\(\sqrt{x^2-2x+1}+x\\ =\sqrt{\left(x-1\right)^2}+x\\ =\left|x-1\right|+x\\ =x-1+x\\ =2x-1\)

`->A`

`(x+2)^3 -(x+1)(x^2-x+1)=1`

`<=> x^3 + 6x^2 + 12x + 8 - x^3 -1-1=0`

`<=> 6x^2 +12x +6=0`

`<=> 6(x^2 +2x+1)=0`

`<=> 6(x+1)^2 =0`

`<=> (x+1)^2 =0`

`<=>x+1=0`

`<=>x=-1`

`2^(x-1)=16`

`=> 2^(x-1)=2^4`

`=> x-1=4`

`=>x=4+1`

`=>x=5`

Vậy `x=5`

Bài `2`

`a, 3/4 +1/4 : x=2/5`

`=> 1/4 : x = 2/5 -3/4`

`=> 1/4 : x = -7/20`

`=> x= 1/4 : (-7/20)`

`=>x=-5/7`

`b,2* 3^(x-1)-7=11`

`=> 2* 3^(x-1) =11+7`

`=> 2*3^(x-1) =18`

`=> 3^(x-1) =18:2`

`=>3^(x-1)=9`

`=> 3^(x-1)=3^2`

`=>x-1=2`

`=>x=2+1`

`=>x=3`

`c, (2x-3)^3 =(-1)^(2023) * 27`

`=> (2x-3)^3 =-1*27`

`=> (2x-3)^3 =-27`

`=>(2x-3)^3 = -3^3`

`=> 2x-3=-3`

`=>2x=-3+3`

`=>2x=0`

`=>x=0:2`

`=>x=0`

`136 -4^2 : x=134`

`=> 136-16:x=134`

`=> 16:x= 136-134`

`=> 16:x=2`

`=>x=16:2`

`=>x=8`

__

`19-5(x-1)=4`

`=> 5(x-1)=19-4`

`=> 5(x-1) =15`

`=> x-1=15:5`

`=>x-1=3`

`=>x=3+1`

`=>x=4`

__

`257-(x^2-1) =177`

`=>x^2-1=257-177`

`=>x^2-1=80`

`=>x^2=80+1`

`=>x^2=81`

`=>x^2=(+- 9)^2`

`=> x= +- 9`

__

`(x^2-17) : 8=2^3`

`=> x^2 -17 = 8 * 8`

`=> x^2 -17= 64`

`=> x^2 =64+17`

`=>x^2=81`

`=>x^2=(+- 9)^2`

`=> x= +- 9`

\(a,\sqrt{4x-20}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9x-45}=4\left(đkxđ:x\ge5\right)\\ \Leftrightarrow\sqrt{4\left(x-5\right)}+\sqrt{x-5}-\dfrac{1}{3}\sqrt{9\left(x-5\right)}=4\\ \Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\\ \Leftrightarrow2\sqrt{x-5}=4\\ \Leftrightarrow\sqrt{x-5}=2\\ \Leftrightarrow x-5=4\\ \Leftrightarrow x=9\left(tm\right)\)

\(b,\sqrt{2x-1}-\sqrt{8x-4}+5=0\left(đkxđ:x\ge\dfrac{1}{2}\right)\\ \Leftrightarrow\sqrt{2x-1}-\sqrt{4\left(2x-1\right)}=-5\\ \Leftrightarrow\sqrt{2x-1}-2\sqrt{2x-1}=-5\\ \Leftrightarrow-\sqrt{2x-1}=-5\\ \Leftrightarrow\sqrt{2x-1}=5\\ \Leftrightarrow2x-1=25\\ \Leftrightarrow2x=26\\ \Leftrightarrow x=13\left(tm\right)\)

\(c,\sqrt{x^2-10x+25}=2\\ \Leftrightarrow\sqrt{\left(x-5\right)^2}=2\\ \Leftrightarrow\left|x-5\right|=2\\ \Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)

\(d,\sqrt{x^2-14x+49}-5=0\\ \Leftrightarrow\sqrt{\left(x-7\right)^2}=5\\ \Leftrightarrow\left|x-7\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}x-7=5\\x-7=-5\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=12\\x=2\end{matrix}\right.\)

\(a,Q=\left(\dfrac{4\sqrt{x}}{x+2\sqrt{x}}+\dfrac{8\sqrt{x}}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\\ =\left(\dfrac{4\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}-\dfrac{8\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2}{\sqrt{x}}\right)\\ =\dfrac{4\sqrt{x}\left(\sqrt{x}-2\right)-8\sqrt{x}\cdot\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\dfrac{\sqrt{x}-1-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{4x-8\sqrt{x}-8x}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\sqrt{x}-1-2\sqrt{x}+4}\)

\(=\dfrac{-4x-8\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}\\ =\dfrac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\\ =\dfrac{-4\sqrt{x}}{3-\sqrt{x}}\)

`b,` Để `Q<4` ta có :

\(\dfrac{-4\sqrt{x}}{3-\sqrt{x}}< 4\\ \Leftrightarrow\dfrac{-4\sqrt{x}}{3-\sqrt{x}}-4< 0\\ \Leftrightarrow\dfrac{-4\sqrt{x}-4\left(3-\sqrt{x}\right)}{3-\sqrt{x}}< 0\\ \Leftrightarrow-4\sqrt{x}-12+4\sqrt{x}< 0\\ \Leftrightarrow-12< 0\left(luon.dung\right)\)

C

A

D

B

C

A

C

Bài `1`

`a,<`

`b,<`

`c,=`

`d,>`

\(\dfrac{5}{\sqrt{7}+\sqrt{2}}-\sqrt{8-2\sqrt{7}}+\sqrt{2}\\ =\dfrac{5\left(\sqrt{7}-\sqrt{2}\right)}{7-2}-\sqrt{7-2\sqrt{7}+1}+\sqrt{2}\\ =\dfrac{5\left(\sqrt{7}-\sqrt{2}\right)}{5}-\sqrt{\left(\sqrt{7}\right)^2-2\cdot\sqrt{7}\cdot1+1^2}+\sqrt{2}\\ =\sqrt{7}-\sqrt{2}-\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{2}\\ =\sqrt{7}-\sqrt{2}-\sqrt{7}+1+\sqrt{2}\\ =1\)