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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
a) VT bạn bình phương rồi B.C.S sẽ được VT<=2
VP=3x^2-12x+12+2=3(x-2)^2+1>=2
Dấu = xảy ra khi x=2
\(\text{Đk: }1,5\le x\le2,5\)
Áp dụng bđt cauchy ta có:
\(\text{VT }\Leftrightarrow\frac{2x-3+1+1-2x+1}{2}=2\)
Mà: \(\text{VP}=3\left(x-2\right)^2+2\ge2\)
\(\text{ĐT}\Leftrightarrow x=2\)
\(\Rightarrow x=2\)
\(1.x^2-4x-2\sqrt{2x-5}+5=0\left(x>=\dfrac{5}{2}\right)\)
\(\text{⇔}2x-5-2\sqrt{2x-5}+1+x^2-6x+9=0\)
\(\text{⇔}\left(\sqrt{2x-5}-1\right)^2+\left(x-3\right)^2=0\)
\(\text{⇔}\sqrt{2x-5}-1=0\) 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...........
\(d.Tuong-tự-nhé-bn\)
b) \(\dfrac{16}{\sqrt{x-3}}+\dfrac{4}{\sqrt{y-1}}+\dfrac{1225}{\sqrt{z-665}}=82-\sqrt{x-3}-\sqrt{y-1}-\sqrt{z-665}\) (*)
Đk: \(\left\{{}\begin{matrix}x>3\\y>1\\z>665\end{matrix}\right.\)
(*) \(\Leftrightarrow\dfrac{16}{\sqrt{x-3}}+\dfrac{4}{\sqrt{y-1}}+\dfrac{1225}{\sqrt{z-665}}=82-\dfrac{x-3}{\sqrt{x-3}}-\dfrac{y-1}{\sqrt{y-1}}-\dfrac{z-665}{\sqrt{z-665}}\)
\(\Leftrightarrow\dfrac{16}{\sqrt{x-3}}+\dfrac{4}{\sqrt{y-1}}+\dfrac{1225}{\sqrt{z-665}}-82+\dfrac{x-3}{\sqrt{x-3}}+\dfrac{y-1}{\sqrt{y-1}}+\dfrac{z-665}{\sqrt{z-665}}=0\)
\(\Leftrightarrow\left(\dfrac{x-3}{\sqrt{x-3}}-\dfrac{8\sqrt{x-3}}{\sqrt{x-3}}+\dfrac{16}{\sqrt{x-3}}\right)+\left(\dfrac{y-1}{\sqrt{y-1}}-\dfrac{4\sqrt{y-1}}{\sqrt{y-1}}+\dfrac{4}{\sqrt{y-1}}\right)+\left(\dfrac{z-665}{\sqrt{z-665}}-\dfrac{70\sqrt{z-665}}{\sqrt{z-665}}+\dfrac{1225}{\sqrt{z-665}}\right)=0\)
\(\Leftrightarrow\dfrac{\left(\sqrt{x-3}-4\right)^2}{\sqrt{x-3}}+\dfrac{\left(\sqrt{y-1}-2\right)^2}{\sqrt{y-1}}+\dfrac{\left(\sqrt{z-665}-35\right)^2}{\sqrt{z-665}}=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-3}-4=0\\\sqrt{y-1}-2=0\\\sqrt{z-665}-35=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=19\\y=5\\z=1890\end{matrix}\right.\)
Kl: x=19, y= 5, z=1890
\(\sqrt{x-2}+\sqrt{y-3}+\sqrt{z-5}=\dfrac{1}{2}\left(x+y+z-7\right)\)
\(\text{⇔}2\sqrt{x-2}+2\sqrt{y-3}+2\sqrt{z-5}=x+y+z-7\)
\(\text{⇔}x-2-2\sqrt{x-2}+1+y-3-2\sqrt{y-3}+1+z-5-2\sqrt{z-5}+1=0\)
\(\text{⇔}\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-1\right)^2+\left(\sqrt{z-5}-1\right)^2=0\)
\(\text{⇔}x=3;y=4;z=6\)
KL..........