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.

ĐK: x>0

Đặt a=1/x ta được: a>0

\(a+\frac{1}{3}=\sqrt{\frac{1}{9}+a\sqrt{\frac{4}{9}+2a^2}}\)

\(\Leftrightarrow a^2+\frac{1}{9}+\frac{2}{3}a=\frac{1}{9}+a\sqrt{\frac{4}{9}+2a^2}\)

<=>\(a^2+\frac{2}{3}a=a\sqrt{\frac{4}{9}+2a^2}\)

<=>\(a.\left(a+\frac{2}{3}\right)=a\sqrt{\frac{4}{9}+2a^2}\)

<=>\(a+\frac{2}{3}=\sqrt{\frac{4}{9}+2a^2}\)

<=>\(a^2+\frac{4}{9}+\frac{4}{3}a=\frac{4}{9}+2a^2\)

<=>\(a^2-\frac{4}{3}a=0\Leftrightarrow a=0\left(loại\right);a=\frac{4}{3}\)

<=>\(x=\frac{3}{4}\)(loại -3/2)

Vậy x=3/4

b) Đặt \(\sqrt{x^2-6x+6}=a\left(a\ge0\right)\)

\(\Rightarrow a^2+3-4a=0\)

=> (a - 3).(a - 1) = 0

=> \(\left[{}\begin{matrix}a=3\\a=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\sqrt{x^2-6x+6}=3\\\sqrt{x^2-6x+6}=1\end{matrix}\right.\)

Bình phương lên giải tiếp nhé!

c) Tương tư câu b nhé

a/ \(x^4+y^4=1\Rightarrow\left\{{}\begin{matrix}x^4\le1\\y^4\le1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left|x\right|\le1\\\left|y\right|\le1\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x^4\ge x^6\\y^4\ge y^6\end{matrix}\right.\) \(\Rightarrow x^6+y^6\le x^4+y^4\le1\)

Dấu "=" xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}x^4=x^6\\y^4=y^6\\x^4+y^4=1\end{matrix}\right.\)

\(\Leftrightarrow\left(x;y\right)=\left(1;0\right);\left(0;1\right);\left(-1;0\right);\left(0;-1\right)\)

b/ \(\Rightarrow x^9+y^4=1.\left(x^4+y^4\right)\)

\(\Rightarrow x^9+y^9=\left(x^5+y^5\right)\left(x^4+y^4\right)\)

\(\Rightarrow x^9+y^9=x^9+y^9+x^5y^4+x^4y^5\)

\(\Rightarrow x^4y^4\left(x+y\right)=0\Rightarrow\left[{}\begin{matrix}xy=0\\x=-y\left(l\right)\end{matrix}\right.\)

\(\Rightarrow\left(x;y\right)=\left(0;1\right);\left(1;0\right)\)

Tớ đã trả lời ở câu hỏi mới nhất r nên xin phép được xóa câu hỏi này nhé

a: ĐKXĐ: x<>-2/3

\(\frac{2x+1}{3x+2}=5\)

=>5(3x+2)=2x+1

=>15x+10=2x+1

=>13x=-9

=>\(x=-\frac{9}{13}\) (nhận)

b: ĐKXĐ: x∉{1;3}

\(\frac{2x^2-5x+2}{x-1}=\frac{2x^2+x+15}{x-3}\)

=>\(\left(2x^2-5x+2\right)\left(x-3\right)=\left(2x^2+x+15\right)\left(x-1\right)\)

=>\(2x^3-6x^2-5x^2+15x+2x-6=2x^3-2x^2+x^2-x+15x-15\)

=>\(-11x^2+17x-6=-x^2+14x-15\)

=>\(-10x^2+3x+9=0\)

=>\(10x^2-3x-9=0\)

=>\(x^2-\frac{3}{10}x-\frac{9}{10}=0\)

=>\(x^2-2\cdot x\cdot\frac{3}{20}+\frac{9}{400}-\frac{9}{400}-\frac{9}{10}=0\)

=>\(\left(x-\frac{3}{20}\right)^2=\frac{9}{400}+\frac{9}{10}=\frac{9}{400}+\frac{360}{400}=\frac{369}{400}\)

=>\(x-\frac{3}{20}=\pm\frac{3\sqrt{41}}{20}\)

=>\(\left[\begin{array}{l}x=\frac{3\sqrt{41}+3}{20}\left(nhận\right)\\ x=\frac{-3\sqrt{41}+3}{20}\left(nhận\right)\end{array}\right.\)

c: ĐKXĐ: x∉{3;-3}

\(\frac{2x+3}{x-3}-\frac{4}{x+3}=\frac{24}{x^2-9}+2\)

=>\(\frac{\left(2x+3\right)\left(x+3\right)-4\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{24+2\left(x^2-9\right)}{\left(x-3\right)\left(x+3\right)}\)

=>(2x+3)(x+3)-4(x-3)=\(24+2x^2-18\)

=>\(2x^2+6x+3x+9-4x+12=2x^2+6\)

=>5x+21=6

=>5x=-15

=>x=-3(loại)

b: Đặt \(\sqrt{x+1}=a;\sqrt{4-x}=b\)

=>\(a+b+ab=5\)

\(\Leftrightarrow\sqrt{a^2+b^2-2ab}=5-ab\)

=>\(a^2+b^2-2ab=\left(5-ab\right)^2\)

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

\(\Leftrightarrow5-2ab=\left(5-ab\right)^2\)

=>5-2ab=25-10ab+a^2b^2

=>a^2b^2-10ab+25-5+2ab=0

=>a^2b^2-8ab+20=0

=>\(\Leftrightarrow\left(a,b\right)\in\varnothing\)

a: \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+1}-3\sqrt{y-1}=-1\\2\sqrt{x+1}+5\sqrt{y-1}=9\end{matrix}\right.\)

=>\(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x+1}-6\sqrt{y-1}=-2\\2\sqrt{x+1}+5\sqrt{y-1}=18\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11\sqrt{y-1}=-20\\\sqrt{x-1}-3\sqrt{y-1}=-1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}y-1=\dfrac{400}{121}\\x-1=\dfrac{2401}{121}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2522}{121}\\y=\dfrac{521}{121}\end{matrix}\right.\)