\(\left(2x-1\right)+\left(x+3\right)-5\left(x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow-5x^2+245+2x-1+x+3=0\)
\(\Leftrightarrow-5x^2+3x+247=0\)
\(\text{Δ}=3^2-4\cdot\left(-5\right)\cdot247=4949\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{3+7\sqrt{101}}{10}\\x_2=\dfrac{3-\sqrt{701}}{10}\end{matrix}\right.\)
(x + 2)^2 + (2x -1)^2-(x - 3)^2 = 36
(2x -1)2 + (x + 3)2 - 5(x + 7)(x - 7) = 0
A, (x-2)^2=4x^2+4x+1 B, 25x^2-9=0 C, (2x-1)^2+(x+3)^2-5(x+7)(x-7)=0 D, (2x+1)(x-4)-2(x-3)^2=8 E, (3x-1)^2=(5-x)^2
tìm x
(2x-1)2+(x+3)2-5(x+7)(x-7) = 0
a)(2x-1)^2+(x+3)^2-5(x+7)(x-7)=0
b)(x+2)(x^2-2x+4)-x(x^2+2)=15
c)(x+3)^3-x(3x+1)^2+(2x-1)(4x^2-2x+1)=28
d)(x^2-1)^3-(x^4+x^2+1)(x^2-1)=0
a)(2x-1)^2+(x+3)^2-5(x+7)(x-7)=0
b)(x+3)^3-x(3x+1)^2+(2x-1)(4x^2-2x+1)=28
c)(x+2)(x^2-2x+4)-x(x^2+2)=15
d)(x^2-1)^3-x^4+x^2+1)(x^2-1)=0
Tìm x:
a) (x - 2) (x2 + 2x + 7) + 2 (x2 - 4) - 5 (x - 2) = 0
b) 3x (x - 1) + (x - 1) = 0
c) 2 (x + 3) x2 - 3x = 0
d) 4x2 - 25 - (2x - 5)(2x + 7) = 0
e) x3 + 27(x + 3)(x - 9) = 0
Tìm x biết
1.(x+3)2-(x+2).(x-2)=4x+17
2.(2x+1)2-(4x-1).(x-3)-15=0
3.(2x+3).(x-1)+(2x-3).(1-x)=0
4.2(5x-8)-3(4x-5)=4(3x-4)+11
5.(3x-1).(2x-7)-(1-3x).(6x-5)=0
1, tìm x :
a) (x+2).(x+3)-(x-2).(x+5)=0
b) (2x+3).(x-4)+(x-5).(x-2)=(x-4).(3x-5)
c) (3x-5).(7-5x)-(5x+2).(2-3x)=3
d)(x-7).(x+7)-(x-4)2-x=10+3
e) (x+1/2)2-(x-1).(x+1)=2x-1
2, Thu gọn :
a) (x+5)2-(3-x)2-2(x-4).(x+4)
b)4(x-1)2+(2x+3)2-8(x-2)(x+2)
Tìm x biết
a) (x-3)2 - 4 = 0
b) (2x-1)2 + (x+3)2 - 5(x+7)(x-7) = 0
