\(2x\sqrt x +12x +22\sqrt x+12 \over (\sqrt x+1)(\sqrt x+2)(\sqrt x+3) \)
Những câu hỏi liên quan
\(2x\sqrt x+12x+22\sqrt x+12 \over(\sqrt x +1)(\sqrt x+2)(\sqrt x+3) \)
a) \(\sqrt{3x^2-5x+7}\)+\(\sqrt{3x^2+x+1}\) = 12x-12
b) \(\sqrt{x^2+33}\)+3 = 2x+\(\sqrt{x^2-12}\)
c) 3x-\(8\sqrt{x+14}\) = \(2\sqrt{2x-3}\) - 28
d) \(x^2\)+\(\sqrt{x+7}\) = 7
\(\sqrt{12-\frac{12}{x^2} }+\sqrt{x^2+\frac{12}{x} }=x^2+\frac{25}{2} \)
\(\frac{\sqrt[3]{10-x}+\sqrt[3]{8-x}}{\sqrt[3]{10-x}-\sqrt[3]{8-x}}=9-x \)
\(4x^2+\sqrt{2x+1}+5=12x\)
Tìm x biết
a) \(\sqrt{-x^2+2x-1}=\sqrt{9-12x+4x^2}\)
b) \(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2-\sqrt{2x-5}}=2\sqrt{2}\)
c)\(x^2+x+12\sqrt{x+1}=36\)
Giải phương trình: a) sqrt{x+3-4sqrt{x+1}}+sqrt{x+8-6sqrt{x-1}}1b) sqrt{x+sqrt{x-11}}+sqrt{x-sqrt{x-11}}4c) sqrt{x+2+3sqrt{2x-5}}+sqrt{x-2sqrt{2x-5}}2sqrt{2}d) sqrt{x-4}+sqrt{6-x}x^2-10x+27e) sqrt{2x+1}+sqrt{17-2x}x^4-8x^3+17x^2-8x+22f) sqrt{3x^2+12x+16}+sqrt{y^2-4y+13}5g) sqrt{x+x^2}+sqrt{x-x^2}x+1ai lmmm giúp tui ikkk
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Giải phương trình:
a) \(\sqrt{x+3-4\sqrt{x+1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)
b) \(\sqrt{x+\sqrt{x-11}}+\sqrt{x-\sqrt{x-11}}=4\)
c) \(\sqrt{x+2+3\sqrt{2x-5}}+\sqrt{x-2\sqrt{2x-5}}=2\sqrt{2}\)
d) \(\sqrt{x-4}+\sqrt{6-x}=x^2-10x+27\)
e) \(\sqrt{2x+1}+\sqrt{17-2x}=x^4-8x^3+17x^2-8x+22\)
f) \(\sqrt{3x^2+12x+16}+\sqrt{y^2-4y+13}=5\)
g) \(\sqrt{x+x^2}+\sqrt{x-x^2}=x+1\)
ai lmmm giúp tui ikkk
\(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\) ( SỬA ĐỀ)
\(\sqrt{x-1-2.2.\sqrt{x-1}+4}+\sqrt{x-1-2.3.\sqrt{x-1}+9}=1\)
\(|x-1-2|+|x-1-3|=1\)
\(|x-3|+|x-4|=1\)
Với \(x\le3\)thì PT thành \(3-x+4-x=1\) \(\Rightarrow-2x=-6\Rightarrow x=3\)(thõa mãn)
Với \(3\le x< 4\)thì PT thành \(x-3+4-x=1\Leftrightarrow0x=0\Rightarrow\)Đúng với mọi x từ \(3\le x< 4\)
Với \(x\ge4\)thì PT thành \(x-3+x-4=1\Leftrightarrow2x=8\Leftrightarrow x=4\)(thõa mãn)
Vậy \(3\le x\le4\)
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4 tháng 8 2021 lúc 11:47
Giải phương trính:
1. \(\sqrt{x^2+x-1}+\sqrt{x-x^2+1}=x^2-x+2\)
2. \(\sqrt{2x-3}+\sqrt{5-2x}=3x^2-12x+14\)
1.
ĐKXĐ: ...
\(x^2-x+2=1\sqrt{x^2+x-1}+1\sqrt{x-x^2+1}\)
\(\Rightarrow x^2-x+2\le\dfrac{1}{2}\left(1+x^2+x-1\right)+\dfrac{1}{2}\left(1+x-x^2+1\right)\)
\(\Rightarrow x^2-2x+1\le0\)
\(\Rightarrow\left(x-1\right)^2\le0\)
\(\Rightarrow x=1\)
Thử lại ta thấy thỏa mãn
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b.
ĐKXĐ: ...
Ta có:
\(VP=3\left(x-2\right)^2+2\ge2\)
\(VT=1\sqrt{2x-3}+1\sqrt{5-2x}\le\dfrac{1}{2}\left(1+2x-3\right)+\dfrac{1}{2}\left(1+5-2x\right)=2\)
\(\Rightarrow VT\le VP\)
Đẳng thức xảy ra khi:
\(\left\{{}\begin{matrix}x-2=0\\1=2x-3\\1=5-2x\end{matrix}\right.\) \(\Leftrightarrow x=2\)
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Tìm x
1, 4x^4+12x^3+12x-47x^2+404x^4+12x^3+12x-47x^2+40
2, x^2+sqrt{x+1}1
3.sqrt{2x^2+8x+6}+sqrt{x^2-1}2x+2
4.x-2sqrt{x-1}-left(x-1right)sqrt{x}+sqrt{x^2-x}0
5.xsqrt[3]{35-x^3}-left(x+sqrt[3]{35-x^3}right)30
6. x^3-12sqrt[3]{2x-1}
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Tìm x
1, \(4x^4+12x^3+12x-47x^2+4=0\)\(4x^4+12x^3+12x-47x^2+4=0\)
2, \(x^2+\sqrt{x+1}=1\)
3.\(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)
4.\(x-2\sqrt{x-1}-\left(x-1\right)\sqrt{x}+\sqrt{x^2-x}=0\)
5.\(x\sqrt[3]{35-x^3}-\left(x+\sqrt[3]{35-x^3}\right)=30\)
6. \(x^3-1=2\sqrt[3]{2x-1}\)
1/\(4x^4+12x^3-47x^2+12x+4=0\)
\(\Leftrightarrow\left(x-2\right)\left(4x^3+20x^2-7x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)\left(2x^2+11x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{2}\\x=\frac{-11\pm\sqrt{105}}{4}\end{matrix}\right.\)
Vậy ....
1, 4x^4+12x^3+12x−47x^2+4=0 nhé
Giai các phương trình sau:
a) \(\sqrt{x-2}+\sqrt{10-x}=x^2-12x+40\)
b) \(\sqrt{x^2+x-1}+\sqrt{x-x^2+1}=x^2-x+2\)
c) \(\sqrt{2x-3}+\sqrt{5-2x}=3x^2-12x+14\)
d) \(x^2-2x+3=\sqrt{2x^2-x}+\sqrt{1+3x-3x^2}\)
a/ ĐKXĐ: \(2\le x\le10\)
\(\Leftrightarrow\sqrt{x-2}+\sqrt{10-x}-x^2+12x-20-20=0\)
Đặt \(\sqrt{x-2}+\sqrt{10-x}=a>0\)
\(\Rightarrow a^2=8+2\sqrt{-x^2+12x-20}\Rightarrow-x^2+12x-20=\frac{\left(a^2-8\right)^2}{4}\)
Phương trình trở thành:
\(a+\frac{\left(a^2-8\right)^2}{4}-20=0\Leftrightarrow a^4-16a^2+4a-16=0\)
\(\Leftrightarrow a^2\left(a-4\right)\left(a+4\right)+4\left(a-4\right)=0\)
\(\Leftrightarrow\left(a-4\right)\left(a^3+4a^2+4\right)=0\)
\(\Leftrightarrow a=4\) (do \(a^3+4a^2+4>0\) \(\) \(\forall a>0\))
\(\Leftrightarrow\sqrt{x-2}+\sqrt{10-x}=4\)
Mà \(\sqrt{x-2}+\sqrt{10-x}\le\sqrt{2\left(x-2+10-x\right)}=4\)
Dấu "=" xảy ra khi và chỉ khi \(x-2=10-x\Leftrightarrow x=6\)
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b/ ĐKXĐ:...
Ta có:
\(VT=1.\sqrt{x^2+x-1}+1.\sqrt{x-x^2+1}\le\frac{1+x^2+x-1}{2}+\frac{1+x-x^2+1}{2}=x+1\)
\(\Rightarrow x^2-x+2\le x+1\)
\(\Leftrightarrow x^2-2x+1\le0\)
\(\Leftrightarrow\left(x-1\right)^2\le0\Rightarrow x=1\)
Vậy pt có nghiệm duy nhất \(x=1\)
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c/ ĐKXĐ: \(\frac{3}{2}\le x\le\frac{5}{2}\)
Ta có:
\(VT=\sqrt{2x-3}+\sqrt{5-2x}\le\sqrt{2\left(2x-3+5-2x\right)}=2\)
\(VP=3\left(x^2-4x+4\right)+2=3\left(x-2\right)^2+2\ge2\)
\(\Rightarrow VP\ge VT\)
Dấu "=" xảy ra khi và chỉ khi:
\(\left\{{}\begin{matrix}\left(x-2\right)^2=0\\2x-3=5-2x\end{matrix}\right.\) \(\Leftrightarrow x=2\)
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Xem thêm câu trả lời
giải phương trình :
a, \(\sqrt{x+1}+x+3=\sqrt{1-x}+3\sqrt{1-x^2}\)
b,\(\left(2x-3\right)\sqrt{3+x}+2x\sqrt{3-x}=6x-8+\sqrt{9-x^2}\)
c, \(2x^2-5x+22=5\sqrt{x^3-11x +20}\)
d, \(x^3-3x^2+2\sqrt{\left(x+2\right)^3}=6x\)