TÌM X,BIẾT:
\(\sqrt{\left(x-2\right)\left(x+1\right)}\) = \(\sqrt{\left(x-2\right)\left(-5x+1\right)}\)
giúp mk vs nha
Tìm x biết
\(a,-\frac{1}{2}\left(3x-1\right)+\frac{3}{4}\left(3-2x\right)=-3\left(\frac{x}{2}-1\right)-\left(\frac{4}{5}\right)^{-1}\)
\(b,\sqrt{9\left(5x-1\right)}-\sqrt{16\cdot\left(5x-1\right)}+\sqrt{36\left(5x-1\right)}=15\)
MÌNH ĐANG CẦN GẤP GIẢI CỤ THỂ GIÚP MÌNH NHA
Tìm điều kiện để căn thức có nghĩa:
1,\(\sqrt{x-2\sqrt{x-1}}\)
2,\(\sqrt{x-1-2\sqrt{x-2}}\)
3,\(\sqrt{\left|2m+1\right|-1}\)
4,\(\sqrt{2\sqrt{2-\left|3n-\sqrt{2}\right|}}\)
5,\(\frac{\sqrt{5x^2-4x-1}}{\left|x^2-3x\right|+\left|x^2-9\right|}\)
6,\(\frac{\sqrt{x^2-x-6}}{\left|x^2-3x\right|+\left|x^2-9\right|}\)
Ai giải giúp mk vs đi ạ!!! Mk camon m.n nhìu :)))
Tìm x biết :
\(\left(2x-1\right)^3+\left(3x+2\right)^3-\left(5x+1\right)^3=0\)
MẤY BẠN GIÚP MK VS Ạ AI NHANH MK VOTE NHA
\(\dfrac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+2\right)}\) \(\dfrac{1}{\left(\sqrt{X}-1\right)\left(3-\sqrt{X}\right)}\)
GIUP MIK VS NHA,CẢM ƠN
`1/((sqrtx-1)(sqrtx+2))-1/((sqrtx-1)(3-sqrtx))`
`=1/((sqrtx-1)(sqrtx+2))+1/((sqrtx-1)(sqrtx-3))`
`=(sqrtx-3+sqrtx+2)/((sqrtx-1)(sqrtx+2)(sqrtx-3))`
`=(2sqrtx-1)/((sqrtx-1)(sqrtx+2)(sqrtx-3))`
1)\(\sqrt{x+3}\) > 2
2) \(\dfrac{1+\sqrt{x}}{\sqrt{x}-2}\)<1
3) \(\left(\sqrt{x}-1\right)\).\(\left(\sqrt{x}-3\right)\)-5=\(\sqrt{x}\) \(\left(\sqrt{x}+2\right)-5\)
tìm x mn giúp mình nha plsss
1: ĐKXĐ: x+3>=0
=>x>=-3
\(\sqrt{x+3}>2\)
=>x+3>4
=>x>4-3=1
2: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >4\end{matrix}\right.\)
\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}< 1\)
=>\(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-1< 0\)
=>\(\dfrac{\sqrt{x}+1-\sqrt{x}+2}{\sqrt{x}-2}< 0\)
=>\(\dfrac{3}{\sqrt{x}-2}< 0\)
=>\(\sqrt{x}-2< 0\)
=>\(\sqrt{x}< 2\)
=>0<=x<4
3: ĐKXĐ: x>=0
\(\left(\sqrt{x}-1\right)\left(\sqrt{x}-3\right)-5=\sqrt{x}\left(\sqrt{x}+2\right)-5\)
=>\(x-4\sqrt{x}+3-5=x+2\sqrt{x}-5\)
=>\(x-4\sqrt{x}-2-x-2\sqrt{x}+5=0\)
=>\(-6\sqrt{x}+3=0\)
=>\(-6\sqrt{x}=-3\)
=>\(\sqrt{x}=\dfrac{1}{2}\)
=>x=1/4(nhận)
Giúp nha :
Tìm x ; y ; z biết :
\(\sqrt{\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2}+3\left(x^2-1\right)\left(y^2-1\right)\left(z^2-1\right)+\left(x-1\right)\left(y-1\right)\left(z-1\right)=0\)
giải phương trình :
a, \(\left(\sqrt{5x-1}+\sqrt{x-1}\right)\left(3x-1-\sqrt{5x^2-6x+1}\right)=4x\)
b, \(2\left(\sqrt{x}-\sqrt{x-1}\right)\left(1+\sqrt{x^2-1}\right)=x\sqrt{x}\)
giải phương trình :
a, \(\left(\sqrt{5x-1}+\sqrt{x-1}\right)\left(3x-1-\sqrt{5x^2}-6x+1\right)=4x\)
b, \(2\left(\sqrt{x}-\sqrt{x-1}\right)\left(1+\sqrt{x^2-1}\right)=x\sqrt{x}\)
giải pt :
a,\(\left(\sqrt{5x-1}+\sqrt{x-1}\right)\left(3x-1-\sqrt{5x^2-6x+1}\right)=4x\)
b,\(2\left(\sqrt{x}-\sqrt{x-1}\right)\left(1+\sqrt{x^2-1}\right)=x\sqrt{x}\)
a, ĐK: \(x\ge1\)
Đặt \(\sqrt{5x-1}=a;\sqrt{x-1}=b\left(a,b\ge0\right)\)
\(pt\Leftrightarrow\left(a+b\right)\left(\dfrac{a^2+b^2}{2}-ab\right)=a^2-b^2\)
\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2=2\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow\left(a+b\right)\left(a-b\right)\left(a-b-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=b+2\end{matrix}\right.\)
TH1: \(a=b\Leftrightarrow\sqrt{5x-1}=\sqrt{x-1}\Leftrightarrow x=0\left(l\right)\)
TH2: \(a=b+2\Leftrightarrow\sqrt{5x-1}=\sqrt{x-1}+2\)
\(\Leftrightarrow5x-1=x-1+4+4\sqrt{x-1}\)
\(\Leftrightarrow4x-4-4\sqrt{x-1}=0\)
\(\Leftrightarrow4x-4-4\sqrt{x-1}+1=1\)
\(\Leftrightarrow\left(2\sqrt{x-1}-1\right)^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}2\sqrt{x-1}-1=1\\2\sqrt{x-1}-1=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=1\\\sqrt{x-1}=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\end{matrix}\right.\)