\(7x^2-10x+14=5\sqrt{x^4+4}\)
giải pt
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PB
2
20 tháng 9 2021
1) \(\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{4-2\sqrt{3}}=\sqrt{3}+1-\sqrt{\left(\sqrt{3}-1\right)^2}=\sqrt{3}+1-\sqrt{3}+1=2\)
2) \(\dfrac{3}{5}\sqrt{25x-50}-\sqrt{x-2}=6\left(đk:x\ge2\right)\)
\(\Leftrightarrow3\sqrt{x-2}-\sqrt{x-2}=6\)
\(\Leftrightarrow2\sqrt{x-2}=6\)
\(\Leftrightarrow\sqrt{x-2}=3\)
\(\Leftrightarrow x-2=9\Leftrightarrow x=11\left(tm\right)\)
LT
giải pt:
a) \(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\)
b) \(2x^4+8=4\sqrt{4+x^4}+4\sqrt{x^4-4}\)
0
DT
1
27 tháng 5 2021
\(A=x^4-7x^3+10x^2+\left(a-1\right)x+b-a\)
\(A=x^4-6x^3+5x^2-x^3+6x^2-5x-x^2+\left(a-1\right)x+b-a\)
\(A=x^2\left(x^2-6x+5\right)-x\left(x^2-6x+5\right)-\left(x^2-\left(a-1\right)x+b-a\right)\)
Ta thấy
\(x^2\left(x^2-6x+5\right)-x\left(x^2-6x+5\right)\) chia hết cho B
\(\Rightarrow-\left(x^2-\left(a-1\right)x+b-a\right)\) phải chia hết cho B
\(\Leftrightarrow\left[{}\begin{matrix}a-1=6\\b-a=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}a=5\\b=0\end{matrix}\right.\)
28 tháng 9 2021
1) \(\Leftrightarrow\sqrt{\left(x+5\right)^2}=4\)
\(\Leftrightarrow\left|x+5\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+5=4\\x+5=-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-9\end{matrix}\right.\)
2) \(ĐK:x\ge2\)
\(\Leftrightarrow\sqrt{x-2}=2\)
\(\Leftrightarrow x-2=4\Leftrightarrow x=6\left(tm\right)\)
3) \(\Leftrightarrow\left(x^2-x+4\right)-\sqrt{x^2-x+4}+\dfrac{1}{4}=\dfrac{9}{4}\)
\(\Leftrightarrow\left(\sqrt{x^2-x+4}-\dfrac{1}{2}\right)^2=\dfrac{9}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}-\dfrac{1}{2}=\dfrac{3}{2}\\\sqrt{x^2-x+4}-\dfrac{1}{2}=-\dfrac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x^2-x+4}=2\\\sqrt{x^2-x+4}=-1\left(VLý\right)\end{matrix}\right.\)
\(\Leftrightarrow x^2-x+4=4\Leftrightarrow x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
4) \(ĐK:x\ge0\)
\(\Leftrightarrow3\sqrt{x}-3=\sqrt{x}+2\)
\(\Leftrightarrow\sqrt{x}=\dfrac{5}{2}\Leftrightarrow x=\dfrac{25}{4}\left(tm\right)\)
19 tháng 9 2021
1) \(\sqrt{5-2x}=6\left(đk:x\le\dfrac{5}{2}\right)\)
\(\Leftrightarrow5-2x=36\)
\(\Leftrightarrow2x=-31\Leftrightarrow x=-\dfrac{31}{2}\left(tm\right)\)
2) \(\sqrt{2-x}=\sqrt{x+1}\left(đk:2\ge x\ge-1\right)\)
\(\Leftrightarrow2-x=x+1\)
\(\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)
3) \(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)
\(\Leftrightarrow\left|2x+1\right|=6\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)
4) \(\sqrt{x^2-10x+25}=x-2\left(đk:x\ge2\right)\)
\(\Leftrightarrow\sqrt{\left(x-5\right)^2}=x-2\)
\(\Leftrightarrow\left|x-5\right|=x-2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-5=x-2\left(x\ge5\right)\\x-5=2-x\left(2\le x< 5\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}5=2\left(VLý\right)\\x=\dfrac{7}{2}\left(tm\right)\end{matrix}\right.\)
ĐK: \(\forall x\in R\)
\(PT\Leftrightarrow\left(7x^2-10x+14\right)^2=25\left(x^4+4\right)\)
\(\Leftrightarrow49x^4-140x^3+296x^2-280x+196=25x^4+100\)
\(\Leftrightarrow24x^4-140x^3+296x^2-280x+96=0\)
\(\Leftrightarrow6x^4-35x^3+74x^2-70x+24=0\)
Với \(x=0\) => Không thỏa mãn phương trình.
Với \(x\ne0\) , chia cả 2 vế cho \(x^2\):
\(\Leftrightarrow6x^2-35x+74-\frac{70}{x}+24x^2=0\)
\(\Leftrightarrow6\left(x^2+\frac{4}{x^2}\right)-35\left(x+\frac{2}{x}\right)+74=0\)
Đặt \(x+\frac{2}{x}=t\) \(\Rightarrow t^2=x^2+\frac{4}{x^2}+4\)
\(\Leftrightarrow6\left(t^2-4\right)-35t+74=0\) \(\Leftrightarrow6t^2-35t+50=0\Rightarrow\left[{}\begin{matrix}t=\frac{10}{3}\\t=\frac{5}{2}\end{matrix}\right.\)
+\(t=\frac{10}{3}\Rightarrow x+\frac{2}{x}=\frac{10}{3}\Rightarrow3x^2-10x+6=0\) \(\Rightarrow x=\frac{5\pm\sqrt{7}}{3}\)
+\(t=\frac{5}{2}\Rightarrow x+\frac{2}{x}=\frac{5}{2}\Rightarrow2x^2-5x+4=0\) (Vô no)
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