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

Câu 1:

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

- Với \(x< -1\Rightarrow\left\{{}\begin{matrix}VT\ge0\\VP< 0\end{matrix}\right.\) pt vô nghiệm

- Nhận thấy \(x=-1\) là 1 nghiệm

- Nếu \(x>-1\) kết hợp ĐKXĐ các căn thức ta được \(x\ge1\), pt tương đương:

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

\(\Leftrightarrow2x+6+x-1+2\sqrt{2\left(x+3\right)\left(x-1\right)}=4x+4\)

\(\Leftrightarrow2\sqrt{2x^2+4x-6}=x-1\)

\(\Leftrightarrow4\left(2x^2+4x-6\right)=\left(x-1\right)^2\)

\(\Leftrightarrow7x^2+18x-25=0\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=-\frac{25}{7}< 0\left(l\right)\end{matrix}\right.\)

Vậy pt có nghiệm \(x=\pm1\)

Câu 2:

ĐKXĐ: \(x\ge1\)

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

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

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

- Nếu \(\sqrt{x-1}-1\ge0\Leftrightarrow x\ge2\) pt trở thành:

\(\sqrt{x-1}+1-\sqrt{x-1}+1=2\Leftrightarrow2=2\) (luôn đúng)

- Nếu \(1\le x< 2\) pt trở thành:

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

Vậy nghiệm của pt là \(x\ge2\)

Câu 3:

Bình phương 2 vế ta được:

\(2x^2+2x+5+2\sqrt{\left(x^2+x+4\right)\left(x^2+x+1\right)}=2x^2+2x+9\)

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

\(\Leftrightarrow\left(x^2+x+4\right)\left(x^2+x+1\right)=4\)

Đặt \(x^2+x+1=a>0\) pt trở thành:

\(a\left(a+3\right)=4\Leftrightarrow a^2+3a-4=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-4\left(l\right)\end{matrix}\right.\)

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

Câu 5:

ĐKXĐ: \(x\ge1\)

\(\sqrt{x-1-4\sqrt{x-1}+4}+\sqrt{x-1-6\sqrt{x-1}+9}=1\)

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

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

Mà \(VT=\left|\sqrt{x-1}-2\right|+\left|3-\sqrt{x-1}\right|\ge\left|\sqrt{x-1}-2+3-\sqrt{x-1}\right|=1\)

\(\Rightarrow VT\ge VP\Rightarrow\) Đẳng thức xảy ra khi và chỉ khi:

\(\left\{{}\begin{matrix}\sqrt{x-1}-2\ge0\\\sqrt{x-1}-3\le0\end{matrix}\right.\) \(\Rightarrow5\le x\le10\)

Vậy nghiệm của pt là \(5\le x\le10\)