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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 \(\orbr{\begin{cases}x=-3\\x=1\end{cases}}\)
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
Bài 1:
a: \(=\left|5-\sqrt{3}\right|-\left|\sqrt{3}-2\right|\)
\(=5-\sqrt{3}-2+\sqrt{3}=3\)
b; \(B=\dfrac{\left(2-\sqrt{3}\right)\cdot\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\cdot\sqrt{52-30\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\left(2-\sqrt{3}\right)\cdot\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)}{\sqrt{2}}\)
\(=\dfrac{6\sqrt{3}+10-9-5\sqrt{3}-6\sqrt{3}+10-9+5\sqrt{3}}{\sqrt{2}}\)
\(=\dfrac{20-18}{\sqrt{2}}=\sqrt{2}\)
c: \(C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3+3-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\left(\sqrt{5}-1\right)}=1\)
d: \(A=\left(\sqrt{5}-1\right)\cdot\sqrt{6+2\sqrt{5}}\)
\(=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=5-1=4\)
Bài 1:
b: \(\Leftrightarrow2+\sqrt{3x-5}=x+1\)
\(\Leftrightarrow\sqrt{3x-5}=x-1\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-2x+1=3x-5\\x>=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x^2-5x+6=0\\x>=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;3\right\}\)
c: \(\Leftrightarrow5x+7=16\left(x+3\right)\)
=>16x+48=5x+7
=>11x=-41
hay x=-41/11
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 \(x=3\)
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
a/ ĐKXĐ: \(x\ge-1\)
\(\sqrt{x+1+2\sqrt{x+1}+1}+\sqrt{x+1-6\sqrt{x+1}+9}=2\sqrt{x+1-2\sqrt{x+1}+1}\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x+1}+1\right)^2}+\sqrt{\left(\sqrt{x+1}-3\right)^2}=2\sqrt{\left(\sqrt{x+1}-1\right)^2}\)
\(\Leftrightarrow\sqrt{x+1}+1+\left|\sqrt{x+1}-3\right|=2\left|\sqrt{x+1}-1\right|\)
- Nếu \(\sqrt{x+1}\ge3\Leftrightarrow x\ge8\) pt trở thành:
\(\sqrt{x+1}+1+\sqrt{x+1}-3=2\sqrt{x+1}-2\)
\(\Leftrightarrow-2=-2\) (đúng)
- Nếu \(\sqrt{x+1}-1\le0\Leftrightarrow-1\le x\le0\) pt trở thành:
\(\sqrt{x+1}+1+3-\sqrt{x+1}=2-2\sqrt{x+1}\)
\(\Leftrightarrow\sqrt{x+1}=-1< 0\) (vô nghiệm)
- Nếu \(0< x< 8\) pt trở thành:
\(\sqrt{x+1}+1+3-\sqrt{x+1}=2\sqrt{x+1}-2\)
\(\Leftrightarrow\sqrt{x+1}=3\Rightarrow x=8\left(l\right)\)
Vậy nghiệm của pt đã cho là \(x\ge8\)
b/ ĐKXĐ: \(x\ge\dfrac{-1}{4}\)
Đặt \(\sqrt{x+\dfrac{1}{4}}=t\ge0\Rightarrow x=t^2-\dfrac{1}{4}\) pt trở thành:
\(t^2-\dfrac{1}{4}+\sqrt{t^2+t+\dfrac{1}{4}}=2\)
\(\Leftrightarrow t^2-\dfrac{1}{4}+\sqrt{\left(t+\dfrac{1}{2}\right)^2}=2\)
\(\Leftrightarrow t^2+t+\dfrac{1}{4}-2=0\)
\(\Leftrightarrow4t^2+4t-7=0\Rightarrow\left[{}\begin{matrix}t=\dfrac{-1+2\sqrt{2}}{2}\\t=\dfrac{-1-2\sqrt{2}}{2}< 0\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x=t^2-\dfrac{1}{4}=\left(\dfrac{-1+2\sqrt{2}}{2}\right)^2-\dfrac{1}{4}=2-\sqrt{2}\)
Vậy pt có nghiệm duy nhất \(x=2-\sqrt{2}\)
\(a,\sqrt{3-x}+\sqrt{2-x}=1\)
\(\Rightarrow\sqrt{3+x}=1-\sqrt{2-x}\)
\(\Rightarrow3+x=1-2\sqrt{2-x}+2-x\)
\(\Rightarrow2x+2\sqrt{2-x}=0\)
\(\Rightarrow x+\sqrt{2-x}=0\)
\(\Rightarrow2-x=\left(-x\right)^2\)
\(\Rightarrow2-x=x^2\)
\(\Rightarrow2-x^2-x=0\)
\(\Rightarrow x^2+x-2=0\)
\(\Rightarrow\orbr{\begin{cases}x+2=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=1\end{cases}}}\)
Vậy....
\(x^2+2x+1-\left(x+1\right)+2\sqrt{x+1}.6-36=0\)
\(\left(x+1\right)^2-\left(\sqrt{x+1}-6\right)^2=0\)
\(\left(x-\sqrt{x+1}+7\right)\left(x+\sqrt{x+1}-5\right)=0\)
\(\left[{}\begin{matrix}x-\sqrt{x+1}+7=0\\x+\sqrt{x+1}-5=0\end{matrix}\right.\)