14-6x+3=0\)
\(\Leftrightarrow-3x=11\)
hay \(x=-\dfrac{11}{3}\)
11 tháng 8 2021
4: Ta có: \(\dfrac{x-2}{4}-\dfrac{2x+3}{3}=\dfrac{x+6}{12}\)
\(\Leftrightarrow3x-6-8x-12=x+6\)
\(\Leftrightarrow-5x-x=6+18\)
hay x=-4
5: Ta có: \(\dfrac{2x-1}{12}-\dfrac{3-x}{18}=\dfrac{-1}{36}\)
\(\Leftrightarrow6x-3+2x-6=-1\)
\(\Leftrightarrow8x=8\)
hay x=1
NH
11 tháng 11 2020
a)(x+2).(x+3)-(x-2).(x+5)=10
( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10
x^2 +3x+2x+6-x^2 -5x+2x+10-10=0
2x+6=0
2x=-6
x=-3
18 tháng 12 2022
a: =18x^3y^2-12x^3y^3+6x^2y^2
b: (-3x+2)(5x^2-1/3x+4)
=-12x^3+x^2-12x+10x^2-2/3x+8
=-12x^3+11x^2-38/3x+8
c: =x^2-x-2+3x-x^2
=2x-2
d: =4x^2+12x+9-4x^2+25-(x-1)(x^2+12)
=12x+34-x^3-12x+x^2+12
=-x^3+x^2+46
27 tháng 5 2021
\(\left(2x^2+1\right)\left(4x-3\right)=\left(x-12\right)\left(2x^2+1\right)\)
<=> 4x - 3 = x - 12 (vì \(2x^2+1\ne0\) với mọi x)
<=> 3x = -9
<=> x = -3
23 tháng 9 2021
d: \(\dfrac{x^4-2x^3+2x-1}{x^2-1}\)
\(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}\)
\(=x^2-2x+1\)
\(=\left(x-1\right)^2\)
NT
1
24 tháng 4 2017
a, (x-3)(2x+2) - (2x+1)(x-3)+12 =0
(x-3)(2x +2-2x-1) +12 = 0
(x-3) . 1 +12=0
x - 3 +12 =0
x = 9
31 tháng 1 2018
\(\Leftrightarrow\left(2x^2+1\right)\left(4x-3\right)-\left(2x^2+1\right)\left(x-12\right)=0\)
\(\Leftrightarrow\left(2x^2+1\right)\left(4x-3-x+12\right)=0\)
\(\Leftrightarrow\left(2x^2+1\right)\left(3x+9\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x^2+1=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x^2=-1\\3x=-9\end{matrix}\right.\Leftrightarrow}x=-3}\)
31 tháng 1 2018
Đề: \(\left(2x^2+1\right).\left(4x-3\right)=\left(2x^2+1\right).\left(x-12\right)\) (1)
\(\left(1\right)\Leftrightarrow\left(2x^2+1\right)\left(4x-3\right)-\left(2x^2+1\right)\left(x-12\right)=0\\ \Leftrightarrow\left(2x^2+1\right)\left(3x+9\right)=0\Rightarrow\left[{}\begin{matrix}3x+9=0\\2x^2+1=0\left(VN\right)\end{matrix}\right.\Rightarrow x=-3\)
29 tháng 7 2021
\(\dfrac{11x}{2x-3}+\dfrac{x-18}{2x-3}\left(ĐKXĐ:x\ne\dfrac{3}{2}\right)\\ =\dfrac{11x+x-18}{2x-3}\\ =\dfrac{12x-18}{2x-3}\\ =\dfrac{6\left(2x-3\right)}{2x-3}\\ =6\)
\(\dfrac{2x+12}{4x^2-9}+\dfrac{2x+5}{4x-6}\left(ĐKXĐ:x\ne\dfrac{3}{2};x\ne\dfrac{-3}{2}\right)\\ =\dfrac{2x+12}{\left(2x-3\right)\left(2x+3\right)}+\dfrac{2x+5}{2\left(2x-3\right)}\\ =\dfrac{4x+24}{2\left(2x-3\right)\left(2x+3\right)}+\dfrac{\left(2x+5\right)\left(2x+3\right)}{2\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{4x+24+4x^2+6x+10x+15}{2\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{4x^2+20x+39}{2\left(2x-3\right)\left(2x+3\right)}\)
\(\dfrac{x}{2x+1}+\dfrac{-1}{4x^2-1}+\dfrac{2-x}{2x-1}\left(ĐKXĐ:x\ne\dfrac{1}{2};x\ne\dfrac{-1}{2}\right)\\ =\dfrac{x\left(2x-1\right)-1+\left(2-x\right)\left(2x+1\right)}{\left(2x+1\right)\left(2x-1\right)}\\ =\dfrac{2x^2-x-1+4x+2-2x^2-x}{\left(2x-1\right)\left(2x+1\right)}\\ =\dfrac{2x+1}{\left(2x+1\right)\left(2x-1\right)}\\ =\dfrac{1}{2x-1}\)
1 tháng 2 2018
\(a,\left(x+1\right)^2=x^2-2x-3\)
\(\Leftrightarrow\left(x+1\right)^2=x^2-3x+x-3\)
\(\Leftrightarrow\left(x+1\right)^2=x\left(x-3\right)+\left(x-3\right)\)
\(\Leftrightarrow\left(x+1\right)^2=\left(x+1\right)\left(x-3\right)\)
\(\Leftrightarrow\left(x+1\right)^2-\left(x+1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+1-x+3\right)=0\)
\(\Leftrightarrow4\left(x+1\right)=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy pt có tập nghiệm S = { - 1 }
b, ĐKXĐ :\(\left\{{}\begin{matrix}1-2x\ne0\\1+2x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{1}{2}\\x\ne-\dfrac{1}{2}\end{matrix}\right.\)
\(\dfrac{3}{1-2x}=\dfrac{2}{1+2x}-\dfrac{3x+12}{1-4x^2}\)
\(\Leftrightarrow\dfrac{3\left(1+2x\right)}{\left(1-2x\right)\left(1+2x\right)}=\dfrac{2\left(1-2x\right)}{\left(1+2x\right)\left(1-2x\right)}-\dfrac{3x+12}{\left(1-2x\right)\left(1+2x\right)}\)
\(\Rightarrow3+6x=2-4x-3x-12\)
\(\Leftrightarrow6x+4x+3x=2-12-3\)
\(\Leftrightarrow13x=-13\)
\(\Leftrightarrow x=-1\) ( t/m )
Vậy pt có tập nghiệm S = { - 1 }
Như thế làm sao mà làm được