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a/ \(-4\le x\le1\)
\(\Leftrightarrow1-x+4+x+2\sqrt{4-3x-x^2}=9\)
\(\Leftrightarrow\sqrt{4-3x-x^2}=2\)
\(\Leftrightarrow-3x-x^2=0\)
\(\Leftrightarrow x\left(x+3\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
b/ ĐKXĐ: \(x\ge-\frac{3}{2}\)
\(\Leftrightarrow x^2+2x+1+\left(2x+3-2\sqrt{2x+3}+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2+\left(\sqrt{2x+3}-1\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=0\\\sqrt{2x+3}-1=0\end{matrix}\right.\) \(\Rightarrow x=-1\)
c/ ĐKXĐ: \(x;y\ge4\)
\(\Leftrightarrow\frac{2\left(x\sqrt{y-4}+y\sqrt{x-4}\right)}{xy}=1\)
\(\Leftrightarrow\frac{2\sqrt{y-4}}{y}+\frac{2\sqrt{x-4}}{x}=1\)
Mặt khác theo BĐT Cauchy:
\(\frac{2\sqrt{y-4}}{y}+\frac{2\sqrt{x-4}}{x}\le\frac{2^2+y-4}{2y}+\frac{2^2+x-4}{2x}=\frac{1}{2}+\frac{1}{2}=1\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}\sqrt{x-4}=2\\\sqrt{y-4}=2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=8\\y=8\end{matrix}\right.\)
a,\(\sqrt{1-x}-1+\sqrt{4+x}-2=0\)\(\Leftrightarrow x\left(\frac{1}{\sqrt{1-x}+1}+\frac{1}{\sqrt{4+x}+2}\right)=0\Rightarrow x=0\)
b,\(\Leftrightarrow x^2+6x+9=2x+3+2\sqrt{2x+3}+1\)
\(\Leftrightarrow\left(x+3\right)^2=\left(\sqrt{2x+3}+1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3-\sqrt{2x+3}-1=0\\x+3+\sqrt{2x+3}+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{2x+3}+2=0\\x+\sqrt{2x+3}+4=0\end{matrix}\right.\)
a,Ta có :\(x=\sqrt[3]{4\left(\sqrt{5}+1\right)}-\sqrt[3]{4\left(\sqrt{5}-1\right)}\)
\(\Rightarrow x^3=4\left(\sqrt{5}+1\right)-4\left(\sqrt{5}-1\right)-3\sqrt[3]{4\left(\sqrt{5}-1\right).4\left(\sqrt{5}+1\right)}.\left(\sqrt[3]{4\left(\sqrt{5}+1\right)}-\sqrt[3]{4\left(\sqrt{5}-1\right)}\right)\)\(\Rightarrow x^3=8-3\sqrt[3]{16\left(\sqrt{5}+1\right)\left(\sqrt{5}-1\right)}.x\)
\(\Rightarrow x^3=8-3\sqrt[3]{64}.x\Rightarrow x^3=8-12x\)\(\Rightarrow x^3-12x+8=0\)
Vậy \(x^3+12x-8=0\)
b,\(\left(x+\sqrt{x^2+3}\right)\left(y+\sqrt{y^2+3}\right)=3\)(1)
Ta có :\(3=\left(x^2+3\right)-x^2=\left(\sqrt{x^2+3}-x\right)\left(\sqrt{x^2+3}+x\right)\)(2)
\(3=\left(y^2+3\right)-y^2=\left(\sqrt{y^2+3}-y\right)\left(\sqrt{y^2+3}+y\right)\) (3)
Từ (1) và (2) ta suy ra :\(y+\sqrt{y^2+3}=\sqrt{x^2+3}-x\)
Từ (1) và (3) ta suy ra :\(x+\sqrt{x^2+3}=\sqrt{y^2+3}-y\)
Cộng 2 đẳng thức trên vế theo vế ta được :
\(x+y+\sqrt{x^2+3}+\sqrt{y^2+3}=\sqrt{x^2+3}+\sqrt{y^2+3}-x-y\)
\(\Leftrightarrow2\left(x+y\right)=0\Leftrightarrow x+y=0\)
Vậy B=0
\(1.x^2-4x-2\sqrt{2x-5}+5=0\left(x>=\dfrac{5}{2}\right)\)
\(\text{⇔}2x-5-2\sqrt{2x-5}+1+x^2-6x+9=0\)
\(\text{⇔}\left(\sqrt{2x-5}-1\right)^2+\left(x-3\right)^2=0\)
\(\text{⇔}\sqrt{2x-5}-1=0\) hoặc \(x-3=0\)
\(\text{⇔}x=3\left(TM\right)\)
KL...........
\(2.x+y+4=2\sqrt{x}+4\sqrt{y-1}\)
\(\text{⇔}x-2\sqrt{x}+1+y-1-4\sqrt{y-1}+4=0\)
\(\text{⇔}\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-2\right)^2=0\)
\(\text{⇔}x=1;y=5\)
KL..........
\(3.\sqrt{x-2}+\sqrt{y-3}+\sqrt{z-5}=\dfrac{1}{2}\left(x+y+z-7\right)\)
\(\text{⇔}2\sqrt{x-2}+2\sqrt{y-3}+2\sqrt{z-5}=x+y+z-7\)
\(\text{⇔}x-2-2\sqrt{x-2}+1+y-3-2\sqrt{y-3}+1+z-5-2\sqrt{z-5}+1=0\)
\(\text{⇔}\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-1\right)^2+\left(\sqrt{z-5}-1\right)^2=0\)
\(\text{⇔}x=1;y=4;z=6\)
KL...........
\(d.Tuong-tự-nhé-bn\)
a)
\(\sqrt{1-x}+\sqrt{4+x}=3\) \(\left(-4\le x\le1\right)\)
Đặt \(\left\{{}\begin{matrix}\sqrt{1-x}=a\\\sqrt{4+x}=b\end{matrix}\right.\)\(\left(a,b\ge0\right)\), ta có hpt:
\(\left\{{}\begin{matrix}a+b=3\left(1\right)\\a^2+b^2=5\end{matrix}\right.\)
\(\left(1\right)\Leftrightarrow a^2+2ab+b^2=9\)
\(\Leftrightarrow5+2ab=9\)
\(\Leftrightarrow ab=2\). Thay \(b=3-a\) vào, ta có:
\(\Rightarrow a\left(3-a\right)=2\)
\(\Leftrightarrow a^2-3a+2=0\)
\(\Leftrightarrow\left(a-1\right)\left(a-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=1\\a=2\end{matrix}\right.\) (nhận)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{1-x}=1\\\sqrt{1-x}=2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\) (nhận)
b)
\(x^2+4x+5=2\sqrt{2x+3}\)
\(\Leftrightarrow\left(x+2\right)^2+1=2\sqrt{2\left(x+2\right)-1}\). Đặt \(x+2=t\), ta có:
\(\Rightarrow t^2+1=2\sqrt{2t-1}\)
\(\Leftrightarrow t^4+2t^2+1=8t-4\)
\(\Leftrightarrow\left(t^2+2t+5\right)\left(t-1\right)^2=0\)
\(\Leftrightarrow t=1\) vì \(t^2+2t+5=\left(t+1\right)^2+4\ge4>0\)
\(\Leftrightarrow x+2=1\)
\(\Leftrightarrow x=-1\)