\(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\left(\dfrac{1-\sqrt{a}}{1-a}\right)^2\) giúp minh với
1\(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right)\cdot\left(\sqrt{a}\cdot\dfrac{4}{\sqrt{a}}\right)\) (với a>0,a≠4)
2\(\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right)\cdot\left(1-\dfrac{1}{\sqrt{a}}\right)\) (với a>0,a≠1)
3\(\left(\dfrac{\sqrt{a}-1}{\sqrt{a}+1}+\dfrac{\sqrt{a}+1}{\sqrt{a}-1}\right)\cdot\left(1-\dfrac{2}{a+1}\right)^2\) (với a>0,a≠1)
Rút gọn các biểru thức sau
1. \(\left(\dfrac{\sqrt{a}-2}{\sqrt{a}+2}-\dfrac{\sqrt{a}+2}{\sqrt{a}-2}\right).\left(\sqrt{a}.\dfrac{4}{\sqrt{a}}\right)=\dfrac{\left(\sqrt{a}-2\right)^2-\left(\sqrt{a}+2\right)^2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.4=\dfrac{a-4\sqrt{a}+4-a-4\sqrt{a}-4}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}.4=\dfrac{-64\sqrt{a}}{a-4}\)Nếu nhân tu thứ 2 của phép tính là \(\sqrt{a}-\dfrac{4}{\sqrt{a}}\) thì kết quả của phép tính là -16 nha bạn
2.\(\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right).\left(1-\dfrac{1}{\sqrt{a}}\right)=\dfrac{1+\sqrt{a}-1+\sqrt{a}}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}.\dfrac{-\left(1-\sqrt{a}\right)}{\sqrt{a}}=\dfrac{-2\sqrt{a}}{\left(1+\sqrt{a}\right)\sqrt{a}}=\dfrac{-2}{1+\sqrt{a}}\)\(\left(a>0,a\ne1\right)\)
Rút gọn:
\(A=\left[1:\left(1-\dfrac{\sqrt{a}}{1+\sqrt{a}}\right)\right]\cdot\left[\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{a\sqrt{a}-a+\sqrt{a}-1}\right]\)
\(A=\left[1:\left(1-\frac{\sqrt{a}}{1+\sqrt{a}}\right)\right]\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{a\sqrt{a}-a+\sqrt{a}-1}\right]\)
\(=\left[1:\left(\frac{1+\sqrt{a}-\sqrt{a}}{1+\sqrt{a}}\right)\right]\left[\frac{1}{\sqrt{a}-1}-\frac{2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\right]\)
\(=\left(1:\frac{1}{1+\sqrt{a}}\right).\frac{a+1-2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)
\(=\left(\sqrt{a}+1\right).\frac{\left(\sqrt{a}-1\right)^2}{\left(\sqrt{a}-1\right)\left(a+1\right)}\)
\(=\frac{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}{a+1}=\frac{a-1}{a+1}\)
chứng minh răng :
a,\(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\cdot\left(1-\dfrac{a-\sqrt{a}}{a-1}\right)=1-a\left(a>hoaăặc=0,a\right)\left(a#1\right)\)b, \(\dfrac{\sqrt{ab}-b}{\sqrt{b}}-\sqrt{\dfrac{a}{b}}< 0\left(a>hoac=0,b>0\right)\)
\(=\dfrac{\sqrt{ab}-b-\sqrt{a}}{\sqrt{b}}\)
Rút gọn:
\(A=\left[1:\left(1-\dfrac{\sqrt{a}}{1+\sqrt{a}}\right)\right]\cdot\left[\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{a\sqrt{a}-a+\sqrt{a}-1}\right]\)
\(A=\left[1:\left(\dfrac{1+\sqrt{a}-\sqrt{a}}{\sqrt{a}+1}\right)\right]\cdot\left(\dfrac{1}{\sqrt{a}-1}-\dfrac{2\sqrt{a}}{\left(\sqrt{a}-1\right)\left(a+1\right)}\right)\)
\(=\dfrac{\sqrt{a}+1}{1}\cdot\dfrac{a+1-2\sqrt{a}}{\left(a+1\right)\left(\sqrt{a}-1\right)}\)
\(=\dfrac{\sqrt{a}+1}{1}\cdot\dfrac{\left(a+1\right)\left(\sqrt{a}-1\right)}{\left(\sqrt{a}-1\right)^2}\)
\(=\dfrac{\left(a+1\right)\left(\sqrt{a}+1\right)}{\sqrt{a}-1}\)
Rút gọn:
\(E=\left(1-a^2\right):\left[\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\right]\)
\(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}=\dfrac{\left(1-a\sqrt{a}\right)\left(1+\sqrt{a}\right)}{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}\right)}=\dfrac{1+\sqrt{a}-a\sqrt{a}-a^2}{1-a}=\dfrac{\left(1-a\right)\left(\sqrt{a}+a+1\right)}{1-a}\)
=> \(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}=a+2\sqrt{a}+1=\left(\sqrt{a}+1\right)^2\)
Tương tự \(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}=\left(\sqrt{a}-1\right)^2\)
biểu thức trong dấu ngoặc vuông = \(\left[\left(\sqrt{a}-1\right).\left(\sqrt{a}+1\right)\right]^2=\left(a-1\right)^2\)
\(E=\dfrac{1-a^2}{\left(a-1\right)^2}\)
Chứng minh;
a, \(\left(\dfrac{1+a+\sqrt{a}}{\sqrt{a}+1}\right)\cdot\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)=1\)
\(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)
=1-a
chứng minh đẳng thức:
a. \(\dfrac{\left(2+\sqrt{a}\right)^2-\left(\sqrt{a}+1\right)^2}{2\sqrt{a}+3}=1\)với a≥0
b.\(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2=1\)với a ≥0
a) \(\dfrac{\left(2+\sqrt{a}\right)^2-\left(\sqrt{a}+1\right)^2}{2\sqrt{a}+3}=\dfrac{\left(2+\sqrt{a}-\sqrt{a}-1\right)\left(2+\sqrt{a}+\sqrt{a}+1\right)}{2\sqrt{a}+3}\)
\(=\dfrac{1.\left(2\sqrt{a}+3\right)}{2\sqrt{a}+3}=1\)
b) \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(a+\sqrt{a}+1\right)}{1-\sqrt{a}}+\sqrt{a}\right).\dfrac{1}{\left(1+\sqrt{a}\right)^2}\)
\(=\left(a+\sqrt{a}+1+\sqrt{a}\right).\dfrac{1}{\left(\sqrt{a}+1\right)^2}=\left(a+2\sqrt{a}+1\right).\dfrac{1}{\left(\sqrt{a}+1\right)^2}\)
\(=\left(\sqrt{a}+1\right)^2.\dfrac{1}{\left(\sqrt{a}+1\right)^2}=1\)
a, \(VT=\dfrac{\left(2+\sqrt{a}\right)^2-\left(\sqrt{a}+1\right)^2}{2\sqrt{a}+3}=\dfrac{a+4\sqrt{a}+4-a-2\sqrt{a}-1}{2\sqrt{a}+3}\)
\(=\dfrac{2\sqrt{a}+3}{2\sqrt{a}+3}=1=VP\)
Vậy ta có đpcm
b, \(VT=\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2\)
\(=\left(1+\sqrt{a}+a+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2=\dfrac{\left(1+\sqrt{a}\right)^2}{\left(1+\sqrt{a}\right)^2}=1=VP\)
Vậy ta có đpcm
a) Ta có: \(\dfrac{\left(2+\sqrt{a}\right)^2-\left(\sqrt{a}+1\right)^2}{2\sqrt{a}+3}\)
\(=\dfrac{a+4\sqrt{a}+4-a-2\sqrt{a}-1}{2\sqrt{a}+3}\)
\(=\dfrac{2\sqrt{a}+3}{2\sqrt{a}+3}=1\)
b) Ta có: \(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right):\left(1+\sqrt{a}\right)^2\)
\(=\left(1+\sqrt{a}+\sqrt{a}+a\right):\left(1+\sqrt{a}\right)^2\)
\(=\left(1+\sqrt{a}\right)^2:\left(1+\sqrt{a}\right)^2=1\)
Cho biểu thức \(A=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right)\cdot\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\)
a. Rút gọn A
b. Tìm \(x\) để \(A>-6\)
c. Tính A khi \(a^2-3=0\)
a) \(A=\left(\dfrac{\sqrt{a}}{2}-\dfrac{1}{2\sqrt{a}}\right).\left(\dfrac{a-\sqrt{a}}{\sqrt{a}+1}-\dfrac{a+\sqrt{a}}{\sqrt{a}-1}\right)\left(đk:a>0,x\ne1\right)\)
\(=\dfrac{a-1}{2\sqrt{a}}.\dfrac{\left(a-\sqrt{a}\right)\left(\sqrt{a}-1\right)-\left(a+\sqrt{a}\right)\left(\sqrt{a}+1\right)}{a-1}\)
\(=\dfrac{a\sqrt{a}-2a+\sqrt{a}-a\sqrt{a}-2a-\sqrt{a}}{2\sqrt{a}}\)
\(=\dfrac{-4a}{2\sqrt{a}}=-2\sqrt{a}\)
b) \(A=-2\sqrt{a}>-6\)
\(\Leftrightarrow\sqrt{a}< 3\Leftrightarrow0\le a< 9\) và \(a\ne1\)
c) \(a^2-3=0\Leftrightarrow a^2=3\Leftrightarrow\sqrt{a}=\sqrt[4]{3}\)
\(\Rightarrow A=-2\sqrt{a}=-2\sqrt[4]{3}\)
Bài 75 (trang 40 SGK Toán 9 Tập 1)
Chứng minh các đẳng thức sau:
a) $\left(\dfrac{2 \sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right) \cdot \dfrac{1}{\sqrt{6}}=-1,5$;
b) $\left(\dfrac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right): \dfrac{1}{\sqrt{7}-\sqrt{5}}=-2$;
c) $\dfrac{a \sqrt{b}+b \sqrt{a}}{\sqrt{a b}}: \dfrac{1}{\sqrt{a}-\sqrt{b}}=a-b$ với $a, b$ dương và $a \neq b$;
d) $\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)=1-a$ với $a \geq 0$ và $a \neq 1$.
a) \(\left(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=\left(\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-\dfrac{\sqrt{216}}{3}\right).\dfrac{1}{\sqrt{6}}=\dfrac{1}{2}-2=-\dfrac{3}{2}\)