Rút gọn Q
\(Q=\frac{\left(x+2\right)^2}{x}.\left(1-\frac{x^2}{x+2}\right)-\frac{x^2+6x+4}{x}\)
mk cần gấp mn giúp với!!
rút gọn
\(\frac{2\left(x+4\right)}{x-3\sqrt{x}-4}+\frac{\sqrt{x}}{\sqrt{x}+1}-\frac{8}{\sqrt{x}-4}\)
mk cần gấp nha bạn
giúp mk với
\(\frac{2.\left(x+4\right)}{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}+\frac{\sqrt{x}.\left(\sqrt{x}-4\right)}{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}-\frac{8.\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-4\right)\left(\sqrt{x}+1\right)}\)
=\(\frac{2x+8+x-4\sqrt{x}-8\sqrt{x}-8}{\left(\sqrt{x}+1\right).\left(\sqrt{x}-4\right)}\)
=\(\frac{3x-12\sqrt{x}}{mc}\)
=\(\frac{3\sqrt{x}.\left(\sqrt{x}-4\right)}{\left(\sqrt{x-4}\right)\left(\sqrt{x}+1\right)}=\frac{3\sqrt{x}}{\sqrt{x}+1}\)
k tk mk cung lam cho
Cho biểu thức Q =\(\frac{\sqrt{x-\sqrt{4\left(x-1\right)}}+\sqrt{x+\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}.\left(1-\frac{1}{x-1}\right)\)
Rút gọn
Các bạn giúp mình với cần gấp nhé ^_^
\(Q=\frac{\sqrt{x-\sqrt{4\left(x-1\right)}}+\sqrt{x+\sqrt{4\left(x-1\right)}}}{\sqrt{x^2-4\left(x-1\right)}}.\left(1-\frac{1}{x-1}\right)\)
\(=\frac{\sqrt{x-1-2\sqrt{x-1}+1}+\sqrt{x-1+2\sqrt{x-1}+1}}{\sqrt{x^2-4x+4}}.\frac{x}{x-1}\)
\(=\frac{\sqrt{\left(\sqrt{x-1}-1\right)^2}+\sqrt{\left(\sqrt{x-1}+1\right)^2}}{\sqrt{\left(x-2\right)^2}}.\frac{x}{x-1}\)
\(=\frac{\left|\sqrt{x-1}-1\right|+\sqrt{x-1}+1}{x-2}.\frac{x}{x-1}\)
Nếu \(x\ge2\) thì
\(Q=\frac{\sqrt{x-1}-1+\sqrt{x-1}+1}{x-2}.\frac{x}{x-1}=\frac{2x\sqrt{x-1}}{\left(x-2\right)\left(x-1\right)}=\frac{2x}{\left(x-2\right)\left(\sqrt{x-1}\right)}\)
Nếu \(x< 2\) thì \(Q=\frac{1-\sqrt{x-1}+\sqrt{x-1}+1}{x-2}.\frac{x}{x-1}=\frac{2x}{\left(x-2\right)\left(x-1\right)}\)
Cảm ơn bạn nhiều nhưng mình thấy \(1-\frac{1}{x-1}=\frac{x-2}{x-1}\) mà bạn sao lại bằng \(\frac{x}{x-1}\)được
Rút gọn biểu thức:
\(\frac{1}{2}x^2.\left(6x-3\right)-x\left(x^2+\frac{1}{2}\right)+\frac{1}{2}\left(x+4\right)\)
Rút gọn biểu thức:
\(\frac{1}{2}x^2.\left(6x-3\right)-x\left(x^2+\frac{1}{2}\right)+\frac{1}{2}\left(x+4\right)\)
\(=3x^3-\frac{3}{2}x^2-x^3-\frac{1}{2}x+\frac{1}{2}x+2\)
\(=2x^3-\frac{3}{2}x^2+2\)
\(Q=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x-2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)
a. Rút gọn Q
b. Tìm x để Q = -1
Tớ đang cần gấp, mong mọi người giúp,cảm ơn trước nhé!
a) ĐKXĐ: x\(\ne\) 0;4
Ta có: Q= \(\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\frac{2}{\sqrt{x}}\right)\)
= \(\frac{4\sqrt{x}\cdot\left(2-\sqrt{x}\right)+8x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}:\frac{\sqrt{x}-1-2\cdot\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
=\(\frac{8\sqrt{x}+4x}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}\)= \(\frac{4\sqrt{x}\cdot\left(2+\sqrt{x}\right)}{2+\sqrt{x}}\cdot\frac{-\sqrt{x}}{3-\sqrt{x}}\)=\(\frac{-4}{3-\sqrt{x}}\)=\(\frac{4}{\sqrt{x}-3}\)
b) Q=-1 => \(\frac{4}{\sqrt{x}-3}=-1\)
<=> \(4=3-\sqrt{x}\)
<=> \(\sqrt{x}=-1\) (vô lí)
Vậy ko tìm được x.
Rút gọn
a) \(\left(\frac{4}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)\)
b) \(\left(\frac{2}{x-2}-\frac{2}{x+2}\right).\frac{x^2+4x+4}{8}\)
c) \(\left(\frac{3x}{1-3x}+\frac{2x}{3x+1}\right):\frac{6x^2+10x}{1-6x+9x^2}\)
Rút gọn \(B=\left(x^4-x+\frac{x-3}{x^3+1}\times\frac{\left(x^3-2x^2+2x-1\right)\left(x+1\right)}{x^9+x^7-3x^2-3}+1-\frac{2\left(x+6\right)}{x^2+1}\right)\times\frac{4x^2+6x+1}{\left(x+3\right)\left(4-x\right)}\)
Cho biểu thức Q= [ \(\frac{x^3-8}{x^2-4}\). - \(\frac{\left(x+2\right)^2}{x^2+2x+4}\)- \(\frac{2x^2+x-15}{x^2+x-6}\) ]: \(\frac{\left(x-1\right)\left(x^2+2x-3\right)}{-7x-21}\)
a. Rút gọn Q b. Tìm x thuộc Z để Q thuộc N c. Tìm x để Q<5
P.s: Giúp với, mình cần gấp
Tìm x thuộc Q
a) 3-4.\(\left|5-6x\right|\) =7
b)\(\left|x+2\right|\) + \(\left|x+\frac{3}{5}\right|\) +\(\left|x+\frac{1}{2}\right|\) =4x
c)\(\left|x+\frac{1}{101}\right|\) +\(\left|x+\frac{2}{101}\right|\) +\(\left|x+\frac{3}{101}\right|\) +...+\(\left|x+\frac{100}{101}\right|\) =101x
lm giúp mk với, mk cần gấp
a ) \(3-4.\left|5-6x\right|=7\)
\(\Leftrightarrow4.\left|5-6x\right|=-4\)
\(\Leftrightarrow\left|5-6x\right|=-1\)
\(\Leftrightarrow\) Không thõa mãn ( vì \(x\ge0\) )
b) Do \(\left|x+2\right|\ge0;\left|x+\frac{3}{5}\right|\ge0;\left|x+\frac{1}{2}\right|\ge0\)
=> \(4x\ge0\)
=> \(x\ge0\)
Lúc này ta có: \(\left(x+2\right)+\left(x+\frac{3}{5}\right)+\left(x+\frac{1}{2}\right)=4x\)
=> \(\left(x+x+x\right)+\left(2+\frac{3}{5}+\frac{1}{2}\right)=4x\)
=> \(3x+\frac{31}{10}=4x\)
=> \(4x-3x=\frac{31}{10}\)
=> \(x=\frac{31}{10}\)
Vậy \(x=\frac{31}{10}\)
c) Do \(\left|x+\frac{1}{101}\right|\ge0;\left|x+\frac{2}{101}\right|\ge0;\left|x+\frac{3}{101}\right|\ge0;...;\left|x+\frac{100}{101}\right|\ge0\)
=> \(101x\ge0\)
=> \(x\ge0\)
Lúc này ta có: \(\left(x+\frac{1}{101}\right)+\left(x+\frac{2}{101}\right)+\left(x+\frac{3}{101}\right)+...+\left(x+\frac{100}{101}\right)=101x\)
=> \(\left(x+x+x+...+x\right)+\left(\frac{1}{101}+\frac{2}{101}+\frac{3}{101}+...+\frac{100}{101}\right)=101x\)
100 số x
=> \(100x+\frac{\left(1+100\right).100:2}{101}=101x\)
=> \(\frac{101.50}{101}=101x-100x\)
=> \(x=50\)
Vậy x = 50