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\(a,\left(đk:x\ge0\right)\)
\(x=0\Rightarrow\sqrt{0+3}+0=0\left(vô-nghiệm\right)\)
\(x>0\)
\(\)\(\sqrt{x+3}+\dfrac{4x}{\sqrt{x+3}}=4\sqrt{x}\Leftrightarrow\dfrac{\sqrt{x+3}}{\sqrt{x}}+\dfrac{4\sqrt{x}}{\sqrt{x+3}}=4\)
\(VT\ge2\sqrt{\dfrac{\sqrt{x+3}}{\sqrt{x}}.\dfrac{4\sqrt{x}}{\sqrt{x+3}}}=4\)
\(dấu"="xảy-ra\Leftrightarrow\dfrac{\sqrt{x+3}}{\sqrt{x}}=\dfrac{4\sqrt{x}}{\sqrt{x+3}}\Leftrightarrow x+3=4x\Leftrightarrow x=1\left(tm\right)\)
\(b.2x^4-5x^3+6x^2-5x+2=0\Leftrightarrow\left(x-1\right)^2\left(2x^2-2x+2\right)\Leftrightarrow\left[{}\begin{matrix}x=1\\2x^2-2x+2=0\left(vô-nghiệm\right)\end{matrix}\right.\)
6.
Đặt \(\left\{{}\begin{matrix}\sqrt{5x^2+6x+5}=a\\4x=b\end{matrix}\right.\)
\(\Rightarrow a\left(a^2+1\right)=b\left(b^2+1\right)\)
\(\Leftrightarrow a^3-b^3+a-b=0\)
\(\Leftrightarrow\left(a-b\right)\left(a^2+b^2+ab+1\right)=0\)
\(\Leftrightarrow a=b\)
\(\Leftrightarrow\sqrt{5x^2+6x+5}=4x\left(x\ge0\right)\)
\(\Leftrightarrow5x^2+6x+5=16x^2\)
\(\Leftrightarrow11x^2-6x-5=0\)
\(\Rightarrow x=1\)
4. Bạn coi lại đề (chính xác là pt này ko có nghiệm thực)
5.
\(\Leftrightarrow x^2+x+6-\left(2x+1\right)\sqrt{x^2+x+6}+6x-6=0\)
Đặt \(\sqrt{x^2+x+6}=t>0\)
\(t^2-\left(2x+1\right)t+6x-6=0\)
\(\Delta=\left(2x+1\right)^2-4\left(6x-6\right)=\left(2x-5\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}t=\frac{2x+1+2x-5}{2}=2x-2\\t=\frac{2x+1-2x+5}{2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2+x+6}=2x-2\left(x\ge1\right)\\\sqrt{x^2+x+6}=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+6=4x^2-8x+4\left(x\ge1\right)\\x^2+x+6=9\end{matrix}\right.\)
ĐK : \(x\ge1\)
\(\Leftrightarrow...\sqrt{\left(x-1\right)\left(x+4\right)}+\sqrt{\left(x-1\right)\left(x+3\right)}-3\sqrt{x+4}-3\sqrt{x+3}=1\)
\(\Leftrightarrow\sqrt{x-1}\cdot\sqrt{x-4}+\sqrt{x-1}\cdot\sqrt{x+3}-3\cdot\left(\sqrt{x+4}+\sqrt{x+3}\right)=1\)
\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x+4}+\sqrt{x+3}\right)-3\left(\sqrt{x+4}+\sqrt{x+3}=1\right)\)
\(\Leftrightarrow\left(\sqrt{x-1}-3\right)\left(\sqrt{x+4}+\sqrt{x+3}\right)=1\)
Làm nốt nhé :)
\(a,PT\Leftrightarrow\left|x+3\right|=3x-6\\ \Leftrightarrow\left[{}\begin{matrix}x+3=3x-6\left(x\ge-3\right)\\x+3=6-3x\left(x< -3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{9}{2}\left(tm\right)\\x=\dfrac{3}{4}\left(ktm\right)\end{matrix}\right.\\ \Leftrightarrow x=\dfrac{9}{2}\\ b,PT\Leftrightarrow\left|x-1\right|=\left|2x-1\right|\\ \Leftrightarrow\left[{}\begin{matrix}x-1=2x-1\\1-x=2x-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{3}\end{matrix}\right.\)
\(c,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=25x^2-20x+4\\ \Leftrightarrow25x^2-15x=0\\ \Leftrightarrow5x\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=\dfrac{3}{5}\left(ktm\right)\end{matrix}\right.\Leftrightarrow x=0\\ d,ĐK:x\le\dfrac{2}{5}\\ PT\Leftrightarrow4-5x=2-5x\\ \Leftrightarrow x\in\varnothing\)
\(\sqrt{\left(2x+y\right)^2-8x+3}-2\sqrt{y}+\sqrt{2x+2y-3}-\sqrt{y}=0\)
\(\Leftrightarrow\dfrac{\left(2x+y\right)^2-4\left(2x+y\right)+3}{\sqrt{\left(2x+y\right)^2-8x+3}+2\sqrt{y}}+\dfrac{2x+y-3}{\sqrt{2x+y-3}+\sqrt{y}}=0\)
\(\Leftrightarrow\dfrac{\left(2x+y-3\right)\left(2x+y-1\right)}{\sqrt{\left(2x+y\right)^2-8x+3}+2\sqrt{y}}+\dfrac{2x+y-3}{\sqrt{2x+y-3}+\sqrt{y}}=0\)
\(\Leftrightarrow2x+y-3=0\)
\(\Leftrightarrow y=3-2x\)
Thế xuống pt dưới:
\(1+\sqrt{5x-4}+\sqrt{2x-1}+6x^2-x-8=0\)
\(\Leftrightarrow\left(\sqrt{5x-4}-1\right)+\left(\sqrt{2x-1}-1\right)+\left(6x^2-x-5\right)=0\)
\(\Leftrightarrow\dfrac{5\left(x-1\right)}{\sqrt{5x-4}+1}+\dfrac{2\left(x-1\right)}{\sqrt{2x-1}+1}+\left(x-1\right)\left(6x+5\right)=0\)
1)
ĐK: \(x\geq 5\)
PT \(\Leftrightarrow \sqrt{4(x-5)}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9(x-5)}=6\)
\(\Leftrightarrow \sqrt{4}.\sqrt{x-5}+3\sqrt{\frac{1}{9}}.\sqrt{x-5}-\frac{1}{3}.\sqrt{9}.\sqrt{x-5}=6\)
\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=6\)
\(\Leftrightarrow 2\sqrt{x-5}=6\Rightarrow \sqrt{x-5}=3\Rightarrow x=3^2+5=14\)
2)
ĐK: \(x\geq -1\)
\(\sqrt{x+1}+\sqrt{x+6}=5\)
\(\Leftrightarrow (\sqrt{x+1}-2)+(\sqrt{x+6}-3)=0\)
\(\Leftrightarrow \frac{x+1-2^2}{\sqrt{x+1}+2}+\frac{x+6-3^2}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow \frac{x-3}{\sqrt{x+1}+2}+\frac{x-3}{\sqrt{x+6}+3}=0\)
\(\Leftrightarrow (x-3)\left(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}\right)=0\)
Vì \(\frac{1}{\sqrt{x+1}+2}+\frac{1}{\sqrt{x+6}+3}>0, \forall x\geq -1\) nên $x-3=0$
\(\Rightarrow x=3\) (thỏa mãn)
Vậy .............
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}+\sqrt{\left(3-x\right)^2}=1\)
\(\Leftrightarrow\left|x-1\right|+\left|3-x\right|=1\)
Mà \(\left|x-1\right|+\left|3-x\right|\ge\left|x-1+3-x\right|=2>1\)
\(\Rightarrow\) Phương trình vô nghiệm
2/ \(\Leftrightarrow\left(2x-3\right)\left(x^2-x+1\right)< 0\) (1)
Do \(x^2-x+1=x^2-x+\frac{1}{4}+\frac{3}{4}=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}>0\)
\(\left(1\right)\Leftrightarrow2x-3< 0\)
\(\Rightarrow x< \frac{3}{2}\)
a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)
\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)
\(\Leftrightarrow3\sqrt{x+5}=6\)
\(\Leftrightarrow x+5=4\)
hay x=-1
b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)
\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)
\(\Leftrightarrow\sqrt{x-1}=17\)
\(\Leftrightarrow x-1=289\)
hay x=290
Điều kiện: \(x\ge\frac{1}{5}\)
Ta có: \(x^3-2x^2+6x+3-4\sqrt{5x-1}\)
\(=\left(x^3-2x^2+x\right)+\left(5x-1-4\sqrt{5x-1}+4\right)\)
\(=x\left(x-1\right)^2+\left(\sqrt{5x-1}-2\right)^2\ge0\)
Làm nốt