a/ ĐKXĐ: \(-\frac{1}{2}\le x\le4\)

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

\(\Leftrightarrow4-x=3x+2+2\sqrt{2x^2+3x+1}\)

\(\Leftrightarrow1-2x=\sqrt{2x^2+3x+1}\) (\(x\le\frac{1}{2}\))

\(\Leftrightarrow4x^2-4x+1=2x^2+3x+1\)

\(\Leftrightarrow2x^2-7x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\frac{7}{2}\left(l\right)\end{matrix}\right.\)

Bài này liên hợp cũng được

b/ ĐKXĐ: ...

\(\Leftrightarrow\sqrt{5x+1}^2-\sqrt{5x+1}\left(\sqrt{14x+7}-\sqrt{2x+3}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x+1=0\Rightarrow x=-\frac{1}{5}\\\sqrt{5x+1}-\sqrt{14x+7}+\sqrt{2x+3}=0\left(1\right)\end{matrix}\right.\)

\(\left(1\right)\Leftrightarrow\sqrt{5x+1}+\sqrt{2x+3}=\sqrt{14x+7}\)

\(\Leftrightarrow7x+4+2\sqrt{10x^2+17x+3}=14x+7\)

\(\Leftrightarrow2\sqrt{10x^2+17x+3}=7x+3\)

\(\Leftrightarrow4\left(10x^2+17x+3\right)=\left(7x+3\right)^2\)

\(\Leftrightarrow...\)

c/ ĐKXĐ: \(x\ge\frac{1}{2}\)

Đặt \(\left\{{}\begin{matrix}\sqrt[3]{2-2x}=a\\\sqrt{2x-1}=b\end{matrix}\right.\) ta được:

\(\left\{{}\begin{matrix}a=1-b\\a^3+b^2=1\end{matrix}\right.\) \(\Rightarrow a^3+\left(1-a\right)^2=1\)

\(\Leftrightarrow a^3+a^2-2a=0\)

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

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

d/ ĐKXĐ: \(x\le\frac{5}{4}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{5-4x}=a\\\sqrt[3]{x+7}=b\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}a+b=3\\a^2+4b^3=33\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}a=3-b\\a^2+4b^3=33\end{matrix}\right.\)

\(\Leftrightarrow\left(3-b\right)^2+4b^3=33\)

\(\Leftrightarrow4b^3+b^2-6b-24=0\)

\(\Leftrightarrow\left(b-2\right)\left(4b^2+9b+12\right)=0\)

\(\Rightarrow b=2\Rightarrow\sqrt[3]{x+7}=2\Rightarrow x=1\)

a/ ĐKXĐ: \(x\ge1\)

Khi \(x\ge1\) ta thấy \(\left\{{}\begin{matrix}VT>0\\VP=1-x\le0\end{matrix}\right.\) nên pt vô nghiệm

b/ \(x\ge1\)

\(\sqrt{\sqrt{x-1}\left(x-2\sqrt{x-1}\right)}+\sqrt{\sqrt{x-1}\left(x+3-4\sqrt{x-1}\right)}=\sqrt{x-1}\)

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

Đặt \(\sqrt{x-1}=a\ge0\) ta được:

\(\sqrt{a\left(a-1\right)^2}+\sqrt{a\left(a-2\right)^2}=a\)

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

\(\Leftrightarrow\left|a-1\right|+\left|a-2\right|=\sqrt{a}\)

- Với \(a\ge2\) ta được: \(2a-3=\sqrt{a}\Leftrightarrow2a-\sqrt{a}-3=0\Rightarrow\left[{}\begin{matrix}\sqrt{a}=-1\left(l\right)\\\sqrt{a}=\frac{3}{2}\end{matrix}\right.\)

\(\Rightarrow a=\frac{9}{4}\Rightarrow\sqrt{x-1}=\frac{9}{4}\Rightarrow...\)

- Với \(0\le a\le1\) ta được:

\(1-a+2-a=\sqrt{a}\Leftrightarrow2a+\sqrt{a}-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-\frac{3}{2}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{x-1}=1\Rightarrow...\)

- Với \(1< a< 2\Rightarrow a-1+2-a=\sqrt{a}\Leftrightarrow a=1\left(l\right)\)

c/ ĐKXĐ: \(x\ge\frac{49}{14}\)

\(\Leftrightarrow\sqrt{14x-49+14\sqrt{14x-49}+49}+\sqrt{14x-49-14\sqrt{14x-49}+49}=14\)

\(\Leftrightarrow\sqrt{\left(\sqrt{14x-49}+7\right)^2}+\sqrt{\left(\sqrt{14x-49}-7\right)^2}=14\)

\(\Leftrightarrow\left|\sqrt{14x-49}+7\right|+\left|7-\sqrt{14x-49}\right|=14\)

Mà \(VT\ge\left|\sqrt{14x-49}+7+7-\sqrt{14x-49}\right|=14\)

Nên dấu "=" xảy ra khi và chỉ khi:

\(7-\sqrt{14x-49}\ge0\)

\(\Leftrightarrow14x-49\le49\Leftrightarrow x\le7\)

Vậy nghiệm của pt là \(\frac{49}{14}\le x\le7\)

d/ ĐKXĐ: \(x\ge\frac{1}{2}\)

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

\(\Leftrightarrow\left|\sqrt{2x-1}-1\right|-2\left|\sqrt{2x-1}-2\right|+3\left|\sqrt{2x-1}-3\right|=4\)

TH1: \(\sqrt{2x-1}\ge3\Rightarrow x\ge5\)

\(\sqrt{2x-1}-1-2\sqrt{2x-1}+4+3\sqrt{2x-1}-9=4\)

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

\(\Leftrightarrow x=13\)

TH2: \(2\le\sqrt{2x-1}< 3\Rightarrow\frac{5}{2}\le x< 5\)

\(\sqrt{2x-1}-1-2\sqrt{2x-1}+4+3\left(3-\sqrt{2x-1}\right)=4\)

\(\Leftrightarrow\sqrt{2x-1}=2\Rightarrow x=\frac{5}{2}\)

TH3: \(1\le\sqrt{2x-1}< 2\Rightarrow1\le x< \frac{5}{2}\)

\(\sqrt{2x-1}-1-2\left(2-\sqrt{2x-1}\right)+3\left(3-\sqrt{2x-1}\right)=4\)

\(\Leftrightarrow4=4\) (luôn đúng)

TH4: \(\frac{1}{2}\le x< 1\)

\(1-\sqrt{2x-1}-2\left(2-\sqrt{2x-1}\right)+3\left(3-\sqrt{2x-1}\right)=4\)

\(\Leftrightarrow\sqrt{2x-1}=1\Rightarrow x=1\left(l\right)\)

Vậy nghiệm của pt là: \(\left[{}\begin{matrix}1\le x\le\frac{5}{2}\\x=13\end{matrix}\right.\)

xin lỗi nhé,tại máy mình bị lỗi nên phải đánh tách  ra :

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

Do \(\left(\sqrt{x^2+x+2}-\sqrt{2x+3}\right)\ge0\)nên \(2x+3\le0\)hay \(x\le\frac{-3}{2}\)

Mà Đk là \(x\ge\frac{-3}{2}\)

\(\Rightarrow x=\frac{-3}{2}\)

Thay lại thì \(x=\frac{-3}{2}\left(L\right)\)

\(\Rightarrow\)pt vô nghiệm

Bài 2 phân tích cái trong căn. tách vế trái thành nt trong căn 

Mình ko biết đúng ko nha:Bài 2 ĐK \(x\ge\frac{-3}{2}\)

Ta có \(2x^3+5x^2+7x+6=\left(2x+3\right)\left(x^2+x+2\right)\)

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

Căn bậc 3 

Phương trình vô tỉ

Khóc lắm bạn ơi *_*

ĐKXĐ: \(x\ge\frac{-1}{2}\)

\(\sqrt{2x+1}+\sqrt[3]{3x-4}=5\Leftrightarrow\left(\sqrt{2x+1}-3\right)+\left(\sqrt[3]{3x-4}-2\right)=0\)

\(\Leftrightarrow\frac{2x+1-9}{\sqrt{2x+1}+3}+\frac{3x-4-8}{\sqrt[3]{\left(3x-4\right)^2}+2\sqrt[3]{3x-4}+4}=0\Leftrightarrow\frac{2\left(x-4\right)}{\sqrt{2x+1}+3}+\frac{3\left(x-4\right)}{\sqrt[3]{\left(3x-4\right)^2}+2\sqrt[3]{3x-4}+4}\)\(\Leftrightarrow\left(x-4\right)\left[\frac{2}{\sqrt{2x+1}+3}+\frac{3}{\sqrt[3]{\left(3x-4\right)^2}+2\sqrt[3]{3x-4}+4}\right]=0\Leftrightarrow x-4=0\)

\(\Leftrightarrow x=4\)

lớp 10 học trường mô đây ?

Đk:\(x\in\left[1;\frac{5}{2}\right]\)

Ta thấy 2 vế luôn dương, bình phương lên đc:

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

\(\Leftrightarrow5-2x=x-1\)

\(\Leftrightarrow3x=6\)

\(\Leftrightarrow x=2\)

Đk:\(\frac{5}{2}\le x\le1\)

2 vế dương bình lên ta có:

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

\(\Leftrightarrow5-2x=x-1\)

\(\Leftrightarrow3x=6\)

\(\Leftrightarrow x=2\)

chết Đk mk sai r`

a. ĐKXĐ: \(x\ge\dfrac{1}{2}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+2x}=a>0\\\sqrt{2x-1}=b\ge0\end{matrix}\right.\)

\(\Rightarrow a+b=\sqrt{3a^2-b^2}\)

\(\Leftrightarrow\left(a+b\right)^2=3a^2-b^2\)

\(\Leftrightarrow a^2-ab-b^2=0\Leftrightarrow\left(a-\dfrac{1+\sqrt{5}}{2}b\right)\left(a+\dfrac{\sqrt{5}-1}{2}b\right)=0\)

\(\Leftrightarrow a=\dfrac{1+\sqrt{5}}{2}b\Leftrightarrow\sqrt{x^2+2x}=\dfrac{1+\sqrt{5}}{2}\sqrt{2x-1}\)

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

\(\Leftrightarrow x^2-\left(\sqrt{5}+1\right)x+\dfrac{3+\sqrt{5}}{2}=0\)

\(\Leftrightarrow\left(x-\dfrac{\sqrt{5}+1}{2}\right)^2=0\)

\(\Leftrightarrow x=\dfrac{\sqrt{5}+1}{2}\)

b. ĐKXĐ: \(x\ge5\)

\(\Leftrightarrow\sqrt{5x^2+14x+9}=\sqrt{x^2-x-20}+5\sqrt{x+1}\)

\(\Leftrightarrow5x^2+14x+9=x^2-x-20+25\left(x+1\right)+10\sqrt{\left(x+1\right)\left(x-5\right)\left(x+4\right)}\)

\(\Leftrightarrow2x^2-5x+2=5\sqrt{\left(x^2-4x-5\right)\left(x+4\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2-4x-5}=a\ge0\\\sqrt{x+4}=b>0\end{matrix}\right.\)

\(\Rightarrow2a^2+3b^2=5ab\)

\(\Leftrightarrow\left(a-b\right)\left(2a-3b\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-4x-5}=\sqrt{x+4}\\2\sqrt{x^2-4x-5}=3\sqrt{x+4}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x-5=x+4\\4\left(x^2-4x-5\right)=9\left(x+4\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)