(x-1)(x-3)(x+5)(x+7)=297
M.n giúp mk ý này vs
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NM
1
1 tháng 4 2020
b) \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}+\frac{x+1}{3}=\frac{x+7}{12}\)
<=> \(\frac{13\left(x+1\right)}{12}-\frac{5x+3}{6}=\frac{x+7}{12}\)
<=> 13(x + 1) - 2(5x + 3) = x + 7
<=> 13x + 13 - 10x - 6 = x + 7
<=> 3x + 7 = x + 7
<=> 3x + 7 - x = 7
<=> 2x + 7 = 7
<=> 2x = 7 - 7
<=> 2x = 0
<=> x = 0
c) 2x + 4(x - 2) = 5
<=> 2x + 4x - 8 = 5
<=> 6x - 8 = 5
<=> 6x = 5 + 8
<=> 6x = 13
<=> x = 13/6
B1
20 tháng 8 2017
X trong (-oo: + oo) (2/5) * x 3/7
= 1 - ((4/7) * x) // - 1 - ((4/7) *
x) (2/5) ) * X + (4/7) * x 3 / 7-1
= 0 34/35 * x-4/7 = 0 // + 4/7 34/35 * x
= 4/7 //: 34/35 X = 4/7/34/35 x
= 10/17 x = 10/17
chắc là sai đó
20 tháng 8 2017
\(\frac{3}{7}\left(x+1\right)=\frac{4}{7}\)
\(x+1=\frac{4}{3}\)
\(x=\frac{1}{3}\)
Vậy \(x=\frac{1}{3}\)
20 tháng 2 2018
a, 2.(4x-3)-3(x+5)+4(x-10)=5(x+2)
2.4x-2.3-3.x+3.5+4x-4.10=5x+5.2
8x-6-3x+15+4x-40=5x-10
8x-3x+4x-5x-6-15-40-10=0
4x-71=0
4x=71
x=71:4
x=71/4
HM
0
17 tháng 2 2018
3. (x+5) - 4.(x+7)=20
3x +15-4x-28 . =20
-x . =20+28-15
-x . =33
x. -33
17 tháng 2 2018
3(x+5)-4(x+7)=20
<=> 3x+15-4x-28=20
<=> -x-13=20
<=> x+13=-20
<=> x=-33
\(\left(x-1\right)\left(x-3\right)\left(x+5\right)\left(x+7\right)=297\)
\(\Rightarrow\left[\left(x-1\right)\left(x+5\right)\right]\left[\left(x-3\right)\left(x+7\right)\right]=297\)
\(\Rightarrow\left[x^2+4x-5\right]\left[x^2+4x-5-16\right]=297\)
Đặt \(x^2+4x-5=t\)
\(\Rightarrow t\left(t-16\right)=297\)
\(\Rightarrow t^2-16t+64=297+64\)
\(\Rightarrow\left(t+8\right)^2=361\)
\(\Rightarrow\left[{}\begin{matrix}t+8=19\\t+8=-19\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}t=11\\t=-27\end{matrix}\right.\)
Ta có : \(\left(x-1\right)\left(x-3\right)\left(x+5\right)\left(x+7\right)=297\)
\(\Leftrightarrow\left[\left(x-1\right)\left(x+5\right)\right]\left[\left(x-3\right)\left(x+7\right)\right]=297\)
\(\Leftrightarrow\left(x^2-x+5x-5\right)\left(x^2-3x+7x-21\right)=297\)
\(\Leftrightarrow\left(x^2+4x-5\right)\left(x^2+4x-21\right)=297\)
\(\Leftrightarrow\left(x^2+4x-13+8\right)\left(x^2+4x-13-8\right)=297\)
\(\Leftrightarrow\left(x^2+4x-13\right)^2-64=297\)
\(\Leftrightarrow\left(x^2+4x-13\right)^2=361\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+4x-13=19\\x^2+4x-13=-19\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+4x+4-17=19\\x^2+4x+4-17=-19\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)^2-17=19\\\left(x+2\right)^2-17=-19\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)^2=36\\\left(x+2\right)^2=-2\left(VL\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=6\\x+2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)