Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bạn tự xét ĐKXĐ nhé ^^
Ta có : \(\sqrt{3x^2-5x+1}-\sqrt{x^2-2}=\sqrt{3\left(x^2-x-1\right)}-\sqrt{x^2-3x+4}\)
\(\Leftrightarrow\left(\sqrt{3x^2-5x+1}-\sqrt{3}\right)-\left(\sqrt{x^2-2}-\sqrt{2}\right)-\left[\sqrt{3\left(x^2-x-1\right)}-\sqrt{3}\right]+\left(\sqrt{x^2-3x+4}-\sqrt{2}\right)=0\)
\(\Leftrightarrow\frac{3x^2-5x+1-3}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{x^2-2-2}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3x^2-3x-3-3}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{x^2-3x+4-2}{\sqrt{x^2-3x+4}+\sqrt{2}}=0\)
\(\Leftrightarrow\frac{\left(x-2\right)\left(3x+1\right)}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{\left(x-2\right)\left(x+2\right)}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3\left(x-2\right)\left(x+1\right)}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{\left(x-2\right)\left(x-1\right)}{\sqrt{x^2-3x+4}+\sqrt{2}}=0\)
\(\Leftrightarrow\left(x-2\right)\left(\frac{3x+1}{\sqrt{3x^2-5x+1}+\sqrt{3}}-\frac{x+2}{\sqrt{x^2-2}+\sqrt{2}}-\frac{3x+3}{\sqrt{3\left(x^2-x-1\right)}+\sqrt{3}}+\frac{x-1}{\sqrt{x^2-3x+4}+\sqrt{2}}\right)=0\)Tới đây bạn tự làm tiếp ^^
Dài quá ^^
\(4\sqrt{x+2}+\sqrt{22-3x}=x^2+8\)
ĐK:\(x\in\left[-2;\frac{22}{3}\right]\)
\(\Leftrightarrow4\sqrt{x+2}-\left(\frac{4}{3}x+\frac{16}{3}\right)+\sqrt{22-3x}-\left(-\frac{1}{3}x+\frac{14}{3}\right)=x^2-x-2\)
\(\Leftrightarrow4\frac{x+2-\left(\frac{1}{3}x+\frac{4}{3}\right)^2}{4\sqrt{x+2}+\frac{4}{3}x+\frac{16}{3}}+\frac{22-3x-\left(-\frac{1}{3}x+\frac{14}{3}\right)^2}{\sqrt{22-3x}+\frac{3}{3}x+\frac{14}{3}}=x^2-x-2\)
\(\Leftrightarrow4\frac{\frac{-x^2-x-2}{9}}{4\sqrt{x+2}+\frac{4}{3}x+\frac{16}{3}}+\frac{\frac{-x^2-x-2}{9}}{\sqrt{22-3x}+\frac{3}{3}x+\frac{14}{3}}-\left(x^2-x-2\right)=0\)
\(\Leftrightarrow-\left(x^2-x-2\right)\left(\frac{4\cdot\frac{1}{9}}{4\sqrt{x+2}+\frac{4}{3}x+\frac{16}{3}}+\frac{\frac{1}{9}}{\sqrt{22-3x}+\frac{3}{3}x+\frac{14}{3}}+1\right)=0\)
Pt trong ngoặc to >0
\(\Rightarrow x^2-x-2=0\Rightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)
ĐKXĐ: \(x\ge1\)
Do \(\sqrt{x-\sqrt{x^2-1}}.\sqrt{x+\sqrt{x^2-1}}=\sqrt{x^2-x^2+1}=1\)
Đặt \(\sqrt{x-\sqrt{x^2-1}}=t\Rightarrow\sqrt{x+\sqrt{x^2-1}}=\dfrac{1}{t}\)
Phương trình trở thành:
\(t+\dfrac{1}{t}=2\Rightarrow t^2-2t+1=0\Rightarrow t=1\)
\(\Rightarrow\sqrt{x-\sqrt{x^2-1}}=1\Leftrightarrow x-\sqrt{x^2-1}=1\)
\(\Leftrightarrow x-1=\sqrt{x^2-1}\)
\(\Rightarrow x^2-2x+1=x^2-1\)
\(\Rightarrow x=1\) (thỏa mãn)
\(2\left(x-4\right)\sqrt{x-2}+\left(x-2\right)\sqrt{x+1}+2\left(x-3\right)=0\)
ĐK:\(x\ge2\)
\(\Leftrightarrow2\left(x-4\right)\left(\sqrt{x-2}-1\right)+\left(x-2\right)\left(\sqrt{x+1}-2\right)-2\left(x-3\right)=0\)
\(\Leftrightarrow2\left(x-4\right)\frac{x-2-1}{\sqrt{x-2}+1}+\left(x-2\right)\frac{x+1-4}{\sqrt{x+1}+2}-2\left(x-3\right)=0\)
\(\Leftrightarrow2\left(x-4\right)\frac{x-3}{\sqrt{x-2}+1}+\left(x-2\right)\frac{x-3}{\sqrt{x+1}+2}-2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{2\left(x-4\right)}{\sqrt{x-2}+1}+\frac{x-2}{\sqrt{x+1}+2}-2\right)=0\)
Suy ra x=3
a)\(\sqrt{3x+1}+2x=\sqrt{x-4}-5\left(ĐKXĐ:x\ge4\right)\)
\(\Leftrightarrow\left(\sqrt{3x+1}-\sqrt{x-4}\right)+\left(2x+5\right)=0\)
\(\Leftrightarrow\frac{3x+1-x+4}{\sqrt{3x+1}+\sqrt{x-4}}+\left(2x+5\right)=0\)
\(\Leftrightarrow\frac{2x+5}{\sqrt{3x+1}+\sqrt{x-4}}+\left(2x+5\right)=0\)
\(\Leftrightarrow\left(2x+5\right)\left(\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1\right)=0\)
a') (tiếp)
\(\Leftrightarrow\orbr{\begin{cases}2x+5=0\\\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-2,5\left(KTMĐKXĐ\right)\\\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1=0\end{cases}}\)
Xét phương trình \(\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1=0\)(1)
Với mọi \(x\ge4\), ta có:
\(\sqrt{3x+1}>0\); \(\sqrt{x-4}\ge0\)
\(\Rightarrow\sqrt{3x+1}+\sqrt{x-4}>0\Rightarrow\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}>0\)
\(\Rightarrow\frac{1}{\sqrt{3x+1}+\sqrt{x-4}}+1>0\)
Do đó phương trình (1) vô nghiệm.
Vậy phương trình đã cho vô nghiệm.
a. ta có
\(x^2+2x-1+4x+2=\left(2x+1\right)\sqrt{x^2+2x+3}\)
\(\Leftrightarrow x^2+2x-1=\left(2x+1\right)\left[\sqrt{x^2+2x+3}-2\right]\Leftrightarrow x^2+2x-1=\left(2x+1\right).\frac{x^2+2x-1}{\sqrt{x^2+2x+3}+2}\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x^2+2x+3}+2=2x+1\\x^2+2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}\sqrt{x^2+2x+3}=2x-1\\x=-1\pm\sqrt{2}\end{cases}}}\)
với \(\sqrt{x^2+2x+3}=2x-1\Leftrightarrow\hept{\begin{cases}x\ge\frac{1}{2}\\x^2+2x+3=4x^2-4x+1\end{cases}\Leftrightarrow x=\frac{3+\sqrt{15}}{3}}\)
b.\(3\sqrt{x-2}-\sqrt{x+6}=2x-6\Leftrightarrow\frac{8\left(x-3\right)}{3\sqrt{x-2}+\sqrt{x+6}}=2\left(x-3\right)\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\3\sqrt{x-2}+\sqrt{x+6}=4\end{cases}}\)
với \(3\sqrt{x-2}+\sqrt{x+6}=4\Leftrightarrow10x-12+6\sqrt{\left(x-2\right)\left(x+6\right)}=16\)
\(\Leftrightarrow3\sqrt{x^2+4x-12}=14-5x\) xét điều kiện rồi bình phương thôi bạn nhé
xét vế trái :
\(\sqrt[]{x-2}+\sqrt{10-x}=< \sqrt{2\left(x-2+10-x\right)}=< 4\)
=>vp=<4
=>\(x^2-12x+40=< 4\)
=>\(\left(x-6\right)^2=< 0\)
=> xảy ra dấu = <=>x=6
vậy pt có nghiệm là 6
Giải bằng liên hợp đúng sở trường của mình rồi ^^
Ta có : \(2\sqrt{x^2-7x+10}=x+\sqrt{x^2-12x+20}\) (ĐKXĐ : \(\orbr{\begin{cases}0\le x\le2\\x\ge10\end{cases}}\) )
\(\Leftrightarrow2\left(\sqrt{x^2-7x+10}-2\right)-\left(\sqrt{x^2-12x+20}-3\right)-\left(x+1\right)=0\)
\(\Leftrightarrow2\left(\frac{x^2-7x+10-4}{\sqrt{x^2-7x+10}+2}\right)-\left(\frac{x^2-12x+20-9}{\sqrt{x^2-12x+20}+3}\right)-\left(x+1\right)=0\)
\(\Leftrightarrow\frac{2\left(x-1\right)\left(x-6\right)}{\sqrt{x^2-7x+10}+2}-\frac{\left(x-1\right)\left(x-11\right)}{\sqrt{x^2-12x+20}+3}-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{2x-12}{\sqrt{x^2-7x+1}+2}-\frac{x-11}{\sqrt{x^2-12x+20}+3}-1\right)=0\)
Đến đây thì dễ rồi ^^
Mình có nhầm một chút xíu ở dòng 3 và 4 nhé ^^