giải các phương trình
6x^3 + x + 4 = 11x^2
x^6 - 14x^4 + 49x^2 = 36
giải các phương trình
6x^3 + x + 4 = 11x^2
x^6 - 14x^4 + 49x^2 = 36
6x^3 + x + 4 = 11x^2
<=>6x3-11x2+x+4=0
<=>6x3+3x2-14x2-7x+8x+4=0
<=>3x2(2x+1)-7x(2x+1)+4(2x+1)=0
<=>(2x+1)(3x2-7x+4)=0
<=>(2x+1)(3x2-3x-4x+4)=0
<=>(2x+1)(3x-4)(x-1)=0
<=>2x+1=0 hoặc 3x-4=0 hoặc x-1=0
<=>x\(\in\){-1/2;1;4/3}
b)x^6 - 14x^4 + 49x^2 = 36
<=>x6-14x4+49x2-36=0
<=>x6-x4-13x4+13x2+36x2-36=0
<=>x4(x2-1)-13x2(x2-1)+36(x2-1)=0
<=>(x2-1)(x4-13x2+36)=0
<=>(x+1)(x-1)(x4-9x2-4x2+36)=0
<=>(x+1)(x-1)[x2(x2-9)-4(x2-9)]=0
<=>(x-1)(x+1)(x2-9)(x2-4)=0
<=>(x-1)(x+1)(x+3)(x-3)(x+2)(x-2)=0
<=>x\(\in\){-3;-2;-1;1;2;3}
phù.mệt
Giải phương trình:
a) 6x^3+x+4=11x^2
b)x6-14x4+49x2=36
Rút gọn a) A=x6-14x4+49x2-36/x4+4x3-x2-16x-12
b) B= (2x4+4)(6x4+4)(10x4+4)(14x4+4)/(44+4)(84+4)(124+4)(164+4)
Giúp mình nha. Tks các bạn
Giải phương trình:
1, \(\sqrt{7x+7}+\sqrt{7x-6}+2\sqrt{49x^2+7x-42}=181-14x\)
2, \(5\sqrt{x}+\dfrac{5}{2\sqrt{x}}=2x+\dfrac{1}{2x}+4\)
Giải BPT
x6 - 14x4 + 49x2 > 36
\(x^6-14x^4+49x^2>36\)
\(\Leftrightarrow x^6-x^5+x^5-x^4-13x^4+13x^3-13x^3+13x^2+36x^2-36x+36x-36>0\)
\(\Leftrightarrow x^5\left(x-1\right)+x^4\left(x-1\right)-13x^3\left(x-1\right)-13x^2\left(x-1\right)+36x\left(x-1\right)+36\left(x-1\right)>0\)
\(\Leftrightarrow\left(x-1\right)\left(x^5+x^4-13x^3-13x^2+36x+36\right)>0\)
\(\Leftrightarrow\left(x-1\right)\left[x^4\left(x+1\right)-13x^2\left(x+1\right)+36\left(x+1\right)\right]>0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^4-13x^2+36\right)>0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^4-9x^2-4x^2+36\right)>0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left[x^2\left(x^2-9\right)-4\left(x^2-9\right)\right]>0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^2-9\right)\left(x^2-4\right) >0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x+3\right)\left(x-3\right)>0\)
Để \(\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x+3\right)\left(x-3\right)>0\)
\(\Rightarrow\left[{}\begin{matrix}x>3\\x< -3\end{matrix}\right.\)
Vậy để \(\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\left(x+3\right)\left(x-3\right)>0\) thì x>3 hoặc x<-3
giải phương trình 3x2 11x √x−2 √2x 3 14x x≥2
Giải phương trình:
1> 12-2(1-x)2=3x-2=2x-3
2> 10x+3-5x=4x+12
3> 11x+42-2x=100-9x-22
4> 2x-(3-5x)=4(x+3)
5> 2(x-3)+5x(x-1)=5x2
6> -6(1,5-2x)=3(-15+2x)
7> 14x-(2x+7)=3x+(12x-13)
8> (x-4)(x+4)-2(3x-2)=(x-4)2
9> 4(x-2)-(x-3)(2x-5)
giải giúp mik với ạ
a, \(12-2\left(1-x\right)^2=\left(3x-2\right)\left(2x-3\right)\)
\(< =>12-2\left(1-2x+x^2\right)=6x^2-9x-4x+6\)
\(< =>12-2+4x-2x^2=6x^2-13x+6\)
\(< =>10+4x-2x^2-6x^2+13x-6=0\)
\(< =>-8x^2+17x+4=0< =>\orbr{\begin{cases}x=\frac{17-\sqrt{417}}{16}\\x=\frac{17+\sqrt{417}}{16}\end{cases}}\)
b, \(10x+3-5x=4x+12< =>5x+3-4x-12=0\)
\(< =>x-9=0< =>x=9\)
c, \(11x+42-2x=100-9x-22< =>9x+42-100+9x+22=0\)
\(< =>18x+64-100=0< =>18x-36=0< =>x=\frac{36}{18}=2\)
d, \(2x-\left(3-5x\right)=4\left(x+3\right)< =>2x-3+5x=4x+12\)
\(< =>7x-3-4x-12=0< =>3x-15=0< =>x=\frac{15}{3}=5\)
e, \(2\left(x-3\right)+5x\left(x-1\right)=5x^2< =>2x-6+5x^2-5=5x^2\)
\(< =>2x-11+5x^2-5x^2=0< =>2x-11=0< =>x=\frac{11}{2}\)
f, \(-6\left(1,5-2x\right)=3\left(-15+2x\right)< =>-6\left(\frac{3}{2}-2x\right)=3\left(2x-15\right)\)
\(< =>-9+12x-6x+45=0< =>6x+36=0< =>x=-6\)
g, \(14x-\left(2x+7\right)=3x+12x-13< =>14x-2x-7=15x-13\)
\(< =>12x-7-15x+13=0< =>-3x+6=0< =>x=-2\)
h, \(\left(x-4\right)\left(x+4\right)-2\left(3x-2\right)=\left(x-4\right)^2\)
\(< =>x^2-16-6x+4=x^2-8x+16\)
\(< =>x^2-6x-12-x^2+8x-16=0\)
\(< =>2x-28=0< =>x=\frac{28}{2}=14\)
q, \(4\left(x-2\right)-\left(x-3\right)\left(2x-5\right)=?\)thiếu đề
a) giải phương trình: 8x-3=5x+12
b) giải bất phương trình sau và biểu diễn tập hợp nghiệm trên trục số: \(\dfrac{8-11x}{4}\)< 13
c) Chứng minh rằng: (\(\dfrac{x}{x^2-36}\)- \(\dfrac{x-6}{x^2+6x}\)): \(\dfrac{2x-6}{x^2+6x}\)+ \(\dfrac{x}{6-x}\)= 1
a:=>3x=15
=>x=5
b: =>8-11x<52
=>-11x<44
=>x>-4
c: \(VT=\left(\dfrac{x^2-\left(x-6\right)^2}{x\left(x+6\right)\left(x-6\right)}\right)\cdot\dfrac{x\left(x+6\right)}{2x-6}+\dfrac{x}{6-x}\)
\(=\dfrac{12x-36}{2x-6}\cdot\dfrac{1}{x-6}-\dfrac{x}{x-6}=\dfrac{6}{x-6}-\dfrac{x}{x-6}=-1\)
Giải phương trình:
a) (x+4)(x+7)(x2+11x+12) = 36(x2+11x+31)
b) (x2+3x-4)3+(2x2- 5x+3) = (3x2 - 2x- 1)3
c) (x+2)/(x+3) - (x+1)/(x-1) = 4/[(x-1)(x+3)]