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1.
\(DK:x\ge2\)
PT
\(\Leftrightarrow\left(2+x\right)\sqrt{x-2}-\left(x+2\right)\left(x-2\right)\)
\(\Leftrightarrow\left(x+2\right)\sqrt{x-2}\left(1-\sqrt{x-2}\right)=0\)
Cho này thì ok ròi nhé
2.
\(DK:x\le\frac{5}{2}\)
Xet \(x\in\left[0;\frac{5}{2}\right]\)
PT
\(\Leftrightarrow x^2-4x=5-2x\)
\(\Leftrightarrow x^2-2x-5=0\)
Ta co:
\(\Delta^`=\left(-1\right)^2-1.\left(-5\right)=6>0\)
\(\Rightarrow\hept{\begin{cases}x_1=1+\sqrt{6}\left(l\right)\\x_2=1-\sqrt{6}\left(l\right)\end{cases}}\)
Xet \(x\le0\)
PT
\(4x-x^2=5-2x\)
\(\Leftrightarrow x^2-6x+5=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=1\left(l\right)\\x=5\left(l\right)\end{cases}}\)
Vay PT vo nghiem
Pt có 2 nghiệm trái dấu khi:
\(ac< 0\Leftrightarrow2\left(m+3\right)< 0\)
\(\Rightarrow m< -3\)
\(|2x^2-3x+4|-|2x-x^2-1|=0\)
\(\Leftrightarrow|2x^2-3x+4|=|2x-x^2-1|\)
\(\Leftrightarrow\orbr{\begin{cases}2x^2-3x+4=2x-x^2-1\\2x^2-3x+4=-2x+x^2+1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x^2-3x+4-2x+x^2+1=0\\2x^2-3x+4+2x-x^2-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x^2-5x+5=0\\x^2-x+3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3\left(x^2-\frac{5}{3}x+\frac{25}{9}-\frac{25}{9}+\frac{5}{3}\right)=0\\x^2-2.x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3\left(x-\frac{5}{3}^2\right)-\frac{10}{3}=0\\\left(x-\frac{1}{2}\right)^2+\frac{11}{4}>0\left(Loai\right)\end{cases}}\)
\(\Leftrightarrow\left(x\sqrt{3}-\frac{5\sqrt{3}}{3}\right)^2-\left(\frac{\sqrt{30}}{3}\right)^2=0\)
\(\Leftrightarrow\left(x\sqrt{3}-\frac{5\sqrt{3}}{3}-\frac{\sqrt{30}}{3}\right)\left(x\sqrt{3}-\frac{5\sqrt{3}}{3}+\frac{\sqrt{30}}{3}\right)=0\)
\(\Leftrightarrow\left(x\sqrt{3}-\frac{\sqrt{30}+5\sqrt{3}}{3}\right)\left(x\sqrt{3}+\frac{\sqrt{30}-5\sqrt{3}}{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x\sqrt{3}-\frac{\sqrt{30}+5\sqrt{3}}{3}=0\\x\sqrt{3}+\frac{\sqrt{30}-5\sqrt{3}}{3}=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{5+\sqrt{10}}{3}\\x=\frac{5-\sqrt{10}}{3}\end{cases}}\)
Vậy ...
\(\left|2x^2-3x+4\right|-\left|2x-x^2-1\right|=0\)
\(\Leftrightarrow\left|2x^2-3x+4\right|=\left|2x-x^2-1\right|\)
\(\Leftrightarrow\orbr{\begin{cases}2x^2-3x+4=2x-x^2-1\\2x^2-3x+4=x^2-2x+1\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}3x^2-5x+5=0\\x^2-x+3=0\end{cases}}\)
\(TH1:3x^2-5x+5=0\)
Ta có: \(\Delta=5^2-4.3.5=-35< 0\)(vô nghiệm)
\(TH2:x^2-x+3=0\)
Ta có: \(\Delta=1^2-4.1.3=-11< 0\)(vô nghiệm)
Vậy pt vô nghiệm
\(x^2-x+\sqrt{x}\left(6-2x\right)-3=0\) (ĐKXĐ : \(3< x\le\frac{1+\sqrt{13}}{2}\))
\(\Leftrightarrow\left(6-2x\right)\left(\sqrt{x}+x-1\right)+3x^2-9x+3=0\)
\(\Leftrightarrow\left(6-2x\right)\left[\left(\sqrt{x}\right)-\left(1-x\right)\right]+3\left(x^2-3x+1\right)=0\)
\(\Leftrightarrow\left(6-2x\right).\frac{x-\left(1-x\right)^2}{\sqrt{x}+1-x}+3\left(x^2-3x+1\right)=0\)
\(\Leftrightarrow\left(6-2x\right)\frac{-x^2+3x-1}{\sqrt{x}+1-x}+3\left(x^2-3x+1\right)=0\)
\(\Leftrightarrow\left(x^2-3x+1\right)\left(\frac{2x-6}{\sqrt{x}+1-x}+3\right)=0\)
Trường hợp 1 : \(x^2-3x+1=0\Leftrightarrow\left[\begin{array}{nghiempt}x=\frac{3+\sqrt{5}}{2}\left(\text{loại}\right)\\x=\frac{3-\sqrt{5}}{2}\left(\text{nhận}\right)\end{array}\right.\)
Trường hợp 2 : \(\frac{2x-6}{\sqrt{x}+1-x}+3=0\) , từ điều kiện \(3< x\le\frac{1+\sqrt{13}}{2}\) ta luôn có \(\frac{2x-6}{\sqrt{x}+1-x}+3>0\)
Vậy phương trình có nghiệm \(x=\frac{3-\sqrt{5}}{2}\)
x+1.2x-1=0
x+2x-1=0
3x-1=0
x-1=0:3
x =0+1
x =1
Đúng thì tick cho mk!
x+1.2x-1=0
x+2x-1 =0
3x-1 =0
3x =0+1
x = 1:3
x =\(\dfrac{1}{3}\)
Sorry bạn nãy làm cuống quá nên nhầm =))