Đk:\(x\ge1\)

\(pt\Leftrightarrow3\left(x-2\right)\sqrt{x-1}\sqrt{x^2+x+1}+18\left(x-1\right)=x\left(x^2+x+1\right)\)

Chia 2 vế của pt cho \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}>0\)ta đc:

\(3\left(x-2\right)\frac{\sqrt{x-1}}{\sqrt{x^2+x+1}}+\frac{18\left(x-1\right)}{x^2+x+1}=x\)

Đặt \(y=\frac{\sqrt{x-1}}{\sqrt{x^2+x+1}}\left(y\ge0\right)\) pt trở thành

\(3\left(x-2\right)y+18y^2-x=0\)

\(\Leftrightarrow\left(3y-1\right)\left(6y+x\right)=0\)

\(\Leftrightarrow3y-1=0\left(y\ge0;x\ge1\Rightarrow6y+x\ge1\right)\)

\(\Leftrightarrow y=\frac{1}{3}\)\(\Leftrightarrow\frac{\sqrt{x-1}}{\sqrt{x^2+x+1}}=\frac{1}{3}\)

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

\(\Leftrightarrow x^2-8x+10=0\)

\(\Leftrightarrow x=4\pm\sqrt{6}\)

Vậy...

\(\sqrt{4x-8}-\sqrt{x-2}=2.\)

ĐK \(x\ge2\)

PT<=> \(2\sqrt{x-2}-\sqrt{x-2}=2\)

<=> \(\sqrt{x-2}=2\)

<=> x-2=4

<=> x=6 (t/m)

Vậ pt có nghiệm x=6

mơn bn nha

bạn đợi mình xíu, mình đang trình bày ^^

Bài giải đây nha bạn https://imgur.com/gallery/NAS59mp

1/

Ta có:  \(\left(1+\sqrt{15}\right)^2\)= 1 + 15 + \(2\sqrt{15}\)= 16 + \(2\sqrt{15}\)

              \(\sqrt{24}^2\)= 24 = 16 + 8

Vì:     \(\sqrt{15}^2\)= 15 < 16 =\(4^2\)

Nên:   \(\sqrt{15}< 4\)

=>       \(2\sqrt{15}< 8\)

=>       \(16+2\sqrt{15}< 24\)

=>      \(\left(1+\sqrt{15}\right)^2< \sqrt{24}^2\)

Vậy     \(1+\sqrt{15}< \sqrt{24}\)

2/

b/    \(3x-7\sqrt{x}=20\)\(\left(x\ge0\right)\)

<=> \(3x-7\sqrt{x}-20=0\)

<=> \(3x-12\sqrt{x}+5\sqrt{x}-20=0\)

<=> \(3\sqrt{x}\left(\sqrt{x}-4\right)+5\left(\sqrt{x}-4\right)=0\)

<=> \(\left(\sqrt{x}-4\right)\left(3\sqrt{x}+5\right)=0\)

<=> \(\sqrt{x}-4=0\)hoặc \(3\sqrt{x}+5=0\)

<=>   \(\sqrt{x}=4\)hoặc \(3\sqrt{x}=-5\)(vô nghiệm)

<=>   \(x=16\)

Vậy S=\(\left\{16\right\}\)

c/    \(1+\sqrt{3x}>3\)

<=> \(\sqrt{3x}>2\)

<=>   \(3x>4\)

<=>  \(x>\frac{4}{3}\)

d/      \(x^2-x\sqrt{x}-5x-\sqrt{x}-6=0\)(\(x\ge0\))

<=>   \(\left(x^2-5x-6\right)-\left(x\sqrt{x}+\sqrt{x}\right)=0\)

<=>   \(\left(x^2-6x+x-6\right)-\left(x\sqrt{x}+\sqrt{x}\right)=0\)

<=>    \([x\left(x-6\right)+\left(x-6\right)]-\sqrt{x}\left(x+1\right)=0\)

<=>   \(\left(x-6\right)\left(x+1\right)-\sqrt{x}\left(x+1\right)=0\)

<=>   \(\left(x+1\right)\left(x-6-\sqrt{x}\right)=0\)

<=>    \(\left(x+1\right)\left(x-3\sqrt{x}+2\sqrt{x}-6\right)=0\) 

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

<=>    \(\left(x+1\right)\left(\sqrt{x}-3\right)\left(\sqrt{x}+2\right)=0\)

<=>     \(x+1=0\)  hoặc \(\sqrt{x}-3=0\)hoặc \(\sqrt{x}+2=0\)

<=>     \(x=-1\)(loại)  hoặc \(x=9\)hoặc \(\sqrt{x}=-2\)(vô nghiệm)

Vậy S={  9 }

Em trục căn thức:

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

<=> \(\frac{-3x+3}{\sqrt{x+3}+2\sqrt{x}}=\frac{-x+1}{\sqrt{2x+2}+\sqrt{3x+1}}\)

=> nhân tử chung là -x + 1 . Tự làm tiếp nhé!

làm như cô thì vẫn cần phải đánh giá rất khó chịu nhé

\(\sqrt{x+3}-2\sqrt{x}=\sqrt{2x+2}-\sqrt{3x+1}\left(ĐKXĐ:x\ge0\right)\)

\(< =>\sqrt{x+3}-\sqrt{2x+2}+\sqrt{3x+1}-2\sqrt{x}=0\)

\(< =>\frac{\sqrt{x+3}^2-\sqrt{2x+2}^2}{\sqrt{x+3}+\sqrt{2x+2}}+\frac{\sqrt{3x+1}^2-4\sqrt{x}^2}{\sqrt{3x+1}+2\sqrt{x}}=0\)

\(< =>\frac{x+3-2x-2}{\sqrt{x+3}+\sqrt{2x+2}}+\frac{3x+1-4x}{\sqrt{3x+1}+2\sqrt{x}}=0\)

\(< =>\frac{1-x}{\sqrt{x+3}+\sqrt{2x+2}}+\frac{1-x}{\sqrt{3x+1}+2\sqrt{x}}=0\)

\(< =>\left(1-x\right)\left(\frac{1}{\sqrt{x+3}+\sqrt{2x+2}}+\frac{1}{\sqrt{3x+1}+2\sqrt{x}}\right)=0< =>x=1\)