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b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)
\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)
\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)
\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)
\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)
1: \(\Leftrightarrow\left(x+4\right)^2+\sqrt{x}-6x-14x=0\)
\(\Leftrightarrow\left(x+4\right)^2+\sqrt{x}-20x=0\)
\(\Leftrightarrow\left(x+4+5\sqrt{x}\right)\left(x+4-4\sqrt{x}\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\)
=>x=4
2: \(\Leftrightarrow\left(x+2\right)^2+6\sqrt{x}+8x-4\sqrt{x}-4=0\)
\(\Leftrightarrow\left(x+2\right)^2+2\sqrt{x}+8x-4=0\)
\(\Leftrightarrow x^2+4x+4+2\sqrt{x}+8x-4=0\)
\(\Leftrightarrow x^2+12x+2\sqrt{x}=0\)
=>x=0
\(\frac{2x-5}{!x-3!}+1>0\Leftrightarrow\frac{2x-5+!x-3!}{!x-3}>0\)
do !x-3!>0 mọi x khác 3=> Bất phương trình tương đương
\(2x-5+!x-3!>0\Leftrightarrow!x-3!>5-2x\)
TH(1) x<3 <=>3-x>5-2x=> x>2
Kết luận(1) \(2< x< 3\)
TH(2) \(x\ge3\Leftrightarrow x-3>5-2x\Rightarrow3x>8\Rightarrow x>\frac{8}{3}\)
Kết luận(2) \(x\ge3\)
(1)và(2) nghiệm của Bpt là: x>2
Lời giải
a) \(\sqrt{\left(x-4\right)^2\left(x+1\right)}>0\Leftrightarrow\left\{{}\begin{matrix}x\ne4\\x+1>0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ne4\\x>-1\end{matrix}\right.\)
b) \(\sqrt{\left(x+2\right)^2\left(x-3\right)}>0\Rightarrow\left\{{}\begin{matrix}x\ne-2\\x-3>0\end{matrix}\right.\) \(\Rightarrow x>3\)
1/ Đặt \(\sqrt[3]{x^2+5x-2}=t\Rightarrow x^2+5x=t^3+2\)
\(t^3+2=2t-2\)
\(\Leftrightarrow t^3-2t+4=0\)
\(\Leftrightarrow\left(t+2\right)\left(t^2-2t+2\right)=0\)
\(\Rightarrow t=-2\)
\(\Rightarrow\sqrt[3]{x^2+5x-2}=-2\)
\(\Leftrightarrow x^2+5x-2=-8\)
\(\Leftrightarrow x^2+5x+6=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
2/ \(\Leftrightarrow2x+11+3\sqrt[3]{\left(x+5\right)\left(x+6\right)}\left(\sqrt[3]{x+5}+\sqrt[3]{x+6}\right)=2x+11\)
\(\Leftrightarrow\sqrt[3]{\left(x+5\right)\left(x+6\right)}\left(\sqrt[3]{x+5}+\sqrt[3]{x+6}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt[3]{x+5}=0\\\sqrt[3]{x+6}=0\\\sqrt[3]{x+5}=-\sqrt[3]{x+6}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-5\\x=-6\\x+5=-x-6\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-5\\x=-6\\x=-\frac{11}{2}\end{matrix}\right.\)
lời giải
a)
\(\left(x+1\right)\left(2x-1\right)+x\le2x^2+3\)
\(\Leftrightarrow2x^2+x-1+x\le2x^2+3\)
\(\Leftrightarrow2x\le4\Rightarrow x\le2\)
\(\)b) \(\left(x+1\right)\left(x+2\right)\left(x+3\right)-x>x^3+6x^2-5\)
\(\left(x^2+3x+2\right)\left(x+3\right)-x>x^3+6x^2-5\)
\(x^3+3x^2+3x^2+9x+2x+6-x>x^3+6x^2-5\)
\(10x+6>-5\Rightarrow x>-\dfrac{11}{10}\)
c)Đkxđ: x≥0
x+√x>(2√x+3)(√x−1)
⇔x+√x>2x+√x−3
⇔x−3>0
⇔x>3. (tmđk).
a/ ĐKXĐ: ...
\(\Leftrightarrow\left(x^2-6x\right)\left(\sqrt{17-x^2}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-6x=0\\\sqrt{17-x^2}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x\left(x-6\right)=0\\x^2=16\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=6\left(l\right)\\x=4\\x=-4\end{matrix}\right.\)
b/ĐKXĐ: \(x\ge-3\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+4=0\\\sqrt{x+3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-4\left(l\right)\\x=-3\end{matrix}\right.\)
c/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ge1\\x\le1\end{matrix}\right.\) \(\Rightarrow x=1\)
Thay \(x=1\) vào pt thấy ko thỏa mãn
Vậy pt vô nghiệm
d/ ĐKXĐ: \(x\ge2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-4x+3=0\\\sqrt{x-2}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=3\left(l\right)\\x=2\end{matrix}\right.\)