\(B=\left(\frac{1}{x-4}-\frac{1}{x+4\sqrt{x}+4}\right).\frac{x+2\sqrt{x}}{\sqrt{x}}\left(vớix>o,x#4\right)\)
rút gọn :
a.\(\sqrt{x+4\sqrt{x-4}}+\sqrt{x-4\sqrt{x-4}}vớix>=8\)
b,\(\sqrt{2x-1+2\sqrt{x^2-x}}+\sqrt{2x-1-2\sqrt{x^2-x}}\)
c,\(\frac{\sqrt{x-2\sqrt{x+1}}}{x+2\sqrt{x+1}}\Rightarrow vớix>=0\)
d,\(\frac{x-1}{\sqrt{y-1}}\cdot\sqrt{\frac{\left(y-2\sqrt{y+1}\right)^2}{\left(x-1\right)^4}}\)
(14,78-a)/(2,87+a)=4/1
14,78+2,87=17,65
Tổng số phần bằng nhau là 4+1=5
Mỗi phần có giá trị bằng 17,65/5=3,53
=>2,87+a=3,53
=>a=0,66.
a,\(\sqrt{x-4+4\sqrt{x-4}+4}\) +\(\sqrt{x-4-4\sqrt{x-4}+4}\)
=\(\sqrt{x-4}+2+\left|\sqrt{x-4}-2\right|\) (vi x>=8)
=\(\sqrt{x-4}+2+\sqrt{x-4}-2=2\sqrt{x-4}\)
b, \(\sqrt{x-1+2\sqrt{x\left(x-1\right)}+x}+\sqrt{x-1-2\sqrt{x\left(x-1\right)}+x}\)
=\(\sqrt{x-1}+\sqrt{x}+\left|\sqrt{x-1}-\sqrt{x}\right|\)
=\(\sqrt{x}+\sqrt{x-1}+\sqrt{x}-\sqrt{x-1}\) =\(2\sqrt{x}\)
c,d sai dau bai hay sao y
1.rút gọn biểu thức sau :
a\(A=\sqrt{4\left(x-1\right)}-\sqrt{9\left(x-1\right)}-\sqrt{16\left(x-1\right)}vớix>=1\)
b.\(B=\frac{2}{x+y}\sqrt{\frac{3\left(x+y\right)^2}{4}}vớix+y>0\)
\(a, A=2\sqrt{x-1}-3\sqrt{x-1}-4\sqrt{x-1}=\left(2-3-4\right)\sqrt{x-1}=-5\sqrt{x-1}\)
\(b, B=\frac{2}{x+y}.\left(x+y\right)\sqrt{\frac{3}{4}}=2\sqrt{\frac{3}{4}}=2.\frac{1}{2}.\sqrt{3}=\sqrt{3}\)
\(B=\left(\frac{a\sqrt{a}+1}{\sqrt{a}+1}\right):\left(a-1\right)+\frac{2a+\sqrt{a}+1}{\sqrt{a}+1}-\frac{\sqrt{a}}{a-1}vớia>1\)
\(C=\left(\frac{X-1}{\sqrt{X}-1}+\frac{\sqrt{X^3}-1}{1-X}\right)-\left(\frac{\left(X-1\right)^2+\sqrt{X}}{\sqrt{X}+1}\right)vớiX>0,X\ne1\)
\(D=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\left(x-1\right)}{\sqrt{x}-1}vớix>0,x\ne1\)
\(B=\frac{-2a\sqrt{a}+2a^2}{\left(\sqrt{a}-\right)\left(a-1\right)}\)
\(C=-x\sqrt{x}+x+\sqrt{x}-1\)
\(D=x-\sqrt{x}+1\)
Mấy cái này chỉ có nhân lên rồi rút gọn thôi ah. Nên mình cho bạn đáp án để kiểm tra lại thôi ah
A=\(\left(1-\frac{4}{\sqrt{x}+1}+\frac{1}{x-1}\right):\frac{x-2\sqrt{x}}{x-1}vớix>0,xkhac1,4\)
Rút gọn biểu thức:
a, \(\sqrt{6+4\sqrt{2}}+\sqrt{6-4\sqrt{2}}\)
b, \(\sqrt{\left(1-\sqrt{3}\right)^2}-\sqrt{12+6\sqrt{3}}\)
c, \(4x-\sqrt{x^2-4x+4}\left(vớix\ge2\right)\)
d, \(\frac{x+6\sqrt{x}+9}{x+9}\)\(\left(vớix\ge0,x\ne9\right)\)
e, \(\frac{\sqrt{x^2+4x+4}}{x+2}\)\(\left(vớix\ne-2\right)\)
mk nhầm dấu sửa lại câu c là \(4x-x+2\)= \(3x+2\)
a, \(\sqrt{\left(\sqrt{2}\right)^2+2\times2\times\sqrt{2}+2^2}\)+ \(\sqrt{2^2+2\times2\times\sqrt{2}+\left(\sqrt{2}\right)^2}\)
= \(\sqrt{\left(\sqrt{2}+2\right)^2}\)+ \(\sqrt{\left(2-\sqrt{2}\right)^2}\)
= \(\sqrt{2}+2+2-\sqrt{2}\)
= 4
c, \(4x-\sqrt{\left(x-2\right)^2}\)
= \(4x-x-2\)
= \(3x-2\)
Cho biểu thức : \(A=\left(\frac{4\sqrt{x}}{2+\sqrt{x}}+\frac{8x}{4-x}\right):\left(\frac{2\sqrt{x}-1}{x-2\sqrt{x}}-\frac{1}{\sqrt{x}}\right)vớix>0,x\ne1,x\ne4\)
a) Rút gọn A
b) Tính giá trị của A biết \(x=3-2\sqrt{2}\)
c) Tìm x để |A| > A
b, \(M=A-B=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\left(\frac{5}{x+\sqrt{x}-6}+\frac{1}{\sqrt{x}-2}\right)\)
\(=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{5}{x+\sqrt{x}-6}-\frac{1}{\sqrt{x}-2}\)
\(=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{x+\sqrt{x}-6}-\frac{5}{x+\sqrt{x}-6}-\frac{1\left(\sqrt{x}+3\right)}{x+\sqrt{x}-6}\)
\(=\frac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-4\sqrt{x}+3\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)\(=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}-4}{\sqrt{x}-2}\)
bạn trung học hay tiểu học vậy
Rút gọn
a) \(\left(\frac{2+\sqrt{a}}{a+2\sqrt{a}+1}-\frac{\sqrt{a}-2}{a-1}\right)\left(\frac{a\sqrt{a}-\sqrt{a}-1}{\sqrt{a}}\right)\)
b) \(\left(\frac{\sqrt{x}+1}{x-4}-\frac{\sqrt{x}-1}{x+4\sqrt{x}+4}\right)\left(\frac{x\sqrt{x}+2x+4\sqrt{x}-8}{\sqrt{x}}\right)\)
A=\(\left(\frac{\sqrt{x}-1}{x-4}-\frac{\sqrt{x+1}}{x+4\sqrt{x+4}}\right)\):\(\frac{x\sqrt{x}}{\left(4-x\right)^2}\)
B=\(\left(\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{x+9\sqrt{x}}{x-9}\right):\frac{x\sqrt{x}}{\left(4-x\right)^2}\)
rút gọn các biểu thức
\(B=\left(\frac{2\sqrt{x}}{\sqrt{x}-3}-\frac{x+9\sqrt{x}}{x-9}\right):\frac{x\sqrt{x}}{\left(4-x\right)^2}\)
\(=\frac{2x+18\sqrt{x}-x-9\sqrt{x}}{x-9}\cdot\frac{\left(4-x\right)^2}{x\sqrt{x}}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+9\right)\left(4-x\right)^2}{x\sqrt{x}\left(x-9\right)}\)
\(=\frac{\left(\sqrt{x}+9\right)\left(4-x\right)^2}{\sqrt{x}\left(x-9\right)}\)