Câu 1 :

Xét điều kiện:\(\hept{\begin{cases}x\ge5\\x\le1\end{cases}}\)(Vô lý) 

Vậy pt vô nghiệm

Câu 2 : 

\(2\sqrt{x+2}+2\sqrt{x+2}-3\sqrt{x+2}=1\)\(\Leftrightarrow\sqrt{x+2}=1\Leftrightarrow x=-1\)

Vậy x=-1

Câu 3 : 

\(\sqrt{3x^2-4x+3}=1-2x\)\(\Leftrightarrow3x^2-4x+3=1+4x^2-4x\)

\(\Leftrightarrow x^2=2\Leftrightarrow x=\sqrt{2}\)

Câu 4 : 

\(4\sqrt{x+1}-3\sqrt{x+1}=4\Leftrightarrow\sqrt{x+1}=4\)

\(\Leftrightarrow x=15\)

Vãi ạ :))

ttpq_Trần Thanh Phương vãi j ?

đặt \(\sqrt{x+5}=a\);\(\sqrt{x+2}=b\)  => ab=\(\sqrt{x^2+7x+10}\) và \(a^2-b^2=3\)

 do đó pt trở thành \(\left(a-b\right)\left(1+ab\right)=a^2-b^2\)

                         \(\left(a-b\right)\left(1+ab\right)-\left(a-b\right)\left(a+b\right)=0\)

                         \(\left(a-b\right)\left(1+ab-a-b\right)=0\) 

đến đây tự giải tiếp nhé

em chưa học , em mới lớp 5 thui

a) \(\left|3x+1\right|=\left|x+1\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+1=x+1\\3x+1=-x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)

c) \(\sqrt{9x^2-12x+4}=\sqrt{x^2}\)

\(\Leftrightarrow\sqrt{\left(3x-2\right)^2}=\sqrt{x^2}\)

\(\Leftrightarrow\left|3x-2\right|=\left|x\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-2=x\\3x-2=-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)

d) \(\sqrt{x^2+4x+4}=\sqrt{4x^2-12x+9}\)

\(\Leftrightarrow\sqrt{\left(x+2\right)^2}=\sqrt{\left(2x-3\right)^2}\)

\(\Leftrightarrow\left|x+2\right|=\left|2x-3\right|\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=2x-3\\x+2=-2x+3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{1}{3}\end{matrix}\right.\)

e) \(\left|x^2-1\right|+\left|x+1\right|=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x^2-1=0\\x+1=0\end{matrix}\right.\)

\(\Leftrightarrow x=-1\)

f) \(\sqrt{x^2-8x+16}+\left|x+2\right|=0\)

\(\Leftrightarrow\sqrt{\left(x-4\right)^2}+\left|x+2\right|=0\)

\(\Leftrightarrow\left|x-4\right|+\left|x+2\right|=0\)

⇒ vô nghiệm

5) \(ĐK:x\ge-\frac{3}{2}\)

\(x^3+4x-\left(2x+7\right)\sqrt{2x+3}=0\)

\(\Leftrightarrow\frac{x^3+4x}{2x+7}=\sqrt{2x+3}\Leftrightarrow\frac{x^3+4x}{2x+7}-3=\sqrt{2x+3}-3\)

\(\Leftrightarrow\frac{\left(x-3\right)\left(x^2+3x+7\right)}{2x+7}=\frac{2\left(x-3\right)}{\sqrt{2x+3}+3}\)

\(\Leftrightarrow\left(x-3\right)\left(\frac{x^2+3x+7}{2x+7}-\frac{2}{\sqrt{2x+3}+3}\right)=0\)

(không có nghiệm thực)

Vậy phương trình có 1 nghiệm duy nhất là 3

1) \(Pt\Leftrightarrow-x^2-3x+10=3\sqrt{x^2+3x}\)( đk: \(x\le-3,x\ge0\)

Đặt \(t=\sqrt{x^2+3x},t\ge0\)

Pt trở thành: \(-t^2-3t+10=0\Leftrightarrow t=2\left(dot\ge0\right)\)

giải \(\sqrt{x^2+3x}=2\Leftrightarrow\orbr{\begin{cases}x=1\\x=-4\end{cases}}\)

b)\(\left(x+3\right)\sqrt{10-x^2}=x^2-x-12\)

Đk:\(-\sqrt{10}\le x\le\sqrt{10}\)

\(pt\Leftrightarrow\left(x+3\right)\sqrt{10-x^2}=\left(x-4\right)\left(x+3\right)\)

\(\Leftrightarrow\left(x+3\right)\sqrt{10-x^2}-\left(x-4\right)\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(\sqrt{10-x^2}-\left(x-4\right)\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x+3}=0\\\sqrt{10-x^2}=x-4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x+3=0\\10-x^2=x^2-8x+16\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\-2x^2+8x-6=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-3\\-\left(x-1\right)\left(x-3\right)=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-3\\\left[{}\begin{matrix}x=1\\x=3\end{matrix}\right.\end{matrix}\right.\)\(\Rightarrow x=-3\) (thỏa)

c)\(\sqrt{\dfrac{x^3+1}{x+3}}+\sqrt{x+3}=\sqrt{x^2-x+1}+\sqrt{x+1}\)

\(\Leftrightarrow\sqrt{\dfrac{\left(x+1\right)\left(x^2-x+1\right)}{x+3}}+\sqrt{x+3}-\sqrt{x^2-x+1}-\sqrt{x+1}=0\)

Đặt \(\sqrt{x^2-x+1}=a;\sqrt{x+1}=b;\sqrt{x+3}=c\left(a,b,c>0\right)\)

\(\Leftrightarrow\dfrac{ab}{c}+c-a-b=0\)

\(\Leftrightarrow\dfrac{\left(a-c\right)\left(b-c\right)}{c}=0\)

\(\Leftrightarrow\left(a-c\right)\left(b-c\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a-c=0\\b-c=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}a=c\\b=c\end{matrix}\right.\)

*)Xét \(a=c\)\(\Rightarrow\sqrt{x^2-x+1}=\sqrt{x+3}\)

\(\Rightarrow x^2-x+1=x+3\Rightarrow x=\dfrac{2\pm\sqrt{12}}{2}\) (thỏa)

*)Xét \(b=c\)\(\Rightarrow\sqrt{x+1}=\sqrt{x+3}\)

\(\Rightarrow x+1=x+3\Rightarrow-2=0\) (loại)

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