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CN
1
KN
5 tháng 10 2020
\(ĐK:x\ge-1\)
\(x^2+x+12\sqrt{x+1}=36\Leftrightarrow\left(x^2+x-12\right)+\left(12\sqrt{x+1}-24\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)+12\left(\sqrt{x+1}-2\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)+12.\frac{x-3}{\sqrt{x+1}+2}=0\Leftrightarrow\left(x-3\right)\left(x+4+\frac{12}{\sqrt{x+1}+2}\right)=0\)
Dễ thấy \(x+4+\frac{12}{\sqrt{x+1}+2}>0\forall x\ge-1\)nên x - 3 = 0 hay x = 3 (tm)
Vậy nghiệm duy nhất của phương trình là 3
MA
1
HM
15 tháng 12 2018
\(x^2+2x+1-\left(x+1\right)+2\sqrt{x+1}.6-36=0\)
\(\left(x+1\right)^2-\left(\sqrt{x+1}-6\right)^2=0\)
\(\left(x-\sqrt{x+1}+7\right)\left(x+\sqrt{x+1}-5\right)=0\)
\(\left[{}\begin{matrix}x-\sqrt{x+1}+7=0\\x+\sqrt{x+1}-5=0\end{matrix}\right.\)
TN
25 tháng 7 2017
a)\(x^2+x+12\sqrt{x+1}=36\)
\(pt\Leftrightarrow x^2+x-12+12\sqrt{x+1}-24=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4\right)+\frac{144\left(x+1\right)-576}{12\sqrt{x+1}+24}=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4\right)+\frac{144\left(x-3\right)}{12\sqrt{x+1}+24}=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+4+\frac{144}{12\sqrt{x+1}+24}\right)=0\)
Dễ thấy: \(x+4+\frac{144}{12\sqrt{x+1}+24}>0\forall x\ge-1\)
\(\Rightarrow x-3=0\Rightarrow x=3\)
b)\(x+\sqrt{x-2}=2\sqrt{x-1}\)
\(pt\Leftrightarrow x-2+\sqrt{x-2}=2\sqrt{x-1}-2\)
\(\Leftrightarrow x-2+\frac{x-2}{\sqrt{x-2}}=2\left(\sqrt{x-1}-1\right)\)
\(\Leftrightarrow x-2+\frac{x-2}{\sqrt{x-2}}-2\cdot\frac{x-1-1}{\sqrt{x-1}+1}=0\)
\(\Leftrightarrow x-2+\frac{x-2}{\sqrt{x-2}}-2\cdot\frac{x-2}{\sqrt{x-1}+1}=0\)
\(\Leftrightarrow\left(x-2\right)\left(1+\frac{1}{\sqrt{x-2}}-\frac{2}{\sqrt{x-1}+1}\right)=0\)
Suy ra x-2=0=>x=2
c)Áp dụng BĐT \(\sqrt{a}+\sqrt{b}\ge\sqrt{a+b}\) ta có:
\(VT=\sqrt{x+3}+\sqrt{1-x}\)
\(\ge\sqrt{x+3+1-x}=\sqrt{4}=2=VP\)
Xảy ra khi \(\orbr{\begin{cases}x=-3\\x=1\end{cases}}\)
9 tháng 7 2018
1) ĐK: \(x\ge-1\)
\(PT\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)
\(\Leftrightarrow12.\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)
\(\Leftrightarrow x=3\text{ hoặc }\frac{12}{\sqrt{x+1}+2}+x+4=0\) (*)
VT của (*) luôn dương với \(x\ge-1\)
=> x = 3
30 tháng 8 2018
Câu b : \(x^2-5x+14=4\sqrt{x+1}\) ( ĐK : \(x\ge-1\) )
\(\Leftrightarrow x^2-5x+14-4\sqrt{x+1}=0\)
\(\Leftrightarrow\left(x^2-6x+9\right)+\left[\left(x+1\right)-4\sqrt{x+1}+4\right]=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(\sqrt{x+1}-2\right)^2=0\)
Do : \(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\\\left(\sqrt{x+1}-2\right)^2\ge0\end{matrix}\right.\Rightarrow\left(x-3\right)^2+\left(\sqrt{x+1}-2\right)^2=0\Leftrightarrow\left\{{}\begin{matrix}\left(x-3\right)^2=0\\\left(\sqrt{x+1}-2\right)^2=0\end{matrix}\right.\Leftrightarrow x=3\)
Vậy \(x=3\)
30 tháng 8 2018
a. Ta có : x2 + x = 36 - 12\(\sqrt{x+1}\)
⇌ x2 + 2x + 1 = 36 - 12\(\sqrt{x+1}\) + x + 1
⇌ (x+1)2 = ( \(\sqrt{x+1}\) -6)2
⇌ (x+1)2 - ( \(\sqrt{x+1}\) -6)2 = 0
còn lại tự làm nha
12 tháng 12 2022
c: \(\Leftrightarrow\sqrt{4x^2\left(x+2\right)}=3x+1\)
\(\Rightarrow\left\{{}\begin{matrix}4x^2\left(x+2\right)=9x^2+6x+1\\x>=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}4x^3+8x^2-9x^2-6x-1=0\\x>=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}4x^3-x^2-6x-1=0\\x>=-\dfrac{1}{3}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4x^3+4x^2-5x^2-5x-x-1=0\\x>=-\dfrac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+1\right)\left(4x^2-5x-1\right)=0\\x>=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow x=\dfrac{5\pm\sqrt{41}}{8}\)
a: \(\Leftrightarrow\sqrt{5+\sqrt{x-1}}=6-x\)
\(\Leftrightarrow5+\sqrt{x-1}=x^2-12x+36\) và x<=6
=>\(\sqrt{x-1}=x^2-12x+31\) và x<=6
=>x-1=(x^2-12x+22+11)^2
=>\(x\in\varnothing\)
H
1 tháng 7 2019
2,\(pt\Leftrightarrow12\left(\sqrt{x+1}-2\right)+x^2+x-12=0\)
\(\Leftrightarrow12\cdot\frac{x-3}{\sqrt{x+1}+2}+\left(x-3\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)=0\)
Vì \(\left(\frac{12}{\sqrt{x+1}+2}+x+4\right)\ge0\left(\forall x>-1\right)\)
\(\Rightarrow x=3\)
QH
1
KB
19 tháng 3 2019
a ) Đặt \(\sqrt{x+1}=a\Rightarrow x+1=a^2\Rightarrow x=a^2-1\)
Ta có : \(x^2+x+12\sqrt{x+1}=36\)
\(\Leftrightarrow x\left(x+1\right)+12a=36\)
\(\Leftrightarrow a^2\left(a^2-1\right)+12a-36=0\)
\(\Leftrightarrow a^4-a^2+12a-36=0\)
\(\Leftrightarrow a^3\left(a-2\right)+2a^2\left(a-2\right)+3a\left(a-2\right)+18\left(a-2\right)=0\)
\(\Leftrightarrow\left(a-2\right)\left(a^3+2a^2+3a+18\right)=0\)
\(\Leftrightarrow\left(a-2\right)\left[a^2\left(a+3\right)-a\left(a+3\right)+6\left(a+3\right)\right]=0\)
\(\Leftrightarrow\left(a-2\right)\left(a+3\right)\left(a^2-a+6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2\\a=-3\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=2\\\sqrt{x+1}=-3\left(VL\right)\end{matrix}\right.\)
\(\Leftrightarrow x+1=4\Leftrightarrow x=3\)
Vậy ...
b ) \(x^4-8x^2+x+12=0\)
\(\Leftrightarrow\left(x^4-8x^2+16\right)+x-4=0\)
\(\Leftrightarrow\left(x^2-4\right)^2+x-4=0\)
Đặt \(4-x^2=a\) , ta có :
\(a^2+x-4=0\) \(\Rightarrow x=4-a^2\)
Ta có : x = \(4-a^2;a=4-x^2\)
\(\Leftrightarrow x-a=x^2-a^2\)
\(\Leftrightarrow\left(x-a\right)\left(1-x-a\right)=0\)
\(\Leftrightarrow\left(x-4+x^2\right)\left(1-x-4+x^2\right)=0\)
\(\Leftrightarrow\left(x^2+x-4\right)\left(x^2+x-3\right)=0\)
\(\Leftrightarrow...\)
ĐKXĐ : \(x\ge-1\)
\(x^2+x+12\sqrt{x+1}=36\)
\(\Leftrightarrow\left(x^2+2x+1\right)-\left(x+1-12\sqrt{x+1}+36\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2-\left(\sqrt{x+1}-6\right)^2=0\)
\(\Leftrightarrow\left(x+\sqrt{x+1}-5\right)\left(x-\sqrt{x+1}+7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+1}=5-x\\\sqrt{x+1}=x+7\end{matrix}\right.\)
Với \(\sqrt{x+1}=5-x\)
<=> \(\left\{{}\begin{matrix}x+1=x^2-10x+25\\-1\le x\le5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-11x+24=0\\-1\le x\le5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-3\right)\left(x-8\right)=0\\-1\le x\le5\end{matrix}\right.\Leftrightarrow x=3\)
Với \(\sqrt{x+1}=x+7\Leftrightarrow\left\{{}\begin{matrix}x+1=x^2+14x+49\\x\ge-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2+13x+48=0\\x\ge-1\end{matrix}\right.\)
\(\Leftrightarrow∄x\)
Vậy tập nghiệm S = {3}