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\(\sqrt{x+8}=\sqrt{3x+2}+\sqrt{x+3}\) dkxd \(\left\{{}\begin{matrix}x\ge-8\\x\ge\\x\ge-\dfrac{2}{3}\end{matrix}\right.-3\)=>x\(\ge\)\(\dfrac{-2}{3}\)
\(x+8=3x+2+x+3+2\sqrt{\left(3x+2\right)\left(x+3\right)}\)
\(x+8=4x+5+2\sqrt{\left(3x+2\right)\left(x+3\right)}\)
\(x+8-4x-5=2\sqrt{\left(3x+2\right)\left(x+3\right)}\)
-3x+3=\(2\sqrt{\left(3x+2\right)\left(x+3\right)}\)
\(\left\{{}\begin{matrix}-3\left(x-3\right)\ge0\\\left(-3x+3\right)^2=4.\left(3x+2\right)\left(x+3\right)\end{matrix}\right.\)
Chắc tới đây bạn làm đc rồi nhỉ
\(\sqrt{12-\frac{12}{x^2}}+\sqrt{x^2-\frac{12}{x^2}}=x^2\)
\(pt\Leftrightarrow\sqrt{12-\frac{12}{x^2}}-3+\sqrt{x^2-\frac{12}{x^2}}-1=x^2-4\)
\(\Leftrightarrow\frac{12-\frac{12}{x^2}-9}{\sqrt{12-\frac{12}{x^2}}+3}+\frac{x^2-\frac{12}{x^2}-1}{\sqrt{x^2-\frac{12}{x^2}}+1}=x^2-4\)
\(\Leftrightarrow\frac{\frac{3x^2-12}{x^2}}{\sqrt{12-\frac{12}{x^2}}+3}+\frac{\frac{x^4-x^2-12}{x^2}}{\sqrt{x^2-\frac{12}{x^2}}+1}-\left(x^2-4\right)=0\)
\(\Leftrightarrow\frac{\frac{3\left(x-2\right)\left(x+2\right)}{x^2}}{\sqrt{12-\frac{12}{x^2}}+3}+\frac{\frac{\left(x-2\right)\left(x+2\right)\left(x^2+3\right)}{x^2}}{\sqrt{x^2-\frac{12}{x^2}}+1}-\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)\left(\frac{\frac{3}{x^2}}{\sqrt{12-\frac{12}{x^2}}+3}+\frac{\frac{x^2+3}{x^2}}{\sqrt{x^2-\frac{12}{x^2}}+1}-1\right)=0\)
SUy ra x=±2
\(\sqrt{x^2}\)+\(\sqrt{x^2+3}\)+\(2x^2\)+3+2\(\sqrt{x^2\left(x^2+3\right)}\)=12
Đặt \(\sqrt{x^2}\)+\(\sqrt{x^2+3}\)=a (a>0)
=> \(2x^2\)+3+2\(\sqrt{x^2\left(x^2+3\right)}\)= \(a^2\)
Chị QA 114 đấy
\(4x^2-4-3x=\sqrt[3]{x^2\left(x^2-1\right)}\)
\(\Leftrightarrow4\left(x-1\right)\left(x+1\right)-3x=\sqrt[3]{x^2\left(x-1\right)\left(x+1\right)}\)
dat \(\left(x-1\right)\left(x+1\right)=y\)
\(4y-3x=\sqrt[3]{x^2y}\)
\(\Leftrightarrow\left(4y-3x\right)^3=x^2y\)
\(\Leftrightarrow64y^3-144y^2x+108yx^2-27x^3=x^2y\)
\(\Leftrightarrow64y^3-144y^2x+107yx^2-27x^3=0\)
\(\Leftrightarrow64y^3-64y^2x-80y^2x+80x^2y+27x^2y-27x^3=0\)
\(\Leftrightarrow\left(y-x\right)\left(64y^2-80xy+27x^2\right)=0\)
de thay \(64y^2-80xy+27x^2=\left(8y\right)^2-2.8y.5x+25x^2+2x^2=\left(8y-5x\right)^2+2x^2>0\)
\(\Rightarrow y=x\)hay \(\left(x-1\right)\left(x+1\right)=x\Rightarrow x^2-x-1=0\)
\(\left(x-\frac{1}{2}\right)^2-\frac{5}{4}=0\Rightarrow\left(x-\frac{1}{2}\right)^2=\frac{5}{4}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{5}+1}{2}\\x=\frac{-\sqrt{5}+1}{2}\end{cases}}\)
câu b tương tự nhé bạn
a) đk: \(x\ge3\)
Ta có: \(\sqrt{x-3}=3x-11\)
\(\Leftrightarrow x-3=9x^2-66x+121\)
\(\Leftrightarrow9x^2-67x+124=0\)
\(\Leftrightarrow\left(9x^2-36x\right)-\left(31x-124\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(9x-31\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\9x-31=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=4\\x=\frac{31}{9}\end{cases}}\)
a, \(\sqrt{x-3}=3x-11\left(đk:x\ge3\right)< =>\sqrt{x-3}-1=3x-12\)
\(< =>\frac{x-4}{\sqrt{x-3}+1}-3\left(x-4\right)=0< =>\left(x-4\right)\left(\frac{1}{\sqrt{x-3}+1}-3\right)=0\)
\(< =>\orbr{\begin{cases}x-4=0\\\frac{1}{\sqrt{x-3}+1}=3\end{cases}}< =>\orbr{\begin{cases}x=4\left(tm\right)\\\sqrt{x-3}+1=\frac{1}{3}\left(vl\right)\end{cases}}\)
ĐK : \(x+1>0=>x\ge-1\)
Đặt \(\sqrt{x+1}=t=>t\ge0=>x+1=t^2=>x=t^2-1=>x^2=t^4-2t^2+1\)
Khi đó ta có \(t^4-2t^2+1+t^2-1+12t-36=0\)
=>\(t^4-t^2+12t-36=0\)
=>\(t^4-2t^3+2t^3-4t^2+3t^2-6t+18t-36=0\)
=>\(t^3\left(t-2\right)+2t^2\left(t-2\right)+3t\left(t-2\right)+18\left(t-2\right)=0\)
=>\(\left(t-2\right)\left(t^3+2t^2+3t+18\right)=0\)
=>\(\hept{\begin{cases}t=2\\t^3+2t^2+3t+18=0\left(loại\right)do\left(t\ge0=>t^3+2t^2+3t+18>0\right)\end{cases}}\)
=>\(t=2=>x+1=4=>x=3\)(thảo mãn đk)
zậy...
x^2 +x + 12√(x+1) =36
<=> x(x+1) + 12√(x+1) =36 (1)
dặt √(x+1) =t =>
{ x+1 =t^2
{ x=t^2 -1
(1) <=> (t^2 -1)(t^2 ) +12t =36
<=> t^4 -t^2 +12t-36 =0
<=> t^4 -(t-6)^2 =0
<=> (t^2 -t+6)(t^2 +t-6) =0
<=>
[ t^2 -t +6 =0 <=> (t-1/2)^2 + 23/4 =0 (vô lý do (t-1/2)^2 ≥ 0 ,23/4 >0 )
[t^2 +t-6 =0 <=> (t+1/2)^2 =25/4 (2)
(2) <=>
[ t+1/2 =5/2 => t =2 => √(x+1) =2 => x=3
[t+1/2 =-5/2 => t=-3 =>√(x+1) =-3 (vô lý do √(x+1) ≥ 0
vậy pt có nghiệm x=3
bạn ơi nhưng đề bài là 12\(\sqrt{x-1}\) mà