Giải phương trình sau:
\(\dfrac{3}{7}\)x-1=\(\dfrac{1}{7}\)x(3x-7)
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19 tháng 2 2022
a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)
\(\Leftrightarrow49-21x+60x+24=84x+1092\)
\(\Leftrightarrow39x-84x=1092-73\)
=>-45x=1019
hay x=-1019/45
b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)
=>21x+63-14=20x+36-49x+63
=>21x+49=-29x+99
=>50x=50
hay x=1
c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)
=>14x+7-15x-6-21x-63=0
=>-22x-64=0
hay x=-32/11
d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)
=>70x-105-30x-45=84x+63-1785
=>40x-150-84x+1722=0
=>-44x+1572=0
hay x=393/11
N
20 tháng 11 2021
\(ĐK:x\ne3\\ PT\Leftrightarrow\dfrac{x^2+3x+2}{x-3}\left(-x-1+x^2-2x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{\left(x+1\right)\left(x+2\right)}{x-3}=0\\x^2-3x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\\x=\dfrac{3+\sqrt{41}}{2}\\x=\dfrac{3-\sqrt{41}}{2}\end{matrix}\right.\)
ND
Nguyễn Đức Trí
VIP
15 tháng 7 2023
1) \(\sqrt[]{3x+7}-5< 0\)
\(\Leftrightarrow\sqrt[]{3x+7}< 5\)
\(\Leftrightarrow3x+7\ge0\cap3x+7< 25\)
\(\Leftrightarrow x\ge-\dfrac{7}{3}\cap x< 6\)
\(\Leftrightarrow-\dfrac{7}{3}\le x< 6\)
27 tháng 1 2022
a, <=> x = -4
b, <=> 6x + 2 = -2x + 5 <=> 8x = 3 <=> x = 3/8
c, <=> 5x + 2x - 2 = 4x + 7 <=> 2x = 9 <=> x = 9 /2
d, <=> 10x^2 - 10x^2 - 15x = 15 <=> x = -1
27 tháng 1 2022
a, <=> x = -4
b, <=> 6x + 2 = -2x + 5 <=> 8x = 3 <=> x = 3/8
c, <=> 5x + 2x - 2 = 4x + 7 <=> 2x = 9 <=> x = 9 /2
d <=> 10x^2 - 10x^2 - 15x = 15 <=> x = -1
27 tháng 1 2022
a: =>-x+2x=3-7
=>x=-4
b: =>6x+2+2x-5=0
=>8x-3=0
hay x=3/8
c: =>5x+2x-2-4x-7=0
=>3x-9=0
hay x=3
d: =>10x2-10x2-15x=15
=>-15x=15
hay x=-1
19 tháng 8 2021
1: Ta có: \(\dfrac{3}{x-3}+\dfrac{4}{x+3}=\dfrac{3x-7}{x^2-9}\)
\(\Leftrightarrow\dfrac{3x+9}{\left(x-3\right)\left(x+3\right)}+\dfrac{4x-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{3x-7}{\left(x-3\right)\left(x+3\right)}\)
Suy ra: \(3x+9+4x-12=3x-7\)
\(\Leftrightarrow4x=-7+12-9=-4\)
hay \(x=-1\left(nhận\right)\)
2: Ta có: \(\dfrac{3}{x-4}-\dfrac{4}{x+4}=\dfrac{3x-4}{x^2-16}\)
\(\Leftrightarrow\dfrac{3x+12}{\left(x-4\right)\left(x+4\right)}-\dfrac{4x-16}{\left(x+4\right)\left(x-4\right)}=\dfrac{3x-4}{\left(x-4\right)\left(x+4\right)}\)
Suy ra: \(3x+12-4x+16=3x-4\)
\(\Leftrightarrow28-4x=-4\)
\(\Leftrightarrow4x=32\)
hay \(x=8\left(tm\right)\)
19 tháng 8 2021
3: Ta có: \(\dfrac{5x^2-12}{x^2-1}+\dfrac{3}{x-1}=\dfrac{5x}{x+1}\)
Suy ra: \(5x^2-12+3x+3=5x^2-5x\)
\(\Leftrightarrow3x-9+5x=0\)
\(\Leftrightarrow8x=9\)
hay \(x=\dfrac{9}{8}\left(nhận\right)\)
NH
20 tháng 3 2023
a)
`2x-3=2-x`
`<=>2x+x=2+3`
`<=>3x=5`
`<=>x=5/3`
b)
`3x+3=7+5x`
`<=>3x-5x=7-3`
`<=>-2x=4`
`<=>x=-2`
c)
`7x-3=3x+13`
`<=>7x-3x=13+3`
`<=>4x=16`
`<=>x=4`
d)
`(5x-2)/3=(5-3x)/2`
`<=>10x-4=15-9x`
`<=>10x+9x=15+4`
`<=>19x=19`
`<=>x=1`
30 tháng 3 2022
a: \(\Leftrightarrow4\left(2x+1\right)-3\left(6x-1\right)=2x+1\)
=>8x+4-18x+3=2x+1
=>-10x+7=2x+1
=>-12x=-6
hay x=1/2
b: \(\Leftrightarrow4x^2-12x+7x-21-x^2=3x^2+6x\)
=>5x-21=6x
=>-x=21
hay x=-21
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):}`
12 tháng 4 2022
g.\(\dfrac{1-3x}{6}+x-1=\dfrac{x+2}{2}\)
\(\Leftrightarrow\dfrac{\left(1-3x\right)+6\left(x-1\right)}{6}=\dfrac{3\left(x+2\right)}{6}\)
\(\Leftrightarrow\left(1-3x\right)+6\left(x-1\right)=3\left(x+2\right)\)
\(\Leftrightarrow1-3x+6x-6=3x+6\)
\(\Leftrightarrow-5=6\left(vô.lí\right)\)
Vậy pt vô nghiệm
12 tháng 4 2022
h.\(\dfrac{3\left(2x+1\right)}{4}-5-\dfrac{3x+2}{10}=\dfrac{2\left(3x-1\right)}{5}\)
\(\Leftrightarrow\dfrac{15\left(2x+1\right)-100-2\left(3x+2\right)}{20}=\dfrac{8\left(3x-1\right)}{20}\)
\(\Leftrightarrow15\left(2x+1\right)-100-2\left(3x+2\right)=8\left(3x-1\right)\)
\(\Leftrightarrow30x+15-100-6x-4=24x-8\)
\(\Leftrightarrow-89=-8\left(vô.lí\right)\)
Vậy pt vô nghiệm
\(\dfrac{3}{7}x-1=\dfrac{1}{7}x\left(3x-7\right)\)
⇔ \(\dfrac{3}{7}x-\dfrac{7}{7}=\dfrac{3}{7}x^2-\dfrac{7}{7}x\)
⇔ \(\dfrac{3}{7}x-\dfrac{7}{7}-\dfrac{3}{7}x^2+\dfrac{7}{7}x=0\)
⇔ \(\dfrac{3}{7}x\left(1-x\right)-\dfrac{7}{7}\left(1-x\right)=0\)
⇔ \(\left(1-x\right)\left(\dfrac{3}{7}x-\dfrac{7}{7}\right)=0\)
⇔ \(\left[{}\begin{matrix}1-x=0\\\dfrac{3}{7}x-\dfrac{7}{7}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{7}{3}\end{matrix}\right.\)
Vậy pt có 2 nghiệm pt : \(x=1;x=\dfrac{7}{3}\)
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