ĐK : x - 8 \(\ne\) 0 <=> x \(\ne\) 8
\(\dfrac{24}{x-8}+2=x-8\)
\(\Leftrightarrow\dfrac{24}{x-8}=x-10\)
\(\Leftrightarrow24=\left(x-8\right)\left(x-10\right)\)
\(\Leftrightarrow x^2-10x-8x+80=24\)
\(\Leftrightarrow x^2-18x+56=0\)
\(\Leftrightarrow x_1=14\left(TM\right),x_2=4\left(TM\right)\)
vậy pt có nghiệm x1 = 14, x2 = 4
Tìm x, biết :
\(x^3+x^2+x=8x^2+8x+8\).
Giải phương trình: \(8\left(x+\dfrac{1}{x}\right)^2+4\left(x^2+\dfrac{1}{x^2}\right)^2-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)^2=\left(x+4\right)^2\)
tìm x
a)x^4-2x^3-25x^2+50x=0
b)x^2(x-1)-4x^2+8x-4=0
c)9x^2-4-2(3x-2)^2=0
d)9x^2+90x+225-(x-7)^2=0
e)x^3-8+(x-2)(x+1)=0
g)(x+1)(x+2)(x+3)(x+4)-24=0
giúp mk vs ah
cho \(\dfrac{x^2-3x}{x.\left(1-3y\right)}=\dfrac{y^2-3x}{y.\left(1-3x\right)}\).CM \(\dfrac{8}{3}+x+y=\dfrac{1}{x}+\dfrac{1}{y}\)
(x2+6x+8)(x2+8x+15)-24
Cho bieu thuc: \(Q=\left(\dfrac{x^2-2x}{2x^2+8}+\dfrac{2x^2}{x^2.\left(x-2\right)}\right).\left(\dfrac{x^2-x-2}{x^2}\right)\)
a, Rut gon bieu thuc Q
b, Tim gia tri ca x de Q co gia tri bang \(\dfrac{1}{4}\)
giải phương trình sau:
a, \(x^4+x^2-2=0\)
b,\(x^4-13x^2+36=0\)
c, \(\dfrac{1}{8}x^3+\dfrac{3}{4}x^4+\dfrac{3}{2}x+=-1\)
Bài 1: Tìm x biết a) x^3 - 4x^2 - x + 4= 0 b) x^3 - 3x^2 + 3x + 1=0 c) x^3 + 3x^2 - 4x - 12=0 d) (x-2)^2 - 4x +8 =0
phân tích đa thức sau thành nhân tử:
a) x^8 + x^4 -2
b)x^2n + 5x^n - 24, n thuộc N*
c) (x^2 + x)^2 -2(x^2 +x ) - 15
(x^2 + x +1)(x^2 +x +2) -12
