Giải phương trình:
a) (4x - 3)3 + ( 3x - 2)3 = ( 7x - 5)3
b) x3 -2x2 - x - 6 = 0
Giải phương trình:
a) 2x2 + 3x - 27 =0
b) -10x2 + x + 3 =0
c) -x3 + x2 + 4 =0
d) x3 - 4x2 - 8x +8 =0
a: =>2x^2+9x-6x-27=0
=>x(2x+9)-3(2x+9)=0
=>(2x+9)(x-3)=0
=>x=3 hoặc x=-9/2
b: =>-10x^2+6x-5x+3=0
=>-2x(5x-3)-(5x-3)=0
=>(5x-3)(-2x-1)=0
=>x=-1/2 hoặc x=5/3
c: =>-x^3+2x^2-x^2+4=0
=>-x^2(x-2)-(x-2)(x+2)=0
=>(x-2)(-x^2-x-2)=0
=>x-2=0
=>x=2
d: =>(x^3+8)-4x(x+2)=0
=>(x+2)(x^2-2x+4)-4x(x+2)=0
=>(x+2)(x^2-6x+4)=0
=>x=-2 hoặc \(x=3\pm\sqrt{5}\)
giải phương trình:
a)(2x-3)(2x+3)=4x(x-5)-3x
b)(2x+1)(4x-3)=4x^2-1
c)3x/x-2+x/5-x-2x^2+5/x^2-7x+10=0
\(a)PT\Leftrightarrow4x^2-9-4x^2+20x+3x=0.\\ \Leftrightarrow23x=9.\\ \Leftrightarrow x=\dfrac{9}{23}.\\ b)PT\Leftrightarrow\left(2x+1\right)\left(4x-3\right)-\left(2x+1\right)\left(2x-1\right)=0.\\\Leftrightarrow\left(2x+1\right)\left(4x-3-2x+1\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(2x-2\right)=0.\\ \Leftrightarrow\left(2x+1\right)\left(x-1\right)=0. \)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}.\\x=1.\end{matrix}\right.\)
Bài 1 : giải phương trình
a) (8x + 3)(2x - 1) = (2x - 1)2
b) (x - 5)2 - 36 = 0
c) (4x - 3)2 - 4(x + 3)2
d) x3 - 3x -2 = 0
e) x3 + 2x2 - 4x - 8 = 0
Bài 1 giải phương trình:
a) (4x2+4x+1)-x2=0
b) x2-2x+1=4
c) x2-5x+6=0
Bài 2: giải phương trình
a) \(\dfrac{2x-5}{x+5}\)= 3
b) \(\dfrac{5}{3x+2}\)= 2x-1
c) \(\dfrac{x^2-6}{x}\)= x+\(\dfrac{3}{2}\)
d) \(\dfrac{1}{x-2}\)+3= \(\dfrac{x-3}{2-x}\)
e) \(\dfrac{3x-2}{x+7}\)=\(\dfrac{6x+1}{2x-3}\)
f) \(\dfrac{x-2}{x+2}\) - \(\dfrac{3}{x-2}\)=\(\dfrac{2\left(x-11\right)}{x^2-4}\)
Bài 1:
a.
$(4x^2+4x+1)-x^2=0$
$\Leftrightarrow (2x+1)^2-x^2=0$
$\Leftrightarrow (2x+1-x)(2x+1+x)=0$
$\Leftrightarrow (x+1)(3x+1)=0$
$\Rightarrow x+1=0$ hoặc $3x+1=0$
$\Rightarrow x=-1$ hoặc $x=-\frac{1}{3}$
b.
$x^2-2x+1=4$
$\Leftrightarrow (x-1)^2=2^2$
$\Leftrightarrow (x-1)^2-2^2=0$
$\Leftrightarrow (x-1-2)(x-1+2)=0$
$\Leftrightarrow (x-3)(x+1)=0$
$\Leftrightarrow x-3=0$ hoặc $x+1=0$
$\Leftrightarrow x=3$ hoặc $x=-1$
c.
$x^2-5x+6=0$
$\Leftrightarrow (x^2-2x)-(3x-6)=0$
$\Leftrightarrow x(x-2)-3(x-2)=0$
$\Leftrightarrow (x-2)(x-3)=0$
$\Leftrightarrow x-2=0$ hoặc $x-3=0$
$\Leftrightarrow x=2$ hoặc $x=3$
2c.
ĐKXĐ: $x\neq 0$
PT $\Leftrightarrow x-\frac{6}{x}=x+\frac{3}{2}$
$\Leftrightarrow -\frac{6}{x}=\frac{3}{2}$
$\Leftrightarrow x=-4$ (tm)
2d.
ĐKXĐ: $x\neq 2$
PT $\Leftrightarrow \frac{1+3(x-2)}{x-2}=\frac{3-x}{x-2}$
$\Leftrightarrow \frac{3x-5}{x-2}=\frac{3-x}{x-2}$
$\Rightarrow 3x-5=3-x$
$\Leftrightarrow 4x=8$
$\Leftrightarrow x=2$ (không tm)
Vậy pt vô nghiệm.
2f.
ĐKXĐ: $x\neq \pm 2$
PT $\Leftrightarrow \frac{(x-2)^2-3(x+2)}{(x+2)(x-2)}=\frac{2(x-11)}{(x-2)(x+2)}$
$\Rightarrow (x-2)^2-3(x+2)=2(x-11)$
$\Leftrightarrow x^2-4x+4-3x-6=2x-22$
$\Leftrightarrow x^2-7x-2=2x-22$
$\Leftrightarrow x^2-9x+20=0$
$\Leftrightarrow (x-4)(x-5)=0$
$\Leftrightarrow x-4=0$ hoặc $x-5=0$
$\Leftrightarrow x=4$ hoặc $x=5$ (tm)
Bài 1: Giải phương trình:
a) 2x2-6=0
b)x3-5x2+6x=0
c) \(\dfrac{3}{x-1}+\dfrac{2}{x^{2^{ }}+x+1}=\dfrac{3x^2}{x^3-1}\)
Bài 2: Tìm MIN của
A=\(\dfrac{2}{-x^2-2x-2}\)
Bài 2:
\(A=\dfrac{2}{-x^2-2x-2}=\dfrac{-2\left(-x^2-2x-2\right)-2x^2-4x-2}{-x^2-2x-2}\) \(=-2+\dfrac{2\left(x+1\right)^2}{-x^2-2x-2}\ge-2\)
Dấu bằng xảy ra \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)
Vậy \(A_{Min}=-2\) khi \(x=-1\)
Bài 1:
a) Ta có: \(2x^2-6=0\)
\(\Leftrightarrow2x^2=6\)
\(\Leftrightarrow x^2=3\)
hay \(x\in\left\{\sqrt{3};-\sqrt{3}\right\}\)
Vậy: \(S=\left\{\sqrt{3};-\sqrt{3}\right\}\)
giải các phương trình:
a)(x2+x+1)2-2x2-2x=5
b)\(\dfrac{1}{\left(x-1\right)\left(x-3\right)}\)+x2-4x+5=0
Giải phương trình:
a) \(\dfrac{3x-2}{x^2-12x+20}-\dfrac{4x+3}{x^2+6x-16}=\dfrac{7x+11}{x^2-2x-80}\)
b) \(\dfrac{2x-5}{x^2+5x-36}-\dfrac{x-6}{x^2+3x-28}=\dfrac{x+8}{x^2+16x+63}\)
a: \(\Leftrightarrow\dfrac{3x-2}{\left(x-2\right)\left(x-10\right)}-\dfrac{4x+3}{\left(x+8\right)\left(x-2\right)}=\dfrac{8x+11}{\left(x-10\right)\left(x+8\right)}\)
=>(3x-2)(x+8)-(4x+3)(x-10)=(8x+11)(x-2)
=>3x^2+24x-2x-16-4x^2+40x-3x+30=8x^2-16x+11x-22
=>-x^2+59x+14-8x^2+5x+22=0
=>-9x^2+54x+36=0
=>x^2-6x-4=0
=>\(x=3\pm\sqrt{13}\)
b: \(\Leftrightarrow\dfrac{2x-5}{\left(x+9\right)\left(x-4\right)}-\dfrac{x-6}{\left(x+7\right)\left(x-4\right)}=\dfrac{x+8}{\left(x+9\right)\left(x+7\right)}\)
=>(2x-5)(x+7)-(x-6)(x+9)=(x+8)(x-4)
=>2x^2+14x-5x-35-x^2-9x+6x+54=x^2+4x-32
=>x^2+6x+19=x^2+4x-32
=>2x=-51
=>x=-51/2
`4x=2+xx+1x<=>4x=2+3x<=>4x-3x=2<=>1x=2<=>x=2`
Giải các phương trình sau:
g/ x(x + 3)(x – 3) – (x + 2)(x2 – 2x + 4) = 0
h/ (3x – 1)(x2 + 2) = (3x – 1)(7x – 10)
i/ (x + 2)(3 – 4x) = x2 + 4x + 4
k/ x(2x – 7) – 4x + 14 = 0
m/ x2 + 6x – 16 = 0
n/ 2x2 + 5x – 3 = 0
\(m,x^2+6x-16=0\)
\(\Leftrightarrow x^2-2x+8x-16=0\)
\(\Leftrightarrow x\left(x-2\right)+8\left(x-2\right)=0\)
\(\Leftrightarrow\left(x+8\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-8\\x=2\end{matrix}\right.\)
\(n,2x^2+5x-3=0\)
\(\Leftrightarrow2x^2-x+6x-3=0\)
\(\Leftrightarrow x\left(2x-1\right)+3\left(2x-1\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\dfrac{1}{2}\end{matrix}\right.\)
\(k,x\left(2x-7\right)-4x+14=0\)
\(\Leftrightarrow2x^2-4x-7x+14=0\)
\(\Leftrightarrow2x\left(x-2\right)-7\left(x-2\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-7=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=2\end{matrix}\right.\)
Giúp vs ạ
Bài 1 giải các bất phương trình sau
a.x2 - x - 6 = 0
b.2x2 - 7x + 5 < 0
c.3x2 - 9x + 6 ≥ 0
d.2x2 - 5x + 3 < 0
Bài 2 Giải phương trình sau
A.√x2 + x + 5 = √2x2 - 4x + 1
B.√11x2 -14x - 12 = √3x2 + 4x - 7
Bài 2:
a: =>2x^2-4x+1=x^2+x+5
=>x^2-5x-4=0
=>\(x=\dfrac{5\pm\sqrt{41}}{2}\)
b: =>11x^2-14x-12=3x^2+4x-7
=>8x^2-18x-5=0
=>x=5/2 hoặc x=-1/4