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19 tháng 2 2019
1) \(\left(5x-4\right)\left(4x+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-4=0\\4x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=4\\4x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{4}{5};\dfrac{3}{2}\right\}\)
2) \(\left(4x-10\right)\left(24+5x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=10\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-24}{5}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{5}{2};\dfrac{-24}{5}\right\}\)
3) \(\left(x-3\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{-1}{2}\end{matrix}\right.\)
Vậy phương trình có tập nghiệm S = \(\left\{3;\dfrac{-1}{2}\right\}\)
NH
23 tháng 9 2018
1,=\(x^2-3x-2x^2+6x=-x^2+3x\)
2,=\(3x^2-x-5+15x=3x^2+14x-5\)
3,=\(5x+15-6x^2-6x=-6x^2-x+15\)
4,=\(4x^2+12x-x-3=4x^2+11x-3\)
4 tháng 9 2022
5: =>(x+5)^3=0
=>x+5=0
=>x=-5
6: =>(2x-3)^2=0
=>2x-3=0
=>x=3/2
7: =>(x-6)(x-10)=0
=>x=10 hoặc x=6
8: \(\Leftrightarrow x^3-12x^2+48x-64=0\)
=>(x-4)^3=0
=>x-4=0
=>x=4
DN
5 tháng 3 2019
\(j,3x^2+7x+2=0\)
\(\Leftrightarrow3x^2+6x+x+2=0\)
\(\Leftrightarrow3x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{1}{3}\\x=-2\end{matrix}\right.\)
Vậy...............................
DN
5 tháng 3 2019
\(m,3x^2+4x-4=0\)
\(\Leftrightarrow3x^2+6x-2x-4=0\)
\(\Leftrightarrow3x\left(x+2\right)-2\left(x+2\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-2=0\\x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{2}{3}\\x=-2\end{matrix}\right.\)
22 tháng 2 2020
\(\left(3x-2\right)\left(x+6\right)\left(x^2+5\right)=0\)
\(TH1:3x-2=0\Leftrightarrow3x=2\Leftrightarrow x=\frac{2}{3}\)
\(TH2:x+6=0\Leftrightarrow x=-6\)
\(TH3:x^2+5=0\Leftrightarrow x^2=5\Leftrightarrow x=\sqrt{5}\)( ns vô nghiệm cx ko sai nha )
\(\left(2x+5\right)^2=\left(3x-1\right)^2\)
\(2x+5=3x-1\)
\(2x-3x=-1-5\)
\(-1x=-6\)
\(x=6\)
2 tháng 9 2022
1: \(\Leftrightarrow-4x^2+3x-4x^2+8x=10\)
=>-8x^2+11x-10=0
=>\(x\in\varnothing\)
2: \(\Leftrightarrow5x^2-15x+5+x-5x^2=x-2\)
=>-14x+5=x-2
=>-15x=-7
=>x=7/15
3: \(\Leftrightarrow12x^2-12x^2+20x=10x-17\)
=>10x=-17
=>x=-17/10
4: \(\Leftrightarrow4x^2-2x+3-4x^2+20x=7x-3\)
=>18x+3=7x-3
=>11x=-6
=>x=-6/11
5: \(\Leftrightarrow-3x+15+5x-5+3x^2=4-x\)
\(\Leftrightarrow3x^2+2x+10-4+x=0\)
=>3x^2+3x+6=0
hay \(x\in\varnothing\)
3 tháng 8 2020
\(5X\left(X-2020\right)+X=2020\)
\(\Leftrightarrow5X^2-10100X+X=2020\)
\(\Leftrightarrow5X^2-10099X=2020\)
\(\Leftrightarrow5X^2-10099X-2020=0\)
\(\Leftrightarrow5X^2-10100X+x-2020=0\)
\(\Leftrightarrow5X\left(X-2020\right)+X-2020=0\)
\(\Leftrightarrow\left(X-2020\right)\left(5X+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=2020\\x=-\frac{1}{5}\end{cases}}\)
3 tháng 8 2020
\(4\left(x-5\right)^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-5\right)\right]^2-\left(2x+1\right)^2=0\)
\(\Leftrightarrow\left[2\left(x-5\right)-2x-1\right]\left[2\left(x-5\right)+2x+1\right]=0\)
\(\Leftrightarrow\left(2x-10-2x-1\right)\left(2x-10+2x+1\right)=0\)
\(\Leftrightarrow-11\left(4x-9\right)=0\)
\(\Leftrightarrow x=\frac{9}{4}\)
TN
5 tháng 3 2020
\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\\ \left(3x+2\right)\left(x^2-1\right)-\left(9x^2-4\right)\left(x+1\right)=0\\ \left(3x+2\right)\left(x+1\right)\left(x-1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0\\ \left(3x+2\right)\left(x+1\right)\left[\left(x-1\right)-\left(3x-2\right)\right]=0\\ \left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0\\ \left(3x+2\right)\left(x+1\right)\left(1-2x\right)=0\\ \left[{}\begin{matrix}3x+2=0\\x+1=0\\1-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-2}{3}\\x=-1\\x=\frac{1}{2}\end{matrix}\right.\)
\(b.x\left(x+3\right)\left(x-3\right)-\left(x+2\right)\left(x^2-2x+4\right)=0\\ x\left(x^2-9\right)-\left(x^3+8\right)=0\\ x^3-9x-x^3-8=0\\ -9x-8=0\\ -9x=8\\ x=\frac{-8}{9}\)
\(c.2x\left(x-3\right)+5\left(x-3\right)=0\\ \left(x-3\right)\left(2x+5\right)=0\\ \left[{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{-5}{2}\end{matrix}\right.\)
\(d.\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\\ \left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\\ \left(3x-1\right)\left[\left(x^2+2\right)-\left(7x-10\right)\right]=0\\ \left(3x-1\right)\left(x^2+2-7x+10\right)=0\\ \left(3x-1\right)\left(x^2-7x+12\right)=0\\ \left(3x-1\right)\left(x^2-4x-3x+12\right)=0\\ \left(3x-1\right)\left[\left(x^2-4x\right)+\left(-3x+12\right)\right]=0\\ \left(3x-1\right)\left[x\left(x-4\right)-3\left(x-4\right)\right]=0\\ \left(3x-1\right)\left(x-4\right)\left(x-3\right)=0\\ \left[{}\begin{matrix}3x-1=0\\x-4=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=4\\x=3\end{matrix}\right.\)
TN
5 tháng 3 2020
\(e.\left(x+2\right)\left(3-4x\right)=x^2+4x+4\\ \left(x+2\right)\left(3-4x\right)=\left(x+2\right)^2\\ \left(x+2\right)\left(3-4x\right)-\left(x+2\right)^2=0\\ \left(x+2\right)\left[\left(3-4x\right)-\left(x+2\right)\right]=0\\ \left(x+2\right)\left(3-4x-x-2\right)=0\\ \left(x+2\right)\left(1-5x\right)=0\left[{}\begin{matrix}x+2=0\\1-5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\frac{1}{5}\end{matrix}\right.\)
\(f.x\left(2x-7\right)-4x+14=0\\ x\left(2x-7\right)-2\left(2x-7\right)=0\\ \left(2x-7\right)\left(x-2\right)=0\\ \left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{7}{2}\\x=2\end{matrix}\right.\)
\(g.3x-15=2x\left(x-5\right)\\ 3\left(x-5\right)=2x\left(x-5\right)\\ 3\left(x-5\right)-2x\left(x-5\right)=0\\ \left(x-5\right)\left(3-2x\right)=0\\ \left[{}\begin{matrix}x-5=0\\3-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\frac{3}{2}\end{matrix}\right.\)
\(h.\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\\ \left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\\ \left(2x+1\right)\left[\left(3x-2\right)-\left(5x-8\right)\right]=0\\ \left(2x+1\right)\left(3x-2-5x+8\right)=0\\ \left(2x+1\right)\left(6-2x\right)=0\\ \left[{}\begin{matrix}2x+1=0\\6-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=3\end{matrix}\right.\)
x + 3x + 4x + 3x + 1 = 0
⇒x + x + 2x + 2x + 2x + 2x + x + 1 = 0
⇒x x + 1 + 2x x + 1 + 2x x + 1 + x + 1 = 0 ⇒ x + 1 x + x + x + x + x + 1 = 0 ⇒ x + 1 x x + 1 + x x + 1 + x + 1 = 0 ⇒ x + 1 x + 1 x + x + 1 = 0 ⇒ x + 1 x + x + 1 = 0 ⇒ x + 1 = 0 vix̀ + x + 1 ≠ 0 ⇒x + 1 = 0 ⇒x = −1 vậy pt có No ......... 3 2x − 3 − 6 x − 3 = 5 4x + 3 − 17 ⇔ 30 10 2x − 3 − 30 5 x − 3 = 30 6 4x + 3 − 30 17.30 ⇔20x − 30 − 5x + 15 = 24x + 18 − 510 ⇔20x − 5x − 24x = 18 − 510 + 30 − 15
⇔− 9x = −477 ⇔x = 53
vậy pt có No........
\(x^4+3x^3+4x^2+3x+1=0\)
\(\Rightarrow x^4+x^3+2x^3+2x^2+2x^2+2x+x+1=0\)
\(\Rightarrow x^3\left(x+1\right)+2x^2\left(x+1\right)+2x\left(x+1\right)+\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(x^3+x^2+x^2+x+x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left[x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)\right]=0\)
\(\Rightarrow\left(x+1\right)\left(x+1\right)\left(x^2+x+1\right)=0\)
\(\Rightarrow\left(x+1\right)^2\left(x^2+x+1\right)=0\)
\(\Rightarrow\left(x+1\right)^2=0\left(vìx^2+x+1\ne0\right)\)
\(\Rightarrow x+1=0\)
\(\Rightarrow x=-1\)
vậy pt có No .........
\(\frac{2x-3}{3}-\frac{x-3}{6}=\frac{4x+3}{5}-17\)
\(\Leftrightarrow\frac{10\left(2x-3\right)}{30}-\frac{5\left(x-3\right)}{30}=\frac{6\left(4x+3\right)}{30}-\frac{17.30}{30}\)
\(\Leftrightarrow20x-30-5x+15=24x+18-510\)
\(\Leftrightarrow20x-5x-24x=18-510+30-15\)
\(\Leftrightarrow-9x=-477\)
\(\Leftrightarrow x=53\)
vậy pt có No........