Em muốn mọi người giải bài nhanh nhưng đến đề bài em cũng chưa ghi đủ?
Lời giải:
ĐKXĐ:
\(x>0; x\neq 4\)
Ta có:
\(\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4\sqrt{x}-4}{\sqrt{x}(\sqrt{x}-2)}.\sqrt{x}=\frac{\sqrt{x}}{\sqrt{x}-2}-\frac{4\sqrt{x}-4}{\sqrt{x}-2}\)
\(=\frac{\sqrt{x}-(4\sqrt{x}-4)}{\sqrt{x}-2}=\frac{4-3\sqrt{x}}{\sqrt{x}-2}=\frac{-2}{\sqrt{x}-2}-3\)
Giải hộ mình với
1 chứng minh đẳng thức:
a) \(\frac{\sqrt{a^2+x^2}+\sqrt{a^2+x^2}}{\sqrt{a^2+x^2}+\sqrt{a^2-x^2}}-\sqrt{\frac{a^4}{x^4}}=\frac{a^2}{x^2}\)với \(\left|a\right|\)>\(\left|x\right|\)
b) \(\left(\frac{5+2\sqrt{6}}{\sqrt{x}+\sqrt{2}}\right)^2-\left(\frac{5-2\sqrt{6}}{\sqrt{3}-\sqrt{6}}\right)^2=4\sqrt{6}\)
2.
A=\(\frac{\sqrt{x}+1}{\sqrt{x}-2}+\frac{2\sqrt{x}}{\sqrt{x}+2}+\frac{2+5\sqrt{x}}{4-x}\)
a) Rút gọn A nếu \(x\ge0\)và \(x\ne4\)
b) Tìm x để A-2
1/Rút gọn
A=\(\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{xy}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\left(x-y\right)\left(\sqrt{x^3+x}\right)}\)(x>0; y>0; x#y)
B= \(\left(\frac{1}{\sqrt{x}+1}-\frac{1}{x+\sqrt{x}}\right):\frac{x-\sqrt{x}+1}{x\sqrt{x}+1}\)( x>0)
C=\(\left(\frac{x+1}{\sqrt{x}}+2\right).\frac{\sqrt{x}}{\left(\sqrt{x}+1\right)\left(x\sqrt{x}+1\right)}\)(x>0)
D=\(\left(\frac{x\sqrt{x}-1}{\sqrt{x}-1}+\sqrt{x}\right):\left(x-1\right)-\frac{2}{\sqrt{x}-1}\)(x>=0; x#1)
giúp em với ạ em đang cần gấp ạ
Bài 1 :1) TÍNH A=\(\left(\sqrt{10}-\sqrt{2}\right)\sqrt{3+\sqrt{5}}\)
2) rút gọn biểu thức B=\(\frac{\left(\frac{\sqrt{x}}{\sqrt{x}+4}+\frac{4}{\sqrt{x}-4}\right)}{\frac{x+16}{\sqrt{x}+2}}\)
LÀM ƠN GIẢI GIÚP
Rút gọn biểu thức:
1) \(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\cdot\left(x-1\right)}{\sqrt{x}-1}\)
2) \(P=\left(\frac{\sqrt{x}-2}{\sqrt{x}-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\cdot\frac{\left(1-x\right)^2}{2}\)
3) \(B=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)
4) \(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right)\div\left(\frac{1}{\sqrt{a}+1}-\frac{2}{a-1}\right)\)
A = (\(\sqrt{x}-\frac{x+2}{x-1}\)) : ( \(\frac{\sqrt{x}}{\sqrt{x}+1}\)- \(\frac{\sqrt{x}-4}{1-x}\))
B= (\(\frac{x\sqrt{x}+1}{x-1}-\frac{x-1}{\sqrt{x}-1}\)) :\(\sqrt{x}+\frac{\sqrt{x}}{\sqrt{x}-1}\))
C=(\(\frac{\sqrt{x}-2}{x-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\)) .(\(\frac{1-x}{\sqrt{2}}\))2
Mọi ng giúp em nhanh với ạ, em cảm ơn ạ.
a, A = \(\frac{5+7\sqrt{5}}{\sqrt{5}}+\frac{11+\sqrt{11}}{1+\sqrt{11}}\)
b, B = \(\left(1-\sqrt{5}\right).\frac{\sqrt{5}+5}{2\sqrt{5}}\)
c, C = \(1+\left(\frac{x+\sqrt{x}}{1+\sqrt{x}}\right).\left(1+\frac{x-\sqrt{x}}{1-\sqrt{x}}\right)\) ( với 0 < bằng x)
Giải giúp ạ- mai mình cần rồi
A= \(\frac{x-4\sqrt{x}+2}{\sqrt{x}-2}\) \(\left(x\ge0;x\ne4\right)\)
B= \(\frac{x\sqrt{x}-1}{x-1}\) \(\left(x\ge0;x\ne1\right)\)
C= \(\frac{x\sqrt{x}-1}{x-\sqrt{x}}-\frac{x\sqrt{x}+1}{x+\sqrt{x}}+\frac{x+1}{\sqrt{x}}\) \(\left(x>0;x\ne1\right)\)
D= \(\sqrt{x+2\sqrt{2x-4}}+\sqrt{x-2\sqrt{2x-4}}\) \(\left(x\ge2\right)\)
E= \(\frac{x+\sqrt{x^2}-2x}{x-\sqrt{x^2-2x}}-\frac{x-\sqrt{x^2-2x}}{x+\sqrt{x^2-2x}}\)
\(D=\left(\frac{x-2\sqrt{x}}{x-4}-1\right):\left(\frac{4-x}{x-\sqrt{x}-6}-\frac{\sqrt{x}-2}{3-\sqrt{x}}-\frac{\sqrt{x}-3}{\sqrt{x}+2}\right)\)
Tìm đkxđ
Rút gọn
Rút gọn:
\(A=\frac{x^2+5x+6+x\sqrt{9-x^2}}{3x-x^2+\left(x+2\right).\sqrt{9-x^2}}\)
\(B=\frac{x^2-5x+6+3\sqrt{x^2-6x+8}}{3x-12+\left(x-3\right).\sqrt{x^2-6x+8}}\)
\(C=\frac{\sqrt{2\sqrt{4-x^2}}.\left(\sqrt{\left(2+x\right)^3}-\sqrt{\left(2-x\right)^3}\right)}{4+\sqrt{4-x^2}}\)
