Giải pt sau:
X^2+2X-2=0
giải pt sau:
\(2^{2x-3}-3.2^{x-2}+1=0\)
\(\Leftrightarrow2^{-3}.2^{2x}-3.2^{-2}.2^x+1=0\)
\(\Leftrightarrow\dfrac{1}{8}2^{2x}-\dfrac{3}{4}2^x+1=0\)
Đặt \(2^x=t>0\)
\(\Rightarrow\dfrac{1}{8}t^2-\dfrac{3}{4}t+1=0\Rightarrow\left[{}\begin{matrix}t=4\\t=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2^x=4\\2^x=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)
GIẢI CÁC PT SAU:
\(\left(x^2+5x\right)^2+2x^2+10x-24=0\)
\(\left(x^2-4x+1\right)^2+2x^2-8x-1=0\)
Lời giải:
1.
PT $\Leftrightarrow (x^2+5x)^2+2(x^2+5x)-24=0$
$\Leftrightarrow t^2+2t-24=0$ (đặt $x^2+5x=t$)
$\Leftrightarrow (t-4)(t+6)=0$
$\Rightarrow t-4=0$ hoặc $t+6=0$
Nếu $t-4=0\Leftrightarrow x^2+5x-4=0$
$\Leftrightarrow x=\frac{-5\pm \sqrt{41}}{2}$
Nếu $t+6=0$
$\Leftrightarrow x^2+5x+6=0$
$\Leftrightarrow (x+2)(x+3)=0\Rightarrow x=-2$ hoặc $x=-3$
2.
PT $\Leftrightarrow (x^2-4x+1)^2+2(x^2-4x+1)-3=0$
$\Leftrightarrow t^2+2t-3=0$ (đặt $x^2-4x+1=t$)
$\Leftrightarrow (t-1)(t+3)=0$
$\Rightarrow t-1=0$ hoặc $t+3=0$
Nếu $t-1=0\Leftrightarrow x^2-4x=0\Leftrightarrow x(x-4)=0$
$\Rightarrow x=0$ hoặc $x=4$
Nếu $t+3=0\Leftrightarrow x^2-4x+4=0$
$\Leftrightarrow (x-2)^2=0\Leftrightarrow x=2$
1. C/m pt sau vô nghiệm
x^4 - 2x^3 + 3x^2 - 2x + 1 =0
2.giải pt
(x^2-4)^2=8x + 1
1. \(x^4-2x^3+3x^2-2x+1=0\)
\(\Leftrightarrow\left(x^4-2x^3+x^2\right)+\left(x^2-2x+1\right)+x^2=0\)
\(\Leftrightarrow x^2\left(x-1\right)^2+\left(x-1\right)^2+x^2=0\)
\(\Leftrightarrow\) (x - 1)2 = 0 và x2 = 0
\(\Leftrightarrow\) x - 1 = 0 và x = 0
\(\Leftrightarrow\) x = 1 và x = 0 (vô lí)
Vậy phương trình vô nghiệm.
2. \(\left(x^2-4\right)^2=8x+1\)
\(\Leftrightarrow x^4-8x^2+16=8x+1\)
\(\Leftrightarrow x^4-8x^2-8x+15=0\)
\(\Leftrightarrow x^4-x^3+x^3-x^2-7x^2+7x-15x+15=0\)
\(\Leftrightarrow x^3\left(x-1\right)+x^2\left(x-1\right)-7x\left(x-1\right)-15\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+x^2-7x-15\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3-3x^2+4x^2-12x+5x-15\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2\left(x-3\right)+4x\left(x-3\right)+5\left(x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x^2+4x+5\right)=0\)
\(\Leftrightarrow\) x - 1 = 0 hoặc x - 3 = 0 hoặc x2 + 4x + 5 = 0
1) x - 1 = 0 \(\Leftrightarrow\) x = 1
2) x - 3 = 0 \(\Leftrightarrow\) x = 3
3) \(x^2+4x+5=0\left(\text{loại vì }x^2+4x+5=\left(x+2\right)^2+1>0\forall x\right)\)
Vậy tập nghiệm của pt là S = {1;3}.
giải pt: x^5 + 2x^4 +3x^3 + 3x^2 + 2x +1=0
giải pt: x^4 + 3x^3 - 2x^2 +x - 3=0
ta có : x^5+2x^4+3x^3+3x^2+2x+1=0
\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0
\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0
\(\Leftrightarrow\)x^4(x+1)+x^3(x+1)+2x^2(x+1)+x(x+1)+(x+1)=0
\(\Leftrightarrow\)(x+1)(x^4+x^3+2x^2+x+1)=0
\(\Leftrightarrow\)(x+1)(x^4+x^3+x^2+x^2+x+1)=0
\(\Leftrightarrow\)(x+1)[x^2(x^2+x+1)+(x^2+x+1)]=0
\(\Leftrightarrow\)(x+1)(x^2+x+1)(x^2+1)=0
VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)
\(\Rightarrow\)x+1=0
\(\Rightarrow\)x=-1
CÒN CÂU B TỰ LÀM (02042006)
b: x^4+3x^3-2x^2+x-3=0
=>x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0
=>(x-1)(x^3+4x^2+2x+3)=0
=>x-1=0
=>x=1
giải pt và bất pt sau:
a.5|2x-1|-3=7
b.(2x+3)(x-2)-x^2+4=0
c. 2x-3/2<1-3x/-5
a, \(5\left|2x-1\right|-3=7\Leftrightarrow5\left|2x-1\right|=10\Leftrightarrow\left|2x-1\right|=2\)
TH1 : \(2x-1=2\Leftrightarrow x=\frac{3}{2}\)
TH2 : \(2x-1=-2\Leftrightarrow x=-\frac{1}{2}\)
b, \(\left(2x+3\right)\left(x-2\right)-x^2+4=0\Leftrightarrow\left(2x+3\right)\left(x-2\right)-\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x+3-x-2\right)=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow x=-1;x=2\)
c, \(\frac{2x-3}{2}< \frac{1-3x}{-5}\Leftrightarrow\frac{2x-3}{2}+\frac{1-3x}{5}< 0\)
\(\Leftrightarrow\frac{10x-15+2-6x}{10}< 0\Rightarrow4x-13< 0\Leftrightarrow x< \frac{13}{4}\)
GIẢI CÁC PT SAU:
\(\sqrt{x^2+5x+1}=\sqrt{x+1}\)
\(\sqrt{x^2+2x+4}=\sqrt{2-x}\)
\(\sqrt{2x+4}-\sqrt{2-x}=0\)
Lời giải:
1. ĐKXĐ: $x\geq \frac{-5+\sqrt{21}}{2}$
PT $\Leftrightarrow x^2+5x+1=x+1$
$\Leftrightarrow x^2+4x=0$
$\Leftrightarrow x(x+4)=0$
$\Rightarrow x=0$ hoặc $x=-4$
Kết hợp đkxđ suy ra $x=0$
2. ĐKXĐ: $x\leq 2$
PT $\Leftrightarrow x^2+2x+4=2-x$
$\Leftrightarrow x^2+3x+2=0$
$\Leftrightarrow (x+1)(x+2)=0$
$\Leftrightarrow x+1=0$ hoặc $x+2=0$
$\Leftrightarrow x=-1$ hoặc $x=-2$
3.
ĐKXĐ: $-2\leq x\leq 2$
PT $\Leftrightarrow \sqrt{2x+4}=\sqrt{2-x}$
$\Leftrightarrow 2x+4=2-x$
$\Leftrightarrow 3x=-2$
$\Leftrightarrow x=\frac{-2}{3}$ (tm)
GIẢI PT SAU:
\(\sqrt{3x^2-2x+6}+3-2x=0\)
\(\sqrt{x+1}+\sqrt{x-1}=4\)
a, ĐKXĐ: ...
\(\sqrt{3x^2-2x+6}+3-2x=0\)
\(\Leftrightarrow\sqrt{3x^2-2x+6}=2x-3\)
\(\Leftrightarrow3x^2-2x+6=4x^2-12x+9\)
\(\Leftrightarrow4x^2-10x+3=0\)
.....
b, ĐKXĐ: ...
\(\sqrt{x+1}+\sqrt{x-1}=4\\ \Leftrightarrow x+1+x-1+2\sqrt{\left(x+1\right)\left(x-1\right)}=16\\ \Leftrightarrow2\sqrt{x^2-1}=16-2x\\ \Leftrightarrow\sqrt{x^2-1}=8-x\\ \Leftrightarrow x^2-1=64-16x+x^2\\ \Leftrightarrow65-16x=0\\ \Leftrightarrow x=\dfrac{65}{16}\)
Giải pt sau
(x^2 + 1)^2 + 3x(x^2 + 1) + 2x^2 = 0
Giải pt sau: x/2x-6 + x/2x+2 - 2x/(x+1)(x-3) =0
cái này dễ đợi tí mình giải cho, gõ đáp số mất khá nhiều thời gian
\(\frac{x}{2x-6}+\frac{x}{2x+2}-\frac{2x}{\left(x+1\right).\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{x}{2.\left(x-3\right)}+\frac{x}{2.\left(x+1\right)}-\frac{2x}{\left(x+1\right).\left(x-3\right)}=0\)
\(\Leftrightarrow\frac{\left(x+1\right).x}{2.\left(x-3\right).\left(x+1\right)}+\frac{x.\left(x-3\right)}{2.\left(x+1\right).\left(x-3\right)}-\frac{4x}{2.\left(x+1\right).\left(x-3\right)}=0\)
tự làm tiếp nha bạn :)))