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16 tháng 6 2023
`@` `\text {Ans}`
`\downarrow`
`a)`
`3x(4x-1) - 2x(6x-3) = 30`
`=> 12x^2 - 3x - 12x^2 + 6x = 30`
`=> 3x = 30`
`=> x = 30 \div 3`
`=> x=10`
Vậy, `x=10`
`b)`
`2x(3-2x) + 2x(2x-1) = 15`
`=> 6x- 4x^2 + 4x^2 - 2x = 15`
`=> 4x = 15`
`=> x = 15/4`
Vậy, `x=15/4`
`c)`
`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`
`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`
`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`
`=> 40x^2 -17x - 1 = 1`
`d)`
`(x+2)(x+2)-(x-3)(x+1)=9`
`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`
`=> 6x + 7 =9`
`=> 6x = 2`
`=> x=2/6 =1/3`
Vậy, `x=1/3`
`e)`
`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`
`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`
`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`
`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`
`=> 12x +8 = 0`
`=> 12x = -8`
`=> x= -8/12 = -2/3`
Vậy, `x=-2/3`
`g)`
`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`
`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`
`=> -3x + 4 =14`
`=> -3x = 10`
`=> x= - 10/3`
Vậy, `x=-10/3`
20 tháng 5 2022
a: =>|7x-9|=5x-3
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(7x-9-5x+3\right)\left(7x-9+5x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{3}{5}\\\left(2x-6\right)\left(12x-12\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;3\right\}\)
b: \(\Leftrightarrow\left|4x+1\right|=8x-x-2=7x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{7}\\\left(7x-2-4x-1\right)\left(7x-2+4x+1\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{2}{7}\\\left(3x-3\right)\left(11x-1\right)=0\end{matrix}\right.\Leftrightarrow x=1\)
c: |17x-5|=|17x+5|
=>17x-5=17x+5 hoặc 17x+5=5-17x
=>x=0
LV
4
LD
13 tháng 2 2022
\(a.\left(2x-3\right)+\left(x+9\right)=0\)
\(3x+6=0\Rightarrow x=-2\)
\(b.10x-2x^2=0\)
\(\Rightarrow10x=2x^2\Rightarrow x=5\)
\(c.2x^2-5x-7=0\)
\(2x^2+2x-7x-7=0\)
\(2x\left(x+1\right)-7\left(x+1\right)=0\)
\(\left(2x-7\right)\left(x+1\right)=0\Rightarrow\orbr{\begin{cases}2x-7=0\\x+1=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=3,5\\x=-1\end{cases}}\)
13 tháng 2 2022
a, Ta có : \(2x-3+x+9=0\Leftrightarrow3x+6=0\Leftrightarrow x=-2\)
b, \(-2x^2+10x=0\Leftrightarrow-2x\left(x-5\right)=0\Leftrightarrow x=0;x=5\)
c, \(2x^2-7x+2x-7=0\Leftrightarrow\left(x+1\right)\left(2x-7\right)=0\Leftrightarrow x=-1;x=\frac{7}{2}\)
M
0
25 tháng 8 2020
a) \(2x+\frac{3}{15}=\frac{7}{5}\)
=> \(2x=\frac{7}{5}-\frac{3}{15}=\frac{21}{15}-\frac{3}{15}=\frac{18}{15}\)
=> \(x=\frac{18}{15}:2=\frac{18}{15}\cdot\frac{1}{2}=\frac{9}{15}\cdot\frac{1}{1}=\frac{9}{15}\)
b) \(x-\frac{2}{9}=\frac{8}{3}\)
=> \(x=\frac{8}{3}+\frac{2}{9}\)
=> \(x=\frac{24}{9}+\frac{2}{9}=\frac{26}{9}\)
c) \(\frac{-8}{x}=\frac{-x}{18}\)
=> x(-x) = (-8).18
=> -x2 = -144
=> x2 = 144(bỏ dấu âm)
=> x = \(\pm\)12
d) \(\frac{2x+3}{6}=\frac{x-2}{5}\)
=> 5(2x + 3) = 6(x - 2)
=> 10x + 15 = 6x - 12
=> 10x + 15 - 6x + 12 = 0
=> 4x + 27 = 0
=> 4x = -27
=> x = -27/4
e) \(\frac{x+1}{22}=\frac{6}{x}\)
=> x(x + 1) = 132
=> x(x + 1) = 11.12
=> x = 11
f) \(\frac{2x-1}{2}=\frac{5}{x}\)
=> x(2x - 1) = 10
=> 2x2 - x = 10
=> 2x2 - x - 10 = 0
tới đây tự làm đi nhé
g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\)
=> (2x - 1)(2x + 1) = 63
=> 4x2 - 1 = 63
=> 4x2 = 64
=> x2 = 16
=> x = \(\pm\)4
h) Tương tự
NC
25 tháng 8 2020
a) \(\frac{2x+3}{15}=\frac{7}{5}\Leftrightarrow10x+15=105\Leftrightarrow10x=90\Rightarrow x=9\)
b) \(\frac{x-2}{9}=\frac{8}{3}\Leftrightarrow3x-6=72\Leftrightarrow3x=78\Rightarrow x=26\)
c) \(\frac{-8}{x}=\frac{-x}{18}\Leftrightarrow x^2=144\Leftrightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)
d) \(\frac{2x+3}{6}=\frac{x-2}{5}\Leftrightarrow10x+15=12x-12\Leftrightarrow2x=27\Rightarrow x=\frac{27}{2}\)
e) \(\frac{x+1}{22}=\frac{6}{x}\Leftrightarrow x^2+x-132=0\Leftrightarrow\left(x-11\right)\left(x+12\right)=0\Leftrightarrow\orbr{\begin{cases}x=11\\x=-12\end{cases}}\)
f) \(\frac{2x-1}{2}=\frac{5}{x}\Leftrightarrow2x^2-x-10=0\Leftrightarrow\left(x-2\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=2\\x=-\frac{5}{2}\end{cases}}\)
g) \(\frac{2x-1}{21}=\frac{3}{2x+1}\Leftrightarrow4x^2=64\Leftrightarrow x^2=16\Rightarrow\orbr{\begin{cases}x=4\\x=-4\end{cases}}\)
h) \(\frac{10x+5}{6}=\frac{5}{x+1}\Leftrightarrow10x^2+15x-25=0\Leftrightarrow5\left(x-1\right)\left(2x+5\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-\frac{5}{2}\end{cases}}\)
10 tháng 7 2017
chào các bạn,có 2 tốt bụng thì tk mik nhé,cần lắm những người như thế
PN
0
K
1
28 tháng 6 2019
:v ( cái đề bài )
Đề : Tìm x để biểu thức có giá trị là nguyên
\(d,\frac{x+4}{x-3}\)
ĐKXĐ : \(x\ne-3\)
Để biểu thức có giá trị là nguyên thì
\(x+4⋮x-3\)
=> \(\left(x-3\right)+7⋮x-3\)
=> \(7⋮x-3\)
=> \(x-3\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\\\left[{}\begin{matrix}x-3=7\\x-3=-7\end{matrix}\right.\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=1+3=4\\x=-1+3=2\end{matrix}\right.\\\left[{}\begin{matrix}x=7+3=10\\x=-7+3=-4\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x\in\left\{4;2;10;-4\right\}\)
e, \(\frac{10x+9}{2x+3}\)
Để biểu thức có giá trị là nguyên thì
\(10x+9⋮2x+3\)
=>\(10x+30-21⋮2x+3\)
=> \(10\left(x+3\right)-21⋮2x+3\)
=> \(-21⋮2x+3\)
=> \(2x+3\inƯ\left(-21\right)=\left\{\pm1;\pm3;\pm7;\pm21\right\}\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}2x+3=1\\2x+3=-1\end{matrix}\right.\\\left[{}\begin{matrix}2x+3=3\\2x+3=-3\end{matrix}\right.\\\left[{}\begin{matrix}2x+3=7\\2x+3=-7\end{matrix}\right.\\\left[{}\begin{matrix}2x+3=21\\2x+3=-21\end{matrix}\right.\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}2x=1-3=-2\\2x=-1-3=-4\end{matrix}\right.\\\left[{}\begin{matrix}2x=3-3=0\\2x=-3-3=-6\end{matrix}\right.\\\left[{}\begin{matrix}2x=7-3=4\\2x=-7-3=-10\end{matrix}\right.\\\left[{}\begin{matrix}2x=21-3=18\\2x=-21-3=-24\end{matrix}\right.\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=-2:2=-1\\x=-4:2=-2\end{matrix}\right.\\\left[{}\begin{matrix}x=0:2=0\\x=-6:2=-3\end{matrix}\right.\\\left[{}\begin{matrix}x=4:2=2\\x=-10:2=-5\end{matrix}\right.\\\left[{}\begin{matrix}x=18:2=9\\x=-24:2=-12\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x\in\left\{-1;-2;0;-3;2;-5;9;12\right\}\)
f, \(\frac{4x+1}{2x-1}\)
ĐKXĐ : \(x\ne\frac{a}{b}\)( phân số tối giản )
Đẻ biểu thức trên có giá trị là nguyên thì
\(4x+1⋮2x-1\)
=> \(\left(4x-1\right)+2⋮2x-1\)
=> \(2⋮2x-1\)
=> \(2x-1\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}2x-1=1\\2x-1=-1\end{matrix}\right.\\\left[{}\begin{matrix}2x-1=2\\2x-1=-2\end{matrix}\right.\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}2x=1+1=2\\2x=-1+1=0\end{matrix}\right.\\\left[{}\begin{matrix}2x=2+1=3\\2x=-2+1=-1\end{matrix}\right.\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2:2=1\left(lấy\right)\\x=0:2=0\left(lấy\right)\end{matrix}\right.\\\left[{}\begin{matrix}x=3:2=\frac{3}{2}\left(loại\right)\\x=-:2=-\frac{1}{2}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x\in\left\{1;0\right\}\)
