giải phương trình sau
1/ 2x( x+3) - 6 (x-3) =0
2/ 2x^2( 2x+3) +(2x+3) =0
3/ (x-2) (x+1) -(x-2) 4x =0
4/ 2x ( x-5) -3x +15=0
5/ 3x(x+4) -2x-8 =0
6/ x^2 (2x-6) + 2x -6 =0
giải phương trình sau
1/ ( x-1) (2x+1) =0
2/ x (2x-1) (3x+15) =0
3/ (2x-6) (3x+4) x=0
4/ (2x-10)(x^2+1)=0
5/ (x^2+3) (2x-1) =0
6/ (3x-1) (2x^2 +1)=0
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: \(\left(2x-6\right)\left(3x+4\right)x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-6=0\\3x+4=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{4}{3}\\x=0\end{matrix}\right.\)
4: Ta có: \(\left(2x-10\right)\left(x^2+1\right)=0\)
mà \(x^2+1>0\forall x\)
nên 2x-10=0
hay x=5
5: Ta có: \(\left(x^2+3\right)\left(2x-1\right)=0\)
mà \(x^2+3>0\forall x\)
nên 2x-1=0
hay \(x=\dfrac{1}{2}\)
6: Ta có: \(\left(3x-1\right)\left(2x^2+1\right)=0\)
mà \(2x^2+1>0\forall x\)
nên 3x-1=0
hay \(x=\dfrac{1}{3}\)
Bài 1: tìm x
1, 2x(3x-1)+1-3x=0
2, x\(^2\)(2x-3)+12-8x=0
3, 25(x-1)\(^2\)-4=0
4, 25x\(^2\)-10x+1=0
5, -4x\(^2\)+\(\dfrac{1}{9}\)=0
6, (x-1)\(^3\)=8
7, (2x-1)\(^3\)+27=0
8, 125+\(\dfrac{1}{8}\)(x-1)\(^3\)=0
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}.`
`@` `\text {Ans}`
`\downarrow`
`5,`
`-4x^2 + 1/9 = 0`
`<=> -4x^2 = 0 - 1/9`
`<=> -4x^2 = -1/9`
`<=> 4x^2 = 1/9`
`<=> x^2 = 1/9 \div 4`
`<=> x^2 = 1/36`
`<=> x^2 = (+-1/6)^2`
`<=> x = +-1/36`
Vậy, `S = {1/36; -1/36}`
`6,`
`(x-1)^3 = 8`
`<=> (x-1)^3 = 2^3`
`<=> x-1=2`
`<=> x = 2 + 1`
`<=> x = 3`
Vậy, `S = {3}`
`7,`
`(2x-1)^3 + 27 = 0`
`<=> (2x - 1)^3 = -27`
`<=> (2x-1)^3 = (-3)^3`
`<=> 2x - 1 = -3`
`<=> 2x = -3 + 1`
`<=> 2x = -2`
`<=> x = -1`
Vậy,` S = {-1}`
`8,`
`125 + 1/8(x-1)^3 = 0`
`<=> 1/8(x-1)^3 = - 125`
`<=> (x-1)^3 = -125 \div 1/8`
`<=> (x-1)^3 = -1000`
`<=> (x-1)^3 = (-10)^3`
`<=> x - 1 = - 10`
`<=> x = -10+1`
`<=> x = -9`
Vậy, `S = {-9}.`
Giải phương trình
1) 2x ( x – 3 ) + 5 ( x – 3 ) = 0
2) ( x2 – 4 ) – ( x – 2 ) ( 3 – 2x ) = 0
3) ( 2x – 1 )2 – ( 2x + 5 )2 = 11
4) ( 2x + 1 )2 ( 3x – 5 ) = 4x2 – 1
5) 3x2 – 5x – 8 = 0
6) ( 2x + 1 )2 ( 3x – 5 ) = 4x2 – 1
7) 3x2 – 5x – 8 = 0
8) \(\left|x-5\right|=3\)
9) \(\left|2x-5\right|=3-x\)
10) \(\left|2x+1\right|=\left|x-1\right|\)
11) \(\dfrac{5x+2}{6}-\dfrac{8x-1}{3}=\dfrac{4x+2}{5}-5\)
12) \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
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: \(\left|2x+1\right|=\left|x-1\right|\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=x-1\\2x+1=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x-x=-1-1\\2x+x=1-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=0\end{matrix}\right.\)
giải phương trình sau
1/ x^2 -3x+2=0
2/ x^2 -6x+5=0
3/ 2x^2 +5x+3 =0
4/ x^2-8x+15=0
5/ x^2 -x-12=0
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) x2 - x - (3x - 3) = 02) x(x - 6) - 7x + 42 = 0
3) x3 - 3x2 + 3x - 1 = 0
4) (2x - 5)2 - (x + 2) 2 = 0
5) x(2x - 9) = 3x(x - 5)
1) x^2-x-(3x-3)=0
⇔ X^2-x-3x+3=0
⇔ x^2-4x+3 =0
⇔x^2-3x-x+3 =0
⇔ x(x-3)-(x-3) =0
⇔(x-1)(x-3) =0
⇔ x-1=0 -> x=1
x-3=0 -> x=3
Vậy tập nghiệm S={ 1;3}
giải phương trình sau
1/ ( x-5)^2 +3(x-5) =0
2/ ( x^2-9) +2 (x-3) =0
3/ ( 2x+1)^2+(x-1)(2x+1)=0
4/ (x-1) (x+3) +( x+3)^2=0
5/ ( x+5)^2 -16x^2 =0
6/ x^3-2x^2-x+2=0
1.
\(\left(x-5\right)^2+3\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-5+3\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=2\end{matrix}\right.\)
2.
\(\left(x^2-9\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
3.
\(\left(2x+1\right)^2+\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x+1+x-1\right)=0\)
\(\Leftrightarrow\left(2x+1\right).3x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=0\\2x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
4.
\(\left(x-1\right)\left(x+3\right)+\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-1+x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
5.
\(\left(x+5\right)^2-16x^2=0\)
\(\Leftrightarrow\left(x+5+4x\right)\left(x+5-4x\right)=0\)
\(\Leftrightarrow\left(5x+5\right)\left(5-3x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x+5=0\\5-3x=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{5}{3}\end{matrix}\right.\)
6.
\(x^3-2x^2-x+2=0\)
\(\Leftrightarrow x^2\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=2\end{matrix}\right.\)
Hãy giải các phương trình sau đây :
1, x2 - 4x + 4 = 0
2, 2x - y = 5
3, x + 5y = - 3
4, x2 - 2x - 8 = 0
5, 6x2 - 5x - 6 = 0
6,( x2 - 2x )2 - 6 (x2 - 2x ) + 5 = 0
7, x2 - 20x + 96 = 0
8, 2x - y = 3
9, 3x + 2y = 8
10, 2x2 + 5x - 3 = 0
11, 3x - 6 = 0
1) Ta có: \(x^2-4x+4=0\)
\(\Leftrightarrow\left(x-2\right)^2=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
Vậy: S={2}
1)4x-20=0 ; 2) 5x+15=0 ; 3) 3x-5=7x+2 ; 4) 4x-(x-1)=2(1+x) ; 5) x2 -2x=0 ; 6) 2(3x-5)-3(x-2)=3(x+4) ; 7) (x+3)(2x-7)=0
8) 5x(x-3)+2x-6=0 ; 9) (3x-1)(2x-1)-(3x-1)(x+2)=0
10)|2x-1|+1=8 ; 11) |x-2|=3x+1 ; 12) |2x|=21-x
Giải các phương trình nha mọi người ^_^
Giải các phương trình sau. 2x-1=2-x ; x-5x-1/6=8-3x/4. ; x/3 - 2x+1/2=x/6 - x ; (2x-5)(x+3)=0. ; (1-7)(2+x)=0
Bạn cần viết đề bằng công thức toán (biểu tượng $\sum$ góc trái khung soạn thảo) đẻ được hỗ trợ tốt hơn. Viết như thế kia rất khó đọc => khả năng bị bỏ qua bài cao.
a: =>3x=3
=>x=1
b: =>12x-2(5x-1)=3(8-3x)
=>12x-10x+2=24-9x
=>2x+2=24-9x
=>11x=22
=>x=2
c: =>2x-3(2x+1)=x-6x
=>-5x=2x-6x-3=-4x-3
=>-x=-3
=>x=3
d: =>2x-5=0 hoặc x+3=0
=>x=5/2 hoặc x=-3
e: =>x+2=0
=>x=-2