\(x-\sqrt{x-15}=17\)

<=> \(\sqrt{x-15}=x-17\) (đk x>=17)

<=> x-15 = (x-17)²

^<=> x² -35x + 304 = 0

<=> x=19 (t/m) hoặc x=16 (loại)

Vậy x=19

P/s:Dấu"[" dùng để thay cho chữ "hoặc" nha!

\(x-\sqrt{x-15}=17\Leftrightarrow x-15-\sqrt{x-15}=2\)

ĐKXĐ: \(x\ge15\)

Đặt \(\sqrt{x-15}=t\Leftrightarrow x=t^2+15\) (1)

PT <=>\(t^2-t=2\Leftrightarrow t\left(t-1\right)=2\Leftrightarrow\left[{}\begin{matrix}t=2\\t-1=2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=2\\t=3\end{matrix}\right.\)

Thay vào (1),được: \(\left[{}\begin{matrix}x=t^2+15=2^2+15=19\left(tm\right)\\x=t^2+15=3^2+15=24\left(L\right)\end{matrix}\right.\)

Vậy x = 19

ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}u=v\\v=\sqrt{17-x^2}\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}u+v+uv=9\\u^2+v^2=17\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}uv=9-\left(u+v\right)\\\left(u+v\right)^2-2uv=17\end{matrix}\right.\)

\(\Rightarrow\left(u+v\right)^2+2\left(u+v\right)-35=0\)

\(\Rightarrow\left[{}\begin{matrix}u+v=5\Rightarrow uv=4\\u+v=-7\Rightarrow uv=16\end{matrix}\right.\)

\(\Rightarrow...\)

Đk:\(x\ge\sqrt{15}\)

Đặt \(\sqrt{x^2-15}=a;\sqrt{x-3}=b\left(a,b>0\right)\)

Thì \(a^2+b^2=x^2+x-18\) khi đó

\(pt\Leftrightarrow a^2+b^2+1=ab+a+b\)

Áp dụng BĐT AM-GM ta có:

\(\left\{{}\begin{matrix}a^2+b^2\ge2\sqrt{a^2b^2}=2ab\\b^2+1\ge2\sqrt{b^2}=2b\\a^2+1\ge2\sqrt{a^2}=2a\end{matrix}\right.\)

Cộng theo vế rồi thu gọn 3 BĐT trên ta có:

\(VT=a^2+b^2+1\ge ab+a+b=VP\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}a^2+b^2=2ab\\b^2+1=2b\\a^2+1=2a\end{matrix}\right.\)\(\Rightarrow a=b=1\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x^2-15}=1\\\sqrt{x-3}=1\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x^2-15=1\\x-3=1\end{matrix}\right.\Rightarrow x=4\left(x\ge\sqrt{15}\right)\)

\(\Leftrightarrow\sqrt{2x^2-x+3}-\left(x+1\right)+\left(x^2+1\right)-\sqrt{21x-17}=0\)

=>\(\dfrac{2x^2-x+3-x^2-2x-1}{\sqrt{2x^2-x+3}+x+1}+\dfrac{x^4+2x^2+1-21x+17}{x^2+1+\sqrt{21x-17}}=0\)

=>x^2-3x+2=0

=>x=1 hoặc x=2

\(ĐK:-5\le x\le3\)

Đặt \(\sqrt{x+5}+\sqrt{3-x}=t\ge0\Leftrightarrow t^2-8=2\sqrt{15-2x-x^2}\), PTTT:

\(t-t^2+8-2=0\\ \Leftrightarrow t^2-t-6=0\\ \Leftrightarrow t=3\left(t\ge0\right)\\ \Leftrightarrow2\sqrt{15-2x-x^2}=3^2-8=1\\ \Leftrightarrow60-8x-4x^2=1\\ \Leftrightarrow4x^2+8x-59=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2+3\sqrt{7}}{2}\left(tm\right)\\x=\dfrac{-2-3\sqrt{7}}{2}\left(tm\right)\end{matrix}\right.\)

Vậy nghiệm pt là ...

đkxđ \(1\le x\le6\)

đặt \(\sqrt{x-1}=a\left(a\ge0\right);\sqrt{6-x}=b\left(b\ge0\right)\)

ta thấy \(a^2+b^2=5\)

ta suy ra hệ phương trình

\(\left\{{}\begin{matrix}a+7b=15\left(1\right)\\a^2+b^2=5\left(2\right)\end{matrix}\right.\)

rút pt (1) thế pt(2) ta có

\(\left(15-7b\right)^2+b^2=5\)

\(\Leftrightarrow50b^2-210b+220=0\)

\(\Leftrightarrow\left[{}\begin{matrix}b=\dfrac{11}{5}\Rightarrow a=-\dfrac{2}{5}\left(l\right)\\b=2\Rightarrow a=1\left(n\right)\end{matrix}\right.\)

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

thay x=2 thấy b=2

vậy pt có nghiệm là x=2

Lời giải:

ĐKXĐ: \(1\leq x\leq 6\)

Ta có:

\(\sqrt{x-`1}+7\sqrt{6-x}=15\)

\(\Leftrightarrow 7\sqrt{6-x}=15-\sqrt{x-1}\)

\(\Rightarrow 49(6-x)=225+x-1-30\sqrt{x-1}\) (bp hai vế)

\(\Leftrightarrow 50x-30\sqrt{x-1}-70=0\)

\(\Leftrightarrow 5x-3\sqrt{x-1}-7=0\)

\(\Leftrightarrow 5(x-1)-3\sqrt{x-1}-2=0\) Đặt \(\sqrt{x-1}=t(t\geq 0)\)

Khi đó: \(5t^2-3t-2=0\Leftrightarrow (t-1)(5t+2)=0\Rightarrow t=1\)

vì $t\geq 0$

Do đó: \(x=t^2+1=2\). Thử lại thấy thỏa mãn

1/ \(\dfrac{5}{3}\le x\le\dfrac{7}{3}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{3x-5}=a>0\\\sqrt{7-3x}=b>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^2+b^2=2\\17-6x=2b^2+3\\6x-7=2a^2+3\end{matrix}\right.\)

Mặt khác theo BĐT Bunhiacốpxki:

\(a+b=\sqrt{3x-5}+\sqrt{7-3x}\le\sqrt{\left(1+1\right)\left(3x-5+7-3x\right)}=2\)

\(\Rightarrow0< a+b\le2\)

Ta được hệ pt:

\(\left\{{}\begin{matrix}a^2+b^2=2\\\left(2b^2+3\right).a+\left(2a^2+3\right)b=2+8ab\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}\left(a+b\right)^2-2ab=2\\2ab^2+3a+2a^2b+3b-8ab-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}2ab=\left(a+b\right)^2-2\\2ab\left(a+b\right)+3\left(a+b\right)-8ab-2=0\end{matrix}\right.\)

\(\Rightarrow\left(\left(a+b\right)^2-2\right)\left(a+b\right)+3\left(a+b\right)-4\left(a+b\right)^2+6=0\)

\(\Leftrightarrow\left(a+b\right)^3-4\left(a+b\right)^2+\left(a+b\right)+6=0\)

\(\Rightarrow\left[{}\begin{matrix}a+b=-1< 0\left(l\right)\\a+b=2\\a+b=3>2\left(l\right)\end{matrix}\right.\)

\(\Rightarrow a+b=2\) , dấu "=" xảy ra khi và chỉ khi:

\(3x-5=7-3x\Rightarrow x=2\)

Vậy pt có nghiệm duy nhất \(x=2\)

2/ ĐKXĐ: \(x\ne\pm2\)

\(\left(\dfrac{x-1}{x+2}\right)^2+4\left(\dfrac{x+1}{x-2}\right)^2-\left(\dfrac{15}{x^2-4}+5\right)=0\)

\(\Leftrightarrow\left(\dfrac{x-1}{x+2}\right)^2+4\left(\dfrac{x+1}{x-2}\right)^2-5.\left(\dfrac{x^2-1}{x^2-4}\right)=0\)

\(\Leftrightarrow\left(\dfrac{x-1}{x+2}\right)^2-\left(\dfrac{x^2-1}{x^2-4}\right)-4\left[\left(\dfrac{x^2-1}{x^2-4}\right)-\left(\dfrac{x+1}{x-2}\right)^2\right]=0\)

\(\Leftrightarrow\left(\dfrac{x-1}{x+2}\right)\left(\dfrac{x-1}{x+2}-\dfrac{x+1}{x-2}\right)-4\left(\dfrac{x+1}{x-2}\right)\left(\dfrac{x-1}{x+2}-\dfrac{x+1}{x-2}\right)=0\)

\(\Leftrightarrow\left(\dfrac{x-1}{x+2}-\dfrac{4\left(x+1\right)}{x-2}\right)\left(\dfrac{x-1}{x+2}-\dfrac{x+1}{x-2}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x-1}{x+2}=\dfrac{4\left(x+1\right)}{x-2}\\\dfrac{x-1}{x+2}=\dfrac{x+1}{x-2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2-3x+2=4\left(x^2+3x+2\right)\\x^2-3x+2=x^2+3x+2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x^2+15x+6=0\\6x=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{-5+\sqrt{17}}{2}\\x=\dfrac{-5-\sqrt{17}}{2}\end{matrix}\right.\)

Đây là phương pháp trừ để hỏng, phương pháp rất đơn giản như sau:

B1: Thử các gt đầu 1;-1;2;-2;3;-3;...... xác định giá trị VT,VP khi ở nghiệm x

B2:GPT

Bài làm 

Thử vào PT ta thấy x=1 là nghiêm pt và VT=VP=4

có đẳng thức sau: \(\sqrt{a}-\sqrt{b}=\frac{a-b}{\sqrt{a}+\sqrt{b}}\)

Trừ cả hai vế PT cho 4 ta có: \(\left(\sqrt{x^2+8}\right)+3x-2-4=\left(\sqrt{x^2+15}\right)-4\)

\(\left(\sqrt{x^2+8}\right)-\sqrt{9}+\left(3x-3\right)=\left(\sqrt{x^2+15}\right)-\sqrt{16}\)

\(\frac{\left(x^2+8-9\right)}{\sqrt{x^2+8}+3}+3\left(x-1\right)=\frac{x^2+15-16}{\left(\sqrt{x^2+15}\right)+4}\)

\(\frac{\left(x-1\right)\left(x+1\right)}{\sqrt{x^2+8}+3}+3\left(x-1\right)=\frac{\left(x-1\right)\left(x+1\right)}{\left(\sqrt{x^2+15}\right)+4}\)

\(\left(x-1\right)\left(\frac{\left(x+1\right)}{\sqrt{x^2+8}+3}+3-\frac{\left(x+1\right)}{\left(\sqrt{x^2+15}\right)+4}\right)=0\)

Giải tiếp ta có x=1 hoặc cái trong ngoặc kia sẽ có nghiêm hoặc vô nghiêm gì đó