Đặt cái BT thứ nhất là √a thì cái BT sau là √(1/a),khi đó phương trình viết lại(a>0)

√a+√(1/a)=7/4;Bình phương 2 vế suy ra:

a+1/a+2=49/16>>>a+1/a=17/16>>>a^2+1=17/16a>>>16A^2+16-17=0(pt vô nghiệm)

Vậy phương trình vô nghiệm

Nhân liên hợp rồi rút gọn thì ta sẽ ra. Tôi nghĩ vậy

Bài 1: 

\(\Leftrightarrow\left(x^2-6x-7\right)^2-\left(3x^2-12x-9\right)^2=0\)

\(\Leftrightarrow\left(3x^2-12x-9-x^2+6x+7\right)\left(3x^2-12x-9+x^2-6x-7\right)=0\)

\(\Leftrightarrow\left(2x^2-6x-2\right)\left(4x^2-18x-16\right)=0\)

\(\Leftrightarrow\left(x^2-3x-1\right)\left(2x^2-9x-8\right)=0\)

hay \(x\in\left\{\dfrac{3+\sqrt{13}}{2};\dfrac{3-\sqrt{13}}{2};\dfrac{9+\sqrt{145}}{4};\dfrac{9-\sqrt{145}}{4}\right\}\)

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: \(x\ge1;x\ne50\)

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

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

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

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

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

d/ ĐKXĐ: \(x\ge\frac{4}{7};x\ne\frac{13}{7}\)

\(\Leftrightarrow\sqrt{x+1}=\sqrt{7x-4}-3\)

\(\Leftrightarrow\sqrt{x+1}+3=\sqrt{7x-4}\)

\(\Leftrightarrow x+10+6\sqrt{x+1}=7x-4\)

\(\Leftrightarrow3\sqrt{x+1}=3x-7\) (\(x\ge\frac{7}{3}\))

\(\Leftrightarrow9\left(x+1\right)=\left(3x-7\right)^2\)

\(\Leftrightarrow...\)

e/ Giống câu b

f/ ĐKXĐ: \(\left[{}\begin{matrix}x\ge1\\x\le-\frac{1}{3}\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}\sqrt{\frac{3x+1}{2x-1}}=a\ge0\\\sqrt{\frac{x-1}{2x-1}}=b\ge0\end{matrix}\right.\) ta được hệ:

\(\left\{{}\begin{matrix}2a-b=2\\a^2+5b^2=4\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=2a-2\\a^2+5b^2=4\end{matrix}\right.\)

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

\(\Leftrightarrow a^2+20\left(a^2-2a+1\right)-4=0\)

\(\Leftrightarrow21a^2-40a+16=0\) \(\Rightarrow\left[{}\begin{matrix}a=\frac{4}{3}\\a=\frac{4}{7}\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{\frac{3x+1}{2x-1}}=\frac{4}{3}\\\sqrt{\frac{3x+1}{2x-1}}=\frac{4}{7}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\frac{3x+1}{2x-1}=\frac{16}{9}\\\frac{3x+1}{2x-1}=\frac{16}{49}\end{matrix}\right.\) \(\Leftrightarrow...\)

a) ĐKXĐ : \(7\le x\le9\)

đặt \(A=\sqrt{x-7}+\sqrt{9-x}\)

\(\Rightarrow A^2=2+2\sqrt{\left(x-7\right)\left(9-x\right)}\le2+\left(x-7\right)+\left(9-x\right)=4\)

\(\Rightarrow A\le2\)

Mà \(x^2-16x+66=\left(x-8\right)^2+2\ge2\)

\(\Rightarrow VT=VP=2\)

do đó : \(x-7=9-x\Leftrightarrow x=8\)( t/m )

b) ĐKXĐ : \(x\le1\)

Ta có : \(\sqrt{1-x}+\sqrt{\left(x-1\right)\left(x-2\right)}-\left|x-2\right|\sqrt{\frac{x-1}{x-2}}=3\)

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

\(\Leftrightarrow\sqrt{1-x}=3\Leftrightarrow x=-8\left(tm\right)\)

2(sin2xcos\(\frac{9\pi}{4}\) + sin\(\frac{9\pi}{4}\)cosx) + 7\(\sqrt{2}\)sinx + \(\sqrt{2}\)( sinx cos\(\frac{11\pi}{2}\)+sin\(\frac{11\pi}{2}\)cosx ) =4\(\sqrt{2}\)

\(\sqrt{2}\)sin2x + \(\sqrt{2}\)cosx +7\(\sqrt{2}\)sinx -\(\sqrt{2}\)cosx =4\(\sqrt{2}\)

2\(\sqrt{2}\)sinxcosx+7\(\sqrt{2}\)sinx - 4\(\sqrt{2}\) =0

PHẦN CÒN LẠI C TỰ LM NỐT NHÉ

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge16\\y\ge9\end{matrix}\right.\)

Từ pt thứ nhất của hệ:

\(\frac{8xy}{x^2+y^2+6xy}+\frac{17}{8}\left(\frac{x}{y}+\frac{y}{x}\right)=\frac{21}{4}\)

\(\Leftrightarrow\frac{8}{\frac{x}{y}+\frac{y}{x}+6}+\frac{17}{8}\left(\frac{x}{y}+\frac{y}{x}\right)=\frac{21}{4}\)

Đặt \(\frac{x}{y}+\frac{y}{x}=t\ge2\)

\(\Rightarrow\frac{8}{6+t}+\frac{17}{8}t=\frac{21}{4}\)

\(\Leftrightarrow\frac{17}{8}t^2+\frac{15}{2}t-\frac{47}{2}=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=2\\t=-\frac{94}{17}< 0\left(l\right)\end{matrix}\right.\)

\(\Leftrightarrow\frac{x}{y}+\frac{y}{x}=2\Leftrightarrow x^2+y^2=2xy\)

\(\Leftrightarrow\left(x-y\right)^2=0\Leftrightarrow x=y\)

Thay xuống pt dưới:

\(\sqrt{x-16}+\sqrt{x-9}=7\)

\(\Leftrightarrow\sqrt{x-16}-3+\sqrt{x-9}-4=0\)

\(\Leftrightarrow\frac{x-25}{\sqrt{x-16}+3}+\frac{x-25}{\sqrt{x-9}+4}=0\)

\(\Leftrightarrow...\)

\(\sqrt{\frac{-6}{1+x}}=5\)

\(\Leftrightarrow\sqrt{\frac{-6}{1+x}}^2=5^2\)

\(\Leftrightarrow\frac{-6}{1+x}=25\)

\(\Leftrightarrow x+1=\frac{-6}{25}\)

\(\Leftrightarrow x=\frac{-6}{25}-1=\frac{-31}{25}\)

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

\(\Leftrightarrow\sqrt{x-49}=2\)

\(\Leftrightarrow x-49=4\Leftrightarrow x=53\)

\(\sqrt{\frac{1}{4}-2a}=3\)

\(\Leftrightarrow\frac{1}{4}-2a=9\)

\(\Leftrightarrow2a=\frac{1}{4}-\frac{36}{4}\)

\(\Leftrightarrow2a=\frac{-35}{4}\)

\(\Leftrightarrow a=\frac{-35}{8}\)