$\sqrt{7x^2+20x-86}+x\sqrt{31-4x-x^2}=x+1$
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$\sqrt{7x^2+20x-86}+x\sqrt{31-4x-x^2}=x+1$
$\sqrt{7x^2+20x-86}+x\sqrt{31-4x-x^2}=x+1$ = $\sqrt{7x^2+20x-86}+x\sqrt{31-4x-x^2}=x+1$
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câu 1:\(\sqrt{7x^2+20x-86}+x\sqrt{31-4x-x^2}=x+1\)
câu 2:\(\sqrt[3]{\frac{12x^2+12x+9}{4}}=x+\sqrt[4]{\frac{4x^3-2}{3}}\)
GPT:
1/ \(\sqrt{7x^2+20x-86}+x\sqrt{31-4x-x^2}=x+1\)
2/ \(\sqrt[3]{\frac{12x^2+12x+9}{4}}=x+\sqrt[4]{\frac{4x^3-2}{3}}\)
1/ \(\sqrt{7x^2+20x-86}+x\sqrt{31-4x-x^2}=x+1\)
2/ \(\sqrt[3]{\frac{12x^2+12x+9}{4}}=x+\sqrt[4]{\frac{4x^3-2}{3}}\)
em mới lớp 10, nên anh chị, thầy cô giải cách nào dễ hiểu giúp em nha
\(\sqrt{x+2\sqrt{x-1}}=2\)
\(\sqrt{4x^2-20x+25}+2x=5\)
\(\sqrt{2x^2-3}=\sqrt{4x-3}\)
\(\sqrt{x^2-x-6}=\sqrt{x-3}\)
\(\sqrt{x^2-x}=\sqrt{3-x}\)
a.
\(\sqrt{x+2\sqrt{x-1}}=2\)
ĐKXĐ: \(x\ge1\)
\(\sqrt{x-1+2\sqrt{x-1}+1}=2\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}=2\)
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|=2\)
\(\Leftrightarrow\sqrt{x-1}+1=2\)
\(\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\)
\(\Leftrightarrow x=2\)
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b.
\(\sqrt{4x^2-20x+25}=5-2x\)
\(\Leftrightarrow\sqrt{\left(2x-5\right)^2}=5-2x\)
\(\Leftrightarrow\left|5-2x\right|=5-2x\)
\(\Leftrightarrow5-2x\ge0\)
\(\Leftrightarrow x\le\dfrac{5}{2}\)
c.
ĐKXĐ: \(x\ge3\)
\(\sqrt{x^2-x-6}=\sqrt{x-3}\)
\(\Rightarrow x^2-x-6=x-3\)
\(\Leftrightarrow x^2-2x-3=0\Rightarrow\left[{}\begin{matrix}x=-1\left(loại\right)\\x=3\end{matrix}\right.\)
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d.
ĐKXĐ: \(\left[{}\begin{matrix}x\le0\\1\le x\le3\end{matrix}\right.\)
\(\sqrt{x^2-x}=\sqrt{3-x}\)
\(\Rightarrow x^2-x=3-x\)
\(\Leftrightarrow x^2=3\)
\(\Rightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\) (thỏa mãn)
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\(\sqrt{4x^2}-20x+25+2x=5\)
\(\sqrt{1-2x}+36x^2=5\)
\(\sqrt{4x^2-20x+25x+2x}=5\)
\(\sqrt{x-2}\sqrt{x-1}=\sqrt{x-1-1}\)
dài v nhg thui cố làm v
a)\(\sqrt{4x^2}-20x+25+2x=5\)
=> \(2x-18x+20=0\)
=> \(-16x+20=0\)
=> \(-4x+5=0\)
=> \(-4x=-5\)
=> \(x=\dfrac{5}{4}\)
vậy........................................................
d) \(\sqrt{x-2}\cdot\sqrt{x-1}=\sqrt{x-1-1}\)
cau này đề sai
ok baby
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gpt \(\sqrt{x^4-x^2+4}+\sqrt{x^4+20x^2+4}=7x\)
Giải phương trình
a) \(\sqrt{12x^2+12x+19}+\sqrt{20x^2+20x+14}=6-4x-4x^2\)
b) \(\left(x+\dfrac{1}{x}\right)-4\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)+6=0\)
b:
ĐKXĐ: x>0
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)^2-2-4\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)+6=0\)
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}-2\right)^2=0\)
\(\Leftrightarrow x+1-2\sqrt{x}=0\)
=>x=1
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Giải phương trình
a) \(\sqrt{12x^2+12x+19}+\sqrt{20x^2+20x+14}=6-4x-4x^2\)
b) \(\left(x+\dfrac{1}{x}\right)-4\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)+6=0\)
a) ta có \(\sqrt{12x^2+12x+19}+\sqrt{20x^2+20x+14}=-4x^2-4x+6\)
\(\Leftrightarrow\sqrt{12\left(x+\dfrac{1}{2}\right)^2+16}+\sqrt{20\left(x+\dfrac{1}{2}\right)^2+9}=-\left(2x+1\right)^2+7\)ta có : \(VT\ge\sqrt{16}+\sqrt{9}=7\) và \(VT\le7\)
\(\Rightarrow VT=VP\) \(\Leftrightarrow x=\dfrac{-1}{2}\) vậy \(x=\dfrac{-1}{2}\)
b) điều kiện \(x>0\)
ta có : \(\left(x+\dfrac{1}{x}\right)-4\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)+6=0\)
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)^2-4\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}\right)+4=0\)
\(\Leftrightarrow\left(\sqrt{x}+\dfrac{1}{\sqrt{x}}-2\right)^2=0\) \(\Leftrightarrow\sqrt{x}+\dfrac{1}{\sqrt{x}}-2=0\)
\(\Leftrightarrow\sqrt{x}+\dfrac{1}{\sqrt{x}}=2\Leftrightarrow\dfrac{x+\sqrt{x}}{\sqrt{x}}=2\Leftrightarrow x+\sqrt{x}=2\sqrt{x}\)
\(\Leftrightarrow x-\sqrt{x}=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}-1\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=0\\\sqrt{x}-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(L\right)\\x=1\left(N\right)\end{matrix}\right.\)
vậy \(x=1\)
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