x-4\(\sqrt[]{\text{x}}=0\)
Những câu hỏi liên quan
1, P=(\(\dfrac{\text{x-1}}{\text{x+3}\sqrt{\text{x-4}}}+\dfrac{\sqrt{\text{x}}+1}{1-\sqrt{\text{x}}}\)) : \(\dfrac{\text{x}+2\sqrt{\text{x}}+1}{x-1}\)+1
a, Rút gọn P
b, Tìm x để P<0
Giải phương trình sau :
\(\sqrt{x}+\sqrt{2x-1}+x^2+x-4=\text{0}\)
ĐK: \(x\ge\dfrac{1}{2}\)
\(pt\Leftrightarrow\sqrt{x}-1+\sqrt{2x-1}-1+x^2+x-2=0\)
\(\Leftrightarrow\dfrac{x-1}{\sqrt{x}+1}+\dfrac{2x-2}{\sqrt{2x-1}+1}+\left(x-1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\sqrt{2x-1}+1}+x+2\right)\left(x-1\right)=0\)
Vì \(\dfrac{1}{\sqrt{x}+1}+\dfrac{2}{\sqrt{2x-1}+1}+x+2>0\) nên \(x-1=0\Leftrightarrow x=1\left(tm\right)\)
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\(\frac{x+\sqrt{x}}{\sqrt{x}}+\frac{x-4}{\sqrt{x}+2}\text{với x > 0}\)
\(\frac{x+\sqrt{x}}{\sqrt{x}}+\frac{x-4}{\sqrt{x}+2}\)
= \(\frac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}+\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}{\sqrt{x}+2}\)
= \(\left(\sqrt{x}+1\right)+\sqrt{x}-2\)
= \(2\sqrt{x}-1\)
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\(\)\(\begin{cases}\left(\sqrt{2016+x^2}+x\right)\left(\sqrt{504+y^2}+y\right)=1008\\x\sqrt{6\text{x}-4\text{x}y+1}=8\text{x}y+6\text{x}+1\end{cases}\)\(\begin{cases}\sqrt{x+1}\left(x+5\right)-3\left(x-y+1\right)\sqrt{y}\left(y+4\right)=0\\x^2+6y+7=2\sqrt{3y+1}+3\sqrt{x+4y+5}\end{cases}\)
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bạn tách từng câu ra mik suy nghĩ từng câu
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B=\(\left(\frac{x\sqrt{x}}{x\text{+}\sqrt{x}\text{+}1}-\frac{1}{x\text{+}\sqrt{x}\text{+}1}\right):\frac{2}{\sqrt{x}\text{+}1}\)
Chứng minh A<0 với mọi 0<x<1
\(\sqrt{3}\text{x}^{^{ }2}-3\sqrt{3}\text{x}+\sqrt{3}+\sqrt{x^4+x^2+1}=0\)
Giải phương trình bằng cách đặt ẩn phụ
Mong mng giúp ạcâu1 rút gọn a)sqrt{4-2sqrt{3}}-sqrt{3}b)dfrac{x^2+2sqrt{2}x+2}{x^2-2}left(xnesqrt{2},xne-sqrt{2}right)c)sqrt{9text{x}^2}-2text{x}left(x 0right)d)sqrt{11+6sqrt{2}}-3+sqrt{2}e)dfrac{x^2-5}{x+sqrt{5}}left(xne-sqrt{5}right)
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Mong mng giúp ạ
câu1 rút gọn
a)\(\sqrt{4-2\sqrt{3}}-\sqrt{3}\)
b)\(\dfrac{x^2+2\sqrt{2}x+2}{x^2-2}\left(x\ne\sqrt{2},x\ne-\sqrt{2}\right)\)
c)\(\sqrt{9\text{x}^2}-2\text{x}\left(x< 0\right)\)
d)\(\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\)
e)\(\dfrac{x^2-5}{x+\sqrt{5}}\left(x\ne-\sqrt{5}\right)\)
\(a,\sqrt{4-2\sqrt{3}}-\sqrt{3}=\sqrt{\sqrt{3^2}-2\sqrt{3}+1}-\sqrt{3}=\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}=\left|\sqrt{3}-1\right|-\sqrt{3}=-1\)
\(b,\dfrac{x^2+2\sqrt{2}x+2}{x^2-2}\left(dk:x\ne\pm\sqrt{2}\right)\\ =\dfrac{x^2+2\sqrt{2}x+\sqrt{2^2}}{x^2-\sqrt{2^2}}\\ =\dfrac{\left(x+\sqrt{2}\right)^2}{\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)}\\ =\dfrac{x+\sqrt{2}}{x-\sqrt{2}}\)
\(c,\sqrt{9x^2}-2x\left(dk:x< 0\right)\\ =\sqrt{3^2}.\sqrt{x^2}-2x\\ =3\left|x\right|-2x\\ =-3x-2x\\ =-5x\)
\(d,\sqrt{11+6\sqrt{2}}-3+\sqrt{2}\\ =\sqrt{\sqrt{2^2}+2.3\sqrt{2}+3^2}-3+\sqrt{2}\\ =\sqrt{\left(\sqrt{2}+3\right)^2}-3+\sqrt{2}\\ =\sqrt{2}+3-3+\sqrt{2}\\ =2\sqrt{2}\)
\(e,\dfrac{x^2-5}{x+\sqrt{5}}\left(dk:x\ne-\sqrt{5}\right)\\ =\dfrac{\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)}{x+\sqrt{5}}\\ =x-\sqrt{5}\)
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giải phương trình sau :
\(\sqrt{x}+\sqrt[4]{x\text{(}1-x\text{)}^2}+\sqrt[4]{\text{(}1-x\text{)}^3}=\sqrt{1-x}+\sqrt[4]{x^3}+\sqrt[4]{x^2.\text{(}1-x\text{)}}\)
giải phương trình
\(\text{x}^2-4=3\sqrt{\text{x}^3-4\text{x}}\)
\(9\text{x}+17=6\sqrt{8\text{x}-1}+4\sqrt{\text{x}+3}\)
\(\sqrt{2\text{x}-1}+\text{x}=\sqrt{\text{x}}+\sqrt{\text{x}^2-\text{x}+1}\)
\(2\sqrt{\text{x}^2-\text{x}+1}+\sqrt{\text{x}^2+\text{x}+1}=\sqrt{\text{x}^4+\text{x}^2+1}+2\)
a: Đặt \(x^2-4=a\)
Pt sẽ là \(a=3\sqrt{xa}\)
\(\Rightarrow a^2=9xa\)
\(\Leftrightarrow a\left(a-9x\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\left(x^2-4-9x\right)=0\)
hay \(x\in\left\{2;-2;\dfrac{9+\sqrt{97}}{2};\dfrac{9-\sqrt{97}}{2}\right\}\)
d: Đặt \(\sqrt{x^2-x+1}=a;\sqrt{x^2+x+1}=b\)
Pt sẽ là 2a+b=ab+2
=>(b-2)(1-a)=0
=>b=2 và 1-a
\(\Leftrightarrow\left\{{}\begin{matrix}x^2+x+1=4\\x^2-x+1=1\end{matrix}\right.\Leftrightarrow x\in\varnothing\)
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