\(3x^2+3x+4-\left(3x+2\right)\sqrt{x^2+3}\ge0\)
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giúp mình giải bpt vs
\(\dfrac{\left|2x-1\right|-x}{2x}>1;\dfrac{2-\left|x-2\right|}{x^2-1}\ge0;\dfrac{\sqrt{x+4}-2}{4-9x^2}\le0;\dfrac{x^2-2x-3}{\sqrt[3]{3x-1}+\sqrt[3]{4-5x}}\ge0;\)\(3x^2-10x+3\ge0;\left(\sqrt{2}-x\right)\left(x^2-2\right)\left(2x-4\right)< 0;\dfrac{1}{x+9}-\dfrac{1}{x}>\dfrac{1}{2};\dfrac{2}{1-2x}\le\dfrac{3}{x+1}\)
Rút gọn:a)2sqrt{3x}-4sqrt{3x}+27-2sqrt{3x}(xge0)b)3sqrt{2x}-5sqrt{8x}+7sqrt{8x}+28left(xge0right)c)frac{2}{x^2-y^2}sqrt{frac{3left(x+yright)^2}{2}}left(xge0,yge0,xne yright)d)frac{2}{2a-1}sqrt{5a^2left(1-4a+4a^2right)}
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Rút gọn:
a)\(2\sqrt{3x}-4\sqrt{3x}\)+\(27-2\sqrt{3x}\)(\(x\ge0\))
b)\(3\sqrt{2x}-5\sqrt{8x}\)+\(7\sqrt{8x}+28\)\(\left(x\ge0\right)\)
c)\(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}\)\(\left(x\ge0,y\ge0,x\ne y\right)\)
d)\(\frac{2}{2a-1}\sqrt{5a^2\left(1-4a+4a^2\right)}\)
a) \(2\sqrt{3x}-4\sqrt{3x}+27-2\sqrt{3x}=27-4\sqrt{3x}\)
b) \(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{8x}+28=3\sqrt{2x}+2\sqrt{8x}+28=3\sqrt{2x}+4\sqrt{2x}+28=7\sqrt{2x}+28\)
c) \(\frac{2}{x^2-y^2}\sqrt{\frac{3\left(x+y\right)^2}{2}}=\frac{2}{\left(x-y\right)\left(x+y\right)}.\frac{\sqrt{3}\left|x+y\right|}{\sqrt{2}}=\frac{\sqrt{6}}{x-y}\)
d) \(\frac{2}{2a-1}\sqrt{5a^2\left(1-4x+4a^2\right)}=\frac{2}{2a-1}\sqrt{5a^2\left(2a-1\right)^2}=\frac{2}{2a-1}.\sqrt{5}\left|a\left(2a-1\right)\right|=2a\sqrt{5}\)
Thiếu ĐKXĐ : ..............
a) Ta có: \(2\sqrt{3x}-4\sqrt{3x}+27-2\sqrt{3x}\)
\(=27-4\sqrt{3x}\)
b) Ta có: \(3\sqrt{2x}-5\sqrt{8x}+7\sqrt{8x}+28\)
\(=3\sqrt{2x}-5.2\sqrt{2x}+7.2\sqrt{2x}+28\)
\(=3\sqrt{2x}-10\sqrt{2x}+14\sqrt{2x}+28\)
\(=7\sqrt{2x}+28\)
c) Ta có: \(\frac{2}{x^2-y^2}.\sqrt{\frac{3\left(x+y\right)^2}{2}}\)
\(=\sqrt{\frac{4}{\left(x-y\right)^2.\left(x+y\right)^2}.\frac{3\left(x+y\right)^2}{2}}\)
\(=\sqrt{\frac{2.3}{\left(x-y\right)^2}}\)
\(=\frac{1}{x-y}.\sqrt{6}\)
d) Ta có: \(\frac{2}{2a-1}.\sqrt{5a^2.\left(1-4a+4a^2\right)}\)
\(=\sqrt{\frac{4}{\left(2a-1\right)^2}.5a^2.\left(2a-1\right)^2}\)
\(=2a.\sqrt{5}\)
\(\left(2x+3\right)^2+\left(3x-2\right)^4=0\) vì \(\left(2x+3\right)^2\ge0;\left(3x-2\right)^4\ge0\) nên\(\Rightarrow\hept{\begin{cases}\left(2x+3\right)^2=0\\\left(3x-2\right)^4=0\end{cases}\Rightarrow\hept{\begin{cases}2x+3=0\\3x-2=0\end{cases}}}\) \(\Rightarrow\hept{\begin{cases}x=-\frac{3}{2}\\x=\frac{2}{3}\end{cases}}\)
Giải các bất phương trình sau:
a)\(\left(x^2+3x-4\right)\left(3-2x\right)\)<0
b) \(\dfrac{x^2+3x+4}{x^2-2}\ge0\)
c) \(\dfrac{x\left(x^2+4x+4\right)}{x^2-1}\ge0\)
a. TH1:
\(\left\{{}\begin{matrix}x^2+3x-4< 0\\3-2x>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 1\\x>-4\end{matrix}\right.\\x>\dfrac{3}{2}\end{matrix}\right.\)
TH2:
\(\left\{{}\begin{matrix}x^2+3x-4>0\\3-2x< 0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1\\x< -4\end{matrix}\right.\\x< \dfrac{3}{2}\end{matrix}\right.\)
Vậy nghiệm của BPT:
\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x< 1\\x>-4\end{matrix}\right.\\x>\dfrac{3}{2}\end{matrix}\right.\) \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>1\\x< -4\end{matrix}\right.\\x< \dfrac{3}{2}\end{matrix}\right.\)
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Giải các bất phương trình sau:a) left(x^2+3x-4right)left(3-2xright) 0 dfrac{x^2+3x+4}{x^2-2}ge0 dfrac{xleft(x^2+4x+4right)}{x^2-1}ge0b) dfrac{3x-2}{2-x}le-xc) dfrac{x-3}{x+1}dfrac{x+4}{x+2}d) dfrac{x+2}{x-2}-dfrac{x+3}{x-2}1e) |2x-3|x+1f) |2x-5|le x+1g) x-4-|x^2+3x-4|0h) left|x^2+4x+3right|left|x^2-4x-5right|
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Giải các bất phương trình sau:
a) \(\left(x^2+3x-4\right)\left(3-2x\right)< 0\)
\(\dfrac{x^2+3x+4}{x^2-2}\ge0\)
\(\dfrac{x\left(x^2+4x+4\right)}{x^2-1}\ge0\)
b) \(\dfrac{3x-2}{2-x}\le-x\)
c) \(\dfrac{x-3}{x+1}>\dfrac{x+4}{x+2}\)
d) \(\dfrac{x+2}{x-2}-\dfrac{x+3}{x-2}>1\)
e) \(|2x-3|>x+1\)
f) \(|2x-5|\le x+1\)
g) \(x-4-|x^2+3x-4|>0\)
h) \(\left|x^2+4x+3\right|>\left|x^2-4x-5\right|\)
Câu 2:
Tìm GTLN của biểu thức sau :
Asqrt{3x-5}+sqrt{7-3x}
Giải
ĐK: frac{5}{3}le xlefrac{7}{3}
Cmr: a+ble2sqrt{ab}left(a,bge0right)(*)
Leftrightarrow a^2+2ab+b^2ge4ab
Leftrightarrow a^2-2ab+b^2ge0
Leftrightarrowleft(a-bright)^2ge0 luôn đúng với a,b0
Dấu xảy ra ab
Ta có Asqrt{3x-5}+sqrt{7-3x}
A^2left[3x-5+7-3x+2sqrt{left(3x-5right)left(7-3xright)}right]2+2sqrt{left(3x-5right)left(7-3xright)}
Áp dụng BĐT (*) ta được:
A^2le2+left(3x-5right)+left(7-3xright)4̸
Right...
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Câu 2:
Tìm GTLN của biểu thức sau :
\(A=\sqrt{3x-5}+\sqrt{7-3x}\)
Giải
ĐK: \(\frac{5}{3}\le x\le\frac{7}{3}\)
Cmr: \(a+b\le2\sqrt{ab}\left(a,b\ge0\right)\)(*)
\(\Leftrightarrow a^2+2ab+b^2\ge4ab\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\) luôn đúng với a,b>0
Dấu "=" xảy ra <=> a=b
Ta có \(A=\sqrt{3x-5}+\sqrt{7-3x}\)
=> \(A^2=\left[3x-5+7-3x+2\sqrt{\left(3x-5\right)\left(7-3x\right)}\right]=2+2\sqrt{\left(3x-5\right)\left(7-3x\right)}\)
Áp dụng BĐT (*) ta được:
\(A^2\le2+\left(3x-5\right)+\left(7-3x\right)=4̸\)
\(\Rightarrow A\le2\)
Vậy MaxA=2 <=> \(\left\{{}\begin{matrix}\frac{5}{3}\le x\le\frac{7}{3}\\3x-5=7-3x\end{matrix}\right.\)=>x=2
Giải phương trình, x>0
\(\frac{\left(x^3+3x^2\sqrt{x^3-3x+6}\right)\left(3x-x^3-2\right)}{2+\sqrt{x^3-3x+6}}=4\left[2\sqrt{\left(x^3-3x+6\right)^3}-\left(x^3-3x+6\right)^2\right]\)
bài này chắc đặt \(\sqrt{x^3-3x+6}\)cho nó gọn thôi
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Giải phương trình, x>0
\(\frac{\left(x^3+3x^2\sqrt{x^3-3x+6}\right)\left(3x-x^3-2\right)}{2+\sqrt{x^3-3x+6}}=4\left[2\sqrt{\left(x^3-3x+6\right)^3}-\left(x^3-3x+6\right)^2\right]\)
3 tháng 1 2021 lúc 22:01
Giải phương trình
\(-3x^2+x+3+\left(\sqrt{3x+2}-4\right)\sqrt{3x-2x^2}+\left(x-1\right)\sqrt{3x+2}=0\)