Lời giải:

ĐK: \(x\geq 1\)

Ta có:

\(16x-13\sqrt{x-1}=9\sqrt{x+1}\)

\(\Leftrightarrow 13(x-\sqrt{x-1})+3(x-3\sqrt{x+1})=0\)

\(\Leftrightarrow 13(x-1-\sqrt{x-1}+\frac{1}{4})+3(x+1-3\sqrt{x+1}+\frac{9}{4})=0\)

\(\Leftrightarrow 13(\sqrt{x-1}-\frac{1}{2})^2+3(\sqrt{x+1}-\frac{3}{2})^2=0\)

Vì \((\sqrt{x-1}-\frac{1}{2})^2; (\sqrt{x+1}-\frac{3}{2})^2\geq 0\)

\(\Rightarrow 13(\sqrt{x-1}-\frac{1}{2})^2+3(\sqrt{x+1}-\frac{3}{2})^2\geq 0\)

Dấu "=" xảy ra khi \(\sqrt{x-1}-\frac{1}{2}=\sqrt{x+1}-\frac{3}{2}=0\Rightarrow x=\frac{5}{4}\) (t.m)

Vậy pt có nghiệm duy nhất $x=\frac{5}{4}$

Bạn ấy chọn điểm rơi x= và mục đích là để làm mất hết ẩn

C2 thêm bớt nhân liên hợp

PT<=>=0

xét pt

Vế trái là hàm nghịch biến vế phải là hằng số nên nghiệm kia là duy nhất

Ta có:

\(\left(x-1\right)+\frac{1}{4}\ge\sqrt{x-1}\)

\(\Leftrightarrow13\left(x-1\right)+\frac{13}{4}\ge13\sqrt{x-1}\)

\(\Leftrightarrow13x-\frac{39}{4}\ge13\sqrt{x-1}\)(1)

Ta lại có

\(\left(x+1\right)+\frac{9}{4}\ge3\sqrt{x+1}\)

\(3\left(x+1\right)+\frac{27}{4}\ge9\sqrt{x+1}\)

\(\Leftrightarrow3x+\frac{39}{4}\ge9\sqrt{x+1}\)(2)

Cộng (1) và (2) vế theo vế được

\(16x\ge13\sqrt{x-1}+9\sqrt{x+1}\)

Dấu = xảy ra khi

\(\hept{\begin{cases}x-1=\frac{1}{4}\\x+1=\frac{9}{4}\end{cases}}\Leftrightarrow x=\frac{5}{4}\)

\(\sqrt{7-x}+\sqrt{x+1}=x^2-6x+13,đkxđ:-1\le x\le7,\Leftrightarrow\left(\sqrt{7-x}+\sqrt{x+1}\right)^2=\left(x^2-6x+13\right)^2\Leftrightarrow7-x+x+1+2\sqrt{\left(7-x\right)\left(x+1\right)}=\left(x^2-6x+13\right)\left(x^2-6x+13\right)\Leftrightarrow8+2\sqrt{7x+8-x^2-x}=x^4-6x^3+13x^2-6x^3+36x^2-78x+13x^2-78x+169\Leftrightarrow8+2\sqrt{-x^2+6x+8}=x^4-12x^3+62x^2-120x+169\Leftrightarrow Bírồi:< \)

\(Chot=7-x\Rightarrow x=7-t\Rightarrow\sqrt{7-x}=\sqrt{7-7+t}=\sqrt{t}và\sqrt{x+1}=\sqrt{7-t+1}=\sqrt{8-t}vàx^2-6x+13=\left(7-t\right)^2-6\left(7-t\right)+13,tacópt:\sqrt{t}+\sqrt{8-t}=49-14t+t^2-42+6t+13\Leftrightarrow\sqrt{t}+\sqrt{8-t}=t^2-8t+20=t^2-2.4.t+16+4=\left(t-4\right)^2+4\Leftrightarrow\left(\sqrt{t}+\sqrt{8-t}\right)^2=\left[\left(t-4\right)^2+4\right]^2\Leftrightarrow t-t+8+2\sqrt{8t-t^2}=...\left(bítiếp\right)\)

\(\sqrt{7-x}+\sqrt{x+1}=x^2-6x+13\left(đk:-1\le x\le7\right)\)

Với a,b>0 ta AD BĐT: \(\sqrt{a}+\sqrt{b}\le2\sqrt{\frac{a+b}{2}}\) (tự CM nha ).Dấu "=" xảy ra<=>a=b (1)

AD bđt (1) có:

\(\sqrt{7-x}+\sqrt{x+1}\le2\sqrt{\frac{7-x+x+1}{2}}\)

\(\le2\sqrt{4}\) =4 (*)

Có x²-6x+13=(x-3)²+4 \(\ge4\) (**)

Từ (*),(**) => Dấu bằng xảy ra \(< =>\left\{{}\begin{matrix}7-x=x+1\\x-3=0\end{matrix}\right.\) \(< =>\left\{{}\begin{matrix}x=3\\x=3\end{matrix}\right.\)\(< =>x=3\)(tm điều kiện của x)

Vậy x=3

ĐK \(-1\le x\le7\)

\(VP=x^2-6x+13=\left(x-3\right)^2+4\ge4\forall-1\le x\le7\)

\((\sqrt{7-x}+\sqrt{x+1})^2\le\left(1+1\right)\left(7-x+x-1\right)=16\)

\(\Rightarrow VT\le\sqrt{16}=4\)

Dấu "= " xảy ra

\(\left\{{}\begin{matrix}x^2-6x+13=4\\\sqrt{7-x}=\sqrt{x+1}\end{matrix}\right.\)

\(\Leftrightarrow x=3\)

Vậy nghiệm của pt là x =3

dễ thấy x \(\ge\)0

bình phương hai vế được :

\(13-\sqrt{13+x}=x^2\)

\(\Rightarrow\sqrt{13+x}+x=13+x-x^2\)

\(\Rightarrow\sqrt{13+x}+x=\left(\sqrt{13+x}+x\right)\left(\sqrt{13+x}-x\right)\)

\(\Rightarrow1=\sqrt{13+x}-x\)

\(\Rightarrow13+x=x^2+2x+1\)

\(\Rightarrow x^2+x-12=0\)

\(\Rightarrow\orbr{\begin{cases}x=3\left(tm\right)\\x=-4\left(kotm\right)\end{cases}}\)

\(ĐK:-1\le x\le1\\ PT\Leftrightarrow13\left(1-2x^2\right)\sqrt{\left(1-x^2\right)\left(1+x^2\right)}+9\left(1+2x^2\right)\sqrt{\left(1+x^2\right)\left(1-x^2\right)}=0\\ \Leftrightarrow\sqrt{1-x^4}\left(13-26x^2+9+18x^2\right)=0\\ \Leftrightarrow\sqrt{1-x^4}\left(22-8x^2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}1-x^4=0\\22-8x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(1+x^2\right)\left(1-x\right)\left(1+x\right)=0\\x^2=\dfrac{22}{8}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x=1\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\\\left[{}\begin{matrix}x=\dfrac{\sqrt{11}}{2}\left(ktm\right)\\x=-\dfrac{\sqrt{11}}{2}\left(ktm\right)\end{matrix}\right.\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)

\(\sqrt{30-x}-\sqrt{x-5}=\sqrt{x-13}\left(1\right)\)

ĐKXĐ: \(13\le x\le30\)

\(\left(1\right)\Leftrightarrow\sqrt{30-x}=\sqrt{x-13}+\sqrt{x-5}\)

\(\Leftrightarrow30-x=x-13+x-5+2\sqrt{\left(x-13\right)\left(x-5\right)}\)

\(\Leftrightarrow2\sqrt{\left(x-13\right)\left(x-5\right)}=48-3x\)

\(\Leftrightarrow\left\{{}\begin{matrix}48-3x\ge0\\4\left(x-13\right)\left(x-5\right)=\left(48-3x\right)^2\end{matrix}\right.\)

+) \(48-3x\ge0\Leftrightarrow3x\le48\Leftrightarrow x\le16\)

+) \(4\left(x-13\right)\left(x-5\right)=\left(48-3x\right)^2\)

\(\Leftrightarrow4x^2-72x+260=2304-288x+9x^2\)

\(\Leftrightarrow5x^2-216x+2044=0\)

△' \(=108^2-2044.5=1444>0\)

\(\Rightarrow\left[{}\begin{matrix}x_1=\frac{108-\sqrt{1444}}{5}\\x_2=\frac{-108-\sqrt{1444}}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x_1=14\\x_2=\frac{-146}{5}\end{matrix}\right.\)

Đối chiếu đk thì chỉ có \(x=14\)thỏa mãn

Vậy pt có nghiệm là \(x=14\)

Gọn nhẹ hơn 1 chút:

ĐKXĐ:...

\(\Leftrightarrow\sqrt{x-13}-1+\sqrt{x-5}-3+4-\sqrt{30-x}=0\)

\(\Leftrightarrow\frac{x-14}{\sqrt{x-13}+1}+\frac{x-14}{\sqrt{x-5}+3}+\frac{x-14}{4+\sqrt{30-x}}=0\)

\(\Leftrightarrow\left(x-14\right)\left(\frac{1}{\sqrt{x-13}+1}+\frac{1}{\sqrt{x-5}+3}+\frac{1}{4+\sqrt{30-x}}\right)=0\)

\(\Leftrightarrow x=14\)

Dễ dàng nhận ra cái ngoặc đằng sau dương

x>=1

\(\Leftrightarrow16x-13\sqrt{x-1}-9\sqrt{x+1}=0\)

\(\Leftrightarrow13\left(x-1-\sqrt{x-1}+\dfrac{1}{4}\right)+3\left(x+1-3\sqrt{x+1}+\dfrac{9}{4}\right)=0\)

\(\Leftrightarrow13\left(\sqrt{x-1}-\dfrac{1}{2}\right)^2+3\left(\sqrt{x+1}-\dfrac{3}{2}\right)^2=0\)

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

x=5/4(tm)