giúp với mọi người

\(\Leftrightarrow\sqrt[3]{4x^2-9x-3}-\sqrt[3]{2x^2-3x-2}=\sqrt[3]{3x^2-2x+2}-\sqrt[3]{x^2+4x+3}\)

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{4x^2-9x-3}=a\\\sqrt[3]{2x^2-3x-2}=b\\\sqrt[3]{3x^2-2x+2}=c\\\sqrt[3]{x^2+4x+3}=d\end{matrix}\right.\) ta được:

\(\left\{{}\begin{matrix}a-b=c-d\\a^3-b^3=c^3-d^3\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a-b=c-d\\\left(a-b\right)\left(a^2+ab+b^2\right)=\left(c-d\right)\left(c^2+cd+d^2\right)\end{matrix}\right.\)

TH1: \(a-b=c-d=0\) \(\Leftrightarrow2x^2-6x-1=0\Leftrightarrow...\)

TH2: \(a-b=c-d\ne0\) \(\Rightarrow a^2+ab+b^2=c^2+cd+d^2\)

\(\Leftrightarrow\left(a-b\right)^2+4ab=\left(c-d\right)^2+4cd\)

\(\Leftrightarrow ab=cd\)

\(\Leftrightarrow\left(4x^2-9x-3\right)\left(2x^2-3x-2\right)=\left(3x^2-2x+2\right)\left(x^2+4x+3\right)\)

\(\Leftrightarrow x\left(5x^3-40x^2+10x+25\right)=0\)

\(\Leftrightarrow5x\left(x-1\right)\left(x^2-7x-5\right)=0\)

\(\Leftrightarrow...\)

Nhận thấy x=0 là nghiệm của PT

Xét x khác 0

\(PT\Leftrightarrow\sqrt[3]{x^2+4x+3}-\sqrt[3]{3x^2-2x+2}=\sqrt[3]{2x^2-3x-2}-\sqrt[3]{4x^2-9x-3}\)\(\Leftrightarrow\frac{-2x^2+6x+1}{\left(\sqrt[3]{x^2+4x+3}\right)^2+\sqrt[3]{\left(x^2+4x+3\right)\left(3x^2-2x+2\right)}+\left(\sqrt[3]{3x^2-2x+2}\right)^2}\)\(=\frac{-2x^2+6x+1}{\left(\sqrt[3]{2x^2-3x-2}\right)^2+\sqrt[3]{\left(2x^2-3x-2\right)\left(4x^2-9x-3\right)}+\left(\sqrt[3]{4x^2-9x-3}\right)}\)

\(\Leftrightarrow\left(-2x^2+6x+1\right)\left(....\right)=0\)(tự viết cái trong ngoặc nhaa :33 dài quá)

\(\Leftrightarrow x=\frac{3\pm\sqrt{11}}{2}\)

Vậy ......

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

3) \(x^2+\sqrt{2x^2+4x+3}=6-2x\Leftrightarrow-\sqrt{2x^2+4x+3}=x^2+2x-6\)\(\Leftrightarrow\left(2x^2+4x+3\right)-15=-2\sqrt{2x^2+4x+3}\)

Đặt \(\sqrt{2x^2+4x+3}=t\)(t > 0) thì phương trình trở thành \(t^2-15=-2t\Leftrightarrow t^2+2t-15=0\Leftrightarrow\left(t+5\right)\left(t-3\right)=0\Leftrightarrow\orbr{\begin{cases}t=-5\left(L\right)\\t=3\left(tm\right)\end{cases}}\)

Với t = 3 thì \(\sqrt{2x^2+4x+3}=3\Leftrightarrow2x^2+4x+3=9\Leftrightarrow2x^2+4x-6=0\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\Leftrightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}\)Vậy phương trình có tập nghiệm S = {1; -3}

Em xin phép làm bài EZ nhất :)

4,ĐK :\(\forall x\in R\)

Đặt \(x^2+x+2=t\) (\(t\ge\dfrac{7}{4}\))

\(PT\Leftrightarrow\sqrt{t+5}+\sqrt{t}=\sqrt{3t+13}\)

\(\Leftrightarrow2t+5+2\sqrt{t\left(t+5\right)}=3t+13\)

\(\Leftrightarrow t+8=2\sqrt{t^2+5t}\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge-8\\\left(t+8\right)^2=4t^2+20t\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\3t^2+4t-64=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left(t-4\right)\left(3t+16\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}t\ge\dfrac{7}{4}\\\left[{}\begin{matrix}t=4\left(tm\right)\\t=-\dfrac{16}{3}\left(l\right)\end{matrix}\right.\end{matrix}\right.\)

\(\Rightarrow x^2+x+2=4\)\(\Leftrightarrow x^2+x-2=0\)

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

Vậy ....

Đk : \(x\ge\frac{3}{4}\)

\(x-\sqrt{4x-3}=2\)

\(x-2=\sqrt{4x-3}\)

\(\Rightarrow\left(x-2\right)^2=\left(\sqrt{4x-3}\right)^2\)

\(x^2-4x+4=4x-3\)

\(x^2-8x+7=0\)

\(\Delta=36\Rightarrow\sqrt{\Delta}=6\)

\(\Rightarrow\)Phương trình có hai nghiệm phân biệt :

\(x_1=1\left(tm\right)\)

\(x_2=7\left(tm\right)\)

\(\sqrt{5x^2-2x\sqrt{5}+1}=\sqrt{6-2\sqrt{5}}\)

\(\Leftrightarrow\)\(5x^2-2x\sqrt{5}+1=6-2\sqrt{5}\)

\(\Leftrightarrow\)\(\left(x\sqrt{5}-1\right)^2=\left(\sqrt{5}-1\right)^2\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x\sqrt{5}-1=\sqrt{5}-1\\x\sqrt{5}-1=1-\sqrt{5}\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=1\\x=\frac{2-\sqrt{5}}{\sqrt{5}}\end{cases}}\)

Vậy...

ĐK:  \(x\ge\frac{3}{4}\)

\(x-\sqrt{4x-3}=2\)

\(\Leftrightarrow\)\(\sqrt{4x-3}=x-2\)

\(\Leftrightarrow\)\(4x-3=x^2-4x+4\)

\(\Leftrightarrow\)\(x^2-8x+7=0\)

\(\Leftrightarrow\)\(\left(x-1\right)\left(x-7\right)=0\)

đến đây tự làm

Mình làm câu còn lại nha :

ĐK : \(x\ge\frac{\sqrt{5}}{5}\)

\(\sqrt{\left(x\sqrt{5}-1\right)^2}=\sqrt{\left(\sqrt{5}-1\right)^2}\)

\(x\sqrt{5}-1=\sqrt{5}-1\)

\(x\sqrt{5}=\sqrt{5}\)

\(x=1\left(tm\right)\)

Đệ biết là có người làm câu c,d nên xin xí câu e :3

ĐK: \(\left\{{}\begin{matrix}x\ge-1\\x\ne2\end{matrix}\right.\)

\(PT\Leftrightarrow5+\sqrt{x+1}=7\left(x-2\right)\)

\(\Leftrightarrow\sqrt{x+1}=7x-19\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{19}{7}\\x+1=49x^2-266x+361\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{19}{7}\\49x^2-267x+360=0\end{matrix}\right.\)

\(\Rightarrow x=3\left(tm\right)\)

a/ \(\Leftrightarrow\left\{{}\begin{matrix}9-2x\ge0\\x^2-4x-12=\left(9-2x\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\le\frac{9}{2}\\3x^2-32x+93=0\end{matrix}\right.\)

Phương trình vô nghiệm

b/ \(\Leftrightarrow\left(x+1\right)\sqrt[3]{15x^2-x-1}-\left(x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(\sqrt[3]{15x^2-x-1}-x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\Rightarrow x=-1\\\sqrt[3]{15x^2-x-1}-x+1=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\sqrt[3]{15x^2-x-1}=x-1\)

\(\Leftrightarrow15x^2-x-1=x^3-3x^2+3x-1\)

\(\Leftrightarrow x^3-18x^2+4x=0\)

\(\Leftrightarrow x\left(x^2-18x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=9\pm\sqrt{77}\\\end{matrix}\right.\)

c/ ĐKXĐ: \(x\ge\frac{1}{2}\)

\(\Leftrightarrow2\left(x-1\right)\sqrt{2x-1}-6\left(x-1\right)=0\)

\(\Leftrightarrow2\left(x-1\right)\left(\sqrt{2x-1}-3\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=1\\2x-1=9\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)

d/ ĐKXĐ: \(1\le x< 3\)

\(\Leftrightarrow\sqrt{-x^2+4x-3}-1=2x-6\)

\(\Leftrightarrow\sqrt{-x^2+4x-3}=2x-5\) (\(x\ge\frac{5}{2}\))

\(\Leftrightarrow-x^2+4x-3=\left(2x-5\right)^2\)

\(\Leftrightarrow5x^2-24x+28=0\)

\(\Rightarrow\left[{}\begin{matrix}x=2< \frac{5}{2}\left(l\right)\\x=\frac{14}{5}\end{matrix}\right.\)

e/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge-1\\x\ne2\end{matrix}\right.\)

\(\Leftrightarrow5+\sqrt{x+1}=7x-14\)

\(\Leftrightarrow\sqrt{x+1}=7x-19\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{19}{7}\\x+1=\left(7x-19\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge\frac{19}{7}\\49x^2-267x+360=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=3\\x=\frac{120}{49}< \frac{19}{7}\left(l\right)\end{matrix}\right.\)

nhiều thế giải ko đổi đâu bạn

vậy trả lời câu a thôi

đkxđ : \(\frac{1}{2}\le x\le7\)

\(x^2-5x+3\sqrt{2x-1}=2\sqrt{14-2x}+5\)

\(\Leftrightarrow\left(x^2-5x\right)+3\left(\sqrt{2x-1}-3\right)=2\left(\sqrt{14-2x}-2\right)\)

\(\Leftrightarrow x\left(x-5\right)+\frac{3.\left(2x-10\right)}{\sqrt{2x-1}+3}+\frac{2.\left(2x-10\right)}{\sqrt{14-2x}+2}=0\)

\(\Leftrightarrow\left(x-5\right)\left(x+\frac{6}{\sqrt{2x-1}+3}+\frac{4}{\sqrt{14-2x}+2}\right)=0\)

\(\Leftrightarrow x=5\)

còn bài a,c lười đánh lắm