Giải PT: x(x+2)(x2+2x+5)=6
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a: =>4x-3x=1-2
=>x=-1
b: =>3x=12
=>x=4
c: =>2(x^2-6)=x(x+3)
=>2x^2-12-x^2-3x=0
=>x^2-3x-12=0
=>\(x=\dfrac{3\pm\sqrt{57}}{2}\)
a: =>4x-3x=1-2
=>x=-1
b: =>3x=12
=>x=4
c: =>2(x^2-6)=x(x+3)
=>2x^2-12=x^2+3x
=>x^2-3x-12=0
=>\(x=\dfrac{3\pm\sqrt{57}}{2}\)
Ta có: \(\dfrac{2x}{x^2-x+1}-\dfrac{x}{x^2+x+1}=\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{2x\left(x^2+x+1\right)-x\left(x^2-x+1\right)}{\left(x^2-x+1\right)\left(x^2+x+1\right)}=\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{2x^3+2x^2+2x-x^3+x^2-x}{\left(x^2-x+1\right)\left(x^2+x+1\right)}=\dfrac{5}{3}\)
\(\Leftrightarrow\dfrac{x^3+3x^2+x}{\left(x^2+1\right)^2-x^2}=\dfrac{5}{3}\)
\(\Leftrightarrow3x^3+9x^2+3x=5\left(x^4+2x^2+1-x^2\right)\)
\(\Leftrightarrow3x^3+9x^2+3x=5x^4+5x^2+5\)
\(\Leftrightarrow5x^4+5x^2+5-3x^3-9x^2-3x=0\)
\(\Leftrightarrow5x^4-3x^3-4x^2-3x+5=0\)
\(\Leftrightarrow5x^4-5x^3+2x^3-2x^2-2x^2+2x-5x+5=0\)
\(\Leftrightarrow5x^3\left(x-1\right)+2x^2\left(x-1\right)-2x\left(x-1\right)-5\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x^3+2x^2-2x-5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(5x^3-5x^2+7x^2-7x+5x-5\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[5x^2\left(x-1\right)+7x\left(x-1\right)+5\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)^2\cdot\left(5x^2+7x+5\right)=0\)
mà \(5x^2+7x+5>0\forall x\)
nên x-1=0
hay x=1
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.
ĐKXĐ: \(x\ge-5\)
\(\Leftrightarrow\left(x+7\right)^2-2\left(x+7\right)\sqrt{x+5}+x+5-16=0\)
\(\Leftrightarrow\left(x+7-\sqrt{x+5}\right)^2-16=0\)
\(\Leftrightarrow\left(x+7-\sqrt{x+5}-4\right)\left(x+7-\sqrt{x+5}+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=x+3\left(x\ge-3\right)\\\sqrt{x+5}=x+11\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=x^2+6x+9\\x+5=x^2+22x+121\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+4=0\\x^2+21x+116=0\left(vn\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-4< -3\left(l\right)\end{matrix}\right.\)
a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
Ta có: \(\dfrac{x+1}{x-2}-\dfrac{5}{x+2}=\dfrac{12}{x^2-4}+1\)
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\dfrac{5\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}=\dfrac{12}{\left(x-2\right)\left(x+2\right)}+\dfrac{x^2-4}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2+3x+2-5x+10=12+x^2-4\)
\(\Leftrightarrow x^2-2x+12-8-x^2=0\)
\(\Leftrightarrow-2x+4=0\)
\(\Leftrightarrow-2x=-4\)
hay x=2(loại)
Vậy: \(S=\varnothing\)
b) Ta có: \(\left|2x+6\right|-x=3\)
\(\Leftrightarrow\left|2x+6\right|=x+3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+6=x+3\left(x\ge-3\right)\\-2x-6=x+3\left(x< -3\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-x=3-6\\-2x-x=3+6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\left(nhận\right)\\x=-3\left(loại\right)\end{matrix}\right.\)
Vậy: S={-3}
Cho pt x²+(a-1)x-6=0 a) Giải pt với a =6 b) Tìm a để pt có 2 nghiệm x1,x2 thoả mãn x1²+x2²-3x1.x2=34
a: \(\Leftrightarrow\left(x+6\right)\left(x-1\right)=0\)
=>x=-6 hoặc x=1
x(x + 2)(x2 + 2x + 5) = 6
\(\Leftrightarrow\)(x2 + 2x)(x2 + 2x + 5) = 6
Đặt x2 + 2x = y; ta có
y(y + 5) = 6
\(\Leftrightarrow\)y2 + 5y - 6 = 0
\(\Leftrightarrow\)y2 - y + 6y - 6 = 0
\(\Leftrightarrow\)y(y - 1)+ 6(y - 1) = 0
\(\Leftrightarrow\)(y - 1)(y + 6) = 0
\(\Leftrightarrow\)\(\orbr{\begin{cases}y-1=0\\y+6=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}y=1\\y=-6\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2+2x=1\\x^2+2x=-6\end{cases}}\)
Mk lm đc thế thôi, bn lm tiếp nha