Điều kiện xác định : \(1\le x\le7\)

Bất phương trình chuyển thành :

\(x-1+2\sqrt{7-x}-2\sqrt{x-1}-\sqrt{\left(x-1\right)\left(7-x\right)}\le0\)

Đặt \(a=\sqrt{x-1};b=\sqrt{7-x}\) ta có :

\(a^2-2a-ab+2b\le0\)

\(\Leftrightarrow\left(a-b\right)\left(a-2\right)\le0\)

\(\Leftrightarrow\left[{}\begin{matrix}a\le b\\a\le2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1\le7-x\\x-1\le4\end{matrix}\right.\)

Sau đó tìm x

a/ ĐKXĐ: \(x\ge-1\)

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

\(\Leftrightarrow2\left(\sqrt{x+1}+1\right)-\sqrt{x+1}=4\)

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

\(\Rightarrow x=3\)

b/ ĐKXĐ: \(x\ge1\)

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

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

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

Nên dấu "=" xảy ra khi và chỉ khi:

\(1\le\sqrt{x-1}\le2\Rightarrow2\le x\le5\)

Vậy nghiệm của pt là \(2\le x\le5\)

c/ ĐKXĐ: \(x\ge1\)

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

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

- Với \(\sqrt{x-1}\ge1\Rightarrow x\ge2\) ta có:

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

\(\Leftrightarrow2=2\) (luôn đúng)

- Với \(1\le x< 2\) ta có:

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

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

Vậy nghiệm của pt là \(x\ge2\)

d/ ĐKXĐ: \(-\le x\le1\)

\(\Leftrightarrow\sqrt{5-4x^2+4\sqrt{1-x^2}}+\sqrt{5-4x^2-4\sqrt{1-x^2}}=2x+2\)

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

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

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

TH1: \(\sqrt{4-4x^2}\ge1\Rightarrow-\frac{\sqrt{3}}{2}\le x\le\frac{\sqrt{3}}{2}\) ta có:

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

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

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

\(\Leftrightarrow5x^2+2x-3=0\Rightarrow\left[{}\begin{matrix}x=-1\left(l\right)\\x=\frac{3}{5}\end{matrix}\right.\)

TH2: \(\left[{}\begin{matrix}-1\le x< -\frac{\sqrt{3}}{2}\\\frac{\sqrt{3}}{2}< x\le1\end{matrix}\right.\) ta có:

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

\(\Leftrightarrow2x=0\Rightarrow x=0\left(l\right)\)

Vậy pt có nghiệm duy nhất \(x=\frac{3}{5}\)

a/ ĐKXĐ: \(-\frac{3}{2}\le x\le4\)

\(\sqrt{2x+3}+\sqrt{4-x}=6x-3\left(x+7-2\sqrt{\left(2x+3\right)\left(4-x\right)}\right)-10\)

\(\Leftrightarrow\sqrt{2x+3}+\sqrt{4-x}=3\left(x+7+2\sqrt{\left(2x+3\right)\left(4-x\right)}\right)-52\)

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

Phương trình trở thành:

\(a=3a^2-52\Leftrightarrow3a^2-a-52=0\Rightarrow\left[{}\begin{matrix}a=-4\left(l\right)\\a=\frac{13}{3}\end{matrix}\right.\)

\(\sqrt{2x+3}+\sqrt{4-x}=\frac{13}{3}\)

Phương trình này vô nghiệm nên ko muốn giải tiếp, bạn bình phương lên và chuyển vế thôi :(

b/ ĐKXĐ: \(-\frac{1}{4}\le x\le1\)

Đặt \(\sqrt{4x+1}+2\sqrt{1-x}=a>0\Rightarrow a^2=5+4\sqrt{-4x^2+3x+1}\)

\(\Rightarrow\sqrt{-4x^2+3x+1}=\frac{a^2-5}{4}\)

Pt trở thành:

\(a+10\left(\frac{a^2-5}{4}\right)=13\)

\(\Leftrightarrow5a^2+2a-51=0\Rightarrow\left[{}\begin{matrix}a=3\\a=-\frac{17}{5}\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{-4x^2+3x+1}=\frac{a^2-5}{4}=1\)

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

c/ \(\Leftrightarrow x^2\left(x^2+2\right)=12-x\sqrt{2x^2+4}\)

\(\Leftrightarrow x^2\left(2x^2+4\right)=24-2x\sqrt{2x^2+4}\)

Đặt \(x\sqrt{2x^2+4}=a\) ta được:

\(a^2=24-2a\Leftrightarrow a^2+2a-24=0\Leftrightarrow\left[{}\begin{matrix}a=4\\a=-6\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x\sqrt{2x^2+4}=4\left(x>0\right)\\x\sqrt{2x^2+4}=-6\left(x< 0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2\left(2x^2+4\right)=16\\x^2\left(2x^2+4\right)=36\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^4+2x^2-8=0\\x^4+2x^2-18=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x^2=2\\x^2=-4\left(l\right)\\x^2=\sqrt{19}-1\\x^2=-\sqrt{19}-1\left(l\right)\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}< 0\left(l\right)\\x=-\sqrt{\sqrt{19}-1}\\x=\sqrt{\sqrt{19}-1}>0\left(l\right)\end{matrix}\right.\)

1.

ĐK: \(-1\le x\le4\)

Đặt \(\sqrt{x+1}+\sqrt{4-x}=t\left(t\ge0\right)\)

\(\Leftrightarrow\sqrt{\left(x+1\right)\left(4-x\right)}=\frac{t^2-5}{2}\)

\(PT\Leftrightarrow t+\frac{t^2-5}{2}=5\Rightarrow t^2+2t-15=0\) \(\Rightarrow\left[{}\begin{matrix}t=3\\t=-5\left(l\right)\end{matrix}\right.\)

\(t=3\Rightarrow\sqrt{-x^2+3x+4}=2\) \(\Leftrightarrow-x^2+3x+4=4\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\) (tm)

2.

ĐK:\(x\ge4\)

Đặt \(\sqrt{x+4}+\sqrt{x-4}=t\left(t\ge0\right)\)

\(\Rightarrow2\sqrt{x^2-16}=t^2-2x\)

\(PT\Leftrightarrow t=2x-12+t^2-2x\)

\(\Leftrightarrow t^2-t-12=0\Rightarrow\left[{}\begin{matrix}t=4\\t=-3\left(l\right)\end{matrix}\right.\) Giải tiếp như trên.

@tran duc huy Bình phương rồi chuyển vế nha.

a/ ĐKXĐ: \(-2\le x\le2\)

Đặt \(x+\sqrt{4-x^2}=a\Rightarrow a^2=4+2x\sqrt{4-x^2}\Rightarrow x\sqrt{4-x^2}=\frac{a^2-4}{2}\)

\(\Rightarrow a-\frac{3\left(a^2-4\right)}{2}=2\)

\(\Leftrightarrow-3a^2+2a+8=0\Rightarrow\left[{}\begin{matrix}a=2\\a=-\frac{4}{3}\end{matrix}\right.\)

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

\(\Rightarrow\left[{}\begin{matrix}\sqrt{4-x^2}=2-x\\3\sqrt{4-x^2}=-4-3x\left(x\le-\frac{4}{3}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4-x^2=x^2-4x+4\\12\left(4-x^2\right)=9x^2+24x+16\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x^2-4x=0\\21x^2+24x-32=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=\frac{-12+4\sqrt{51}}{2}\left(l\right)\\x=\frac{-12-4\sqrt{51}}{2}\end{matrix}\right.\)

Mấy câu còn lại và bài kia tầm 30ph nữa sẽ làm, bận chút xíu việc

b/ ĐKXĐ: \(-2\le x\le2\)

\(\Leftrightarrow\left(2\sqrt{4-x^2}+4+4\right)\left(\sqrt{x+2}+\sqrt{2-x}\right)-5=0\)

Đặt \(\sqrt{x+2}+\sqrt{2-x}=a>0\Rightarrow a^2=4+2\sqrt{4-x^2}\)

\(\Rightarrow\left(a^2+4\right)a-5=0\)

\(\Leftrightarrow a^3+4a-5=0\Leftrightarrow\left(a-1\right)\left(a^2+a+5\right)=0\)

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

\(\Leftrightarrow4+2\sqrt{4-x^2}=1\Rightarrow2\sqrt{4-x^2}=-3\)

Vậy pt vô nghiệm

Thật ra bài này có thể biện luận vô nghiệm ngay từ đầu:

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

\(2\left(\sqrt{4-x^2}+4\right)\ge2.4=8\)

\(\Rightarrow VT>8.2-5=11>0\) nên pt vô nghiệm

a)\(ĐK:-3\le x\le6\)

\(PT\Leftrightarrow\sqrt{x+3}+\sqrt{6-x}=3\)

\(\Leftrightarrow x+3+6-x+2\sqrt{\left(x+3\right)\left(6-x\right)}=9\)

\(\Leftrightarrow\sqrt{\left(x+3\right)\left(6-x\right)}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=6\end{matrix}\right.\left(tm\right)\)

b/ ĐKXĐ: \(x\ge7\)

\(\sqrt{3x-2}=1+\sqrt{x-7}\)

\(\Leftrightarrow3x-2=x-6+2\sqrt{x-7}\)

\(\Leftrightarrow x+2=\sqrt{x-7}\)

\(\Leftrightarrow x^2+4x+4=x-7\)

\(\Leftrightarrow x^2+3x+11=0\) (vô nghiệm)

c/ ĐKXĐ: \(x\ge1;x\ne50\)

\(1-\sqrt{3x+1}=\sqrt{x-1}-7\)

\(\Leftrightarrow\sqrt{x-1}+\sqrt{3x+1}=8\)

\(\Leftrightarrow4x+2\sqrt{3x^2-2x-1}=64\)

\(\Leftrightarrow\sqrt{3x^2-2x-1}=32-2x\) (\(x\le16\))

\(\Leftrightarrow3x^2-2x-1=\left(32-2x\right)^2\)

Dk 1<x<2

√x^2 -x -2<x+2

5x+6>0

X > -6/5

Bpt vô nghiệm