Có sai đề không vậy bạn?
Mình nghĩ đúng đề phải là x^2-x+2 chứ.
https://hoc24.vn/hoi-dap/question/480187.html
Có sai đề không vậy bạn?
Mình nghĩ đúng đề phải là x^2-x+2 chứ.
https://hoc24.vn/hoi-dap/question/480187.html
Giải phương trình :
\(a,\sqrt{x+2+2\sqrt{x+1}}+\sqrt{x+10-6\sqrt{x+1}}=2\sqrt{x+2-2\sqrt{x+1}}\)
\(b,\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=x+\frac{3}{2}\)
giải các pt
1, \(\sqrt{2x^2+8x+6}+\sqrt{x^2-1}=2x+2\)
2, \(\sqrt{x+2\sqrt{x-1}}-\sqrt{x-2\sqrt{x-1}}=2\)
3, \(\sqrt{x^2+x+4}+\sqrt{x^2+x+1}=\sqrt{2x^2+2x+9}\)
4, \(2x^2+\sqrt{x^2-4x+12}=4x+8\)
5, \(\sqrt{x+3-4\sqrt{x-1}}+\sqrt{x+8-6\sqrt{x-1}}=1\)
Rút gọn các biểu thức sau
a,\(A=\left(\dfrac{1}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
b,\(B=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}+\dfrac{3\sqrt{x}-1}{x-\sqrt{x}+1}-\dfrac{2x\sqrt{x}-2x+2\sqrt{x}-3}{x\sqrt{x}+1}\)
c,\(C=\left(1-\dfrac{x+3\sqrt{x}}{x-9}\right):\left(\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{3+\sqrt{x}}-\dfrac{9-x}{x+\sqrt{x}-6}\right)\)
d,\(D=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{x+9}{9-x}\right):\left(\dfrac{3\sqrt{x}+1}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\right)\)
e,\(E=\dfrac{15\sqrt{x}-11}{x+2\sqrt{x}-3}+\dfrac{3\sqrt{x}-2}{1-\sqrt{x}}-\dfrac{2\sqrt{x}+3}{\sqrt{x}+3}\)
Rút gọn:
1) \(\dfrac{1}{\sqrt{3}+1}+\dfrac{1}{\sqrt{3}-1}-2\sqrt{3}\)
\(P=\left(\dfrac{1}{\sqrt{x}-1}-\dfrac{1}{\sqrt{x}}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-1}\right)\)
2) \(\sqrt{3-2\sqrt{2}}+\dfrac{1}{\sqrt{2}-1}\)
\(M=\left(\dfrac{\sqrt{a}}{\sqrt{a}-2}+\dfrac{\sqrt{a}}{\sqrt{a}+2}\right).\dfrac{a-4}{\sqrt{4a}}\)
\(N=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+3}+\dfrac{\sqrt{x}-3}{2-\sqrt{x}}+\dfrac{\sqrt{x}-2}{x+\sqrt{x}-6}\right)\)
\(Q=\left(1-\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}+\dfrac{\sqrt{x}+3}{2-\sqrt{x}}\right)\)
Làm chi tiết giúp mình với vì mình yếu phần này lắm
Giải phương trình :
a,\(\sqrt{x-2-2\sqrt{x-3}}-\sqrt{x+1-4\sqrt{x-3}}=2\)
b,\(\sqrt{2x-1+2\sqrt{x^2-x}}+\sqrt{2x-1-2\sqrt{x^2-x}}=5\) với \(x\frac{>}{ }1\)
c,\(\sqrt{x+6-4\sqrt{x+2}}+\sqrt{x+11-6\sqrt{x+2}}=1\)
Rút gọn biểu thức:
a) \(A=\left(\frac{3x-3\sqrt{x}-3}{x+\sqrt{x}-2}+\frac{1}{\sqrt{x}-1}-\frac{1}{\sqrt{x}+2}\right):\frac{1}{\sqrt{x}+2}\left(x\ge0,x\ne1\right)\)
b) \(B=\frac{x\sqrt{x}-3}{x-2\sqrt{x}-3}-\frac{2\left(\sqrt{x-3}\right)}{\sqrt{x}+1}+\frac{\sqrt{x}+3}{3-\sqrt{x}}\left(x>0,x\ne9\right)\)
c) \(C=\frac{2\sqrt{x}-9}{x-5+6}-\frac{\sqrt{x}+3}{\sqrt{x}-2}-\frac{2\sqrt{x}+1}{3-\sqrt{x}}\left(x\ge0,x\ne4,x\ne9\right)\)
\(\sqrt{\dfrac{x+2}{4}}+\sqrt{25x+50}-2\sqrt{x+2}=14\) ; \(\sqrt{2x+3}=x\) ; \(\sqrt{25x^2+20x+4}=1\) ; \(\sqrt{\dfrac{x+1}{2x-1}}=2\) ; \(\dfrac{\sqrt{x-2}}{\sqrt{3x+1}}=6\)
Tìm x
Giải các pt sau:
1, \(\sqrt{x^2+x+1}=2x+\sqrt{x^2-x+1}\)
2, \(2x^2+2x+6=2x\sqrt{x^2-x+1}+4\sqrt{3x+1}\)
3, \(\left(\sqrt{x+3}-\sqrt{x}\right)\left(1+\sqrt{x^2+3x}\right)=3\)
4, \(\sqrt{2x^2-1}+\sqrt{x^2-3x-2}=\sqrt{2x^2-2x+3}+\sqrt{x^2-x+2}\)
5, \(13\sqrt{x-1}+9\sqrt{x+1}=16x\)
\[D=\left ( \frac{1}{3\sqrt{x}-6} +\frac{1}{x-2\sqrt{x}}\right )\left ( \frac{1}{6} +\frac{1}{2\sqrt{x}}\right )\\ D=\left ( \frac{1}{3\left ( \sqrt{x}-2 \right )} +\frac{1}{\sqrt{x}\left ( \sqrt{x}-2 \right )}\right ).\frac{\sqrt{x}+3}{6\sqrt{x}}\\ D=\frac{\sqrt{x}+3}{3\sqrt{x}\left ( \sqrt{x}-2 \right )}.\frac{\sqrt{x}+3}{6\sqrt{x}}\\ D=\frac{\left ( \sqrt{x}+3 \right )^{2}}{18x\left ( \sqrt{x}-2 \right )}\\ D=\frac{x+6\sqrt{x}+9}{18x\sqrt{x}-36x}\]
A/ Đúng
B/ Sai
Thu gọn
\(B=\left(\sqrt{x}-1\right)\left(\sqrt{x}+4\right)-\left(\sqrt{x}-3\right)^2+\left(2\sqrt{x}+1\right)^2\)
C = \(\frac{x\sqrt{x}+1}{x-\sqrt{x}+1}+\frac{x-4}{x-2}-\frac{x+2\sqrt{x}+1}{\sqrt{x}+1}\)