a) đk: \(1\le x\le5\)

 \(\sqrt[4]{5-x}+\sqrt[4]{x-1}=\sqrt{2}\)

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

<=> \(5-x+x-1+4\sqrt[4]{5-x}^3.\sqrt[4]{x-1}+6\sqrt[4]{5-x}^2.\sqrt[4]{x-1}^2+4\sqrt[4]{5-x}.\sqrt[4]{x-1}^3=4\)

<=> \(\sqrt[4]{\left(5-x\right)\left(x-1\right)}.\left(2\sqrt[4]{5-x}^2+3\sqrt[4]{5-x}.\sqrt[4]{x-1}+2\sqrt[4]{x-1}^2\right)=0\)

<=> \(\left[{}\begin{matrix}\sqrt[4]{\left(5-x\right)\left(x-1\right)}=0\left(2\right)\\2\sqrt[4]{5-x}^2+3\sqrt[4]{\left(5-x\right)\left(x-1\right)}+2\sqrt[4]{x-1}^2=0\left(1\right)\end{matrix}\right.\)

Giải (2) <=> \(\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\left(tm\right)\)

Giải (1) : Đặt \(\sqrt[4]{5-x}=a;\sqrt[4]{x-1}=b\)(đk : a, b \(\ge\)0)

Khi đó, ta có: \(2a^2+3ab+2b^2=0\)

<=> 2(a² + 3/2ab + 9/16b²) + \(\dfrac{7}{8}b^2=0\)

<=> \(2\left(a+\dfrac{3}{4}b\right)^2+\dfrac{7}{8}b^2=0\)

<=> \(\left\{{}\begin{matrix}a+\dfrac{3}{4}b=0\\b=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}a=0\\b=0\end{matrix}\right.\)

<=> \(\left\{{}\begin{matrix}\sqrt[4]{x-1}=0\\\sqrt[4]{5-x}=0\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)(vô lí)

Bài 1: 

Để M có nghĩa thì \(\left\{{}\begin{matrix}x+4\ge0\\2-x\ge0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-4\\x\le2\end{matrix}\right.\Leftrightarrow-4\le x\le2\)

Số giá trị nguyên thỏa mãn điều kiện là:

\(\left(2+4\right)+1=7\)

\(\sqrt{x-5+4\sqrt{x-5}+4}+\sqrt{x-5-4\sqrt{x-5}+4}=2\left(x-17\right)\)

⇔ \(\sqrt{\left(\sqrt{x-5}+4\right)^2}+\sqrt{\left(\sqrt{x-5}-4\right)^2}=2\left(x-17\right)\)

⇔ \(\left|\sqrt{x-5}+4\right|+\left|\sqrt{x-5}-4\right|=2\left(x-17\right)\) (1)

Do \(x\ge17\) nên từ (1) suy ra \(2\sqrt{x-5}=2\left(x-17\right)\)

⇔ \(x-5-\sqrt{x-5}-12=0\)

⇔ \(\left[{}\begin{matrix}\sqrt{x-5}=4\\\sqrt{x-5}=-3\end{matrix}\right.\)⇔ \(x=21\) (t/m)

`a)\sqrt{9-4sqrt5}-sqrt5`

`=sqrt{5-2.2sqrt5+4}-sqrt5`

`=sqrt{(sqrt5-2)^2}-sqrt5`

`=|\sqrt5-2|-sqrt5`

`=sqrt5-2-sqrt5=-2`

`b)\sqrt{7-4sqrt3}+sqrt{4-2sqrt3}`

`=\sqrt{4-2.2sqrt3+3}+\sqrt{3-2sqrt3+1}`

`=sqrt{(2-sqrt3)^2}+sqrt{(sqrt3-1)^2}`

`=|2-sqrt3|+|sqrt3-1|`

`=2-sqrt3+sqrt3-1=1`

`c)(x-49)/(sqrtx-7)(x>=0,x ne 49)`

`=((sqrtx-7)(sqrtx+7))/(sqrtx-7)`

`=sqrtx+7`

`d)\sqrt{4+2\sqrt3}-\sqrt{13+4sqrt3}`

`=\sqrt{3+2sqrt3+1}-\sqrt{12+2.2sqrt3+1}`

`=sqrt{(sqrt3+1)^2}-\sqrt{(2sqrt3+1)^2}`

`=sqrt3+1-2sqrt3-1=-sqrt3`

`e)2+sqrt{17-4sqrt{9+4sqrt{45}}}`(câu này hơi sai)

2:

a: =căn 17-4-căn 17=-4

b: =5-2căn 3-2căn 3=5-4căn 3

1:

a: =>|x+1|=-x

=>x<=0 và (x+1)^2=x^2

=>x<=0 và (x+1+x)(x+1-x)=0

=>x=-1/2

a) ĐKXĐ: 1\(\le x\le7\)

phương trình <=> \(x-1-2\sqrt{x-1}+2\sqrt{7-x}-\sqrt{\left(7-x\right)\left(x-1\right)}=0\\ \Leftrightarrow\sqrt{x-1}\left(\sqrt{x-1}-2\right)-\sqrt{7-x}\left(\sqrt{x-1}-2\right)=0\\ \Leftrightarrow\left(\sqrt{x-1}-2\right)\left(\sqrt{x-1}-\sqrt{7-x}\right)=0\\\Leftrightarrow\left[{}\begin{matrix}x-1=4\\x-1=7-x\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=4\end{matrix}\right.\left(thoả.mãn\right) \)

Vậy S={5,4} là tập nghiệm của phương trình

b) PT <=> \(2x^2-6x+4=\sqrt[2]{\left(x+2\right)\left(x^2-2x+4\right)}\)

Đặt \(\sqrt[2]{x+2}=y,\sqrt[2]{x^2-2x+4}=z\) (y,z>=0)

=> z^2-y^2=x^2-3x+2

pt<=> 2z^2-2y^2=3yz <=> (2z+y)(z-2y)=0

đến đó tự làm tự đặt dkxd

c) Đặt 2 cái căn là a,b => 2a+b=8

và 2a^3 -b^3=1

Thế b=8-2a. pt<=> 2a^3 -(8-2a)^3=1. Đến đó tự giải

\( 1)\sqrt[3]{{12 - x}} + \sqrt[3]{{14 + x}} = 2\\ \Leftrightarrow 12 - x + 3\sqrt[3]{{{{\left( {12 - x} \right)}^2}.\left( {14 + x} \right)}} + 3\sqrt[3]{{\left( {12 - x} \right){{\left( {14 + x} \right)}^2}}} + 14 + x = 8\\ \Leftrightarrow 3\sqrt[3]{{\left( {12 - x} \right)\left( {14 + x} \right)}}\left( {\sqrt[3]{{12 - x}} + \sqrt[3]{{14 + x}}} \right) = - 18\\ \Leftrightarrow 3\sqrt[3]{{\left( {12 - x} \right)\left( {14 + x} \right)}}.2 = - 18\\ \Leftrightarrow \sqrt[3]{{\left( {12 - x} \right)\left( {14 + x} \right)}} = - 3\\ \Leftrightarrow \left( {12 - x} \right)\left( {14 + x} \right) = {\left( { - 3} \right)^3}\\ \Leftrightarrow 168 - 2x - {x^2} = - 27\\ \Leftrightarrow {x^2} + 2x - 195 = 0\\ \Leftrightarrow \left[ \begin{array}{l} x = - 15\\ x = 13 \end{array} \right. \)

Vậy...

1.

Đặt\(\left\{{}\begin{matrix}u=\sqrt[3]{12-x}\\v=\sqrt[3]{14+x}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u^3=12-x\\v^3=14+x\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}u^3+v^3=26\\u+v=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}\left(u+v\right)\left(u^2-uv+v^2\right)=26\\u+v=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u^2-uv+v^2=13\\v=2-u\end{matrix}\right.\)

\(\Rightarrow u^2-u\left(2-u\right)+\left(2-u\right)^2=13\) \(\Leftrightarrow3u^2-6u-9=0\) \(\Rightarrow\left[{}\begin{matrix}u=3\Rightarrow v=-1\\u=-1\Rightarrow v=3\end{matrix}\right.\) Tìm x.

2.ĐK: \(-40\le x\le57\)

Đặt \(\left\{{}\begin{matrix}\sqrt[4]{57-x}=u\\\sqrt[4]{x+40}=v\end{matrix}\right.\) \(\left(u,v\ge0\right)\) \(\Rightarrow\left\{{}\begin{matrix}u^4=57-x\\v^4=x+40\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}u+v=5\\u^4+v^4=97\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}u^2+v^2=25-2uv\\\left(u^2+v^2\right)^2-2u^2v^2=97\end{matrix}\right.\) \(\Rightarrow\left(25-2uv\right)^2-2u^2v^2=97\)

\(\Leftrightarrow2u^2v^2-100uv+528=0\) \(\Rightarrow\left[{}\begin{matrix}uv=44\\uv=6\end{matrix}\right.\) Kết hợp \(u+v=5\) giải 2 trường hợp.

3.

ĐK: \(-\sqrt{17}\le x\le\sqrt{17}\)

Đặt \(x+\sqrt{17-x^2}=t\) \(\Rightarrow\frac{t^2-17}{2}=x\sqrt{17-x^2}\)

\(PT\Leftrightarrow t+\frac{t^2-17}{2}=9\) \(\Leftrightarrow t^2+2t-35=0\Rightarrow\left[{}\begin{matrix}t=5\\t=-7\end{matrix}\right.\) Giải tiếp.