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a)ĐK:\(\begin{cases}25x^2-9 \ge 0\\5x+3 \ge 0\\\end{cases}\)
`<=>` \(\begin{cases}(5x-3)(5x+3) \ge 0\\5x+3 \ge 0\\\end{cases}\)
`<=>` \(\begin{cases}\left[ \begin{array}{l}x\ge \dfrac35\\x \le -\dfrac35\end{array} \right.\\\end{cases}\)
`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\x \ge \dfrac35\end{array} \right.\)
`pt<=>\sqrt{5x+3}(\sqrt{5x-3}-2)=0`
`<=>` \(\left[ \begin{array}{l}5x+3=0\\\sqrt{5x-3}=2\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\5x-3=4\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=-\dfrac35\\x=7/5\end{array} \right.\)
`b)sqrt{x-3}/sqrt{2x+1}=2`
ĐK:\(\begin{cases}x-3 \ge 0\\2x+1>0\\\end{cases}\)
`<=>x>=3`
`pt<=>sqrt{x-3}=2sqrt{2x+1}`
`<=>x-3=8x+4`
`<=>7x=7`
`<=>x=1(l)`
`c)sqrt{x^2-2x+1}+sqrt{x^2-4x+4}=3`
`<=>sqrt{(x-1)^2}+sqrt{(x-2)^2}=3`
`<=>|x-1|+|x-2|=3`
`**x>=2`
`pt<=>x-1+x-2=3`
`<=>2x=6`
`<=>x=3(tm)`
`**x<=1`
`pt<=>1-x+2-x=3`
`<=>3-x=3`
`<=>x=0(tm)`
`**1<=x<=2`
`pt<=>x-1+2-x=3`
`<=>=-1=3` vô lý
Vậy `S={0,3}`
\(a,PT\Leftrightarrow x\sqrt{3}=x+2\\ \Leftrightarrow3x^2=x^2+4x+4\\ \Leftrightarrow2x^2-4x-4=0\Leftrightarrow x^2-2x-2=0\\ \Delta=4+8=12\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2-2\sqrt{3}}{2}=1-\sqrt{3}\\x=\dfrac{2+2\sqrt{3}}{2}=1+\sqrt{3}\end{matrix}\right.\)
\(b,ĐK:x\ge\dfrac{2}{3}\\ PT\Leftrightarrow3x-2=7-4\sqrt{3}\\ \Leftrightarrow3x=9-4\sqrt{3}\\ \Leftrightarrow x=\dfrac{9-4\sqrt{3}}{3}\left(tm\right)\)
\(c,ĐK:x\ge-1\\ PT\Leftrightarrow\left(x+1-4\sqrt{x+1}+4\right)+\left(x^2-6x+9\right)=0\\ \Leftrightarrow\left(\sqrt{x+1}-2\right)^2+\left(x-3\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}\sqrt{x+1}=2\\x-3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x+1=4\\x=3\end{matrix}\right.\Leftrightarrow x=3\left(tm\right)\)
\(a,PT\Leftrightarrow\left|x+3\right|=3x-6\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3x-6\left(x\ge-3\right)\\x+3=6-3x\left(x< -3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\left(tm\right)\\x=\dfrac{3}{4}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{9}{2}\\ b,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\1-x=2x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
\(c,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=25x^2-20x+4\\ \Leftrightarrow25x^2-15x=0\\ \Leftrightarrow5x\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\\ d,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow x\in\varnothing\)
\(a,ĐK:\left\{{}\begin{matrix}x\ge5\\x\le3\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
Vậy pt vô nghiệm
\(b,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow0x=2\Leftrightarrow x\in\varnothing\)
\(c,ĐK:x\ge-\dfrac{3}{2}\\ PT\Leftrightarrow x^2+4x+5-2\sqrt{2x+3}=0\\ \Leftrightarrow\left(2x+3-2\sqrt{2x+3}+1\right)+\left(x^2+2x+1\right)=0\\ \Leftrightarrow\left(\sqrt{2x+3}-1\right)^2+\left(x+1\right)^2=0\\ \Leftrightarrow\left\{{}\begin{matrix}2x+3=1\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\x=-1\end{matrix}\right.\Leftrightarrow x=-1\left(tm\right)\\ d,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\x-1=1-2x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
a.
ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1}+\sqrt{x^3+x^2+x+1}=1+\sqrt{\left(x-1\right)\left(x^3+x^2+x+1\right)}\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x^3+x^2+x+1}-1\right)-\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-1}-1\right)\left(\sqrt{x^3+x^2+x+1}-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x^3+x^2+x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^3+x^2+x=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
b.
ĐKXĐ: \(x\ge-1\)
\(x^2-6x+9+x+1-4\sqrt{x+1}+4=0\)
\(\Leftrightarrow\left(x-3\right)^2+\left(\sqrt{x+1}-2\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\\sqrt{x+1}-2=0\end{matrix}\right.\)
\(\Leftrightarrow x=3\)
c.
ĐKXĐ: \(-2\le x\le\dfrac{4}{5}\)
\(VT=2x+3\sqrt{4-5x}+1.\sqrt{x+2}\)
\(VT\le2x+\dfrac{1}{2}\left(9+4-5x\right)+\dfrac{1}{2}\left(1+x+2\right)=8\)
Dấu "=" xảy ra khi và chỉ khi \(x=-1\)
a) ĐK:\(x\ge4\)
\(\sqrt{x-1}+\sqrt{x-4}=\sqrt{x+4}\Leftrightarrow x-1+x-4+2\sqrt{\left(x-1\right)\left(x-4\right)}=x+4\Leftrightarrow9-x=2\sqrt{x^2-5x+4}\left(ĐK:x\le9\right)\Leftrightarrow\left(9-x\right)^2=4\left(x^2-5x+4\right)\Leftrightarrow81-18x+x^2=4x^2-20x+16\Leftrightarrow3x^2-2x-65=0\Leftrightarrow3x^2-15x+13x-65=0\Leftrightarrow3x\left(x-5\right)+13\left(x-5\right)=0\Leftrightarrow\left(x-5\right)\left(3x+13\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x-5=0\\3x+13=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\left(tm\right)\\x=-\dfrac{13}{3}\left(ktm\right)\end{matrix}\right.\)
Vậy S={5}
b)\(\sqrt[3]{2x-1}+\sqrt[3]{x-1}=1\Leftrightarrow\sqrt[3]{2x-1}-1+\sqrt[3]{x-1}=0\Leftrightarrow\dfrac{2x-1-1}{\left(\sqrt[3]{2x-1}\right)^2+2.\sqrt[3]{2x-1}+1}+\dfrac{x-1}{\left(\sqrt[3]{x-1}\right)^2}=0\Leftrightarrow\left(x-1\right)\left[\dfrac{2}{\left(\sqrt[3]{2x-1}+2.\sqrt[3]{2x-1}+1\right)}+\dfrac{1}{\left(\sqrt[3]{x-1}\right)^2}\right]=0\)(1)
Dễ thấy \(\dfrac{2}{\left(\sqrt[3]{2x-1}+2.\sqrt[3]{2x-1}+1\right)}+\dfrac{1}{\left(\sqrt[3]{x-1}\right)^2}>0\)
Vậy (1)\(\Leftrightarrow x-1=0\Leftrightarrow x=1\)
Vậy S={1}
c) ĐK:\(\left[{}\begin{matrix}x\le-4\\x\ge-1\end{matrix}\right.\)
\(5\sqrt{x^2+5x+8}=x^2+5x+4\left(2\right)\)
Đặt a=x2+5x+4(a\(\ge0\))
(2)\(\Leftrightarrow5\sqrt{a+4}=a\Leftrightarrow25\left(a+4\right)=a^2\Leftrightarrow a^2-25a-100=0\Leftrightarrow\)\(\left[{}\begin{matrix}a=\dfrac{25+5\sqrt{41}}{2}\left(tm\right)\\a=\dfrac{25-5\sqrt{41}}{2}\left(ktm\right)\end{matrix}\right.\)\(\Leftrightarrow a=\dfrac{25+5\sqrt{41}}{2}\Leftrightarrow\dfrac{25+5\sqrt{41}}{2}=x^2+5x+4\Leftrightarrow25+5\sqrt{41}=2x^2+10x+8\Leftrightarrow2x^2+10x-17-5\sqrt{41}=0\Leftrightarrow\)\(\left[{}\begin{matrix}x=3,045972466\left(tm\right)\\x=-8,045972466\left(tm\right)\end{matrix}\right.\)
Vậy S={-8,045972466;3,045972466}
c) ĐK:\(\left[{}\begin{matrix}x\le-4\\x\ge-1\end{matrix}\right.\)
\(5\sqrt{x^2+5x+28}=x^2+5x+4\left(1\right)\)
Đặt a=x2+5x+4(a\(\ge0\))
Vậy \(\left(1\right)\Leftrightarrow5\sqrt{a+24}=a\Leftrightarrow25\left(a+24\right)=a^2\Leftrightarrow a^2-25a-600=0\Leftrightarrow a^2-40a+15a-600=0\Leftrightarrow a\left(a-40\right)+15\left(a-40\right)=0\Leftrightarrow\left(a-40\right)\left(a+15\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}a-40=0\\a+15=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}a=40\left(tm\right)\\a=-15\left(ktm\right)\end{matrix}\right.\)
Vậy ta có a=40\(\Leftrightarrow x^2+5x+4=40\Leftrightarrow x^2+5x-36=0\Leftrightarrow x^2-4x+9x-36=0\Leftrightarrow x\left(x-4\right)+9\left(x-4\right)=0\Leftrightarrow\left(x-4\right)\left(x+9\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}x-4=0\\x+9=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\left(tm\right)\\x=-9\left(tm\right)\end{matrix}\right.\)
Vậy S={-9;4}