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12 tháng 7 2019
a) \(\frac{x}{x+1}=\frac{1}{2}\)
=> 2x = x + 1
=> 2x - x = 1
=> x = 1
b) \(\frac{x}{2}=\frac{x}{3}\)
=> 3x = 2x
=> 3x - 2x = 0
=> x = 0
c) \(\frac{x+1}{2}=\frac{x+1}{2017}\)
=> \(2017\left(x+1\right)=2\left(x+1\right)\)
=> 2017x + 2017 = 2x + 2
=> 2017x - 2x = 2 - 2017
=> 2015x = -2015
=> x = -2015 : 2015
=> x = -1
i) \(\frac{3}{x}=\frac{x}{2017}\)
=> x2 = 2017.3
=> x2 = 6051
=> \(\orbr{\begin{cases}x=\sqrt{6051}\\x=-\sqrt{6051}\end{cases}}\)
còn lại tự lm
12 tháng 7 2019
\(a,\frac{x}{x+1}=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}.\left(x+1\right)\)
\(\Rightarrow x=\frac{1}{2}x+\frac{1}{2}\)
\(\Rightarrow x-\frac{1}{2}x=\frac{1}{2}\)
\(\Rightarrow\frac{1}{2}x=\frac{1}{2}\)
\(\Rightarrow x=1\)
\(b,\frac{x}{2}=\frac{x}{3}\)
\(\Rightarrow x=\frac{x}{3}.2\)
\(\Rightarrow x=\frac{2x}{3}\)
\(\Rightarrow3x=2x\)
\(\Rightarrow x=0\)
\(c,\frac{x+1}{2}=\frac{x+1}{2017}\)
\(\Rightarrow x+1=\frac{x+1}{2017}.2\)
\(\Rightarrow x+1=\frac{2x+2}{2017}\)
\(\Rightarrow2017x+2017=2x+2\)
\(\Rightarrow2017x-2x=2-2017\)
\(\Rightarrow2015x=-2015\)
\(\Rightarrow x=-1\)
\(i,\frac{3}{x}=\frac{x}{2017}\)
\(\Rightarrow x=3:\frac{x}{2017}\)
\(\Rightarrow x=\frac{6051}{x}\)
\(\Rightarrow x^2=6051\)
\(\Rightarrow x=\sqrt{6051}\)
\(o,\frac{x}{3}=\frac{x+1}{2}\)
\(\Rightarrow x=\frac{x+1}{2}.3\)
\(\Rightarrow x=\frac{3x+3}{2}\)
\(\Rightarrow2x=3x+3\)
\(\Rightarrow-x=3\)
\(\Rightarrow x=-3\)
\(m,\frac{x+1}{2}=\frac{x+2}{3}\)
\(\Rightarrow x+1=\frac{x+2}{3}.2\)
\(\Rightarrow x+1=\frac{2x+4}{3}\)
\(\Rightarrow3x+3=2x+4\)
\(\Rightarrow x=1\)
\(p,\frac{x+1}{2}=x\)
\(\Rightarrow2x=x+1\)
\(\Rightarrow x=1\)
\(m,\frac{2}{x}=\frac{x}{8}\)
\(\Rightarrow x=2:\frac{x}{8}\)
\(\Rightarrow x=\frac{16}{x}\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x=4\)
\(Q,\frac{x^2}{2}=\frac{8}{x^2}\)
\(\Rightarrow x^2=\frac{8}{x^2}.2\)
\(\Rightarrow x^2=\frac{16}{x^2}\)
\(\Rightarrow x^4=16\)
\(\Rightarrow x=2\)
\(r,\frac{x^3}{2}=\frac{32}{x}\)
\(\Rightarrow x^3=\frac{32}{x}.2\)
\(\Rightarrow x^3=\frac{64}{x}\)
\(\Rightarrow x^4=64\)
\(\Rightarrow x=\sqrt[4]{64}\)
22 tháng 6 2017
1, \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=5\)
\(\Leftrightarrow4x^2+12x+9-4x^2-1=5\)
\(\Leftrightarrow12x=-3\)
\(\Leftrightarrow x=\dfrac{-1}{4}\)
Vậy \(x=\dfrac{-1}{4}\)
2, \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+5\right)=20\)
\(\Leftrightarrow x^3+27-x^3-5x=20\)
\(\Leftrightarrow5x=7\)
\(\Leftrightarrow x=\dfrac{7}{5}\)
Vậy...
5, \(x^2-9+5\left(x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)+5\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-3+5\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-2\end{matrix}\right.\)
Vậy...
Q
22 tháng 6 2017
1) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=5\) (1)
\(\Leftrightarrow4x^2+12x+9-\left(4x^2-1\right)=5\)
\(\Leftrightarrow4x^2+12x+9-4x^2+1=5\)
\(\Leftrightarrow12x+10=5\)
\(\Leftrightarrow12x=5-10\)
\(\Leftrightarrow12x=-5\)
\(\Leftrightarrow x=-\dfrac{5}{12}\)
Vậy tập nghiệm phương trình (1) là \(S=\left\{-\dfrac{5}{12}\right\}\)
2) \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+5\right)=20\) (2)
\(\Leftrightarrow x^3+27-x^3-5x=20\)
\(\Leftrightarrow27-5x=20\)
\(\Leftrightarrow-5x=20-27\)
\(\Leftrightarrow-5x=-7\)
\(\Leftrightarrow x=\dfrac{7}{5}\)
Vậy tập nghiệm phương trình (2) là \(S=\left\{\dfrac{7}{5}\right\}\)
3) \(\left(x+2\right)^3-x\left(x^2+6x\right)=15\) (3)
\(\Leftrightarrow x^3+6x^2+12x+8-x^3-6x^2=15\)
\(\Leftrightarrow12x+8=15\)
\(\Leftrightarrow12x=15-8\)
\(\Leftrightarrow12x=7\)
\(\Leftrightarrow x=\dfrac{7}{12}\)
Vậy tập nghiệm phương trình (3) là \(S=\left\{\dfrac{7}{12}\right\}\)
4) \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+10\right)\left(x-1\right)=7\) (4)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x\left(x+10\right)\right)=7\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2-10x\right)=7\)
\(\Leftrightarrow\left(x-1\right)\left(-9x+1\right)=7\)
\(\Leftrightarrow-9x^2+x+9x-1=7\)
\(\Leftrightarrow-9x^2+10-1=7\)
\(\Leftrightarrow-9x^2+10x-1-7=0\)
\(\Leftrightarrow-9x^2+10x-8=0\)
\(\Leftrightarrow9x^2-10x+8=0\)
\(\Leftrightarrow x\notin R\)
5) \(x^2-9+5\left(x+3\right)=0\) (5)
\(\Leftrightarrow x^2-9+5x+15=0\)
\(\Leftrightarrow x^2+5x+6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-5+1}{2}\\x=\dfrac{-5-1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
Vậy tập nghiệm phương trình (5) là \(S=\left\{-3;-2\right\}\)
22 tháng 11 2022
a: TH1: x<-5
Pt sẽ là \(-x-5+3-x=9\)
=>-2x-2=9
=>-2x=11
=>x=-11/2(nhận)
TH2: -5<=x<3
Pt sẽ là x+5+3-x=9
=>8=9(loại)
TH3: x>=3
Pt sẽ là x+5+x-3=9
=>2x+2=9
=>x=7/2(nhận)
d: TH1: x<-2
Pt sẽ là \(2\left(-x-2\right)+4-x=22\)
=>-2x-4+4-x=22
=>-3x=22
=>x=-22/3(nhận)
TH2: \(-2< =x< 4\)
Pt sẽ là 2(x+2)+4-x=22
=>2x+4+4-x=22
=>x+8=22
=>x=14(loại)
TH3: x>=4
Pt sẽ là 2x+4+x-4=22
=>3x=22
=>x=22/3(nhận)
6 tháng 8 2022
a: =>(3/2-2x):2/3=1/6
=>3/2-2x=1/6x2/3=2/18=1/9
=>2x=25/18
hay x=25/36
b: \(\Leftrightarrow2x-2x+\dfrac{5}{2}-2=x-\dfrac{1}{4}\)
=>x-1/4=1/2
=>x=3/4
c: \(\Leftrightarrow2x-\dfrac{2}{3}-\dfrac{1}{3}x+\dfrac{1}{4}x=0\)
=>23/12x=2/3
=>x=8/23
17 tháng 6 2018
1) \(\left(x+2\right)^3-\left(x+6\right)^2-\left(x+1\right)\left(x^2-x+1\right)\)
\(=x^3+3.x^2.2+3.x.2^2+2^3-\left(x^2+2.x.6+6^2\right)-\left(x^3+1\right)\)
\(=x^3+6x^2+12x+8-x^2-12x-36-x^3-1\)
\(=5x^2-29\)
2. \(\left(x-3\right)\left(x^2+3x+9\right)-\left(x+3\right)^2\)
\(=x^3-3^3-\left(x^2+2.x.3+3^2\right)\)
\(=x^3-27-x^2-6x-9\)
\(=x^3-x^2-6x-36\)
\(=x^3-2x^2+3x^2-6x-36\)
\(=x^2\left(x-2\right)+3x\left(x-2\right)-36\)
\(=x\left(x-2\right)\left(x+3\right)-36\)
3. \(\left(2x-1\right)^2+2x^2.\left(x-2\right)\left(x+3\right)^2\)
\(=4x^2-4x+1+2x^2\left(x-2\right)\left(x^2+6x+9\right)\)
\(=4x^2-4x+1+2x^2\left(x^3+6x^2+9x-2x^2-12x-18\right)\)
\(=4x^2-4x+1+2x^2\left(x^3+4x^2-3x-18\right)\)
\(=4x^2-4x+1+2x^5+8x^4-6x^3-36x^2\)
\(=2x^5+8x^4-6x^3-32x^2-4x+1\)
....
P/s: Không chắc lắm
(x-2)(x+3)(x-2)
= phân tích thành nhân tử (x-2)^2(x+3)
(𝑥−2)(𝑥+3)(𝑥−2)
= 𝑥(𝑥−2)(𝑥+3)−2(𝑥−2)(𝑥+3)
= (𝑥+3)⋅𝑥2−2𝑥(𝑥+3)−2(𝑥−2)(𝑥+3)
= 𝑥3+3𝑥2−2𝑥(𝑥+3)−2(𝑥−2)(𝑥+3)
= 𝑥3+3𝑥2−2𝑥2−6𝑥−2(𝑥−2)(𝑥+3)
= 𝑥2+1𝑥2−6𝑥−2(𝑥−2)(𝑥+3)
= 𝑥2+𝑥2−6𝑥−2(𝑥−2)(𝑥+3)
= 𝑥3+𝑥2−6𝑥−2𝑥(𝑥+3)+4(𝑥+3)
= 𝑥3+𝑥2−6𝑥−2𝑥2−6𝑥+4(𝑥+3)
= 𝑥3+𝑥2−6𝑥−2𝑥2−6𝑥+4𝑥+12
= 𝑥3+𝑥2−6𝑥−2𝑥2−2𝑥+12
= 𝑥3−1𝑥2−6𝑥−2𝑥+12
= 𝑥3−𝑥2−8𝑥+12
Cre : google