x(2 + x)(7 - x) = 0
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\2+x=0\\7-x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-2\\x=7\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là S = {0;-2;7}
a,(x+3)^2+(x-2).(x+2)-2(x-1)^2=7
b,(x+2)^2-(x-2)(x+2)=0
c,(3x-2)^2-(2x-1)^2=0
d,x^2-4x+3=0
tìm x biết:
a, 4x\(^2\)-25-(2x-5)(2x+7)=0
b, 2x\(^3\)+3x\(^2\)+2x+3=0
c, x\(^3\) +27+(x+3)(x-9)=0
d, x\(^2\)(x+7)-4(x+7)=0
Tìm x, biết:
1) 2x (x-3) + 5x - 15 = 0
2) x ( 2x - 7) - 4x + 14 = 0
3) x2 - 12x + 36 = 0
4) (x + 3) (x2 - 3x + 9) - x (x -1)(x + 1) -27 = 0
tim x biet:
a,x+x\(^2\)=0
b,x+1-(x+1)\(^2\)=0
c,15y(4y-9)-3(4y-9)=0
d,8(25z+7)-27z(25z+7)=0
Bài 1 :Giải phương trình
6) x2+x-12 =0
7) x4+2x3-2x2+2x-3=0
8) (x-1)( x2+5x-2)-x3+1=0
9) x2+(x+2)(11x-7)=4
Tìm x (x+2)(x2-2x-4)-x(x2+2)=15. (x-2)3-(x-3)(x2+3x+9)+6(x+1)2+49. 25x2-9=0. (x+4)2-(x+1)(x-1)=16. (2x+1)2+(x+3)2-5(x-7)(x-7)=0. (x+1)(x+2)(x+5)(x+7)=24. Giúp mình làm câu này
3) \(\frac{x-2}{x-5}\) \(-\frac{5}{x^2-5x}=\frac{1}{x}\)
\(\Leftrightarrow\) \(\frac{x-2}{x-5}-\frac{5}{x.\left(x-5\right)}=\frac{1}{x}\)
\(\Leftrightarrow\frac{\left(x-2\right).\left(x+5\right)}{x.\left(x-5\right)}-\frac{5}{x.\left(x-5\right)}=\frac{1.\left(x+5\right)}{x.\left(x-5\right)}\)
\(\Leftrightarrow x^2+5x-2x-10-5=1x+5\)
\(\Leftrightarrow x^2+5x-2x-1x-10-5-5\) = 0
\(\Leftrightarrow\) \(x^2+2x-20=0\)
\(\Leftrightarrow x^2+2x-10x-20=0\)
\(\Leftrightarrow\) (x\(^2\) + 2x) - (10x + 20) = 0
\(\Leftrightarrow\) x.(x + 2) - 10.(x + 2) = 0
\(\Leftrightarrow\)
4) \(\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x^2+7x}\)
\(\Leftrightarrow\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x\left(x+7\right)}\)
\(\Leftrightarrow\frac{\left(x-4\right).\left(x+7\right)}{x.\left(x+7\right)}-\frac{1.\left(x+7\right)}{x.\left(x+7\right)}=\frac{-7}{x.\left(x+7\right)}\)
\(\Leftrightarrow\) \(x^2+7x-4x-28-x-7=-7\)
\(\Leftrightarrow x^2+7x-4x-x-28-7+7=0\)
\(\Leftrightarrow\) x\(^2\) + 2x - 28 = 0
\(\Leftrightarrow\) x\(^2\) + 2x - 14x - 28 = 0
\(\Leftrightarrow\) (x\(^2\) + 2x) - (14x + 28) = 0
\(\Leftrightarrow\) x.(x + 2) - 14.(x + 2) = 0
\(\Leftrightarrow\) (x - 14) = 0 hoặc (x + 2) = 0
\(\Leftrightarrow\) x = 4 (Nhận) hoặc x = -2 (Loại)
5) \(\frac{x+2}{x-2}+\frac{x-2}{x+2}=\frac{8x}{x^2-4}\)
\(\Leftrightarrow\) \(\frac{\left(x+2\right).\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}+\frac{\left(x-2\right).\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{8x}{\left(x-2\right).\left(x+2\right)}\)
\(\Leftrightarrow x^2+2x+2x+4+x^2-2x-2x+4=8x\)
\(\Leftrightarrow\) \(x^2+x^2+2x+2x-2x-2x-8x+4+4=0\)
\(\Leftrightarrow2x^2-8x+8=0\)
\(\Leftrightarrow\) 2x\(^2\) - 2x - 8x + 8 = 0
\(\Leftrightarrow\) 2x(x - 1) - 8(x - 1) = 0
\(\Leftrightarrow\) 2x - 8 = 0 hoặc x - 1 = 0
\(\Leftrightarrow\) 2x = 8 hoặc x = 1
\(\Leftrightarrow\) x = 4 (Nhận) hoặc x = 1 (Nhận)
Vậy S = {4; 1}
6) \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4}{x^2-1}\)
\(\Leftrightarrow\) \(\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}=\frac{4}{\left(x-1\right).\left(x+1\right)}\)
\(\Leftrightarrow\) x\(^2\) + x + x + 1 - x\(^2\) + x + x - 1 = 4
\(\Leftrightarrow\) 4x - 4 = 0
\(\Leftrightarrow\) 4 (x - 1) =0
\(\Leftrightarrow\) x - 1 = 0 / 4 = 0
\(\Leftrightarrow\) x = 1 (Nhận)
Vậy S = {1}
7) \(\frac{x+1}{x-1}+\frac{-4x}{x^2-1}=\frac{x-1}{x+1}\)
\(\Leftrightarrow\) \(\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}+\frac{-4x}{\left(x-1\right).\left(x+1\right)}=\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x+1\right)}\)
\(\Leftrightarrow x^2+x+x+1-4x=x^2-x-x+1\)
\(\Leftrightarrow\) 0
Vậy S ={\(\varnothing\)}
1. (4x-10).(24+5x)=0
2 .(2x-5).(3x-2)=0
3. (2x-1).(3x+1)=0
4. x.(\(x\)2-1)=0
5.(5x+3).(\(x^2\)+4).(x-1)=0
6.(x-1).(x+2).(x+3)=0
7.(x-1).(x+5).(-3x+8)=0
Tìm x biêt :
a. ( x + 3)2 + ( x - 2).( x + 2) - 2. ( x - 1)2 = 7
b. 36x2 - 4g = 0
c. ( x + 3) . ( x2 - 3x + g) - x . ( x - 1 ) . ( x + 1) - 27 = 0
d. x2 - 4x + 3 = 0
e. x . ( 2x -1 ) - ( x- 2) . ( 2x + 3) = 0
g. ( x - 1 ) . ( x + 2) - x - 2 = 0
mong jup dỡ ạ
