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1: Ta có: \(2x\left(x+3\right)-6\left(x-3\right)=0\)
\(\Leftrightarrow2x^2+6x-6x+18=0\)
\(\Leftrightarrow2x^2+18=0\left(loại\right)\)
2: Ta có: \(2x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
hay \(x=-\dfrac{3}{2}\)
3: Ta có: \(\left(x-2\right)\left(x+1\right)-4x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(1-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
4: Ta có: \(2x\left(x-5\right)-3x+15=0\)
\(\Leftrightarrow\left(x-5\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)
5: Ta có: \(3x\left(x+4\right)-2x-8=0\)
\(\Leftrightarrow\left(x+4\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=\dfrac{2}{3}\end{matrix}\right.\)
6: Ta có: \(x^2\left(2x-6\right)+2x-6=0\)
\(\Leftrightarrow2x-6=0\)
hay x=3
1/ x2-3x+2=0
⇒ (x2-2x)-(x-2)=0
⇒ x(x-2)-(x-2)=0
⇒ (x-1)(x-2)=0
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
2) x2-6x+5=0
⇒x2-6x+9-4=0
⇒(x2-6x+9)-22=0
⇒(x-3)2-22=0
⇒(x-3-2)(x-3+2)=0
⇒(x-5)(x-1)=0
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
3) 2x2+5x+3=0
⇒ (2x2+2x)+(3x+3)=0
⇒ 2x(x+1)+3(x+1)=0
⇒ (x+1)(2x+3)=0
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\2x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-1\\x=-1,5\end{matrix}\right.\)
4) x2-8x+15=0
⇒ (x2-8x+16)-1=0
⇒ (x-4)2-12=0
⇒ (x-4-1)(x-4+1)=0
⇒ (x-5)(x-3)=0
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=5\end{matrix}\right.\)
5) x2-x-12=0
⇒ (x2-4x)+(3x-12)=0
⇒ x(x-4)+3(x-4)=0
⇒ (x-4)(x+3)=0
\(\Rightarrow\left[{}\begin{matrix}x+3=0\\x-4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-3\\x=4\end{matrix}\right.\)
1: Ta có: \(x^2-3x+2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
2: Ta có: \(x^2-6x+5=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)
3: Ta có: \(2x^2+5x+3=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-\dfrac{3}{2}\end{matrix}\right.\)
4: Ta có: \(x^2-8x+15=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=5\end{matrix}\right.\)
5: Ta có: \(x^2-x-12=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-3\end{matrix}\right.\)
1/ ( x-1) (2x+1) =0
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=1\\x=-0,5\end{matrix}\right.\)
2/ x (2x-1) (3x+15) =0
\(\Rightarrow\left[{}\begin{matrix}x=0\\2x-1=0\\3x+15=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=0,5\\x=-5\end{matrix}\right.\)
3/ (2x-6) (3x+4).x=0
\(\Rightarrow\left[{}\begin{matrix}2x-6=0\\3x+4=0\\x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\\x=0\end{matrix}\right.\)
4/ (2x-10)(x2+1)=0
\(\Rightarrow\left[{}\begin{matrix}2x-10=0\\x^2+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x^2=-1\left(loại\right)\end{matrix}\right.\)
5/ (x2+3) (2x-1) =0
\(\Rightarrow\left[{}\begin{matrix}x^2+3=0\\2x-1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x^2=-3\left(loại\right)\\x=0,5\end{matrix}\right.\)
6/ (3x-1) (2x2 +1)=0
\(\Rightarrow\left[{}\begin{matrix}3x-1=0\\2x^2+1=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x^2=-0,5\left(loại\right)\end{matrix}\right.\)
1: Ta có: \(\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
2: Ta có: \(x\left(2x-1\right)\left(3x+15\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\2x-1=0\\3x+15=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\\x=-5\end{matrix}\right.\)
3: Ta có: \(\left(2x-6\right)\left(3x+4\right)x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\3x+4=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\\x=0\end{matrix}\right.\)
1) Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)
2) Ta có: \(\left(x^2-4\right)-\left(x-2\right)\left(3-2x\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)+\left(x-2\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)
3) Ta có: \(\left(2x-1\right)^2-\left(2x+5\right)^2=11\)
\(\Leftrightarrow4x^2-4x-1-4x^2-20x-25=11\)
\(\Leftrightarrow-24x=11+1+25=37\)
hay \(x=-\dfrac{37}{24}\)
5) Ta có: \(3x^2-5x-8=0\)
\(\Leftrightarrow3x^2+3x-8x-8=0\)
\(\Leftrightarrow3x\left(x+1\right)-8\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{8}{3}\end{matrix}\right.\)
8) Ta có: \(\left|x-5\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)
10) Ta có: \(\left|2x+1\right|=\left|x-1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=x-1\\2x+1=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-x=-1-1\\2x+x=1-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=0\end{matrix}\right.\)
5: =>4x^2-1/9=0
=>(2x-1/3)(2x+1/3)=0
=>x=1/6 hoặc x=-1/6
6: =>x-1=2
=>x=3
7:=>(2x-1)^3=-27
=>2x-1=-3
=>2x=-2
=>x=-1
8: =>1/8(x-1)^3=-125
=>(x-1)^3=-1000
=>x-1=-10
=>x=-9
3: =>(5x-5)^2-4=0
=>(5x-7)(5x-3)=0
=>x=3/5 hoặc x=7/5
4: =>(5x-1)^2=0
=>5x-1=0
=>x=1/5
1: =>(3x-1)(2x-1)=0
=>x=1/3 hoặc x=1/2
2: =>x^2(2x-3)-4(2x-3)=0
=>(2x-3)(x^2-4)=0
=>(2x-3)(x-2)(x+2)=0
=>x=3/2;x=2;x=-2
`@` `\text {Answer}`
`\downarrow`
`1,`
\(2x\left(3x-1\right)+1-3x=0\)
`<=> 2x(3x - 1) - 3x + 1 = 0`
`<=> 2x(3x - 1) - (3x - 1) = 0`
`<=> (2x - 1)(3x-1) = 0`
`<=>`\(\left[{}\begin{matrix}2x-1=0\\3x-1=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}2x=1\\3x=1\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy, `S = {1/2; 1/3}`
`2,`
\(x^2\left(2x-3\right)+12-8x=0\)
`<=> x^2(2x - 3) - 8x + 12 =0`
`<=> x^2(2x - 3) - (8x - 12) = 0`
`<=> x^2(2x - 3) - 4(2x - 3) = 0`
`<=> (x^2 - 4)(2x - 3) = 0`
`<=>`\(\left[{}\begin{matrix}x^2-4=0\\2x-3=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x^2=4\\2x=3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x^2=\left(\pm2\right)^2\\x=\dfrac{3}{2}\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=\pm2\\x=\dfrac{3}{2}\end{matrix}\right.\)
Vậy, `S = {+-2; 3/2}`
`3,`
\(25\left(x-1\right)^2-4=0\)
`<=> 25(x-1)(x-1) - 4 = 0`
`<=> 25(x^2 - 2x + 1) - 4 = 0`
`<=> 25x^2 - 50x + 25 - 4 = 0`
`<=> 25x^2 - 15x - 35x + 21 = 0`
`<=> (25x^2 - 15x) - (35x - 21) = 0`
`<=> 5x(5x - 3) - 7(5x - 3) = 0`
`<=> (5x - 7)(5x - 3) = 0`
`<=>`\(\left[{}\begin{matrix}5x-7=0\\5x-3=0\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}5x=7\\5x=3\end{matrix}\right.\)
`<=>`\(\left[{}\begin{matrix}x=\dfrac{7}{5}\\x=\dfrac{3}{5}\end{matrix}\right.\)
Vậy, `S = {7/5; 3/5}`
`4,`
\(25x^2-10x+1=0\)
`<=> 25x^2 - 5x - 5x + 1 = 0`
`<=> (25x^2 - 5x) - (5x - 1) = 0`
`<=> 5x(5x - 1) - (5x - 1) = 0`
`<=> (5x - 1)(5x-1)=0`
`<=> (5x-1)^2 = 0`
`<=> 5x - 1 = 0`
`<=> 5x = 1`
`<=> x = 1/5`
Vậy,` S = {1/5}.`
1.
\(x^2-5x+6=0\\ \Rightarrow x^2-2x-3x+6=0\\ \Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\\ \Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\\ \Rightarrow\left(x-2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
2.
\(\left(x+4\right)^2-\left(3x-1\right)^2=0\\ \Rightarrow\left(x+4-3x+1\right)\left(x+4+3x-1\right)=0\\ \Rightarrow\left(-2x+5\right)\left(4x+3\right)=0\\ \Rightarrow\left[{}\begin{matrix}-2x+5=0\\4x+3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
3.
\(x^2-2x+24=0\\ \Rightarrow\left(x^2-2x+1\right)+23=0\\ \Rightarrow\left(x-1\right)^2+23=0\)
Vì (x-1)2≥0
23>0
\(\Rightarrow\left(x-1\right)^2+23>0\)
Vậy x vô nghiệm
4.
\(9x^2-4=0\\ \Rightarrow\left(3x-4\right)\left(3x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}3x-4=0\\3x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\dfrac{4}{3}\\x=-\dfrac{4}{3}\end{matrix}\right.\)
5.
\(x^2+2x-8=0\\ \Rightarrow\left(x^2+2x+1\right)-9=0\\ \Rightarrow\left(x+1\right)^2-3^2=0\\ \Rightarrow\left(x-2\right)\left(x+4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)
a) 3x(4x-3)-2x(5-6x)=0
\(\Leftrightarrow12x^2-9x-10x+12x^2=0\)
\(\Leftrightarrow24x^2-19x=0\)
\(\Leftrightarrow x\left(24x-19\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\24x-19=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\24x=19\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{19}{24}\end{matrix}\right.\)
Vậy x=0 hoặc x=\(\dfrac{19}{24}\)
Answer:
\(3x^2-4x=0\)
\(\Rightarrow x\left(3x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)
\(\left(x^2-5x\right)+x-5=0\)
\(\Rightarrow x\left(x-5\right)+\left(x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)
\(x^2-5x+6=0\)
\(\Rightarrow x^2-2x-3x+6=0\)
\(\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)
\(\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
\(5x\left(x-3\right)-x+3=0\)
\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Rightarrow\left(5x-1\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)
\(x^2-2x+5=0\)
\(\Rightarrow\left(x^2-2x+1\right)+4=0\)
\(\Rightarrow\left(x-1\right)^2=-4\) (Vô lý)
Vậy không có giá trị \(x\) thoả mãn
\(x^2+x-6=0\)
\(\Rightarrow x^2+3x-2x-6=0\)
\(\Rightarrow x.\left(x+3\right)-2\left(x+3\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)