Bài 1: Giải phương trình: 5x3 + 6x2 + 12x + 8 =0
Bài 2: Giải phương trình: x5 = x4 + x3 + x2 + x + 2
Bài 3: Giải pt: ( x2 + 11x+12 )( x2 + 9x + 20)(x2 + 13x + 42)=36(x2 + 11x + 30)(x2 + 11x + 31)
Giải các phương trình sau:
1/x2+5x+6 + 1/x2+7x+12 + 1/x2+9x+20 + 1/x2+11x+30 = 1/8
\(\Leftrightarrow\dfrac{1}{\left(x+2\right)\left(x+3\right)}+\dfrac{1}{\left(x+3\right)\left(x+4\right)}+...+\dfrac{1}{\left(x+5\right)\left(x+6\right)}=\dfrac{1}{8}\)
=>\(\dfrac{1}{x+2}-\dfrac{1}{x+3}+\dfrac{1}{x+3}-\dfrac{1}{x+4}+...+\dfrac{1}{x+5}-\dfrac{1}{x+6}=\dfrac{1}{8}\)
=>1/x+2-1/x+6=1/8
=>\(\dfrac{x+6-x-2}{\left(x+2\right)\left(x+6\right)}=\dfrac{1}{8}\)
=>x^2+8x+12=32
=>x^2+8x-20=0
=>(x+10)(x-2)=0
=>x=-10 hoặc x=2
giải pt
1/x2+9x+20 + 1/x2+11x+30 + 1/x2+13x+42=1/18
help ae ưiiiiii
ĐKXĐ: \(x\ne\left\{-4;-5;-6;-7\right\}\)
\(\dfrac{1}{x^2+9x+20}+\dfrac{1}{x^2+11x+30}+\dfrac{1}{x^2+13x+42}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{\left(x+4\right)\left(x+5\right)}+\dfrac{1}{\left(x+5\right)\left(x+6\right)}+\dfrac{1}{\left(x+6\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+6}+\dfrac{1}{x+6}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{1}{x+4}-\dfrac{1}{x+7}=\dfrac{1}{18}\)
\(\Leftrightarrow\dfrac{3}{\left(x+4\right)\left(x+7\right)}=\dfrac{1}{18}\)
\(\Rightarrow\left(x+4\right)\left(x+7\right)=54\)
\(\Leftrightarrow x^2+11x-26=0\)
\(\Leftrightarrow x^2-2x+13x-26=0\)
\(\Leftrightarrow x\left(x-2\right)+13\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-13\end{matrix}\right.\)
Bài 1: Giải các phương trình dưới đây
1) x2 - 9 = (x - 3)(5x +2)
2) x3 - 1 = (x - 1)(x2 - 2x +16)
3) 4x2 (x - 1) - x + 1 = 0
4) x3 + 4x2 - 9x - 36 = 0
5) (3x + 5)2 = (x - 1)2
6) 9 (2x + 1)2 = 4 (x - 5)2
7) x2 + 2x = 15
8) x4 + 5x3 + 4x2 = 0
9) (x2 - 4) - (x - 2)(3 - 2x) = 0
10) (3x + 2)(x2 - 1) = (9x2 - 4) (x + 1)
11) (3x - 1)(x2 + 2) = (3x - 1)(7x - 10)
12) (2x2 + 1) (4x - 3) = (x - 12)(2x2 + 1)
1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\)
hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)
2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)
hay \(x\in\left\{1;5\right\}\)
3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)\left(2x+1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)
4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{-4;3;-3\right\}\)
5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)
\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)
\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)
hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)
1.
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(5x-2\right)\)
\(\Leftrightarrow x+3=5x-2\)
\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)
2.
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)
\(\Leftrightarrow x^2+x+1=x^2-2x+16\)
\(\Leftrightarrow3x=15\Leftrightarrow x=5\)
3.
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2};x=-\dfrac{1}{2}\end{matrix}\right.\)
7.
\(\Leftrightarrow x^2+2x-15=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
8.\(\Leftrightarrow x^4+x^3+4x^3+4x^2=0\)
\(\Leftrightarrow x^3\left(x+1\right)+4x^2\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^3+4x^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=0;x=-4\end{matrix}\right.\)
9.\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=\left(x-2\right)\left(3-2x\right)\)
\(\Leftrightarrow x+2=3-2x\)
\(\Leftrightarrow3x=1\Leftrightarrow x=\dfrac{1}{3}\)
BÀI 1. Giải các phương trình sau bằng công thức nghiệm hoặc (công thức nghiện thu gọn). 1) x2 - 11x + 38 = 0 ; 2) 6x2 + 71x + 175 = 0 ; 3) 5x2 - 6x + 27 = 0 ; 4) - 30x2 + 30x - 7,5 = 0 ; 5) 4x2 - 16x + 17 = 0 ; 6) x2 + 4x - 12 = 0 ;
1, \(\Delta=\left(-11\right)^2-4.1.38=121-152=-31< 0\)
\(\Rightarrow\) pt vô nghiệm
2, \(\Delta=71^2-4.6.175=5041-4200=841\)
\(x_1=\dfrac{-71+\sqrt{841}}{2.6}=\dfrac{-71+29}{12}=\dfrac{-42}{12}=-\dfrac{7}{2}\)
\(x_2=\dfrac{-71-\sqrt{841}}{2.6}=\dfrac{-71-29}{12}=\dfrac{-10}{12}=-\dfrac{25}{3}\)
3, \(\Delta=\left(-3\right)^2-5.27=9-135=-126< 0\)
⇒ pt vô nghiệm
4, \(\Delta=15^2-\left(-30\right)\left(-7,5\right)=225-225=0\)
\(\Rightarrow x_1=x_2=\dfrac{-30}{2.\left(-30\right)}=\dfrac{1}{2}\)
5, \(\Delta'=\left(-8\right)^2-4.17=64-68=-4\)
⇒ pt vô nghiệm
6, \(\Delta=4^2-4.1.\left(-12\right)=16+48=64\)
\(x_1=\dfrac{-4+\sqrt{64}}{2.1}=\dfrac{-4+8}{2}=\dfrac{4}{2}=2\)
\(x_2=\dfrac{-4-\sqrt{64}}{2.1}=\dfrac{-4-8}{2}=\dfrac{-12}{2}=-6\)
Giải các phương trình sau:
a, (9x2 - 4)(x + 1) = (3x +2)(x2 - 1)
b, (x - 1)2 - 1 + x2 = (1 - x)(x + 3)
c, (x2 - 1)(x + 2)(x - 3) = (x - 1)(x2 - 4)(x + 5)
d, x4 + x3 + x + 1 = 0
e, x3 - 7x + 6 = 0
f, x4 - 4x3 + 12x - 9 = 0
g, x5- 5x3 + 4x = 0
h, x4 - 4x3 + 3x2 + 4x - 4 = 0
a, \(\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)-\left(3x+2\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(\left(9x^2-4\right)-\left(\left(3x+2\right)\left(x-1\right)\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-\left(3x^2-x-2\right)\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(9x^2-4-3x^2+x+2\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+1\right)=0;3x^2+x-2=0\)
=> x=-1
với \(3x^2+x-2=0\)
ta sử dụng công thức bậc 2 suy ra : \(x=\dfrac{2}{3};x=-1\)
Vậy ghiệm của pt trên \(S\in\left\{-1;\dfrac{2}{3}\right\}\)
b: \(\Leftrightarrow x^2-2x+1-1+x^2=x+3-x^2-3x\)
\(\Leftrightarrow2x^2-2x=-x^2-2x+3\)
\(\Leftrightarrow3x^2=3\)
hay \(x\in\left\{1;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x-3\right)-\left(x-1\right)\left(x-2\right)\left(x+2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left[\left(x+1\right)\left(x-3\right)-\left(x-2\right)\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2-2x-3-x^2-3x+10\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(-5x+7\right)=0\)
hay \(x\in\left\{1;-2;\dfrac{7}{5}\right\}\)
Giải phương trình:
a)x2-11x+15=-15
b)2x-3x+10=x
c)x3-4=4
d)x4+x3-x2-x=0
\(a.x^2-11x+15=-15.\Leftrightarrow x^2-11x+30=0.\)
\(\Leftrightarrow\left(x-6\right)\left(x-5\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=6.\\x=5.\end{matrix}\right.\)
\(b.2x-3x+10=x.\Leftrightarrow-2x+10=0.\Leftrightarrow x=5.\)
\(c.x^3-4=4.\Leftrightarrow x^3=8.\Leftrightarrow x^3=2^3.\Rightarrow x=2.\)
\(d.x^4+x^3-x^2-x=0.\Leftrightarrow x^2\left(x^2+x\right)-\left(x^2+x\right)=0.\Leftrightarrow\left(x^2-1\right)\left(x^2+x\right)=0.\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)x\left(x+1\right)=0.\Leftrightarrow\left(x-1\right)\left(x+1\right)^2x=0.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0.\\x+1=0.\\x=0.\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=-1.\\x=0.\end{matrix}\right.\)
7) x4+2x3-2x2+2x-3=0
8) (x-1)( x2+5x-2)-x3+1=0
9) x2+(x+2)(11x-7)=4
(GIẢI PHƯƠNG TRÌNH)
\(x^4+2x^3-2x^2+2x-3=0\\ \Leftrightarrow x^4+3x^3-x^3-3x^2+x^2+3x-x-3=0\\ \Leftrightarrow x^3\left(x+3\right)-x^2\left(x+3\right)+x\left(x+3\right)-\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^3-x^2+x-1\right)=0\\ \Leftrightarrow\left(x+3\right)\left[x^2\left(x-1\right)+\left(x-1\right)\right]=0\\ \Leftrightarrow\left(x+3\right)\left(x-1\right)\left(x^2+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x-1=0\\x^2+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\left(\text{vì }x^2+1\ge1>0\right)\)
Vậy ...
\(\left(x-1\right)\left(x^2+5x-2\right)-x^3+1=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x-1\right)\left(x^2+x+1\right)=0\\ \Leftrightarrow\left(x-1\right)\left[\left(x^2+5x-2\right)-\left(x^2+x+1\right)\right]=0\\ \Leftrightarrow\left(x-1\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\4x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy ...
\(x^2+\left(x+2\right)\left(11x-7\right)=4\\ \Leftrightarrow x^2-4+\left(x+2\right)\left(11x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-2\right)+\left(11x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(x-2+11x-7\right)=0\\ \Leftrightarrow\left(x+2\right)\left(12x-9\right)=0\\ \Leftrightarrow3\left(x+2\right)\left(4x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+2=0\\4x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy ...
a) Ta có: \(x^2-11x-26=0\)
nên a=1; b=-11; c=-26
Áp dụng hệ thức Viet, ta được:
\(x_1+x_2=\dfrac{-b}{a}=\dfrac{-\left(-11\right)}{1}=11\)
và \(x_1x_2=\dfrac{c}{a}=\dfrac{-26}{1}=-26\)
bài 1 giải các bất phương trình sau
a, -x2 +5x-6 ≥ 0
b, x2-12x +36≤0
c, -2x2 +4x-2≤0
d, x2 -2|x-3| +3x ≥ 0
e, x-|x+3| -10 ≤0
bài 2 xét dấu các biểu thức sau
a,<-x2+x-1> <6x2 -5x+1>
b, x2-x-2/ -x2+3x+4
c, x2-5x +2
d, x-< x2-x+6 /-x2 +3x+4 >
Bài 1:
a: \(\Leftrightarrow x^2-5x+6< =0\)
=>(x-2)(x-3)<=0
=>2<=x<=3
b: \(\Leftrightarrow\left(x-6\right)^2< =0\)
=>x=6
c: \(\Leftrightarrow x^2-2x+1>=0\)
\(\Leftrightarrow\left(x-1\right)^2>=0\)
hay \(x\in R\)
giải phương trình sau:
a. (9x2-4)(x+1) = (3x+2) (x2-1)
b. (x-1)2-1+x2 = (1-x)(x+3)
c. (x2-1)(x+2)(x-3) = (x-1)(x2-4)(x+5)
d. x4+x3+x+1=0
e. x3-7x+6 = 0
f. x4-4x3+12x-9 = 0
g. x5-5x3+4x = 0
h. x4-4x3+3x2+4x-4 = 0
m.n jup vs