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a) \(3x-2\sqrt{x-1}=4\) (ĐK: x ≥ 1)
\(\Rightarrow3x-2\sqrt{x-1}-4=0\)
\(\Rightarrow3x-6-2\sqrt{x-1}+2=0\)
\(\Rightarrow3\left(x-2\right)-2\left(\sqrt{x-1}-1\right)=0\)
\(\Rightarrow3\left(x-2\right)-2.\dfrac{x-2}{\sqrt{x-1}+1}=0\)
\(\Rightarrow\left(x-2\right)\left[3-\dfrac{2}{\sqrt{x-1}+1}\right]=0\)
*TH1: x = 2 (t/m)
*TH2: \(3-\dfrac{2}{\sqrt{x-1}+1}=0\)
\(\Rightarrow3=\dfrac{2}{\sqrt{x-1}+1}\)
\(\Rightarrow3\sqrt{x-1}+3=2\)
\(\Rightarrow3\sqrt{x-1}=-1\) (vô lí)
Vậy S = {2}
b) \(\sqrt{4x+1}-\sqrt{x+2}=\sqrt{3-x}\) (ĐK: \(-\dfrac{1}{4}\le x\le3\) )
\(\Rightarrow\sqrt{4x+1}-3-\sqrt{x+2}+2-\sqrt{3-x}+1=0\)
\(\Rightarrow\dfrac{4x-8}{\sqrt{4x+1}+3}-\dfrac{x-2}{\sqrt{x+2}+2}+\dfrac{x-2}{\sqrt{3-x}+1}=0\)
\(\Rightarrow\left(x-2\right)\left(\dfrac{4}{\sqrt{4x+1}+3}-\dfrac{1}{\sqrt{x+2}+2}+\dfrac{1}{\sqrt{3-x}+1}\right)=0\)
=> x = 2
\(a,3x-2\sqrt{x-1}=4\left(x\ge1\right)\\ \Leftrightarrow-2\sqrt{x-1}=4-3x\\ \Leftrightarrow4\left(x-1\right)=16-24x+9x^2\\ \Leftrightarrow9x^2-28x+20=0\\ \Leftrightarrow\left(x-2\right)\left(9x-10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\left(tm\right)\\x=\dfrac{10}{9}\left(tm\right)\end{matrix}\right.\)
\(b,\sqrt{4x+1}-\sqrt{x+2}=\sqrt{3-x}\left(-\dfrac{1}{4}\le x\le3\right)\\ \Leftrightarrow4x+1+x+2-2\sqrt{\left(4x+1\right)\left(x+2\right)}=3-x\\ \Leftrightarrow-2\sqrt{\left(4x+1\right)\left(x+2\right)}=2-6x\\ \Leftrightarrow\sqrt{4x^2+9x+2}=3x-1\\ \Leftrightarrow4x^2+9x+2=9x^2-6x+1\\ \Leftrightarrow5x^2-15x-1=0\\ \Leftrightarrow\Delta=225+20=245\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15-\sqrt{245}}{10}=\dfrac{15-7\sqrt{5}}{10}\left(ktm\right)\\x=\dfrac{15+\sqrt{245}}{10}=\dfrac{15+7\sqrt{5}}{10}\left(tm\right)\end{matrix}\right.\Leftrightarrow x=\dfrac{15+7\sqrt{5}}{10}\)
Bài 2:
a: Ta có: \(\sqrt{\sqrt{5}-x\sqrt{3}}=\sqrt{8+2\sqrt{15}}\)
\(\Leftrightarrow\sqrt{5}-x\sqrt{3}=8+2\sqrt{15}\)
\(\Leftrightarrow x\sqrt{3}=\sqrt{5}-8-2\sqrt{15}\)
\(\Leftrightarrow x=\dfrac{\sqrt{15}-8\sqrt{3}-6\sqrt{5}}{3}\)
b: Ta có: \(\sqrt{2+\sqrt{\sqrt{x}+3}}=3\)
\(\Leftrightarrow\sqrt{\sqrt{x}+3}=7\)
\(\Leftrightarrow\sqrt{x}=46\)
hay x=2116
Đặt \(A=ab\sqrt{ab}+bc\sqrt{bc}+ac\sqrt{ac}=1.\\ \)( cho đỡ phải đánh máy nhiều )
Ta có : \(\frac{a^6}{a^3+b^3}=a^3-\frac{a^3b^3}{a^3+b^3}\ge a^3-\frac{a^3b^3}{2\sqrt{a^3b^3}}=a^3-\frac{ab\sqrt{ab}}{2}\left(1\right).\)
( do a,b> 0 nên \(a^3+b^3\ge2\sqrt{a^3b^3}\Rightarrow\frac{a^3b^3}{a^3+b^3}\le\frac{a^3b^3}{2\sqrt{a^3b^3}}\))
chứng minh tương tự ta có :
\(\frac{b^6}{b^6+c^6}\ge b^3-\frac{bc\sqrt{bc}}{2}\left(2\right).\); \(\frac{c^6}{c^3+a^3}\ge c^3-\frac{ca\sqrt{ca}}{2}\left(3\right).\)
cộng vế với vế các bđt (1) (2), (3) ta được :
\(P\ge a^3+b^3+c^3-\frac{A}{2}\left(4\right).\)
Áp dụng BĐT Cô si ( AM - GM ) : \(\frac{a^3+b^3}{2}\ge\sqrt{a^3b^3}=ab\sqrt{ab}.\)( làm tương tự 2 lần nữa với a^3, b^3 , c^3 rồi cộng vế với vế ta được )
=> \(a^3+b^3+c^3\ge ab\sqrt{ab}+bc\sqrt{bc}+ca\sqrt{ca}=A\left(5\right).\)
Thay (5) vào (4) ta được :
\(P\ge A-\frac{A}{2}=\frac{A}{2}=\frac{1}{2}.\)
Vậy Pmin = 1/2 khi a = b = c = \(\frac{1}{\sqrt[3]{3}}.\)
a, ĐKXĐ: \(x\ge-\dfrac{1}{3}\)
\(\Leftrightarrow\dfrac{3}{2}.2\sqrt{1+3x}-\dfrac{5}{3}.3\sqrt{1+3x}-\dfrac{1}{4}.4\sqrt{1+3x}=1\\ \Leftrightarrow3\sqrt{1+3x}-5\sqrt{1+3x}-\sqrt{1+3x}=1\\ \Leftrightarrow-3\sqrt{1+3x}=1\\ \Leftrightarrow\sqrt{1+3x}=-\dfrac{1}{3}\left(vô.lí\right)\)
b, \(\Leftrightarrow\sqrt{\left(x-\dfrac{1}{2}\right)^2}=3\\ \Leftrightarrow\left|x-\dfrac{1}{2}\right|=3\\ \Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=3\\x-\dfrac{1}{2}=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
a) ĐKXĐ: \(x\ge-\dfrac{1}{3}\)
\(pt\Leftrightarrow3\sqrt{3x+1}-5\sqrt{3x+1}-\sqrt{3x+1}=1\)
\(\Leftrightarrow-3\sqrt{3x+1}=1\Leftrightarrow\sqrt{3x+1}=-\dfrac{1}{3}\left(VLý\right)\)
Vậy \(S=\varnothing\)
b) \(pt\Leftrightarrow\sqrt{\left(x-\dfrac{1}{2}\right)^2}=3\Leftrightarrow\left|x-\dfrac{1}{2}\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{1}{2}=3\\x-\dfrac{1}{2}=-3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-\dfrac{5}{2}\end{matrix}\right.\)
\(a,PT\Leftrightarrow x\sqrt{3}=x+2\\ \Leftrightarrow3x^2=x^2+4x+4\\ \Leftrightarrow2x^2-4x-4=0\Leftrightarrow x^2-2x-2=0\\ \Delta=4+8=12\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2-2\sqrt{3}}{2}=1-\sqrt{3}\\x=\dfrac{2+2\sqrt{3}}{2}=1+\sqrt{3}\end{matrix}\right.\)
\(b,ĐK:x\ge\dfrac{2}{3}\\ PT\Leftrightarrow3x-2=7-4\sqrt{3}\\ \Leftrightarrow3x=9-4\sqrt{3}\\ \Leftrightarrow x=\dfrac{9-4\sqrt{3}}{3}\left(tm\right)\)
\(c,ĐK:x\ge-1\\ PT\Leftrightarrow\left(x+1-4\sqrt{x+1}+4\right)+\left(x^2-6x+9\right)=0\\ \Leftrightarrow\left(\sqrt{x+1}-2\right)^2+\left(x-3\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+1}=2\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1=4\\x=3\end{matrix}\right.\Leftrightarrow x=3\left(tm\right)\)
1/ab(a+b)+bc(b+c)+ca(c+a)+2abc
= a^2b + ab^2 + b^2c + bc^2 + ca(c+a) + 2abc
= ab^2 + b^2c + a^2b + bc^2 + 2abc + ca(c+a)
=b^2(a+c) + b(a^2 + c^2 + 2ac) + ca(c+a)
=b^2(a+c) + b(a+c)^2 + ca(c+a)
=(c+a)[b^2 + b(a+c) + ca]
=(c+a)[b^2 + ab + bc + ca]
=(c+a)[b(b+a) + c(b+a)]
=(c+a)(b+c)(b+a)
2/VP=-(x-4)
pt trở thành \(\sqrt[3]{16-x^3}=-\left(x-4\right)\)
<=>x=2