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HP
4 tháng 2 2022
\(i.\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\dfrac{4x^2+4x+1}{5}-\dfrac{x^2-2x+1}{3}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow\dfrac{12x^2+12x+3}{15}-\dfrac{5x^2-10x+5}{15}=\dfrac{7x^2-14x-5}{15}\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)
\(\Leftrightarrow36x=-3\)
\(\Leftrightarrow x=-\dfrac{1}{12}\)
2
11 tháng 1 2023
Bài `1:`
`h)(3/4x-1)(5/3x+2)=0`
`=>[(3/4x-1=0),(5/3x+2=0):}=>[(x=4/3),(x=-6/5):}`
______________
Bài `2:`
`b)3x-15=2x(x-5)`
`<=>3(x-5)-2x(x-5)=0`
`<=>(x-5)(3-2x)=0<=>[(x=5),(x=3/2):}`
`d)x(x+6)-7x-42=0`
`<=>x(x+6)-7(x+6)=0`
`<=>(x+6)(x-7)=0<=>[(x=-6),(x=7):}`
`f)x^3-2x^2-(x-2)=0`
`<=>x^2(x-2)-(x-2)=0`
`<=>(x-2)(x^2-1)=0<=>[(x=2),(x^2=1<=>x=+-2):}`
`h)(3x-1)(6x+1)=(x+7)(3x-1)`
`<=>18x^2+3x-6x-1=3x^2-x+21x-7`
`<=>15x^2-23x+6=0<=>15x^2-5x-18x+6=0`
`<=>(3x-1)(5x-1)=0<=>[(x=1/3),(x=1/5):}`
`j)(2x-5)^2-(x+2)^2=0`
`<=>(2x-5-x-2)(2x-5+x+2)=0`
`<=>(x-7)(3x-3)=0<=>[(x=7),(x=1):}`
`w)x^2-x-12=0`
`<=>x^2-4x+3x-12=0`
`<=>(x-4)(x+3)=0<=>[(x=4),(x=-3):}`
2
11 tháng 1 2023
`m)(1-x)(5x+3)=(3x-7)(x-1)`
`<=>(1-x)(5x+3)+(1-x)(3x-7)=0`
`<=>(1-x)(5x+3+3x-7)=0`
`<=>(1-x)(8x-4)=0<=>[(x=1),(x=1/2):}`
`p)(2x-1)^2-4=0`
`<=>(2x-1-2)(2x-1+2)=0`
`<=>(2x-3)(2x+1)=0<=>[(x=3/2),(x=-1/2):}`
`r)(2x-1)^2=49`
`<=>(2x-1-7)(2x-1+7)=0`
`<=>(2x-8)(2x+6)=0<=>[(x=4),(x=-3):}`
`t)(5x-3)^2-(4x-7)^2=0`
`<=>(5x-3-4x+7)(5x-3+4x-7)=0`
`<=>(x+4)(9x-10)=0<=>[(x=-4),(x=10/9):}`
`u)x^2-10x+16=0`
`<=>x^2-8x-2x+16=0`
`<=>(x-2)(x-8)=0<=>[(x=2),(x=8):}`
ND
19 tháng 3 2020
Bạn bạn nhân phân phối (3x-1)(x-2) và (3x-1)(7x-10)
Sau đó chuyển vế sao cho về phương trình bậc 2
Sau đó giải pt bậc hai là ra
19 tháng 3 2020
Ta có : (3x -1 ) . ( x + 2 ) = ( 3x-1 ) .( 7x - 10)
<=>3.x2 + 6x -x -2 = 21x2 -30x - 7x +10
<=> 3x2 + 5x -2 = 21x2 -37x + 10
<=> 3x2 +5x - 3 - 21x2 +37x -10 = 0
<=> -18x2 + 42x -12 = 0
<=> 3x2 -7x +2 = 0
<=> 3x2 -x -6x + 2 = 0
<=> x. ( 3x -1 ) -2.(3x -1 ) = 0
<=> (3x -1 ) . ( x - 2 ) = 0
<=> \(\orbr{\begin{cases}3x-1=0\\x-2=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{1}{3}\\x=2\end{cases}}\)
Tập nghiệm của phương trình là : { \(\frac{1}{3}\); 2}
AT
19 tháng 3 2020
( 3x - 1)( x + 2) = ( 3x - 1)(7x - 10)
<=>( 3x - 1)( x + 2) - ( 3x - 1)(7x - 10) = 0
<=> ( 3x - 1)( x + 2 - 7x + 10) = 0
<=>( 3x - 1)( -6x + 12) = 0
\(\Leftrightarrow\orbr{\begin{cases}3x-1=0\\-6x+12=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=2\end{cases}}}\)
Vậy.....
19 tháng 3 2020
\(\left(3x-1\right)\left(x+2\right)=\left(3x-1\right)\left(7x-10\right)\)
\(3x^2+5x-2=21x^2-37x+10\)
\(3x^2+5x-2-21x^2+37x-10=0\)
\(-18x^2+42x-12=0\)
\(-6\left(3x-1\right)\left(x-2\right)=0\)
\(-6\ne0\)
\(\left(3x-1\right)\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x=1\\x=2\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=2\end{cases}}}\)
21 tháng 1 2022
1, <=> 3x-1-2x=6 <=> x =7
2, (2x-1)(7-x)=x^2-7x
<=> 14x -2x^2-7+x=x^2-7x
<=> -3x^2+ 15x - 7 = -7x
<=> -3x^2 +23x - 7 =0
<=> \(x=\dfrac{23\pm\sqrt{445}}{6}\)
NV
Nguyễn Việt Lâm
Giáo viên
21 tháng 1 2022
1.
\(\Leftrightarrow3x-1-2x=6\)
\(\Leftrightarrow x-1=6\)
\(\Leftrightarrow x=7\)
2.
\(\Leftrightarrow\left(2x-1\right)\left(7-x\right)+7x-x^2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(7-x\right)+x\left(7-x\right)=0\)
\(\Leftrightarrow\left(7-x\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}7-x=0\\3x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{1}{3}\end{matrix}\right.\)
30 tháng 1 2018
a) \(2x^3 + 6x^2 = x^2 +3x\)
\(\Leftrightarrow2x^2\left(x+3\right)=x\left(x+3\right)\)
\(\Leftrightarrow2x^2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x^2-x\right)=0\)
\(\Leftrightarrow\left(x+3\right).x\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x=0\\2x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=0\\x=\dfrac{1}{2}\end{matrix}\right.\)
S = \(\left\{-3;0;\dfrac{1}{2}\right\}\)
b) \((3x-1) (x^2 +2 ) = (3x-1) (7x - 10)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)
S = \(\left\{\dfrac{1}{3};3;4\right\}\)
D
22 tháng 1 2022
\(\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\)
⇔ \(\dfrac{3\left(2x+1\right)^2}{15}-\dfrac{5\left(x-1\right)^2}{15}=\dfrac{7x^2-14x-5}{15}\)
⇔ \(3\left(2x+1\right)^2-5\left(x-1\right)^2=7x^2-14x-5\)
⇔ \(3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)=7x^2-14x-5\)
⇔ \(12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)
⇔ \(7x^2+22x-2=7x^2-14x-5\) ⇔ \(36x+3=0\) ⇔ x=\(\dfrac{-1}{12}\)
22 tháng 1 2022
\(\Leftrightarrow3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)=7x^2-14x-5\)
\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)
\(\Leftrightarrow36x=-3\)
hay x=-1/12
TN
16 tháng 1 2017
a)\(2x^3=x^2+2x-1\)
\(\Rightarrow2x^3-x^2-2x+1=0\)
\(\Rightarrow x^2\left(2x-1\right)-\left(2x-1\right)=0\)
\(\Rightarrow\left(x^2-1\right)\left(2x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x+1\right)\left(2x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=\pm1\\x=\frac{1}{2}\end{cases}}\)
b)\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-x\right)\)
\(\Rightarrow\left(3x-1\right)\left(x^2+2\right)-6x\left(3x-1\right)=0\)
\(\Rightarrow\left(3x-1\right)\left(x^2+2-6x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\\Delta_{x^2-6x+2=0}=\left(-6\right)^2-4\cdot1\cdot2=28\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x_{2,3}=\frac{6\pm\sqrt{28}}{2}\end{cases}}\)
\(\left(3x+1\right)^2=\left(15-7x\right)^2\)
\(\Leftrightarrow\left(3x+1\right)^2-\left(15-7x\right)^2=0\)
\(\Leftrightarrow\left(3x+1+15-7x\right)\left(3x+1-15+7x\right)=0\)
\(\Leftrightarrow\left(16-4x\right)\left(10x-14\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}16-4x=0\\10x-14=0\end{cases}\Leftrightarrow\orbr{\begin{cases}4x=16\\10x=14\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=4\\x=\frac{7}{5}\end{cases}}}\)
Vậy ...............