RÚT GỌN CÁC BIỂU THỨC SAU
M = \(\dfrac{a-1}{\sqrt{a}-1}\)
N =\(\dfrac{a-1}{\sqrt{a}+1}\)
P=\(\dfrac{a\sqrt{a}-1}{\sqrt{a}-1}\)
Q=\(\dfrac{a\sqrt{a}+1}{\sqrt{a}+1}\)
Rút gọn các biểu thức sau:
\(C=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{1}}\right):\dfrac{\sqrt{a}+1}{a-1}\)
ĐK: `a \ne 1 ; a>=0`
`C=(\sqrta/(\sqrta-1)-\sqrta/(a-1)) : (\sqrta+1)/(a-1)`
`=(\sqrta/(\sqrta-1) - \sqrta/((\sqrta-1)(\sqrta+1))) . (a-1)/(\sqrta+1)`
`= (\sqrta(\sqrta+1)-\sqrta)/((\sqrta-1)(\sqrta+1)) . ((\sqrta-1)(\sqrta+1))/(\sqrta+1)`
`= a/(\sqrta+1)`
Ta có: \(C=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-1}\right):\dfrac{\sqrt{a}+1}{a-1}\)
\(=\dfrac{a+\sqrt{a}-\sqrt{a}}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}:\dfrac{1}{\sqrt{a}-1}\)
\(=\dfrac{a}{\sqrt{a}+1}\)
Rút gọn các biểu thức sau đây:
a) $M=5 \sqrt{\dfrac{1}{5}}+\dfrac{5}{2} \sqrt{\dfrac{4}{5}}-3 \sqrt{5}$;
b) $N=3 \sqrt{\dfrac{1}{2}}+\sqrt{4,5}-\sqrt{12,5}$;
c) $P=\sqrt{\dfrac{1}{3}}+\sqrt{1 \dfrac{1}{5}}+4 \sqrt{3}$ :
d) $Q=2 \sqrt{a}-a \sqrt{\dfrac{4}{a}}+a^{2} \sqrt{\dfrac{9}{a^{3}}}$.
a) M=-căn 5
b) N=căn 2/2
c) P=5 căn 3
d) Q=3 căn a
M=-√5
N=√2/2
P= 3√30 +65√3 / 15
Q=3√a
M=-√5
N=√2/2
P= 3√30 +65√3 / 15
Q=3√a
Câu 1: Rút gọn biểu thức: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)(với a \(\ge\) 0;a \(\ne\)1)
Câu 2: Rút gọn biểu thức: \(M=\left(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}+1\right)\left(1+\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)(với a\(\ge\)0; a\(\ne\)1)
Câu 2:
Ta có: \(M=\left(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}+1\right)\left(1+\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\left(\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}+1\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)
\(=1-a\)
Câu 1:
Ta có: \(A=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1}{\sqrt{a}+1}\right)^2\)
\(=\left(\sqrt{a}+1\right)^2\cdot\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)
\(=1\)
Rút gọn các biểu thức sau:
\(C=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{1}}\right):\dfrac{\sqrt{a}+1}{a-1}\)
\(D=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(E=\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1+\dfrac{x-\sqrt{x}}{1-\sqrt{x}}\right)\)
Làm ơn giúp mình với ạ!!mình đang cần gấp lắm!!
Bài 1 : Cho biểu thức A=(\(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{a}})\):\(\dfrac{\sqrt{a}+1}{a-1}\) với a>0 , a≠1
a) Rút gọn b.thức A
b) Tìm các giá trị của a để A<0
Bài 2 : Rút gọn các b.thức :
A =\((\dfrac{3\sqrt{x}+6}{x-4}+\dfrac{\sqrt{x}}{\sqrt{x}-2}):\dfrac{x-9}{\sqrt{x}-3}\) với x ≥ 0 , x ≠ 4 , x ≠ 9
B = \(3\sqrt{8}-\sqrt{50}-\sqrt{(\sqrt{2}-1)^2}\)
C = \(\dfrac{2}{x-1}\times\sqrt{\dfrac{x^2-2x+1}{4x^2}}\) với 0 < x < 1
D = \((\dfrac{1-a\sqrt{a
}}{1-\sqrt{a}}+\sqrt{a})(\dfrac{1-\sqrt{a}}{1-a})^2\) với a ≥ 0 , a ≠ 1
( giúp hộ em với ạ , em đang cần gấp ạ )
Rút gọn biểu thức:
\(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left[\sqrt{a}-\dfrac{1}{\sqrt{a}}\right]\left[\dfrac{\sqrt{a}+1}{\sqrt{a}-1}+\dfrac{\sqrt{a}-1}{\sqrt{a}+1}\right]\)
Ta có: \(\dfrac{a\sqrt{a}-1}{a-\sqrt{a}}-\dfrac{a\sqrt{a}+1}{a+\sqrt{a}}+\left(\sqrt{a}-\dfrac{1}{\sqrt{a}}\right)\left(\dfrac{\sqrt{a}+1}{\sqrt{a}-1}+\dfrac{\sqrt{a}-1}{\sqrt{a}+1}\right)\)
\(=\dfrac{a+\sqrt{a}+1-a+\sqrt{a}-1}{\sqrt{a}}+\dfrac{a-1}{\sqrt{a}}\cdot\dfrac{a+2\sqrt{a}+1+a-2\sqrt{a}+1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}\)
\(=2+\dfrac{2a+2}{\sqrt{a}}\)
\(=\dfrac{2a+2\sqrt{a}+2}{\sqrt{a}}\)
Cho 2 biểu thức M = \(3\sqrt{3}-\sqrt{12}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
N = \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\) với a>0 và a≠1
a, Rút gọn biểu thức M
b, Tìm các giá trị của a để giá trị của biểu thức M bằng 2 lần giá trị của biểu thức N
a) \(M=3\sqrt{3}-\sqrt{12}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(M=3\sqrt{3}-2\sqrt{3}-\left|\sqrt{3}-1\right|\)
\(M=\sqrt{3}-\sqrt{3}+1\)
\(M=1\)
b) Ta có:
\(N=\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{a}-1}\right):\dfrac{\sqrt{a}+1}{a-2\sqrt{a}+1}\)
\(N=\left(\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}+\dfrac{\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\dfrac{\sqrt{a}+1}{\left(\sqrt{a}-1\right)^2}\)
\(N=\left(\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\right)\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\)
\(N=\dfrac{\left(\sqrt{a}+1\right)\cdot\left(\sqrt{a}-1\right)^2}{\sqrt{a}\left(\sqrt{a}-1\right)\cdot\left(\sqrt{a}+1\right)}\)
\(N=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Theo đề ta có: \(M=2N\)
Khi: \(1=2\cdot\left(\dfrac{\sqrt{a}-1}{\sqrt{a}}\right)\)
\(\Leftrightarrow1=\dfrac{2\sqrt{a}-2}{\sqrt{a}}\)
\(\Leftrightarrow\sqrt{a}=2\sqrt{a}-2\)
\(\Leftrightarrow2\sqrt{a}-\sqrt{a}=2\)
\(\Leftrightarrow\sqrt{a}=2\)
\(\Leftrightarrow a=4\left(tm\right)\)
Cho biểu thức \(M=(\dfrac{1}{\sqrt{a}-1}-\dfrac{1}{\sqrt{a}}):(\dfrac{\sqrt{a}+1}{\sqrt{a}-2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-1})\) ( với a>0; a \(\ne\) 1, a \(\ne\) 4)
a. Rút gọn M
b. Tìm a để M<\(\dfrac{1}{6}\)
\(a,M=\dfrac{\sqrt{a}-\sqrt{a}+1}{\sqrt{a}\left(\sqrt{a}-1\right)}:\dfrac{a-1-a+4}{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}\\ M=\dfrac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)\left(\sqrt{a}-2\right)}{3}=\dfrac{\sqrt{a}-2}{3\sqrt{a}}\\ b,M< \dfrac{1}{6}\Leftrightarrow\dfrac{\sqrt{a}-2}{3\sqrt{a}}-\dfrac{1}{6}< 0\\ \Leftrightarrow\dfrac{2\sqrt{a}-4-\sqrt{a}}{6\sqrt{a}}< 0\Leftrightarrow\dfrac{\sqrt{a}-4}{6\sqrt{a}}< 0\\ \Leftrightarrow\sqrt{a}-4< 0\left(6\sqrt{a}>0\right)\\ \Leftrightarrow a< 16\\ \Leftrightarrow0< a< 16\left(kết.hợp.ĐKXĐ\right)\)
Rút gọn các biểu thức sau:
\(A=\left(\dfrac{1}{\sqrt{a}+2}+\dfrac{1}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}}{a-4}\)
\(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right).\dfrac{\sqrt{x}-1}{x^2}\)
\(C=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{x-9}\right).\dfrac{2\sqrt{x}+6}{\sqrt{x}-1}\)
\(D=\left(\dfrac{5\sqrt{x}-6}{x-9}-\dfrac{2}{\sqrt{x}+3}\right):\left(1+\dfrac{6}{x-9}\right)\)
\(E=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{9+x}{9-x}\right).\left(3\sqrt{x}-x\right)\)
help
Rút gọn các biểu thức sau:
\(A=\left(\dfrac{1}{\sqrt{a}+2}+\dfrac{1}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}}{a-4}\)
\(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right).\dfrac{\sqrt{x}-1}{x^2}\)
\(C=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{x-9}\right).\dfrac{2\sqrt{x}+6}{\sqrt{x}-1}\)
\(D=\left(\dfrac{5\sqrt{x}-6}{x-9}-\dfrac{2}{\sqrt{x}+3}\right):\left(1+\dfrac{6}{x-9}\right)\)
\(E=\left(\dfrac{\sqrt{x}}{3+\sqrt{x}}+\dfrac{9+x}{9-x}\right).\left(3\sqrt{x}-x\right)\)
help
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{a}+2}+\dfrac{1}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}}{a-4}\)
\(=\dfrac{\sqrt{a}-2+\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\cdot\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\sqrt{a}}\)
=2
b) Ta có: \(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}-1}{x^2}\)
\(=\dfrac{4x-1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{x^2}\)
\(=\dfrac{4x-1}{x^2}\)