Tìm x để 3(2x-5)=5(7x+8)
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DL
2
1 tháng 9 2019
a) 2x - 5 = 3 + 2x - 7x
=> 2x - 2x + 7x = 3 +5
=> 7x = 8
=> x = 8/7
b) \(\left(2x-1\right)^2=\left(2x-1\right)^5\)
=> \(\left(2x-1\right)^2-\left(2x-1\right)^5=0\)
=> \(\left(2x-1\right)^2\left[1-\left(2x-1\right)^3\right]=0\)
=> \(\orbr{\begin{cases}\left(2x-1\right)^2=0\\1-\left(2x-1\right)^3=0\end{cases}}\)
=> \(\orbr{\begin{cases}2x-1=0\\\left(2x-1\right)^3=1\end{cases}}\)
=> \(\orbr{\begin{cases}2x=1\\2x-1=1\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\2x=2\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}\)
QN
1
30 tháng 1 2022
a: \(\Leftrightarrow3x^3-2x^2+15x^2-10x+3x-2+7⋮3x-2\)
\(\Leftrightarrow3x-2\in\left\{1;-1;7;-7\right\}\)
hay \(x\in\left\{3;1\right\}\)
b: \(\Leftrightarrow2x^5-7x^3+4x^4-14x^2+14x^2-49x+49x-44⋮2x^2-7\)
\(\Leftrightarrow2401x^2-1936⋮2x^2-7\)
\(\Leftrightarrow4802x^2-3872⋮2x^2-7\)
\(\Leftrightarrow2x^2-7\inƯ\left(12935\right)\)
\(\Leftrightarrow2x^2-7\in\left\{1;5;13;65;199;995;2587;12935;-1;-5\right\}\)
\(\Leftrightarrow2x^2\in\left\{8;72;2\right\}\)
hay \(x\in\left\{2;-2;6;-6;1;-1\right\}\)
19 tháng 12 2020
a)
ĐKXĐ: \(x\ne-4\)
Để A nguyên thì \(3x+21⋮x+4\)
\(\Leftrightarrow3x+12+9⋮x+4\)
mà \(3x+12⋮x+4\)
nên \(9⋮x+4\)
\(\Leftrightarrow x+4\inƯ\left(9\right)\)
\(\Leftrightarrow x+4\in\left\{1;-1;3;-3;9;-9\right\}\)
\(\Leftrightarrow x\in\left\{-3;-5;-1;-7;5;-13\right\}\)(nhận)
Vậy: Để A nguyên thì \(x\in\left\{-3;-5;-1;-7;5;-13\right\}\)
b) ĐKXĐ: \(x\ne\dfrac{1}{2}\)
Để B nguyên thì \(2x^3-7x^2+7x+5⋮2x-1\)
\(\Leftrightarrow2x^3-x^2-6x^2+3x+4x-2+7⋮2x-1\)
\(\Leftrightarrow x^2\left(2x-1\right)-3x\left(2x-1\right)+2\left(2x-1\right)+7⋮2x-1\)
\(\Leftrightarrow\left(2x-1\right)\left(x^2-3x+2\right)+7⋮2x-1\)
mà \(\left(2x-1\right)\left(x^2-3x+2\right)⋮2x-1\)
nên \(7⋮2x-1\)
\(\Leftrightarrow2x-1\inƯ\left(7\right)\)
\(\Leftrightarrow2x-1\in\left\{1;-1;7;-7\right\}\)
\(\Leftrightarrow2x\in\left\{2;0;8;-6\right\}\)
hay \(x\in\left\{1;0;4;-3\right\}\)(nhận)
Vậy: \(x\in\left\{1;0;4;-3\right\}\)
22 tháng 1 2021
a) ĐKXĐ: \(x\notin\left\{-1;0\right\}\)
Ta có: \(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=2\)
\(\Leftrightarrow\dfrac{x\left(x+3\right)}{x\left(x+1\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{x\left(x+1\right)}=\dfrac{2x\left(x+1\right)}{x\left(x+1\right)}\)
Suy ra: \(x^2+3x+x^2-3x+2=2x^2+2x\)
\(\Leftrightarrow2x^2+2-2x^2-2x=0\)
\(\Leftrightarrow-2x+2=0\)
\(\Leftrightarrow-2x=-2\)
hay x=1(nhận)
Vậy: S={1}
b) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)
Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)
\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)
\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)
\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)
\(\Leftrightarrow-56x-1=0\)
\(\Leftrightarrow-56x=1\)
hay \(x=-\dfrac{1}{56}\)(nhận)
Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)
c) ĐKXĐ: \(x\ne-\dfrac{2}{3}\)
Ta có: \(\dfrac{5}{3x+2}=2x-1\)
\(\Leftrightarrow5=\left(3x+2\right)\left(2x-1\right)\)
\(\Leftrightarrow6x^2-3x+4x-2-5=0\)
\(\Leftrightarrow6x^2+x-7=0\)
\(\Leftrightarrow6x^2-6x+7x-7=0\)
\(\Leftrightarrow6x\left(x-1\right)+7\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\6x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\6x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{7}{6}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{1;-\dfrac{7}{6}\right\}\)
d) ĐKXĐ: \(x\ne\dfrac{2}{7}\)
Ta có: \(\left(2x+3\right)\cdot\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)
\(\Leftrightarrow\left(2x+3\right)\cdot\left(\dfrac{3x+8+2-7x}{2-7x}\right)-\left(x-5\right)\left(\dfrac{3x+8+2-7x}{2-7x}\right)=0\)
\(\Leftrightarrow\left(2x+3-x+5\right)\cdot\dfrac{-4x+6}{2-7x}=0\)
\(\Leftrightarrow\left(x+8\right)\cdot\left(-4x+6\right)=0\)(Vì \(2-7x\ne0\forall x\) thỏa mãn ĐKXĐ)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\-4x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\-4x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\left(nhận\right)\\x=\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(S=\left\{-8;\dfrac{3}{2}\right\}\)
7 tháng 5 2017
a)
f(x)= -x5 -7x4 -2x3+ x2 + 4x + 8
g(x)=x5 +7x4+2x3+3x2 - 3x -8
f(x)+g(x) =0 -0 -0 + 4x2 +x+0
g(x)=x5 +7x4+2x3+3x2 - 3x -8
f(x)= -x5 -7x4 -2x3+ x2 + 4x + 8
g(x)-f(x) =2x5+14x4+4x3+2x2-7x -16
b)
Bậc:5
Hệ số cao nhất:2
hệ số tự do:16
c)
Để đt h(x) có nghiệm thì
4x2+x=0
->4x.x+x=0
->(4x+1)x=0
->th1:x=0 -> x=0
4x+1=0 -> x=-1/4
Vậy đt h(x) có nghiệm là x=0 hoặc x=-1/4
Lần sau bn viết rõ hơn nhé
mik dich mún lòi mắt
30 tháng 1 2021
Để \(A⋮B\) thì \(10x^2-7x-5⋮2x-3\)
\(\Leftrightarrow10x^2-15x+8x-12+7⋮2x-3\)
\(\Leftrightarrow5x\left(2x-3\right)+4\left(2x-3\right)+7⋮2x-3\)
\(\Leftrightarrow\left(2x-3\right)\left(5x+4\right)+7⋮2x-3\)
mà \(\left(2x-3\right)\left(5x+4\right)⋮2x-3\)
nên \(7⋮2x-3\)
\(\Leftrightarrow2x-3\inƯ\left(7\right)\)
\(\Leftrightarrow2x-3\in\left\{1;-1;7;-7\right\}\)
\(\Leftrightarrow2x\in\left\{4;2;10;-4\right\}\)
hay \(x\in\left\{2;1;5;-2\right\}\)(nhận)
Vậy: Khi \(x\in\left\{2;1;5;-2\right\}\) thì \(A⋮B\)
MH
30 tháng 1 2021
Điều kiện: \(B\ne0\Leftrightarrow2x-3\ne0\Leftrightarrow x\ne\dfrac{3}{2}\).
Ta có: \(\dfrac{A}{B}=\dfrac{10x^2-7x-5}{2x-3}=\dfrac{10x^2-15x+8x-12+7}{2x-3}\\ =\dfrac{5x\left(2x-3\right)+4\left(2x-3\right)+7}{2x-3}=5x+4+\dfrac{7}{2x-3}\)
Để \(A⋮B\) thì \(\left(2x-3\right)\inƯ\left(7\right)=\left\{\pm1;\pm7\right\}\)
Nếu \(2x-3=1\Leftrightarrow2x=4\Leftrightarrow x=2\) (Thỏa mãn)
Nếu \(2x-3=-1\Leftrightarrow2x=2\Leftrightarrow x=1\) (Thỏa mãn)
Nếu \(2x-3=7\Leftrightarrow2x=10\Leftrightarrow x=5\) (Thỏa mãn)
Nếu \(2x-3=-7\Leftrightarrow2x=-4\Leftrightarrow x=-2\) (Thỏa mãn).
Vậy tập các giá trị \(x\) thỏa mãn là \(\left\{1;\pm2;5\right\}\).
5 tháng 1 2017
a, 3 - 2x = 3 . (5 - x) + 4
3 - 2x = 15 - 3x + 4
-2x + 3x = 15 + 4 - 3
x = 16
b, 4 - (7x + 2017) = 6 . (5 - x) - 2017
4 - 7x - 2017 = 30 - 6x - 2017
-7x + 6x = 30 - 2017 - 4 + 2017
-x = 26
x = -26
c, 15 - x . (x + 1) = 4 - x^2 + 2x
15 - x^2 - x = 4 - x^2 + 2x
-x^2 - x + x^2 - 2x = 4 - 15
-3x = -11
x = 11/3
d, -4 . (x - 5) + 2016 = 3 . (8 - x) - (2x - 2016)
-4x + 20 + 2016 = 24 - 3x - 2x + 2016
-4x + 3x +2x = 24 + 2016 - 20 - 2016
x = 4
đúng 100%
3(2x-5)=5(7x+8)
=> 6x-15=35x+40
=> 6x-35x=15+40
=> -29x=55
=> x=\(\dfrac{-55}{29}\)
vậy x=\(\dfrac{-55}{29}\)