tìm x
1, \(x.\left(x-4\right)-\left(x^2-8\right)=0\)
2, \(x^3-3x^2+3x-1=0\)
3, \(x^3-6x^2+12x-8=0\)
4, \(8x^3-12x^2+6x-1=0\)
các bạn giúp mk vs ạ
Giải các phương trình sau:
a \(x^2-11=0\)
b \(x^2-12x+52=0\)
c \(x^2-3x-28=0\)
d \(x^2-11x+38=0\)
e \(6x^2+71x+175=0\)
f \(x^2-\left(\sqrt{2}+\sqrt{8}\right)x+4=0\)
g\(\left(1+\sqrt{3}\right)x^2-\left(2\sqrt{3}+1\right)x+\sqrt{3}=0\)
a.
$x^2-11=0$
$\Leftrightarrow x^2=11$
$\Leftrightarrow x=\pm \sqrt{11}$
b. $x^2-12x+52=0$
$\Leftrightarrow (x^2-12x+36)+16=0$
$\Leftrightarrow (x-6)^2=-16< 0$ (vô lý)
Vậy pt vô nghiệm.
c.
$x^2-3x-28=0$
$\Leftrightarrow x^2+4x-7x-28=0$
$\Leftrightarrow x(x+4)-7(x+4)=0$
$\Leftrightarrow (x+4)(x-7)=0$
$\Leftrightarrow x+4=0$ hoặc $x-7=0$
$\Leftrightarrow x=-4$ hoặc $x=7$
d.
$x^2-11x+38=0$
$\Leftrightarrow (x^2-11x+5,5^2)+7,75=0$
$\Leftrightarrow (x-5,5)^2=-7,75< 0$ (vô lý)
Vậy pt vô nghiệm
e.
$6x^2+71x+175=0$
$\Leftrightarrow 6x^2+21x+50x+175=0$
$\Leftrightarrow 3x(2x+7)+25(2x+7)=0$
$\Leftrightarrow (3x+25)(2x+7)=0$
$\Leftrightarrow 3x+25=0$ hoặc $2x+7=0$
$\Leftrightarrow x=-\frac{25}{3}$ hoặc $x=-\frac{7}{2}$
f.
$x^2-(\sqrt{2}+\sqrt{8})x+4=0$
$\Leftrightarrow x^2-\sqrt{2}x-2\sqrt{2}x+4=0$
$\Leftrightarrow x(x-\sqrt{2})-2\sqrt{2}(x-\sqrt{2})=0$
$\Leftrightarrow (x-\sqrt{2})(x-2\sqrt{2})=0$
$\Leftrightarrow x-\sqrt{2}=0$ hoặc $x-2\sqrt{2}=0$
$\Leftrightarrow x=\sqrt{2}$ hoặc $x=2\sqrt{2}$
g.
$(1+\sqrt{3})x^2-(2\sqrt{3}+1)x+\sqrt{3}=0$
$\Leftrightarrow (1+\sqrt{3})x^2-(1+\sqrt{3})x-(\sqrt{3}x-\sqrt{3})=0$
$\Leftrightarrow (1+\sqrt{3})x(x-1)-\sqrt{3}(x-1)=0$
$\Leftrightarrow (x-1)[(1+\sqrt{3})x-\sqrt{3}]=0$
$\Leftrightarrow x-1=0$ hoặc $(1+\sqrt{3})x-\sqrt{3}=0$
$\Leftrightarrow x=1$ hoặc $x=\frac{3-\sqrt{3}}{2}$
Giải phương trình
\(x^4+x^3-12x^2=0\)
\(x^4-3x^3+3x^2-x=0\)
\(6x^4+5x^3-38x^2+5x+6=0\)
\(\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24=0\)
\(\left(x-1\right)^3+\left(2x+3\right)^3=27x^3+8\)
x(x + 1). (x - 1 ). ( x + 2) = 24
a: \(\Leftrightarrow x^2\left(x^2+x-12\right)=0\)
\(\Leftrightarrow x^2\left(x+4\right)\left(x-3\right)=0\)
hay \(x\in\left\{0;-4;3\right\}\)
d: \(\left(x^2+5x\right)^2-2\left(x^2+5x\right)-24=0\)
\(\Leftrightarrow\left(x^2+5x-6\right)\left(x^2+5x+4\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-1\right)\left(x+1\right)\left(x+4\right)=0\)
hay \(x\in\left\{-6;1;-1;-4\right\}\)
f: \(x\left(x+1\right)\left(x-1\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)
\(\Leftrightarrow\left(x^2+x\right)^2-2\left(x^2+x\right)-24=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)
hay \(x\in\left\{-3;2\right\}\)
Tìm x biết:
\(\left(2x-3\right)^2-\left(3x+2\right)^2=5x\left(2-x\right)\)
\(8x^3-12x^2+6x-1=0\)
a/ \(\left(2x-3\right)^2-\left(3x+2\right)^2=5x\left(2-x\right)\)
<=> \(\left(2x-3-3x-2\right)\left(2x-3+3x+2\right)=5x\left(2-x\right)\)
<=> \(\left(-x-5\right)\left(5x-1\right)=5x\left(2-x\right)\)
<=> \(-5x^2-25x+x+5=10x-5x^2\)
<=> \(10x+25x-x=5\)
<=> \(34x=5\)
<=> \(x=\frac{5}{34}\)
b/ pt <=> \(2^3x^3-3.2^2.x^2.1+3.2.x.1^2-1^3=0\)
<=> \(\left(2x-1\right)^3=0\)
<=> 2 x - 1 = 0
<=> x = 1/2.
a,\(8x^3-12x^2+6x-5=0\Leftrightarrow8\left(x^3-\frac{3}{2}x^2+\frac{3}{4}x-\frac{1}{8}\right)-4=0\)
\(\Leftrightarrow8\left(x-\frac{1}{2}\right)^3=4\Leftrightarrow\left(x-\frac{1}{2}\right)^3=\frac{1}{2}\Leftrightarrow x=\frac{1}{\sqrt[3]{2}}+\frac{1}{2}\)
Tìm x:
a) \(3x\left(3x-8\right)-9x^2+8=0\)
b)\(6x-15-x\left(5-2x\right)=0\)
c) \(x^3-16x=0\)
d) \(2x^2+3x-5=0\)
e) \(3x^2-x\left(3x-6\right)=36\)
f) \(\left(x+2\right)^2-\left(x-5\right)\left(x+1\right)=17\)
g) \(\left(x-4\right)^2-x\left(x+6\right)=9\)
h) \(4x\left(x-1000\right)-x+1000=0\)
i) \(x^2-36=0\)
j) \(x^2y-2+x+x^2-2y+xy=0\)
k) \(x\left(x+1\right)-\left(x-1\right).\left(2x-3\right)=0\)
l) \(3x^3-27x=0\)
tìm x biét
a, \(\left(x+2\right)^2-9=0\)
b, \(\left(x+2\right)^2-x^2+4=0\)
c, \(\left(x-3\right)^2=\left(2-3x\right)^2\)
d, \(x^3-6x^2+12x-8=0\)
e. \(\left(x +1\right)^3+\left(x-2\right)^3=0\)
a, (a, (x + 2)2 - 9 = 0
⇒ (x + 2)2 = 0 + 9 = 9
⇒ (x + 2)2 = \(\left(\pm3\right)^2\)
⇒ x + 2 = \(\pm3\)
\(\Rightarrow\left\{{}\begin{matrix}x+2=3\\x+2=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3-2\\x=-3-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
Vậy x ∈ {1; -5}
b, \(\left(x+2\right)^2-x^2+4=0\)
⇒ x2 + 4x + 4 - x2 + 4 =0
⇒ 4x + 8 = 0
⇒ 4 (x + 2) = 0
⇒ x + 2 = 0
⇒ x = 0 - 2
⇒ x = -2
Vậy x = -2
c, (x - 3)2 = (2 - 3x)2
⇒ (x - 3)2 - (2 - 3x)2 = 0
⇒ x2 - 6x + 9 - 4 + 12x - 9x2 = 0
⇒ 6x - 8x2 + 5 = 0
⇒2 \(\left(3x-4x^2+\dfrac{5}{2}\right)\)= 0
⇒ 3x - 4x2 + \(\dfrac{5}{2}\) = 0
⇒ - (4x2- 3x + \(\dfrac{9}{16}+\dfrac{31}{16}\)) = 0
⇒ - (4x2 - 3x + \(\dfrac{9}{16}\)) - \(\dfrac{31}{16}\) = 0
⇒ - (2x - \(\dfrac{3}{4}\))2 = \(\dfrac{31}{16}\) (vô lí)
Vậy x ∈ ∅
e: \(\Leftrightarrow\left(x+1+x-2\right)\left(x^2+2x+1-x^2+x+2+x^2-4x+4\right)=0\)
=>2x-1=0
=>x=1/2
a: \(\Leftrightarrow\left(x+2+3\right)\left(x+2-3\right)=0\)
=>(x+5)(x-1)=0
=>x=1 hoặc x=-5
b: \(\Leftrightarrow x^2+4x+4-x^2+4=0\)
=>4x+8=0
=>x=-2
c: \(\Leftrightarrow\left(3x-2-x+3\right)\left(3x-2+x-3\right)=0\)
=>(2x+1)(4x-5)=0
=>x=5/4 hoặc x=-1/2
Giải các phương trình
1,\(x\left(x-1\right)=2\left(x-1\right)\)
2, \(\left(x+2\right)\left(2x-3\right)=x^2-4\)
3, \(x^2+3x+2=0\)
4, \(5x^2+5x+3=0\)
5, \(x^3+x^2-12x=0\)
1, x(x-1)=2(x-1)
<=> x(x-1)-2(x-1)=0
<=> (x-2)(x-1)=0
<=>x=2 hoặc x=1
vậy ...
2, (x+2)(2x-3)=x^2 -4
<=>(x+2)(2x-3)=(x-2)(x+2)
<=> (x+2)(2x-3)-(x-2)(x+2)=0
<=> (x+2)(2x-3-x+2)=0
<=> x=-2 hoặc x=1
vây...
3,x^2 +3x +2=0
<=> x^2 +x+2x+2=0
<=>(x+2)(x+1)=0
<=> x=-2 hoặc x=-1
vậy ...
5, x^3+x^2-12x =0
<=> x(x^2+x-12)=0
<=>x(x^2-3x+4x-12)=0
<=>x(x+4)(x-3)=0
<=> x=0 hoặc x=-4 hoặc x=3
vậy ...
Giải các phương trình sau:
a \(x^4=5x^2+2x-3\)
b \(x^4=6x^2+12x+10\)
c \(3x^3+3x^2+3x=-1\)
d \(8x^3-12x^2+6x-5=0\)
Giải phương trình:
a, \(2x^3+3x^2+6x+5=0\) b, \(4x^4+12x^3+5x^2-6x-15=0\) c, \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)=40\)2x3 + 3x2 + 6x + 5 = 02
<=> 2x3 + x2 + 5x + 2x2 + x + 5 = 0
<=> x(2x2 + x + 5) + (2x2 + x + 5) = 0
<=> (2x2 + x + 5)(x + 1) = 0
<=> x + 1 = 0 (vì 2x2 + x + 5 \(\ge\) 4,875 > 0 \(\forall\) x)
<=> x = - 1
Vậy tập nghiệm của pt là \(S=\left\{-1\right\}\)
b) 4x4 + 12x3 + 5x2 - 6x - 15 = 0
<=> 4x4 + 10x3 + 2x3 + 5x2 - 6x - 15 = 0
<=> 2x3(2x + 5) + x2(2x + 5) - 3(2x + 5) = 0
<=> (2x + 5)(2x3 + x2 - 3) = 0
<=> (2x + 5)(2x3 - 2x2 + 3x2 - 3) = 0
<=> (2x + 5)(x - 1)(2x2 + 3x + 3) = 0
<=> (2x + 5)(x - 1)[x2 + (x + 3/2)2 + 3/4]= 0
Mà x2 + (x + 3/2)2 + 3/4 > 0\(\forall x\)
\(\Rightarrow\left[\begin{matrix}2x+5=0\\x-1=0\end{matrix}\right.\)\(\Leftrightarrow\left[\begin{matrix}x=-\frac{5}{2}\\x=1\end{matrix}\right.\)
Vậy ...
c) (x + 1)(x + 2)(x + 3)(x + 4)(x + 5) = 40
Khai triển hết ra ta được:
x5 + 15x4 + 85x3 + 225x2 + 274x + 80 = 0 (*) (đây là phương trình đối xứng bậc 5)
<=> (x + 1)(x4 + 14x3 + 71x2 + 154x + 80) = 0
=> x = -1 hoặc x4 + 14x3 + 71x2 + 154x + 80 = 0
Bây giờ ta cần giải pt x4 + 14x3 + 71x2 + 154x + 80 = 0 (đây là pt đối xứng bậc chẵn)
Dễ thấy x = 0 không là nghiệm của pt
Chia cả 2 vế của pt cho x2 (x2 khác 0)
Tới đây tự lm tiếp nhé!